Solução parcial Goldstein Mecanica Classica

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<p>See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/358249174</p><p>Partial Solutions Manual Herbert B. Goldstein 3RD ED.</p><p>Preprint · January 2022</p><p>CITATIONS</p><p>0</p><p>READS</p><p>24,940</p><p>1 author:</p><p>Jhordan Silveira de Borba</p><p>Universidade Federal do Rio Grande do Sul</p><p>18 PUBLICATIONS   0 CITATIONS</p><p>SEE PROFILE</p><p>All content following this page was uploaded by Jhordan Silveira de Borba on 31 January 2022.</p><p>The user has requested enhancement of the downloaded file.</p><p>https://www.researchgate.net/publication/358249174_Partial_Solutions_Manual_Herbert_B_Goldstein_3RD_ED?enrichId=rgreq-86994239523ec0781f71890b579960fd-XXX&enrichSource=Y292ZXJQYWdlOzM1ODI0OTE3NDtBUzoxMTE4NDU4NTg2NjM2MjkzQDE2NDM2NzI3MTU2Mzg%3D&el=1_x_2&_esc=publicationCoverPdf</p><p>https://www.researchgate.net/publication/358249174_Partial_Solutions_Manual_Herbert_B_Goldstein_3RD_ED?enrichId=rgreq-86994239523ec0781f71890b579960fd-XXX&enrichSource=Y292ZXJQYWdlOzM1ODI0OTE3NDtBUzoxMTE4NDU4NTg2NjM2MjkzQDE2NDM2NzI3MTU2Mzg%3D&el=1_x_3&_esc=publicationCoverPdf</p><p>https://www.researchgate.net/?enrichId=rgreq-86994239523ec0781f71890b579960fd-XXX&enrichSource=Y292ZXJQYWdlOzM1ODI0OTE3NDtBUzoxMTE4NDU4NTg2NjM2MjkzQDE2NDM2NzI3MTU2Mzg%3D&el=1_x_1&_esc=publicationCoverPdf</p><p>https://www.researchgate.net/profile/Jhordan-Silveira-De-Borba?enrichId=rgreq-86994239523ec0781f71890b579960fd-XXX&enrichSource=Y292ZXJQYWdlOzM1ODI0OTE3NDtBUzoxMTE4NDU4NTg2NjM2MjkzQDE2NDM2NzI3MTU2Mzg%3D&el=1_x_4&_esc=publicationCoverPdf</p><p>https://www.researchgate.net/profile/Jhordan-Silveira-De-Borba?enrichId=rgreq-86994239523ec0781f71890b579960fd-XXX&enrichSource=Y292ZXJQYWdlOzM1ODI0OTE3NDtBUzoxMTE4NDU4NTg2NjM2MjkzQDE2NDM2NzI3MTU2Mzg%3D&el=1_x_5&_esc=publicationCoverPdf</p><p>https://www.researchgate.net/institution/Universidade_Federal_do_Rio_Grande_do_Sul?enrichId=rgreq-86994239523ec0781f71890b579960fd-XXX&enrichSource=Y292ZXJQYWdlOzM1ODI0OTE3NDtBUzoxMTE4NDU4NTg2NjM2MjkzQDE2NDM2NzI3MTU2Mzg%3D&el=1_x_6&_esc=publicationCoverPdf</p><p>https://www.researchgate.net/profile/Jhordan-Silveira-De-Borba?enrichId=rgreq-86994239523ec0781f71890b579960fd-XXX&enrichSource=Y292ZXJQYWdlOzM1ODI0OTE3NDtBUzoxMTE4NDU4NTg2NjM2MjkzQDE2NDM2NzI3MTU2Mzg%3D&el=1_x_7&_esc=publicationCoverPdf</p><p>https://www.researchgate.net/profile/Jhordan-Silveira-De-Borba?enrichId=rgreq-86994239523ec0781f71890b579960fd-XXX&enrichSource=Y292ZXJQYWdlOzM1ODI0OTE3NDtBUzoxMTE4NDU4NTg2NjM2MjkzQDE2NDM2NzI3MTU2Mzg%3D&el=1_x_10&_esc=publicationCoverPdf</p><p>UNIVERSIDADE FEDERAL DO RIO GRANDE DO SUL</p><p>INSTITUTO DE FÍSICA</p><p>SOLUTIONS MANUAL HERBERT B. GOLDSTEIN 3RD ED.</p><p>LIST 1 OF CLASSICAL MECHANICS</p><p>CLASS: FIP00001</p><p>TEACHER: MARIA BEATRIZ DE LEONE GAY DUCATI</p><p>PORTO ALEGRE, RS</p><p>2021/01</p><p>Contents</p><p>1 Exercise 1 - Edgard Kretschmann 3</p><p>2 Exercise 2 - Francisco A. F. Almeida 5</p><p>3 Exercise 3 - Francisco A. F. Almeida 7</p><p>4 Exercise 4 - Rodrigo Weber P. 9</p><p>5 Exercise 5 - Jenifer A. de Matos 11</p><p>6 Exercise 6 - Jackson Galvão 17</p><p>7 Exercise 7 - Jackson Galvão 19</p><p>8 Exercise 8 - Gustavo C. G. Kessler 23</p><p>9 Exercise 9 - Rodrigo Weber P. 25</p><p>10 Exercise 10 - Jhordan S. de Borba 27</p><p>11 Exercise 11 - Jhordan S. de Borba 31</p><p>12 Exercise 12 - Jefferson S. Martins 35</p><p>13 Exercise 13 - Geovane Naysinger 37</p><p>14 Exercise 14 - Edgard Kretschmann 41</p><p>15 Exercise 15 - Guilherme S. Y. Giardini 43</p><p>15.1 a) Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43</p><p>15.2 a) Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49</p><p>15.3 b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56</p><p>15.4 c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59</p><p>16 Exercise 16 - Lucas Bourscheidt 63</p><p>17 Exercise 17 - Jenifer A. de Matos 65</p><p>18 Exercise 18 - Gustavo C. G. Kessler 69</p><p>iii</p><p>CONTENTS 1</p><p>19 Exercise 19 - Douglas O. Novaes 71</p><p>20 Exercise 20 - Jefferson S. Martins 75</p><p>21 Exercise 21 - Lucas Bourscheidt 77</p><p>22 Exercise 22 - Douglas O. Novaes 81</p><p>23 Exercise 23 - Guilherme S. Y. Giardini 85</p><p>23.1 Non Conservative Euler-Lagrange Equation . . . . . . . . . . . . . . . . . 85</p><p>23.2 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87</p><p>24 Exercise 24 - Geovane Naysinger 91</p><p>24.1 Questão 1.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91</p><p>24.1.1 a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91</p><p>24.1.2 b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93</p><p>24.1.3 c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94</p><p>24.1.4 d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95</p><p>Bibliography 99</p><p>Exercise 1 - Edgard Kretschmann</p><p>The equation for the kinetic energy of a particle is given by</p><p>T =</p><p>m</p><p>2</p><p>vv (1.1)</p><p>Taking the time derivative of the kinetic energy it comes that</p><p>dT</p><p>dt</p><p>= mv</p><p>dv</p><p>dt</p><p>So</p><p>dT</p><p>dt</p><p>= v</p><p>d(mv)</p><p>dt</p><p>Once that</p><p>F =</p><p>d(mv)</p><p>dt</p><p>Then</p><p>dT</p><p>dt</p><p>= Fv (1.2)</p><p>And if mass isn’t constant the expression d(mT)/dt becomes</p><p>d(mT )</p><p>dt</p><p>= T</p><p>dm</p><p>dt</p><p>+m</p><p>dT</p><p>dt</p><p>Using the definition (1) in the equation above , we will have that:</p><p>d(mT )</p><p>dt</p><p>= m</p><p>dm</p><p>dt</p><p>vv +m2v</p><p>dv</p><p>dt</p><p>Using the chain rule, we can obtain that</p><p>d(mT )</p><p>dt</p><p>= mv</p><p>(</p><p>d(mv)</p><p>dt</p><p>)</p><p>And from this</p><p>d(mT )</p><p>dt</p><p>= Fp</p><p>3</p><p>Exercise 2 - Francisco A. F. Almeida</p><p>Prove that the magnitude R of the position vector for the center of mass from an</p><p>arbitrary origin is given by the equation</p><p>M2R2 =M</p><p>∑</p><p>i,j</p><p>miri</p><p>2 − 1</p><p>2</p><p>∑</p><p>i,j</p><p>mimjrij</p><p>2</p><p>Solution:</p><p>M ~R =</p><p>∑</p><p>i</p><p>miri</p><p>M2R2 = (M ~R) · (M ~R)</p><p>M2R2 =</p><p>∑</p><p>i</p><p>mi~ri ·</p><p>∑</p><p>j</p><p>mj ~rj</p><p>M2R2 =</p><p>∑</p><p>i,j</p><p>mimj(~ri · ~rj)</p><p>According to the graphic above, we have:</p><p>(~ri − ~rj) · (~ri − ~rj) = ~ri · ~ri − ~ri · ~rj − ~rj · ~ri + ~rj · ~rj</p><p>rij</p><p>2 = ri</p><p>2 − 2~ri · ~rj + rj</p><p>2</p><p>5</p><p>6 CHAPTER 2. EXERCISE 2 - FRANCISCO A. F. ALMEIDA</p><p>FIGURE 2.1</p><p>2~ri · ~rj = ri</p><p>2 + rj</p><p>2 − rij2</p><p>~ri · ~rj =</p><p>ri</p><p>2 + rj</p><p>2 − rij2</p><p>2</p><p>M2R2 =</p><p>∑</p><p>i,j</p><p>mimj</p><p>(</p><p>ri</p><p>2 + rj</p><p>2 − rij2</p><p>2</p><p>)</p><p>M2R2 =</p><p>1</p><p>2</p><p>∑</p><p>i,j</p><p>mimjri</p><p>2 +</p><p>1</p><p>2</p><p>∑</p><p>i,j</p><p>mimjrj</p><p>2 − 1</p><p>2</p><p>∑</p><p>i,j</p><p>mimjrij</p><p>2</p><p>M2R2 =</p><p>1</p><p>2</p><p>M</p><p>∑</p><p>i,j</p><p>miri</p><p>2 +</p><p>1</p><p>2</p><p>M</p><p>∑</p><p>i,j</p><p>mjrj</p><p>2 − 1</p><p>2</p><p>∑</p><p>i,j</p><p>mimjrij</p><p>2</p><p>As the i and j indexes are arbitrary, we have:</p><p>M2R2 =M</p><p>∑</p><p>i,j</p><p>miri</p><p>2 − 1</p><p>2</p><p>∑</p><p>i,j</p><p>mimjrij</p><p>2</p><p>Exercise 3 - Francisco A. F. Almeida</p><p>Suppose a system of two particles is know to obey the equations of motion, Eqs.</p><p>(1.22) and (1.26). From the equations of the motion of the individual particles show</p><p>that the internal forces between particles satisfy both the weak and the strong laws</p><p>of action and reaction. The argument may be generalized to a system with arbitrary</p><p>number of particles, thus proving the converse of the arguments leading to Eqs. (1.22)</p><p>and (1.26).</p><p>Solution: For individual particles, we have:</p><p>~̇p1 = ~F1</p><p>e</p><p>+ ~F21</p><p>~̇p2 = ~F2</p><p>e</p><p>+ ~F12</p><p>Adding the two equations above, we have:</p><p>~̇p1 + ~̇p2 = ~F1</p><p>e</p><p>+ ~F2</p><p>e</p><p>+ ~F21 + ~F12 (3.1)</p><p>The system need to satisfy</p><p>M</p><p>d2</p><p>dt2</p><p>~R =</p><p>∑</p><p>i</p><p>Fi</p><p>e</p><p>Thus</p><p>��M</p><p>d2</p><p>dt2</p><p>(</p><p>1</p><p>��M</p><p>∑</p><p>i</p><p>mi~ri</p><p>)</p><p>= ~F1</p><p>(e)</p><p>+ ~F2</p><p>(e)</p><p>d2</p><p>dt2</p><p>(m1~r1 +m2~r2) = ~F1</p><p>(e)</p><p>+ ~F2</p><p>(e)</p><p>d</p><p>dt</p><p>[</p><p>d</p><p>dt</p><p>(m1~r1 +m2~r2)</p><p>]</p><p>= ~F1</p><p>(e)</p><p>+ ~F2</p><p>(e)</p><p>7</p><p>8 CHAPTER 3. EXERCISE 3 - FRANCISCO A. F. ALMEIDA</p><p>d</p><p>dt</p><p>(</p><p>m1</p><p>d</p><p>dt</p><p>~r1 +m2</p><p>d</p><p>dt</p><p>~r2</p><p>)</p><p>= ~F1</p><p>(e)</p><p>+ ~F2</p><p>(e)</p><p>d</p><p>dt</p><p>(m1 ~v1 +m2 ~v2) = ~F1</p><p>(e)</p><p>+ ~F2</p><p>(e)</p><p>d</p><p>dt</p><p>(m1 ~v1) +</p><p>d</p><p>dt</p><p>(m2 ~v2) = ~F1</p><p>(e)</p><p>+ ~F2</p><p>(e)</p><p>d</p><p>dt</p><p>~p1 +</p><p>d</p><p>dt</p><p>~p2 = ~F1</p><p>(e)</p><p>+ ~F2</p><p>(e)</p><p>~p1 + ~p2 = ~F1</p><p>(e)</p><p>+ ~F2</p><p>(e)</p><p>(3.2)</p><p>Comparing 3.1 with 3.2, follow:</p><p>~F21 + ~F12 = 0</p><p>~F21 = − ~F12</p><p>weak law of action and reaction.</p><p>How the system satisfies (1.26), so:</p><p>d~L</p><p>dt</p><p>= ~N (e) +</p><p>∑</p><p>i</p><p>~Ni</p><p>d~L</p><p>dt</p><p>= ~N (e) +</p><p>(</p><p>~r1 × ~F21 + ~r2 × ~F12</p><p>)</p><p>How the system to satisfy d~L</p><p>dt</p><p>= ~N (e), so:</p><p>~r1 × ~F21 + ~r2 × ~F12 = 0</p><p>but</p><p>~F21 = − ~F12</p><p>so</p><p>~r1 × ~F21 + ~r2 ×−( ~F21) = 0</p><p>~r1 × ~F21 − ~r2 × ~F21 = 0</p><p>(~r1 − ~r2)× ~F21 = 0</p><p>Therefore ~F21 and ~F12 are on the line joining the particles. This is the strong law of</p><p>action and reaction.</p><p>Exercise 4 - Rodrigo Weber P.</p><p>The equations of constraint for the rolling disk, Eqs. (1.39), are special cases of</p><p>general linear differential</p><p>pθ = Lθ −</p><p>∑</p><p>i</p><p>n · ri ×∇viU</p><p>Which is what we would like to demonstrate. As for the electromagnetic force, writing</p><p>the Lorentz force in the cgs system:</p><p>F = q</p><p>(</p><p>−∇φ+</p><p>1</p><p>c</p><p>∇ (v ·A)− 1</p><p>c</p><p>dA</p><p>dt</p><p>)</p><p>9</p><p>So we can demonstrate that this force on the charge can be derived from the following</p><p>velocity-dependent potential:</p><p>U = q� �</p><p>q</p><p>c</p><p>(v � A ) (2.3)</p><p>From:</p><p>Qj =</p><p>d</p><p>dt</p><p>�</p><p>@U</p><p>@_qj</p><p>�</p><p>�</p><p>@U</p><p>@qj</p><p>(2.4)</p><p>We can see that:</p><p>dAi</p><p>dt</p><p>=</p><p>d</p><p>dt</p><p>�</p><p>@</p><p>@vi</p><p>(v � A )</p><p>�</p><p>Since the derivative will only take the i component of the product. And since the scalar</p><p>potential � , the chargeq and c are all speed independent, we can also write that:</p><p>�</p><p>dAi</p><p>dt</p><p>=</p><p>d</p><p>dt</p><p>�</p><p>@</p><p>@vi</p><p>(c� � v � A )</p><p>�</p><p>In this way then, replacing 2.3 in 2.4:</p><p>Fx =</p><p>d</p><p>dt</p><p>�</p><p>@U</p><p>@_x</p><p>�</p><p>�</p><p>@U</p><p>@x</p><p>=</p><p>d</p><p>dt</p><p>�</p><p>@</p><p>@_x</p><p>�</p><p>q� �</p><p>q</p><p>c</p><p>(v � A )</p><p>� �</p><p>�</p><p>@</p><p>@x</p><p>�</p><p>q� �</p><p>q</p><p>c</p><p>(v � A )</p><p>�</p><p>=</p><p>�</p><p>q</p><p>c</p><p>d</p><p>dt</p><p>�</p><p>@</p><p>@_x</p><p>(c� � v � A )</p><p>�</p><p>� q</p><p>@�</p><p>@x</p><p>+</p><p>q</p><p>c</p><p>@</p><p>@x</p><p>(v � A )</p><p>�</p><p>So:</p><p>Fx = q</p><p>�</p><p>�</p><p>@�</p><p>@x</p><p>+</p><p>1</p><p>c</p><p>@</p><p>@x</p><p>(v � A ) �</p><p>1</p><p>c</p><p>dAx</p><p>dt</p><p>�</p><p>As desired. So if we have a set of particles, we have:</p><p>r v i U = r v i</p><p>X</p><p>i</p><p>qi � i �</p><p>qi</p><p>c</p><p>(v i � A i )</p><p>!</p><p>=</p><p>qi</p><p>c</p><p>A i</p><p>Where f i = f (x i ). As we wanted to demonstrate, we can write the canonical moment</p><p>as:</p><p>p� = L � �</p><p>X</p><p>i</p><p>n � r i �</p><p>qi</p><p>c</p><p>A i</p><p>Exercise 3 - Geovane Naysing er</p><p>FIGURE 3.1</p><p>First, the distance between two points A and B in the space can be write as</p><p>SAB =</p><p>∫ B</p><p>A</p><p>ds (3.1)</p><p>where ds must be write as</p><p>ds =</p><p>√</p><p>dx2 + dy2 + dz2</p><p>and with a simple algebra, we have</p><p>ds =</p><p>√</p><p>1 +</p><p>(</p><p>dy</p><p>dx</p><p>)2</p><p>+</p><p>(</p><p>dz</p><p>dx</p><p>)2</p><p>dx.</p><p>For now, we can de�ne the function (functional!) f as</p><p>f ≡</p><p>√</p><p>1 +</p><p>(</p><p>dy</p><p>dx</p><p>)2</p><p>+</p><p>(</p><p>dz</p><p>dx</p><p>)2</p><p>. (3.2)</p><p>Then, we can rewrite (3.1) as</p><p>SAB =</p><p>∫ xB</p><p>xA</p><p>√</p><p>1 +</p><p>(</p><p>dy</p><p>dx</p><p>)2</p><p>+</p><p>(</p><p>dz</p><p>dx</p><p>)2</p><p>dx</p><p>11</p><p>12 CHAPTER 3. EXERCISE 3 - GEOVANE NAYSINGER</p><p>and applying the Hamilton principle δS = 0, we have the Euler equation</p><p>∂f</p><p>∂q</p><p>− d</p><p>dx</p><p>(</p><p>∂f</p><p>∂q̇</p><p>)</p><p>= 0, (3.3)</p><p>For now, we replace f in the Euler equation and do some calculating. First, we already</p><p>see that ∂f</p><p>∂q</p><p>= 0, so the only thing to do is the second term. Then,</p><p>d</p><p>dx</p><p>(</p><p>∂f</p><p>∂q̇</p><p>)</p><p>=</p><p>d</p><p>dx</p><p>(1 +</p><p>(</p><p>dy</p><p>dx</p><p>)2</p><p>+</p><p>(</p><p>dz</p><p>dx</p><p>)2</p><p>)− 1</p><p>2</p><p>dy</p><p>dx</p><p> = 0</p><p>(</p><p>1 +</p><p>(</p><p>dy</p><p>dx</p><p>)2</p><p>+</p><p>(</p><p>dz</p><p>dx</p><p>)2</p><p>)− 1</p><p>2</p><p>dy</p><p>dx</p><p>= a</p><p>and</p><p>d</p><p>dx</p><p>(1 +</p><p>(</p><p>dy</p><p>dx</p><p>)2</p><p>+</p><p>(</p><p>dz</p><p>dx</p><p>)2</p><p>)− 1</p><p>2</p><p>dz</p><p>dx</p><p> = 0</p><p>(</p><p>1 +</p><p>(</p><p>dy</p><p>dx</p><p>)2</p><p>+</p><p>(</p><p>dz</p><p>dx</p><p>)2</p><p>)− 1</p><p>2</p><p>dz</p><p>dx</p><p>= b</p><p>where a and b are constants. So, we do the ratio between these two equations as</p><p>(</p><p>1 +</p><p>(</p><p>dy</p><p>dx</p><p>)2</p><p>+</p><p>(</p><p>dz</p><p>dx</p><p>)2)− 1</p><p>2 dy</p><p>dx(</p><p>1 +</p><p>(</p><p>dy</p><p>dx</p><p>)2</p><p>+</p><p>(</p><p>dz</p><p>dx</p><p>)2)− 1</p><p>2 dz</p><p>dx</p><p>=</p><p>a</p><p>b</p><p>dy/dx</p><p>dz/dx</p><p>=</p><p>a</p><p>b</p><p>dy</p><p>dz</p><p>=</p><p>a</p><p>b</p><p>dy =</p><p>a</p><p>b</p><p>dz</p><p>integrating we can write</p><p>y =</p><p>a</p><p>b</p><p>z + c (3.4)</p><p>where c is a constant and this (3.4) is a equation of a straight line. We can do the same</p><p>procedure for x and rewrite other equations for x-y and x-z spaces for example.</p><p>Exercise 4 - Douglas Oliveira Novaes</p><p>Question: Show that the geodesics of a spherical surface are great circles, i.e.,</p><p>circles whose centers lie at the center of the sphere.</p><p>Response : An in�nitesimal displacement through Cartesian space is given by:</p><p>ds2 = dx2 + dy2 + dz2 (4.1)</p><p>where, s is the arc of displacement over the sphere, whose components in spherical</p><p>coordinates can be written as:</p><p>x = r sin θ cosφ</p><p>y = r sin θ sinφ</p><p>z = r cos θ</p><p>where, r is the sphere radius; a constant for any movement on the sphere.</p><p>13</p><p>14 CHAPTER 4. EXERCISE 4 - DOUGLAS OLIVEIRA NOVAES</p><p>Then, the components of the in�nitesimal displacement (4.1) are:</p><p>dx =</p><p>@x</p><p>@r</p><p>dr +</p><p>@x</p><p>@�</p><p>d� +</p><p>@x</p><p>@�</p><p>d�</p><p>dy =</p><p>@y</p><p>@r</p><p>dr +</p><p>@y</p><p>@�</p><p>d� +</p><p>@y</p><p>@�</p><p>d�</p><p>dz =</p><p>@z</p><p>@r</p><p>dr +</p><p>@z</p><p>@�</p><p>d� +</p><p>@z</p><p>@�</p><p>d�</p><p>where the �rst term of each equation is zero because the restriction of movement on</p><p>the sphere implies that the radius r of any path between two points has the same value</p><p>and is constant. The same for the last term of z-component because it does not vary in</p><p>� -direction.</p><p>Then,</p><p>dx = r cos� cos�d� � r sin� sin�d�</p><p>dy = r cos� sin�d� + r sin� cos�d�</p><p>dz = � r sin�d�</p><p>(4.2)</p><p>Replacing (4.2) in (4.1) come</p><p>ds2 = r 2 cos2 � cos2 � d� 2 � r 2 sin2 � sin2 �</p><p>| {z }</p><p>d� 2 �</p><p>( ( ( ( ( ( ( ( ( ( ( ( ( ( (</p><p>2r 2 cos� sin� cos� sin�d�d�</p><p>+ r 2 cos2 � sin2 � d� 2 + r 2 sin2 � cos2 �</p><p>| {z }</p><p>d� 2 +</p><p>( ( ( ( ( ( ( ( ( ( ( ( ( ( (</p><p>2r 2 cos� sin� cos� sin�d�d�</p><p>+ r 2 sin2 �d� 2</p><p>= r 2 cos2 �| {z } d� 2 + r 2 sin2 �d� 2 + r 2 sin2 �| {z } d� 2</p><p>Then,</p><p>ds = r</p><p>q</p><p>d� 2 + sin2 �d� 2</p><p>Without loss of generality, one can assign � = �= 2, id est, the two points lie on</p><p>the equator of the sphere. Hence,</p><p>ds = r</p><p>p</p><p>_� 2 + 1 d�</p><p>Another simpler way to get this expression is to attack it directly, by recognizing</p><p>that an in�nitesimal arc in any direction can be represented as</p><p>ds =</p><p>p</p><p>r 2d� 2 + r 2d� 2</p><p>Hence, factoring r and d� 2 one can write</p><p>ds = r</p><p>p</p><p>_� 2 + 1 d�</p><p>15</p><p>The total length of a curve going between angle φ1 and φ2 is</p><p>I =</p><p>∫ φ2</p><p>φ1</p><p>ds = r</p><p>∫ φ2</p><p>φ1</p><p>√</p><p>θ̇2 + 1 dφ where θ̇ =</p><p>dθ</p><p>dφ</p><p>To �nd the maximum circle, one have minimize this integral using the Lagrange</p><p>equation</p><p>∂f</p><p>∂θ</p><p>− d</p><p>dφ</p><p>∂f</p><p>∂θ̇</p><p>= 0</p><p>whose result of derivatives is</p><p>∂f</p><p>∂θ</p><p>= 0</p><p>∂f</p><p>∂θ̇</p><p>=</p><p>θ̇√</p><p>(θ̇2 + 1)</p><p>Then, replacing in Lagrange equation</p><p>d</p><p>dφ</p><p>θ̇√</p><p>(θ̇2 + 1)</p><p>= 0</p><p>then,</p><p>θ̇√</p><p>(θ̇2 + 1)</p><p>= a where a = constant</p><p>θ̇2 = a2(θ̇2 + 1)</p><p>θ̇2 − a2θ̇2 = a2</p><p>θ̇2(1− a2) = a2</p><p>θ̇2 =</p><p>a2</p><p>(1− a2)</p><p>dθ</p><p>dφ</p><p>=</p><p>a√</p><p>(1− a2)</p><p>= b = another constant</p><p>Integrating it</p><p>(θ2 − θ1) = b(φ2 − φ1)</p><p>16 CHAPTER 4. EXERCISE 4 - DOUGLAS OLIVEIRA NOVAES</p><p>Since b is the angular coef�cient between θ̂ and φ̂, and its value is constant, this</p><p>means that the path between the two points lies on a maximum circle on the sphere.</p><p>Another way to interpret this result is to consider θ1 = θ2. So,</p><p>0 = b(φ2 − φ1)</p><p>Since φ2 6= φ1, b is necessarily equal to zero, which result in a path that describes a</p><p>maximum circle on the sphere.</p><p>Exercise 5 - Jackson Galvªo</p><p>A particle is subjected to the potential V (x) = −F (x), where F is a constant. The</p><p>particle travels from x = 0 to x = a in time interval t0. Assume the motion of the</p><p>particle can be expressed in the form x(t) = A + Bt + Ct2. Found the values A,B and</p><p>C such that the action is a minimum.</p><p>First we obtain the derivative of x(t) = x, such that</p><p>ẋ = B + 2Ct (5.1)</p><p>ẍ = 2C (5.2)</p><p>The Lagragian’s equation of motion is</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ẋ</p><p>)</p><p>− ∂L</p><p>∂x</p><p>= 0 (5.3)</p><p>Where,</p><p>L = T − V (x)</p><p>17</p><p>18 CHAPTER 5. EXERCISE 5 - JACKSON GALVˆO</p><p>such that we write as,</p><p>L =</p><p>1</p><p>2</p><p>mẋ2 − Fx (5.4)</p><p>Therefore, we have</p><p>d</p><p>dt</p><p>[</p><p>∂</p><p>∂ẋ</p><p>(</p><p>1</p><p>2</p><p>mẋ2</p><p>)]</p><p>− ∂</p><p>∂x</p><p>(−Fx) = 0 (5.5)</p><p>then,</p><p>d</p><p>dt</p><p>[mẋ] + F = 0 (5.6)</p><p>What take us</p><p>ẍ =</p><p>F</p><p>m</p><p>(5.7)</p><p>In this way, we have C value to Eq. (5.2)</p><p>C =</p><p>F</p><p>2m</p><p>(5.8)</p><p>With this, we can rewrite the Eq. (5.1) how</p><p>ẋ = B +</p><p>F</p><p>m</p><p>t (5.9)</p><p>19</p><p>which leads us to</p><p>∫ a</p><p>0</p><p>dx =</p><p>∫ t0</p><p>0</p><p>(</p><p>B +</p><p>F</p><p>m</p><p>t</p><p>)</p><p>dt (5.10)</p><p>what result in</p><p>a = Bt0 +</p><p>F</p><p>2m</p><p>t20 (5.11)</p><p>Therefore, B have the value</p><p>B =</p><p>(</p><p>a</p><p>t0</p><p>− F</p><p>2m</p><p>t0</p><p>)</p><p>(5.12)</p><p>For the A value, our expected is that x(t = 0) = 0. So, we have</p><p>x(0) = A = 0 (5.13)</p><p>.</p><p>Thus, we obtain the values A,B and C in the respective equations (5.8), (5.12) and</p><p>(5.13), which leads us �nally form to x = x(t) by the Eq. (5.14)</p><p>x(t) =</p><p>(</p><p>a</p><p>t0</p><p>− F</p><p>2m</p><p>t0</p><p>)</p><p>t+</p><p>F</p><p>2m</p><p>t2 (5.14)</p><p>Exercise 6 - Francisco Augusto Fer-</p><p>reira Almeida</p><p>Find the Euler-Lagrange equation describing the brachistochrone curve for a</p><p>particle moving inside a spherical Earth of uniform density. Obtain a �rst integral</p><p>for this differential equation by analogy to the Jacobi integral h. With the help of this</p><p>integral, show that the desired curve is a hypocycloid (the curve described by a point</p><p>on a circle rolling on the inside of a larger circle). Obtain an expression for the time</p><p>of travel along the brachistochrone between two points on Earth’s surface. How long</p><p>would it take to go from New York to Los Angeles (assumed to be 4800 km apart on</p><p>the surface) along a brachistochrone tunnel (assuming no friction) and how far below</p><p>the surface would the deepest point of the tunnel be?</p><p>Solution</p><p>FIGURE 6.1</p><p>21</p><p>22 CHAPTER 6. EXERCISE 6 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>~r = r cos θ î+ r sin θ ĵ (6.1)</p><p>ds2 = (dr cos θ − r sin θ dθ)2 + (dr sin θ + r cos θ dθ)2 (6.2)</p><p>ds2 = (dr)2 cos2 θ + r2 sin2 θ(dθ)2 − 2r cos θ sin θdrdθ</p><p>+ (dr)2 sin2 θ + r2 cos2 θ(dθ)2 + 2r cos θ sin θdrdθ</p><p>ds2 = (dr)2 + r2(dθ)2 (6.3)</p><p>By energy conservation:</p><p>(1/2)mv2 +m(g/R)r2 = 0 +m(g/r)R2 (6.4)</p><p>v =</p><p>√</p><p>2(g/R)(R2 − r2) (6.5)</p><p>The least time is given by:</p><p>∆t =</p><p>∫ t2</p><p>t1</p><p>(1/v) ds (6.6)</p><p>∆t =</p><p>∫ t2</p><p>t1</p><p>√ (</p><p>dr</p><p>dθ</p><p>)2</p><p>+ r2</p><p>2(g/R)(R2 − r2)</p><p>dθ (6.7)</p><p>∆t =</p><p>∫ t2</p><p>t1</p><p>√</p><p>R</p><p>2g</p><p>√(</p><p>dr</p><p>dθ</p><p>)2</p><p>+ r2</p><p>R2 − r2</p><p>dθ (6.8)</p><p>Where</p><p>f =</p><p>√</p><p>R</p><p>2g</p><p>√(</p><p>dr</p><p>dθ</p><p>)2</p><p>+ r2</p><p>R2 − r2</p><p>(6.9)</p><p>23</p><p>since the potential derives from forces that depend only on position, we can</p><p>write:</p><p>d</p><p>dθ</p><p>(∑</p><p>j</p><p>∂L</p><p>∂q̇j</p><p>q̇j − L</p><p>)</p><p>+</p><p>∂L</p><p>∂θ</p><p>= 0 (6.10)</p><p>or</p><p>d</p><p>dθ</p><p>(</p><p>∂f</p><p>∂ṙ</p><p>· ṙ − f</p><p>)</p><p>+</p><p>∂f</p><p>∂θ</p><p>= 0 (6.11)</p><p>The �rst portion of the above sum is equal to zero (Jacobi integral), thus:</p><p>(</p><p>∂f</p><p>∂ṙ</p><p>· ṙ − f</p><p>)</p><p>= constant (6.12)</p><p>√</p><p>R</p><p>2g</p><p>· 1</p><p>2</p><p>[</p><p>2ṙ</p><p>(</p><p>√</p><p>R2 − r2)(</p><p>√</p><p>ṙ2 + r2)</p><p>]</p><p>ṙ −</p><p>√</p><p>R</p><p>2g</p><p>√</p><p>ṙ2 + r2</p><p>R2 − r2</p><p>= constant (6.13)</p><p>√</p><p>R</p><p>2g</p><p>[</p><p>ṙ2 − (ṙ2 + r2)</p><p>(</p><p>√</p><p>R2 − r2)(</p><p>√</p><p>ṙ2 + r2)</p><p>]</p><p>= constant (6.14)</p><p>[</p><p>ṙ2 − (ṙ2 + r2)</p><p>(</p><p>√</p><p>R2 − r2)(</p><p>√</p><p>ṙ2 + r2)</p><p>]</p><p>= constant (6.15)</p><p>[</p><p>r2</p><p>(</p><p>√</p><p>R2 − r2)(</p><p>√</p><p>ṙ2 + r2)</p><p>]</p><p>= constant (6.16)</p><p>calculating this constant for the minimum position r = ro of the particle, where</p><p>ṙ = 0, we have:</p><p>ro</p><p>2</p><p>(</p><p>√</p><p>R2 − ro2)(</p><p>√</p><p>ro2)</p><p>= constant (6.17)</p><p>ro</p><p>(</p><p>√</p><p>R2 − ro2)</p><p>= constant (6.18)</p><p>24 CHAPTER 6. EXERCISE 6 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>Replacing 6.18 in 6.16, we have:</p><p>[</p><p>r2</p><p>(</p><p>√</p><p>R2 − r2)(</p><p>√</p><p>ṙ2 + r2)</p><p>]</p><p>=</p><p>ro</p><p>(</p><p>√</p><p>R2 − ro2)</p><p>(6.19)</p><p>ṙ2 + r2 =</p><p>(R− ro2)r4</p><p>ro2(R2 − r2)</p><p>(6.20)</p><p>ṙ2 =</p><p>(R2 − ro2)r4 − r2ro2(R2 − r2)</p><p>ro2(R2 − r2)</p><p>(6.21)</p><p>ṙ2 =</p><p>R2r4 − ro2r4 − r2ro2R2 + r4ro</p><p>2</p><p>ro2(R2 − r2)</p><p>(6.22)</p><p>ṙ =</p><p>dr</p><p>dθ</p><p>= R</p><p>r</p><p>ro</p><p>√</p><p>r2 − ro2</p><p>R2 − r2</p><p>(6.23)</p><p>dθ =</p><p>1</p><p>R</p><p>ro</p><p>r</p><p>√</p><p>R2 − r2</p><p>r2 − ro2</p><p>dr (6.24)</p><p>∫ θ</p><p>0</p><p>dθ =</p><p>∫ r</p><p>ro</p><p>1</p><p>R</p><p>ro</p><p>r</p><p>√</p><p>R2 − r2</p><p>r2 − ro2</p><p>dr (6.25)</p><p>Rθ =</p><p>∫ r</p><p>ro</p><p>ro</p><p>r</p><p>√</p><p>R2 − r2</p><p>r2 − ro2</p><p>dr (6.26)</p><p>If u = ( r</p><p>2−ro2</p><p>R2−r2 )1/2, we have:</p><p>du =</p><p>1</p><p>2</p><p>1</p><p>u</p><p>[</p><p>2r</p><p>R2 − r2</p><p>+</p><p>(r2 − ro22r</p><p>(R2 − r2)2</p><p>]</p><p>dr (6.27)</p><p>du =</p><p>r</p><p>u</p><p>[</p><p>1</p><p>R2 − r2</p><p>+</p><p>u2</p><p>R2 − r2</p><p>]</p><p>dr (6.28)</p><p>25</p><p>Going back to (6.26), we have:</p><p>Rθ =</p><p>∫ u</p><p>0</p><p>ro</p><p>r</p><p>1</p><p>u</p><p>(</p><p>u</p><p>r</p><p>R2 − r2</p><p>1 + u2</p><p>du</p><p>)</p><p>(6.29)</p><p>Let’s do the following development:</p><p>R2 − r2</p><p>r2</p><p>=</p><p>(R2 − r2)(R2 − ro2)</p><p>r2(R2 − ro2)</p><p>(6.30)</p><p>R2 − r2</p><p>r2</p><p>=</p><p>(R2 − r2)(R2 − ro2)</p><p>r2R2 − r2ro2 +R2ro2 −R2ro2</p><p>(6.31)</p><p>R2 − r2</p><p>r2</p><p>=</p><p>(R2 − r2)(R2 − ro2)</p><p>ro2(R2 − r2) +R2(r2 − ro2)</p><p>(6.32)</p><p>R2 − r2</p><p>r2</p><p>=</p><p>(R2 − ro2)</p><p>ro2 +R2u2</p><p>(6.33)</p><p>R2 − r2</p><p>r2</p><p>=</p><p>1− (ro/R)2</p><p>(ro/R)2 + u2</p><p>(6.34)</p><p>Going back to (6.29), we have:</p><p>Rθ = ro</p><p>∫ u</p><p>0</p><p>1− (ro/R)2</p><p>[(ro/R)2 + u2](1 + u2)</p><p>du (6.35)</p><p>Rθ = ro</p><p>∫ u</p><p>0</p><p>[(</p><p>1</p><p>(ro/R)2 + u2</p><p>)</p><p>− 1</p><p>1 + u2</p><p>]</p><p>du (6.36)</p><p>Rθ =</p><p>ro</p><p>(ro/R)</p><p>tan−1</p><p>u</p><p>(ro/r)</p><p>− rotan−1u (6.37)</p><p>Rθ = R</p><p>[</p><p>tan−1</p><p>(</p><p>8u</p><p>ro</p><p>)</p><p>− tan−10</p><p>]</p><p>− ro(tan−1u− 0) (6.38)</p><p>26 CHAPTER 6. EXERCISE 6 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>θ = tan−1</p><p>R</p><p>ro</p><p>√</p><p>r2 − r2o</p><p>R2 − r2</p><p>− ro</p><p>R</p><p>tan−1</p><p>√</p><p>r2 − r2o</p><p>R2 − r2</p><p>(6.39)</p><p>If we do</p><p>√</p><p>r2−r2o</p><p>R2−r2 = tanφ</p><p>2</p><p>=</p><p>sinφ</p><p>2</p><p>cosφ</p><p>2</p><p>, we have:</p><p>(r2 − r2o)cos2</p><p>φ</p><p>2</p><p>= (R2 − r2)sin2φ</p><p>2</p><p>(6.40)</p><p>r2</p><p>(</p><p>cos2</p><p>φ</p><p>2</p><p>+ sin2φ</p><p>2</p><p>)</p><p>= R2sin2φ</p><p>2</p><p>+ r2ocos</p><p>2φ</p><p>2</p><p>(6.41)</p><p>r2 = R21</p><p>2</p><p>(1− cosφ) + r2o</p><p>1</p><p>2</p><p>(r + cosφ) (6.42)</p><p>r2 =</p><p>1</p><p>2</p><p>(R2 + r2o)−</p><p>1</p><p>2</p><p>(R2 − r2o)cosφ (6.43)</p><p>Besides that</p><p>θ = tan−1</p><p>(</p><p>R</p><p>ro</p><p>tan</p><p>φ</p><p>2</p><p>)</p><p>− ro</p><p>R</p><p>φ</p><p>2</p><p>(6.44)</p><p>The hypocloid is generated by a circle that rotates within a larger circle. By the</p><p>law of cosines, we have:</p><p>r2 = (R− α)2 + α2 − 2α(R− α)cosφ (6.45)</p><p>If α = 1</p><p>2</p><p>(R− ro)</p><p>So</p><p>R− α =</p><p>1</p><p>2</p><p>(R + ro) (6.46)</p><p>27</p><p>r2 =</p><p>1</p><p>4</p><p>(R2 + r2o − 2Rro) +</p><p>1</p><p>4</p><p>(R2 + r2o + 2Rro)−</p><p>1</p><p>2</p><p>(R2 − r2o)cosφ (6.47)</p><p>r2 =</p><p>1</p><p>2</p><p>(R2 + ro)−</p><p>1</p><p>2</p><p>(R2 − r2o)cosφ (6.48)</p><p>We can test the relationship between a and α and φ.</p><p>tan(θ − θo) =</p><p>αsinφ</p><p>(r − α)− αcosφ</p><p>(6.49)</p><p>tan(θ − θo) =</p><p>1</p><p>2</p><p>(R− ro)sinφ</p><p>1</p><p>2</p><p>(R + ro)− 1</p><p>2</p><p>(R− ro)cosφ</p><p>(6.50)</p><p>tan(θ − θo) =</p><p>(R− ro)sinφ</p><p>R(1− cosφ) + ro(1 + cosφ)</p><p>(6.51)</p><p>tan(θ − θo) =</p><p>(R− ro)2sin(φ/2)cos(φ/2)</p><p>R(2sin2(φ/2)) + ro(2cos2(φ/2))</p><p>(6.52)</p><p>tan(θ − θo) =</p><p>(R− ro)tan(φ/2)</p><p>Rtan2(φ/2) + ro</p><p>(6.53)</p><p>tan(θ − θo + φ/2) =</p><p>tan(θ − θo) + tan(φ/2)</p><p>1− tan(θ − θo)tan(φ/2)</p><p>(6.54)</p><p>tan(θ − θo + φ/2) =</p><p>R</p><p>ro</p><p>tan(φ/2) (6.55)</p><p>When the inner circle moves an arc Rθo over the larger circle, the point that</p><p>describes the hypocycloid has described an arc equivalent to Rθ</p><p>Rθ = αφ =</p><p>1</p><p>2</p><p>(R− ro)φ (6.56)</p><p>28 CHAPTER 6. EXERCISE 6 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>θo =</p><p>(</p><p>1− ro</p><p>R</p><p>)</p><p>(</p><p>φ</p><p>2</p><p>) (6.57)</p><p>θ = tan−1</p><p>(</p><p>R</p><p>ro</p><p>tan(φ/2)</p><p>)</p><p>− ro</p><p>R</p><p>φ</p><p>2</p><p>(6.58)</p><p>Which corresponds to the lei found before, this means that the trajectory effect</p><p>corresponds to a hypocycloid in which the radius of the inner circle is α = 1</p><p>2</p><p>(r − ro)</p><p>Finally</p><p>∆t =</p><p>√</p><p>R</p><p>2g</p><p>∫ θ</p><p>0</p><p>√</p><p>(dr/dθ)2 + r2</p><p>R2 − r2</p><p>dθ (6.59)</p><p>∆t =</p><p>√</p><p>R</p><p>2g</p><p>∫ r</p><p>ro</p><p>√</p><p>1 + r2(dθ/dr)2</p><p>R2 − r2</p><p>dr (6.60)</p><p>Where</p><p>dθ</p><p>dr</p><p>= R</p><p>r</p><p>ro</p><p>√</p><p>r2 − r2o</p><p>R2 − r2</p><p>(6.61)</p><p>dθ</p><p>dr</p><p>=</p><p>ro</p><p>Rr</p><p>√</p><p>R2 − r2</p><p>r2 − r2o</p><p>(6.62)</p><p>Thus</p><p>∆t =</p><p>√</p><p>R</p><p>2g</p><p>∫ r</p><p>ro</p><p>√</p><p>1 + r2 r2o</p><p>R2r2</p><p>R2−r2</p><p>r2−r2o</p><p>R2 − r2</p><p>dr (6.63)</p><p>∆t =</p><p>√</p><p>R2 − r2o</p><p>2gR</p><p>∫ r</p><p>ro</p><p>√</p><p>rdr√</p><p>(R2 − r2)(r2 − r2o)</p><p>(6.64)</p><p>29</p><p>Remembering that</p><p>tan(φ/2) =</p><p>√</p><p>r2 − r2o√</p><p>R2 − r2</p><p>(6.65)</p><p>Thus</p><p>∆t =</p><p>√</p><p>R2 − r2o</p><p>2gR</p><p>∫ φ</p><p>0</p><p>dφ</p><p>2</p><p>(6.66)</p><p>∆t =</p><p>√</p><p>R2 − r2o</p><p>2gR</p><p>φ</p><p>2</p><p>(6.67)</p><p>The total angle φ to travel from one side to the other should be 2π</p><p>∆t =</p><p>√</p><p>R2 − r2o</p><p>2gR</p><p>π (6.68)</p><p>∆t =</p><p>√</p><p>R</p><p>2g</p><p>√</p><p>1− (ro/R)2π (6.69)</p><p>On the other hand:</p><p>θ =</p><p>( s</p><p>2πR</p><p>)</p><p>2π (6.70)</p><p>It will be the angular separation between cities, where s is the separation over</p><p>land.</p><p>∆t =</p><p>√(</p><p>2− s</p><p>πR</p><p>)( s</p><p>2πg</p><p>)</p><p>π (6.71)</p><p>With S=4800km and R=6378 km</p><p>Exercise 7 - Lucas Bourscheidt</p><p>Solution: The function that gives the minimum surface of revolution is the cate-</p><p>nary:</p><p>x = a cosh</p><p>(</p><p>y − b</p><p>a</p><p>)</p><p>where a and b are integration constants determined by the values of x and y at the</p><p>endpoints of the catenary. For the symmetric case x2 = x1 we have:</p><p>cosh</p><p>(</p><p>y2 − b</p><p>a</p><p>)</p><p>= cosh</p><p>(</p><p>y1 − b</p><p>a</p><p>)</p><p>= cosh</p><p>(</p><p>−y2 − b</p><p>a</p><p>)</p><p>= cosh</p><p>(</p><p>y2 + b</p><p>a</p><p>)</p><p>From this, we conclude that b = 0. So:</p><p>x = a cosh</p><p>(y</p><p>a</p><p>)</p><p>Applying this result tho the point (x2, y2) we have:</p><p>x2</p><p>a</p><p>= cosh</p><p>(y2</p><p>a</p><p>)</p><p>= cosh</p><p>(</p><p>y2</p><p>x2</p><p>x2</p><p>a</p><p>)</p><p>31</p><p>32 CHAPTER 7. EXERCISE 7 - LUCAS BOURSCHEIDT</p><p>That is:</p><p>k = cosh (αk) (A)</p><p>Graphically, we found that for values of α smaller than α0 ≈ 0,66 the linear function</p><p>f(k) = k intercept the hyperbolic function g(k) = cosh (αk) twice, so in these cases two</p><p>values of k are possible. The situation with α = 0,5 (�gure 1) illustrates this.</p><p>Figure 1</p><p>For values of α greater than α0, f(k) do not intercept the function g(k) at all, so</p><p>no real value of k can be found. See the �gure 2 below, which corresponds to the case</p><p>with α = 0,8.</p><p>33</p><p>Figure 2</p><p>Finally, in �gure 3 the case with α = 0,66 ≈ α0 is presented. Note that now the</p><p>function f(k) is tangent to g(k), which means that only one value of k is possible.</p><p>34 CHAPTER 7. EXERCISE 7 - LUCAS BOURSCHEIDT</p><p>Figure 3</p><p>Finding the value of α0 iteratively</p><p>If α = α0, a single value of k (= k0) is possible. At this point the function f(k)</p><p>is tangent to g(k) and, therefore, it is legitimate to equate the derivatives of f(k0) and</p><p>g(k0). Thus, taking the derivative of equation (A) with respect to k, we �nd:</p><p>1</p><p>α0</p><p>= sinh (α0k0) (B)</p><p>Furthermore:</p><p>k0 = cosh (α0k0) (C)</p><p>Taking the square of (B) and (C) and subtracting the �rst result from the second:</p><p>k20 −</p><p>1</p><p>α2</p><p>0</p><p>= cosh2(α0k0)− sinh2(α0k0) = 1</p><p>Isolating α0 we �nd:</p><p>α0 =</p><p>1√</p><p>k20 − 1</p><p>(D)</p><p>Inserting (D) into (C) we �nd:</p><p>k0 = cosh</p><p>(</p><p>k0√</p><p>k20 − 1</p><p>)</p><p>(E)</p><p>35</p><p>When solved iteratively, equation (E) gives k0 ≈ 1,81 (for example, starting with k0 =</p><p>2, after 6 iterations k0 starts to oscillate between 1.809 and 1.811). With this result,</p><p>according to (D), we �nd α0 ≈ 0,66.</p><p>Exercise 8 - Guilherme Shoiti Yoshi-</p><p>matsu Giardini</p><p>The broken-segment solution described in the text (cf. p. 42), in which the area</p><p>of revolution is only that of the end circles of radius y1 and y2, respectively, is known as</p><p>the Goldschmidt solution. For the symmetric situation discussed in Exercise 7, obtain an</p><p>expression for the ratio of the area generated by the catenary solutions to that given by</p><p>the Goldschmidt solution. Your result should be a function only of the parameters k and</p><p>alpha.</p><p>Show that for suf�ciently large values of α at least one of the catenaries gives an</p><p>area below that of the Goldschmidt solution. On the other hand, show that if α = α0, the</p><p>Goldschmidt solution gives a lower area than the catenary.</p><p>37</p><p>38 CHAPTER 8. EXERCISE 8 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>For us to solve the enunciated exercise above, we �rst have to obtain a formula</p><p>for the area of an arbitrary surface of revolution.</p><p>We know that an area Acat can be obtained by the addition of small contribu-</p><p>tions of area being revolved around an axis, resulting in a double integral. But it can be</p><p>simpli�ed because the calculation in question involves a revolution (cylindrical sym-</p><p>metry). In other words, to �nd the area of revolution, given a height y, we can multiply</p><p>the circumference length with the small contributions of varying height.</p><p>dAcat = 2πx ds (8.1)</p><p>the equation above illustrates a small area contribution, where 2π x is the cir-</p><p>cumference length and ds =</p><p>√</p><p>dx2 + dy2 is the variation in the circumference radius</p><p>and height.</p><p>39</p><p>Then the total area becomes</p><p>Acat = 2π</p><p>∫ (x2,y2)</p><p>(x1,y1)</p><p>x</p><p>√</p><p>dx2 + dy2 .</p><p>We can further manipulate the equation knowing that</p><p>dy =</p><p>∂y</p><p>∂x</p><p>dx , (8.2)</p><p>such that</p><p>40 CHAPTER 8. EXERCISE 8 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>Acat = 2π</p><p>∫ (x2,y2)</p><p>(x1,y1)</p><p>x</p><p>√</p><p>dx2 + dy2</p><p>= 2π</p><p>∫ (x2,y2)</p><p>(x1,y1)</p><p>x</p><p>√</p><p>dx2 +</p><p>(</p><p>∂y</p><p>∂x</p><p>dx</p><p>)2</p><p>= 2π</p><p>∫ (x2,y2)</p><p>(x1,y1)</p><p>x</p><p>√</p><p>1 +</p><p>(</p><p>∂y</p><p>∂x</p><p>)2</p><p>dx</p><p>= 2π</p><p>∫ (x2,y2)</p><p>(x1,y1)</p><p>x</p><p>√</p><p>1 + ẏ2dx .</p><p>Now, if we want to minimize the surface of rotation, we assume that the path of</p><p>y(x) is arbitrary in a way that it minimizes the surface of rotation. We may say that</p><p>y(x, α) = y(0) + g(x)α (8.3)</p><p>where g(x1, y1) = g(x2, y2) = 0 is an arbitrary function that vanishes in the end-</p><p>points of our chosen path and α is a variable whose objective is to determine the path</p><p>in y that will result in an minimal valued f(y, ẏ, x) = 2π x</p><p>√</p><p>1 + ẏ2.</p><p>Now, we derive Acat with respect to α and �nd the minimum value of surface</p><p>possible</p><p>dAcat</p><p>dα</p><p>=</p><p>d</p><p>dx</p><p>∫ (x2,y2)</p><p>(x1,y1)</p><p>f(y, ẏ, x)dx</p><p>=</p><p>∫ (x2,y2)</p><p>(x1,y1)</p><p>∂f</p><p>∂y</p><p>∂y</p><p>∂α</p><p>+</p><p>∂f</p><p>∂ẏ</p><p>∂ẏ</p><p>∂α</p><p>dx = 0 .</p><p>Above, we equal the derivative to zero to �nd the extrema of the function</p><p>f(y, ẏ, x). To proceed, we rewrite ∂f</p><p>∂ẏ</p><p>∂ẏ</p><p>∂α</p><p>as</p><p>∫ (x2,y2)</p><p>(x1,y1)</p><p>∂f</p><p>∂ẏ</p><p>∂ẏ</p><p>∂α</p><p>=</p><p>∫ (x2,y2)</p><p>(x1,y1)</p><p>∂f</p><p>∂ẏ</p><p>∂2y</p><p>∂α∂x</p><p>and (8.4)</p><p>via the partial fractions method, rewrite it again as</p><p>41</p><p>∫ (x2,y2)</p><p>(x1,y1)</p><p>∂f</p><p>∂ẏ</p><p>∂2y</p><p>∂α∂x</p><p>=</p><p>∂f</p><p>∂ẏ�</p><p>�</p><p>�∂y</p><p>∂α</p><p>∣∣∣∣∣</p><p>(x2,y2)</p><p>(x1,y1)</p><p>−</p><p>∫ (x2,y2)</p><p>(x1,y1)</p><p>d</p><p>dx</p><p>(</p><p>∂f</p><p>∂ẏ</p><p>)</p><p>∂y</p><p>∂α</p><p>dx , (8.5)</p><p>where u = ∂f</p><p>∂ẏ</p><p>, du = d</p><p>dx</p><p>(</p><p>∂f</p><p>∂ẏ</p><p>)</p><p>dx and dv = ∂2y</p><p>∂x∂α</p><p>, v = ∂y</p><p>∂α</p><p>. Also it is important to</p><p>stress that ∂y</p><p>∂α</p><p>∣∣(x2,y2)</p><p>(x1,y1)</p><p>cancels because y does not depend on α at the extremities (x1, y1)</p><p>and (x2, y2).</p><p>Going back to the surface integral derivative, we obtain</p><p>dAcat</p><p>dα</p><p>=</p><p>∫ (x2,y2)</p><p>(x1,y1)</p><p>[</p><p>∂f</p><p>∂y</p><p>− d</p><p>dx</p><p>(</p><p>∂f</p><p>∂ẏ</p><p>)]</p><p>∂y</p><p>∂α</p><p>dx = 0 .</p><p>Now, because ∂y</p><p>∂α</p><p>= g(x) is a non zero arbitrary function, the only way that the</p><p>resulting integral above goes to zero is such that</p><p>∂f</p><p>∂y</p><p>− d</p><p>dx</p><p>(</p><p>∂f</p><p>∂ẏ</p><p>)</p><p>= 0 . (8.6)</p><p>Knowing that, we may �nd y as a function of x such that the surface of rotation</p><p>is minimal. We start by noting again that f(y, ẏ, t) = 2π x</p><p>√</p><p>1 + ẏ2, then</p><p>∂f</p><p>∂y</p><p>=</p><p>∂</p><p>∂y</p><p>(</p><p>2π x</p><p>√</p><p>1 + ẏ2</p><p>)</p><p>= 0 and (8.7)</p><p>∂f</p><p>∂ẏ</p><p>=</p><p>∂</p><p>∂ẏ</p><p>(</p><p>2π x</p><p>√</p><p>1 + ẏ2</p><p>)</p><p>= 2πx</p><p>2ẏ</p><p>2</p><p>√</p><p>1 + ẏ2</p><p>= 2π</p><p>xẏ√</p><p>1 + ẏ2</p><p>, (8.8)</p><p>now we derive the remaining term with respect to x</p><p>42 CHAPTER 8. EXERCISE 8 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>d</p><p>dx</p><p>�</p><p>@f</p><p>@_y</p><p>�</p><p>= 2�</p><p>d</p><p>dx</p><p>x _y</p><p>p</p><p>1 + _y2</p><p>!</p><p>.</p><p>Now we put (8.9) and (8.7) together to obtain</p><p>d</p><p>dx</p><p>x _y</p><p>p</p><p>1 + _y2</p><p>!</p><p>= 0 , (8.9)</p><p>this equation is rather simple, because it results in a zero derivative, then its</p><p>integral is any constant that we will name a.</p><p>x _y</p><p>p</p><p>1 + _y2</p><p>= a , (8.10)</p><p>here, the resulting equation to y becomes</p><p>_y =</p><p>a</p><p>p</p><p>x2 � a2</p><p>integrating ... (8.11)</p><p>y =</p><p>Z</p><p>a</p><p>p</p><p>x2 � a2</p><p>dx</p><p>=</p><p>Z</p><p>a</p><p>a</p><p>q</p><p>x2</p><p>a2 � 1</p><p>dx</p><p>=</p><p>Z</p><p>1</p><p>q</p><p>x2</p><p>a2 � 1</p><p>dx . (8.12)</p><p>Now we make a change of variables from x2</p><p>a2 to sec(� ) such that x</p><p>a = sec(� ) and</p><p>dx = a tan (� )</p><p>cos(� ) d� .</p><p>y =</p><p>Z</p><p>1</p><p>p</p><p>sec2(� ) � 1</p><p>dx</p><p>=</p><p>Z</p><p>a</p><p>p</p><p>sec2(� ) � 1</p><p>tan(� )</p><p>cos(� )</p><p>d� , (8.13)</p><p>we know that sec2(� ) � 1 = tan2(� ), due to this, we can write</p><p>43</p><p>y =</p><p>∫</p><p>a</p><p>tan(θ)</p><p>tan(θ)</p><p>cos(θ)</p><p>dθ</p><p>=</p><p>∫</p><p>a</p><p>cos(θ)</p><p>dθ</p><p>= a ln (tan(θ) + sec(θ)) + c</p><p>= a ln</p><p>(√</p><p>x2 − a2 +</p><p>x</p><p>a</p><p>)</p><p>+ c .</p><p>Now we use the identity that ln</p><p>(√</p><p>x2 − a2 + x</p><p>a</p><p>)</p><p>= arccosh</p><p>(</p><p>x</p><p>a</p><p>)</p><p>, rearranging, we</p><p>get</p><p>x = a cosh</p><p>(</p><p>y − c</p><p>a</p><p>)</p><p>. (8.14)</p><p>In exercise 2.7 (Herbert Goldstein 3rd Ed.), it is said that x1 = x2 and y2 = −y1 ></p><p>0, then we can infer that c in (8.14) is zero, because the curve is symmetrical with respect</p><p>to the y axis. Reescaling the variables according also to exercise 2.7, with k = x2</p><p>a</p><p>and</p><p>β = y2</p><p>x2</p><p>, we get that</p><p>k =</p><p>x2</p><p>a</p><p>= cosh</p><p>(</p><p>y2</p><p>a</p><p>x2</p><p>x2</p><p>)</p><p>= cosh</p><p>(</p><p>y2</p><p>x2</p><p>x2</p><p>a</p><p>)</p><p>(8.15)</p><p>= cosh (βk) . (8.16)</p><p>To �nd the minimum of the resulting transcendental function with respect to β,</p><p>we derive it with respect to k.</p><p>dk</p><p>dk</p><p>=</p><p>d</p><p>dk</p><p>(cosh (β k))</p><p>1 = β sinh(β k)</p><p>sinh(β k) =</p><p>1</p><p>β</p><p>, (8.17)</p><p>noting that cosh2(x)− sinh2(x) = 1, the equation becomes</p><p>44 CHAPTER 8. EXERCISE 8 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>cosh2(β k)− sinh2(β k) = 1</p><p>k2 − 1</p><p>β2</p><p>= 1 and (8.18)</p><p>isolating β, it becomes β0 where</p><p>β0 =</p><p>√</p><p>1</p><p>k2 − 1</p><p>. (8.19)</p><p>Solving the transcendental equation for</p><p>k0 = cosh(β0k0)</p><p>= cosh</p><p>(√</p><p>k20</p><p>k20 − 1</p><p>)</p><p>, (8.20)</p><p>we get an approximate value for k0 ≈ 1,81 and β0 ≈ 0,66.</p><p>Now we proceed to �nd the area of the catenary function. In the beginning of</p><p>this exercise, it was stated that</p><p>Acat = 2π</p><p>∫ (x2,y2)</p><p>(x2,y1)</p><p>x</p><p>√</p><p>1 + ẏ2dx . (8.21)</p><p>but we may also rewrite x as a function of y instead</p><p>Acat = 2π</p><p>∫ (x2,y2)</p><p>(x2,y1)</p><p>x</p><p>√</p><p>1 + ẏ2dx (8.22)</p><p>= 2π</p><p>∫ (x2,y2)</p><p>(x2,y1)</p><p>a cosh</p><p>(y</p><p>a</p><p>)√(∂x</p><p>∂y</p><p>)2</p><p>+ 1 dy , (8.23)</p><p>where</p><p>∂x</p><p>∂y</p><p>=</p><p>∂</p><p>∂y</p><p>(</p><p>a cosh</p><p>(y</p><p>a</p><p>))</p><p>= sinh</p><p>(y</p><p>a</p><p>)</p><p>, (8.24)</p><p>45</p><p>then</p><p>Acat = 2π</p><p>∫ (x2,y2)</p><p>(x2,y1)</p><p>a cosh</p><p>(y</p><p>a</p><p>)√</p><p>sinh2</p><p>(y</p><p>a</p><p>)</p><p>+ 1 dy</p><p>= 2π</p><p>∫ (x2,y2)</p><p>(x2,y1)</p><p>a cosh2</p><p>(y</p><p>a</p><p>)</p><p>dy . (8.25)</p><p>Now we write the hyperbolic cossine in its exponential form such that</p><p>cosh2</p><p>(y</p><p>a</p><p>)</p><p>=</p><p>1</p><p>4</p><p>[</p><p>e2y/a + e−2y/a + 1</p><p>]</p><p>=</p><p>1</p><p>2</p><p>[</p><p>cosh</p><p>(</p><p>2y</p><p>a</p><p>)</p><p>+</p><p>1</p><p>2</p><p>]</p><p>, (8.26)</p><p>this way</p><p>Acat = a π</p><p>∫ (x2,y2)</p><p>(x2,y1)</p><p>cosh</p><p>(</p><p>2y</p><p>a</p><p>)</p><p>+</p><p>1</p><p>2</p><p>dy</p><p>= a π</p><p>[</p><p>a</p><p>2</p><p>sinh</p><p>(</p><p>2y</p><p>a</p><p>)</p><p>+</p><p>y</p><p>2</p><p>]y2</p><p>y1</p><p>(8.27)</p><p>= a π</p><p>[</p><p>a</p><p>2</p><p>sinh</p><p>(</p><p>2y</p><p>a</p><p>)</p><p>+</p><p>y</p><p>2</p><p>]y2</p><p>−y2=y1</p><p>(8.28)</p><p>= π a2</p><p>[</p><p>sinh</p><p>(</p><p>2y2</p><p>a</p><p>)</p><p>+</p><p>y2</p><p>a</p><p>]</p><p>. (8.29)</p><p>Because the Goldschmidt solution is degenerated (p. 42 Herbert Goldstein 3rd</p><p>Ed.), its revolution surface is just the sum of the two circles with radii x1 and x2 re-</p><p>spectively. But as stated in exercise 2.7, x1 = x2, such that the area of revolution of the</p><p>Goldschmidt solution becomes</p><p>AGold = π</p><p>(</p><p>x21 + x22</p><p>)</p><p>= 2 πx22 . (8.30)</p><p>This way, the ratio between the areas of the catenary solution and Goldschmits’</p><p>one is</p><p>46 CHAPTER 8. EXERCISE 8 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>AGold</p><p>Acat</p><p>=</p><p>2π x22</p><p>π a2</p><p>(</p><p>sinh</p><p>(</p><p>2y2</p><p>a</p><p>)</p><p>+ y2</p><p>a</p><p>)</p><p>=</p><p>2 k2</p><p>sinh(2β k) + βk</p><p>. (8.31)</p><p>Now, because k varies with β, we must obtain the equation for k with respect</p><p>only to itself</p><p>k = cosh(β k)</p><p>β =</p><p>arccos(k)</p><p>k</p><p>, (8.32)</p><p>then, we can rewrite the equation above (assuming that k > 1) as</p><p>β =</p><p>ln(k +</p><p>√</p><p>k2 − 1)</p><p>k</p><p>, (8.33)</p><p>making the ratio of areas</p><p>AGold</p><p>Acat</p><p>=</p><p>2 k2</p><p>sinh(2 ln(k +</p><p>√</p><p>k2 − 1)) + ln(k +</p><p>√</p><p>k2 − 1)</p><p>=</p><p>4 k2</p><p>e2ln(k+</p><p>√</p><p>k2−1) − e−2ln(k+</p><p>√</p><p>k2−1) + 2 ln(k +</p><p>√</p><p>k2 − 1)</p><p>. (8.34)</p><p>For values of β >> 1 that implicates into k >> 1, we can ignore a series of terms</p><p>and assume an approximate value for the areas ratio</p><p>AGold</p><p>Acat</p><p>k>>1</p><p>≈ k2</p><p>eln(k+</p><p>√</p><p>k2−1) − e−ln(k+</p><p>√</p><p>k2−1) + ln(k +</p><p>√</p><p>k2 − 1)</p><p>k>>1</p><p>≈ k2</p><p>(k +</p><p>√</p><p>k2 − 1) + (k +</p><p>√</p><p>k2 − 1) + (k +</p><p>√</p><p>k2 − 1)</p><p>k>>1</p><p>≈ k2</p><p>6k</p><p>. (8.35)</p><p>It is easy to see that k2 > k, then</p><p>47</p><p>AGold</p><p>Acat</p><p>k>>1</p><p>> 1 , (8.36)</p><p>which implies that</p><p>AGold</p><p>k>>1</p><p>> Acat . (8.37)</p><p>On the other hand if β = β0 ≈ 0,66 and consequently k = k0 ≈ 1,81, we have</p><p>AGold</p><p>Acat</p><p>≈ 2 1.812</p><p>sinh (2 0.66 1.81) + 0.66 1.81</p><p>≈ 0.990 , (8.38)</p><p>which implies</p><p>that Acat</p><p>β=β0</p><p>> AGold. Which ends our proof (show that for β >> 1,</p><p>AGold > Acat and for β = β0, Acat > AGold).</p><p>Exercise 9 - Rodrigo Weber Pereira</p><p>A chain or rope of inde�nite length passes freely over pulleys at height y1 and y2</p><p>above the plane surface of Earth, with a horizontal distance x2−x1 between them. If the</p><p>chain or rope has a uniform linear mass density, show that the problem of �nding the</p><p>curve assumed between the pulleys is identical with that of the problem of minimum</p><p>surface of revolution. (The transition to the Goldschmidt solution as the heights y1 and</p><p>y2 are changed makes for a striking lecture demonstration. See Exercise 8.)</p><p>The linear mass density is constant, so ρ = dm</p><p>ds</p><p>= m</p><p>l</p><p>, where m is the total mass</p><p>of the rope and l is it’s length, supposedly constant. The differential element of length</p><p>can be written ass:</p><p>ds2 = dx2 + dy2 ⇒ ds =</p><p>√</p><p>1 + y′2dx ; y′ =</p><p>dy</p><p>dx</p><p>(9.1)</p><p>From mass density equation, we get:</p><p>dm</p><p>ρ</p><p>= ds⇒ l =</p><p>1</p><p>ρ</p><p>∫ m</p><p>0</p><p>dm =</p><p>∫ x2</p><p>x1</p><p>√</p><p>1 + y′2dx⇒ l =</p><p>∫ x2</p><p>x1</p><p>√</p><p>1 + y′2dx (9.2)</p><p>So, we get the following equation of constraint:</p><p>l =</p><p>∫ x2</p><p>x1</p><p>√</p><p>1 + y′2dx (9.3)</p><p>We now use a basic principle of analytical dynamics: a system will always tend to be</p><p>in the position of least (total) potential energy. For the rope, the total potential energy</p><p>can be calculated as follows:</p><p>V =</p><p>N∑</p><p>i=1</p><p>migyi =</p><p>∫ x2</p><p>x1</p><p>gydm (9.4)</p><p>Using dm = ρds, and ds =</p><p>√</p><p>1 + y′2dx, we get:</p><p>V = gρ</p><p>∫ x2</p><p>x1</p><p>y</p><p>√</p><p>1 + y′2dx (9.5)</p><p>We want to minimize V taking into account the constraint l = cte. But to do that is</p><p>49</p><p>50 CHAPTER 9. EXERCISE 9 - RODRIGO WEBER PEREIRA</p><p>equivalent of �nding the minimum surface revolution of the curve formed by the rope</p><p>around the x-axis. To calculate the surface area, we just integrate the small area element</p><p>dA by multiplying the length element ds by the perimeter of a circle of radius y around</p><p>the x-axis (2πy):</p><p>S =</p><p>∫ x2</p><p>x1</p><p>dA =</p><p>∫ x2</p><p>x1</p><p>2πyds = 2π</p><p>∫ x2</p><p>x1</p><p>y</p><p>√</p><p>1 + y′2dx (9.6)</p><p>To minimize V and S we would take into account only the function witch is inside the</p><p>integral: y</p><p>√</p><p>1 + y′2, so both problems are equivalent, as we wished to show.</p><p>Exercise 10 - Edgard Kretschmann</p><p>We will use our reference frame with a y-axis pointing downwards By the</p><p>guessed equation of y(t) we can write the parameter a in terms of the parameter b.</p><p>y0 = a,t0 + b,t20 (10.1)</p><p>a =</p><p>y0</p><p>t0</p><p>− b,t0 (10.2)</p><p>Since we know the Lagrangean of this system we can write it in terms of t</p><p>L =</p><p>m</p><p>2</p><p>, ˙y(t) +mgy(t) (10.3)</p><p>L =</p><p>m</p><p>2</p><p>,(a+ 2bt)2 +mg(at+ bt2) (10.4)</p><p>Now, we have to extremize the integral S</p><p>S =</p><p>∫ t0</p><p>0</p><p>Ldt (10.5)</p><p>So</p><p>S =</p><p>∫ t0</p><p>0</p><p>m</p><p>2</p><p>,(a+ 2bt)2 +mg(at+ bt2)dt (10.6)</p><p>Computing the integral:</p><p>S =</p><p>m</p><p>2</p><p>(a2t0 + 2abt20 +</p><p>4b2</p><p>3</p><p>t30) +mg(</p><p>at20</p><p>2</p><p>+</p><p>bt30</p><p>3</p><p>) (10.7)</p><p>Now we have to �nd for which values of b, we extremize the integral S. (Remember</p><p>that we know how a depends on b)</p><p>dS</p><p>db</p><p>=</p><p>m</p><p>2</p><p>(2a</p><p>da</p><p>db</p><p>t0 + 2</p><p>da</p><p>db</p><p>bt20 + 2at0 +</p><p>8b</p><p>3</p><p>t30) +mg(</p><p>da</p><p>db</p><p>t20</p><p>2</p><p>+</p><p>t30</p><p>3</p><p>) (10.8)</p><p>From (2):</p><p>da</p><p>db</p><p>= −t0 (10.9)</p><p>51</p><p>52 CHAPTER 10. EXERCISE 10 - EDGARD KRETSCHMANN</p><p>And applying (2) and (9) on (8):</p><p>dS</p><p>db</p><p>=</p><p>m</p><p>2</p><p>(−2(</p><p>y0</p><p>t0</p><p>− b,t0)t20 − 2bt30 + 2(</p><p>y0</p><p>t0</p><p>− b,t0)t0 +</p><p>8b</p><p>3</p><p>t30) +mg(−t</p><p>3</p><p>0</p><p>2</p><p>+</p><p>t30</p><p>3</p><p>) (10.10)</p><p>This expression will be 0, when b=g/2</p><p>And then, applying this result on (2), using the given result</p><p>t0 =</p><p>√</p><p>2y0</p><p>g</p><p>(10.11)</p><p>We now have that a=0</p><p>Exercise 11 - Gustavo Cor tazzi Garcia</p><p>Kessler</p><p>53</p><p>Exercise 12 - Jenifer Andrade De Matos</p><p>The term generalized mechanics has come to designate a variety of classical me-</p><p>chanics in which the Lagrangian contains time derivatives of qi higher than the �rst.</p><p>problems for which x = f(x, ẋ, ẍ, t) have been referred to as "jerky" mechanics. Such</p><p>equations of motion have interesting applications in chaos theory (cf. Chapter 11). By</p><p>applying the methods of the calculus of variations, show that if there is a Lagrangian</p><p>of the form L(qi, q̇i, q̈i, t), and Hamilton principle holds with the zero variation of both</p><p>qi and q̇i at the end points, then the corresponding Euler-Lagrange equations are</p><p>d2</p><p>dt2</p><p>(</p><p>∂L</p><p>∂q̈i</p><p>)</p><p>− d</p><p>dt</p><p>(</p><p>∂L</p><p>∂q̇i</p><p>)</p><p>+</p><p>∂L</p><p>∂qi</p><p>= 0, i = 1, 2,,, n,</p><p>Apply this result to the Lagrangian</p><p>L = −m</p><p>2</p><p>qq̈ − k</p><p>2</p><p>q2,</p><p>Do you recognize the equations of motion?</p><p>Let´s apply the Hamilton principle to a Lagrangian of the form, L(qi, q̇i, q̈i, t):</p><p>I =</p><p>∫ t2</p><p>t1</p><p>L(qi, q̇i, q̈i, t)dt</p><p>55</p><p>56 CHAPTER 12. EXERCISE 12 - JENIFER ANDRADE DE MATOS</p><p>δI = δ</p><p>∫ t2</p><p>t1</p><p>L(qi, q̇i, q̈i, t)dt = 0</p><p>For I as a function of the parameter α, each coordinate can be written as:</p><p>q1(t, α) = q1(t, 0) + α,η1(t)</p><p>q2(t, α) = q2(t, 0) + α,η2(t)</p><p>...</p><p>Then the extreme solution will be</p><p>(q1(t, 0), q2(t, 0), etc)</p><p>Hence the variation of I can be written as:</p><p>δI =</p><p>∂I</p><p>∂α</p><p>dα =</p><p>∫ t2</p><p>t1</p><p>∑</p><p>i</p><p>(</p><p>∂L</p><p>∂qi</p><p>∂qi</p><p>∂α</p><p>dα +</p><p>∂L</p><p>∂q̇i</p><p>∂q̇i</p><p>∂α</p><p>dα +</p><p>∂L</p><p>∂q̈i</p><p>∂q̈i</p><p>∂α</p><p>dα</p><p>)</p><p>dt</p><p>Let´s do each part of the integral separately:</p><p>∫ t2</p><p>t1</p><p>∂L</p><p>∂q̇i</p><p>∂q̇i</p><p>∂α</p><p>dαdt (12.1)</p><p>dα</p><p>∫ t2</p><p>t1</p><p>∂L</p><p>∂q̇i</p><p>∂2qi</p><p>∂α∂t</p><p>dt (12.2)</p><p>Integrating by parts:</p><p>u =</p><p>∂L</p><p>∂q̇i</p><p>, dv =</p><p>∂2qi</p><p>∂α∂t</p><p>dt (12.3)</p><p>Then [</p><p>∂L</p><p>∂q̇i</p><p>∂qi</p><p>∂α</p><p>]t2</p><p>t1</p><p>−</p><p>∫ t2</p><p>t1</p><p>∂qi</p><p>∂α</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂q̇i</p><p>)</p><p>dt (12.4)</p><p>57</p><p>But, [</p><p>∂L</p><p>∂q̇i</p><p>∂qi</p><p>∂α</p><p>]t2</p><p>t1</p><p>= 0 (12.5)</p><p>Because all the curves pass by the extremes of the function</p><p>Now, let´s apply integration by parts to the other term:</p><p>∫ t2</p><p>t1</p><p>∂L</p><p>∂q̈i</p><p>∂q̈i</p><p>∂α</p><p>dt =</p><p>∫ t2</p><p>t1</p><p>∂q</p><p>∂α</p><p>d2</p><p>dt2</p><p>(</p><p>∂L</p><p>∂q̈i</p><p>)</p><p>dt (12.6)</p><p>Back to the complete integral:</p><p>δI =</p><p>∫ t2</p><p>t1</p><p>∑</p><p>i</p><p>(</p><p>∂L</p><p>∂qi</p><p>− d</p><p>dt</p><p>∂L</p><p>∂q̇i</p><p>+</p><p>d2</p><p>dt2</p><p>∂L</p><p>∂q̈i</p><p>)</p><p>∂q</p><p>∂α</p><p>dαdt = 0 (12.7)</p><p>Since, δ I = 0</p><p>Then we must have that:</p><p>∂L</p><p>∂qi</p><p>− d</p><p>dt</p><p>∂L</p><p>∂q̇i</p><p>+</p><p>d2</p><p>dt2</p><p>∂L</p><p>∂q̈i</p><p>= 0 for i = 1, 2, 3,,n (12.8)</p><p>Now, let´s apply this equation to the folowing Lagrangian:</p><p>L = −m</p><p>2</p><p>qq̈ − k</p><p>2</p><p>q2,</p><p>We have,</p><p>∂L</p><p>∂q</p><p>= −m</p><p>2</p><p>q̈ − kq (12.9)</p><p>∂L</p><p>∂q̇</p><p>= 0 (12.10)</p><p>∂L</p><p>∂q̈</p><p>= −m</p><p>2</p><p>q (12.11)</p><p>58 CHAPTER 12. EXERCISE 12 - JENIFER ANDRADE DE MATOS</p><p>d2</p><p>dt2</p><p>(</p><p>∂L</p><p>∂q̈</p><p>)</p><p>= −m</p><p>2</p><p>q̈ (12.12)</p><p>So we �nd that</p><p>−m</p><p>2</p><p>q̈ − kq − m</p><p>2</p><p>q̈ = 0 (12.13)</p><p>mq̈ + kq = 0 (12.14)</p><p>And this is the equation of a simple harmonic oscillator.</p><p>Exercise 13 - Jhordan Silveira De</p><p>Borba</p><p>A heavy particle is placed at the top of a vertical hoop. Calculate the reaction</p><p>of the hoop on the particle by means of the Lagrange’s undetermined multipliers and</p><p>Lagrange’s equations. Find the height at which the particle falls off.</p><p>Answer :</p><p>As we have the particle at the top of a hoop, we can describe the movement</p><p>bidimensionally, for that we can use coordinates as follows:</p><p>While under the hoop, we have the constraint that r = R is constant. then we</p><p>have</p><p>f = r −R = 0</p><p>59</p><p>60 CHAPTER 13. EXERCISE 13 - JHORDAN SILVEIRA DE BORBA</p><p>Or writing in the in�nitesimal form:</p><p>ardr + aθdθ = 0</p><p>Then we have ar = 1 and aθ = 0, this is evident since integrating</p><p>∫</p><p>dr = r + C = 0.</p><p>Euler Lagrange’s equation for r is then:</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ṙ</p><p>)</p><p>− ∂L</p><p>∂r</p><p>= Qr</p><p>And:</p><p>Qr = −λ1r</p><p>[</p><p>∂f1</p><p>∂r</p><p>]</p><p>= −λ1r</p><p>The speci�c form of the generalized force deserves a little more discussion, which I will</p><p>return to at the end. So −λ1 is the normal reaction force of the hoop on the particle.</p><p>To make it easier, let’s just denote Qr = −λ1r = λ. Using r = (r sin θ, r cos θ), we have</p><p>ṙ =</p><p>(</p><p>ṙ sin θ + r cos θθ̇, ṙ cos θ − r sin θθ̇</p><p>)</p><p>, and then:</p><p>ṙ2 =</p><p>(</p><p>ṙ sin θ + r cos θθ̇</p><p>)2</p><p>+</p><p>(</p><p>ṙ cos θ − r sin θθ̇</p><p>)2</p><p>=ṙ2</p><p>(</p><p>cos2 θ + sin2 θ</p><p>)</p><p>+ r2</p><p>(</p><p>cos2 θ + sin2 θ</p><p>)</p><p>θ̇2 + 2rṙ cos θ sin θθ̇ (1− 1)</p><p>=ṙ2 + r2θ̇2</p><p>So the kinetic energy is:</p><p>T =</p><p>1</p><p>2</p><p>mṙ2 =</p><p>1</p><p>2</p><p>m</p><p>(</p><p>ṙ2 + r2θ̇2</p><p>)</p><p>And the height is given by h = r cos θ. So the potential energy is:</p><p>U = mgh = mgr cos θ</p><p>Thus the Lagragreean is:</p><p>L = T − U =</p><p>1</p><p>2</p><p>m</p><p>(</p><p>ṙ2 + r2θ̇2</p><p>)</p><p>−mgr cos θ</p><p>Then:</p><p>∂L</p><p>∂ṙ</p><p>= mṙ,</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ṙ</p><p>)</p><p>= mr̈,</p><p>∂L</p><p>∂r</p><p>= mrθ̇2 −mg cos θ</p><p>Then:</p><p>λ =</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ṙ</p><p>)</p><p>− ∂L</p><p>∂r</p><p>= mr̈ −mrθ̇2 +mg cos θ</p><p>Since r = R is constant, we just have:</p><p>λ = −mRθ̇2 +mg cos θ (13.1)</p><p>From energy conservation, the initial potential energy must equal the total energy at</p><p>61</p><p>any given time, that is:</p><p>U0 = T + U</p><p>mgR =</p><p>1</p><p>2</p><p>mR2θ̇2 +mgR cos θ</p><p>(13.2)</p><p>Then:</p><p>mRθ̇2 = 2m (g − g cos θ) (13.3)</p><p>This same result can be</p><p>obtained by solving the Euler-Lagrange equation forθ:</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂θ̇</p><p>)</p><p>− ∂L</p><p>∂θ</p><p>= −λ1�</p><p>[</p><p>∂f1</p><p>∂θ</p><p>]</p><p>(13.4)</p><p>d</p><p>dt</p><p>(</p><p>mR2θ̇</p><p>)</p><p>− (mgR sin θ) = 0 (13.5)</p><p>mR2θ̈ = mgr sin θ (13.6)</p><p>And if we differentiate the equation 13.3:</p><p>d</p><p>dt</p><p>(</p><p>mRθ̇2</p><p>)</p><p>=</p><p>d</p><p>dt</p><p>(2m (g − g cos θ))</p><p>mR2θ̇θ̈ = −2m</p><p>(</p><p>−g sin θθ̇</p><p>)</p><p>mR2θ̈ = mgR sin θ</p><p>We retrieve the differential equation obtained in 13.6. Then replacing 13.3 in λ in the</p><p>equation 13.1, we have:</p><p>λ = −2m (g − g cos θ) +mg cos θ</p><p>So the force is:</p><p>λ = mg (3 cos θ − 2)</p><p>And the particle falls when λ = 0, so the angle θc is given by:</p><p>mg (3 cos θ − 2) = 0→ θc = arccos</p><p>(</p><p>2</p><p>3</p><p>)</p><p>≈ 48,19�</p><p>So the height we are looking for is:</p><p>H = R cos</p><p>(</p><p>arccos</p><p>(</p><p>2</p><p>3</p><p>))</p><p>=</p><p>2</p><p>3</p><p>R</p><p>The height at which the particle falls is then given by:</p><p>H =</p><p>2</p><p>3</p><p>R</p><p>Returning to the discussion of generalized force, as the main bibliography (and from</p><p>where this question was copied) Goldstein et al. (2002) was used. The generalized force</p><p>is given by equation 2.25 from the book:</p><p>Qk = −</p><p>m∑</p><p>� =1</p><p>{</p><p>λ�</p><p>[</p><p>∂f�</p><p>∂qk</p><p>− d</p><p>dt</p><p>(</p><p>∂f�</p><p>∂q̇k</p><p>)]</p><p>− dλ�</p><p>dt</p><p>∂f�</p><p>∂q̇k</p><p>}</p><p>(13.7)</p><p>62 CHAPTER 13. EXERCISE 13 - JHORDAN SILVEIRA DE BORBA</p><p>When the constraints do not depend on speed, it reduces to:</p><p>Qk = −</p><p>m∑</p><p>α=1</p><p>λα</p><p>∂fα</p><p>∂qk</p><p>(13.8)</p><p>As used. Or if it is semi-holonomic depending only on the speed it can be written</p><p>simply as:</p><p>Qk = −</p><p>m∑</p><p>α=1</p><p>µα</p><p>∂fα</p><p>∂q̇k</p><p>(13.9)</p><p>Being µα = −dλ�</p><p>dt</p><p>any function, as it is frequently found in the literature. Remember-</p><p>ing that we can only determine the magnitude of force of Qk needed to produce the</p><p>constraint, but the signal is an arbitrary choice, so we need to understand the physics</p><p>involved to determine directions.</p><p>However, this more generic form of the generalized force was not written, as it</p><p>was looking for the reference indicated by the book: J. Ray, Amer. J. Phys.34 (406-8),</p><p>1996 in addition to containing an error, as volume 34 is from 1966 (RAY, 1966b), an</p><p>erratum (RAY, 1966a) was found which states that the method originally proposed for</p><p>dealing with non-holonomic constraint is incorrect. Speci�cally the generalization of</p><p>Hamilton’s principle to non-holonomic systems that takes the following form:</p><p>δ</p><p>∫ t2</p><p>t1</p><p>(L+ λαfα) dt = 0 (13.10)</p><p>for constraints of the form fα (q, q̇, t) = 0, and results in the equation</p><p>d</p><p>dt</p><p>∂L</p><p>∂q̇k</p><p>− ∂L</p><p>∂qk</p><p>=</p><p>∂ (λαfa)</p><p>∂qk</p><p>− d</p><p>dt</p><p>∂ (λαfa)</p><p>∂q̇k</p><p>(13.11)</p><p>which gives rise to the generalized force used by the book, is an incorrect generaliza-</p><p>tion.</p><p>Exercise 14 - Douglas Oliveira Novaes</p><p>Question: A uniform hoop of mass m and radius r rolls without slipping on a</p><p>�xed cylinder of radius R as shown in the �gure. The only external force is that of</p><p>gravity. If the smaller cylinder starts rolling from rest on top of the bigger cylinder,</p><p>use the method of Lagrange multipliers to �nd the point at which the hoop falls off the</p><p>cylinder.</p><p>Response : The system can be represented according �gure below.</p><p>The point at which the hoop falls off the cylinder is the same which the contact</p><p>forces vanish. These forces are constraints forces.</p><p>63</p><p>64 CHAPTER 14. EXERCISE 14 - DOUGLAS OLIVEIRA NOVAES</p><p>The Lagrange equations with constraints are:</p><p>∂L</p><p>∂qj</p><p>− d</p><p>dt</p><p>∂L</p><p>∂q̇j</p><p>+</p><p>∑</p><p>k</p><p>λk(t)</p><p>∂fk</p><p>∂qj</p><p>= 0</p><p>Let r, R and ρ be the radii involving the problem, and, the relationship between</p><p>them as</p><p>ρ = r +R</p><p>The constraint equations are</p><p>f1 = ρ− r −R = 0</p><p>f2 = rφ̇− ρθ̇ = 0</p><p>f3 = f2 with λ3 = λ2</p><p>where φ and θ are the angles described on the hoop and static cylinder respectively. The</p><p>�rst constraint is holonomic, because it depends on the coordinates. The second, which</p><p>is the same as third, constraint is semi-holonomic, i.e., it depends on the velocities but</p><p>it can be integrated into a holonomic constraint.</p><p>The kinetic energy of hoop is</p><p>T =</p><p>1</p><p>2</p><p>mv2 +</p><p>1</p><p>2</p><p>Iω2</p><p>where I = mr2 is the moment of inertia of the hoop and ω = φ̇ is its angular frequency.</p><p>Then</p><p>T =</p><p>1</p><p>2</p><p>m(ρ̇2 + ρ2θ̇2) +</p><p>1</p><p>2</p><p>mr2φ̇2</p><p>and the potential energy is given by</p><p>U = mgy = mgρ cos θ</p><p>Thus, the Lagrangian is written as if there were no constraints like that</p><p>L =</p><p>1</p><p>2</p><p>m(ρ̇2 + ρ2θ̇2) +</p><p>1</p><p>2</p><p>mr2φ̇2 −mgρ cos θ</p><p>Since</p><p>∂f1</p><p>∂ρ</p><p>= 1</p><p>∂f2</p><p>∂θ</p><p>= −ρ ∂f3</p><p>∂φ</p><p>= r</p><p>65</p><p>Solving the derivatives of Lagrange equations for ρ and replacing them</p><p>∂L</p><p>∂ρ</p><p>−</p><p>�</p><p>�</p><p>�</p><p>�></p><p>0</p><p>d</p><p>dt</p><p>∂L</p><p>∂ρ̇</p><p>+ λ1</p><p>∂f1</p><p>∂ρ</p><p>= 0</p><p>mρθ̇2 −mg cos θ + λ1 = 0</p><p>λ1 = mg cos θ −mρθ̇2</p><p>• λ1 corresponds to the ρ-coordinate and is the normal force on the hoop. Thus, the</p><p>condition one is looking for is λ1 = 0. But it remains to know θ̇2.</p><p>Then, for θ</p><p>∂L</p><p>∂θ</p><p>− d</p><p>dt</p><p>∂L</p><p>∂θ̇</p><p>+ λ2</p><p>∂f2</p><p>∂θ</p><p>= 0</p><p>mg�ρ sin θ −����:0</p><p>2mρρ̇ −mρ�2 θ̈ − λ2�ρ = 0</p><p>λ2 = mg sin θ −mρθ̈</p><p>• λ2 corresponds to the rotation of the hoop on the cylinder (in either θ or φ), and is</p><p>thus the tangent friction that make the �no-skipping� condition possible.</p><p>Finally, for φ</p><p>�</p><p>�</p><p>��7</p><p>0</p><p>∂L</p><p>∂φ</p><p>− d</p><p>dt</p><p>∂L</p><p>∂φ̇</p><p>+ λ2</p><p>∂f2</p><p>∂φ</p><p>= 0</p><p>−mr�2φ̈+ λ2�r = 0</p><p>λ2 = mrφ̈</p><p>Matching the λ2 found,</p><p>��mrφ̈ = ��mg sin θ −��mρθ̈</p><p>Since, rφ̇− ρθ̇ = 0, the second derivative yields φ̈ = (ρ/r)θ̈. Then, solving to θ̈</p><p>�r</p><p>ρ</p><p>�r</p><p>θ̈ + ρθ̈ = g sin θ</p><p>θ̈ =</p><p>g sin θ</p><p>2ρ</p><p>Multiplying both sides by θ̇</p><p>θ̇</p><p>d</p><p>dt</p><p>θ̇ = θ̇</p><p>g sin θ</p><p>2ρ</p><p>d</p><p>dt</p><p>θ̇2</p><p>�2</p><p>=</p><p>gθ̇ sin θ</p><p>�2ρ</p><p>θ̇2 = −g</p><p>ρ</p><p>cos θ + c</p><p>66 CHAPTER 14. EXERCISE 14 - DOUGLAS OLIVEIRA NOVAES</p><p>For θ̇ = 0 and θ = 0 as initial conditions</p><p>0 = −g</p><p>ρ</p><p>+ c</p><p>c =</p><p>g</p><p>ρ</p><p>Then</p><p>θ̇2 =</p><p>g</p><p>ρ</p><p>(1− cos θ)</p><p>Rewriting λ1, and remembering that the hoop loses contact with the �xed cylin-</p><p>der when λ1 = 0, come</p><p>λ1 = mg cos θ −m�ρ</p><p>g</p><p>�ρ</p><p>(1− cos θ)</p><p>��mg cos θ = ��mg (1− cos θ)</p><p>2 cos θ = 1</p><p>Hence, the angle corresponding the point in which the hoop falls off the cylinder</p><p>is</p><p>θ = 60°</p><p>Exercise 15 - Jenifer Andrade De Matos</p><p>A form of the Wheatstone impedance bridge has, in addition to the usual four</p><p>resistances, an inductance in one arm and a capacitance in the opposite arm. Set up L</p><p>and F for the unbalanced bridge, with the charges in the elements as coordinates. Using</p><p>the Kirchhoff junction conditions as constraints on the currents, obtain the Lagrange</p><p>equations of motion, and show that eliminating the λ´s reduces these to the usual</p><p>network equations.</p><p>According to the problem, we want to �nd the equations of the following circuit:</p><p>FIGURE 15.1. A Wheatstone impedance bridge with extra inductance and capacitance</p><p>67</p><p>68 CHAPTER 15. EXERCISE 15 - JENIFER ANDRADE DE MATOS</p><p>Let´s write the Lagrangian to this RLC circuit:</p><p>L = T − V (15.1)</p><p>L =</p><p>1</p><p>2</p><p>L1Q̇1</p><p>2 − Q2</p><p>2</p><p>2C</p><p>+ ε0Q0 (15.2)</p><p>This circuit dissipates energy, so let´s write a dissipation function:</p><p>F =</p><p>1</p><p>2</p><p>(R1 +R4) Q̇2</p><p>1 +</p><p>1</p><p>2</p><p>(R2 +R3) Q̇2</p><p>2 (15.3)</p><p>In this circuit there is also a constraint that determines that the coordinates Q0,</p><p>Q1 and Q2 aren´t completely independent:</p><p>Q̇0 − Q̇1 − Q̇2 = 0 (15.4)</p><p>So we must use the Lagrange Multipliers and we have:</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂q̇k</p><p>)</p><p>− ∂L</p><p>∂qk</p><p>+</p><p>∂F</p><p>∂q̇k</p><p>= Qk (15.5)</p><p>For Q0</p><p>∂L</p><p>∂Q̇0</p><p>= 0,</p><p>∂L</p><p>∂Q0</p><p>= ε0,</p><p>∂F</p><p>∂Q̇0</p><p>= 0 (15.6)</p><p>For Q1</p><p>∂L</p><p>∂Q̇1</p><p>= L1Q̇1,</p><p>d</p><p>dt</p><p>∂L</p><p>∂Q̇1</p><p>= L1Q̈1,</p><p>∂L</p><p>∂Q1</p><p>= 0,</p><p>∂F</p><p>∂Q̇1</p><p>= (R1 +R4) Q̇1 (15.7)</p><p>69</p><p>For Q2</p><p>∂L</p><p>∂Q̇2</p><p>= 0,</p><p>∂L</p><p>∂Q2</p><p>= −Q2</p><p>C</p><p>,</p><p>∂F</p><p>∂Q̇2</p><p>= (R2 +R3) Q̇2 (15.8)</p><p>So we can write the equation for, Q0, Q1 and Q2</p><p>−ε0 = λ1a10 (15.9)</p><p>L1Q̈1 + (R1 +R4) Q̇1 = λ1a11 (15.10)</p><p>Q2</p><p>C</p><p>+ (R2 +R3) Q̇2 = λ1a12 (15.11)</p><p>From (15.4), we can write that:</p><p>a10 = 1 a11 = −1, a12 = −1 (15.12)</p><p>So we can rewrite (15.9, 15.10 and 15.11)</p><p>−ε0 = λ1 (15.13)</p><p>L1Q̈1 + (R1 +R4) Q̇1 = −λ1 (15.14)</p><p>Q2</p><p>C</p><p>+ (R2 +R3) Q̇2 = −λ1 (15.15)</p><p>If we add (15.13) and (15.14), we �nd:</p><p>−ε0 + L1Q̈1 + (R1 +R4) Q̇1 = 0 (15.16)</p><p>70 CHAPTER 15. EXERCISE 15 - JENIFER ANDRADE DE MATOS</p><p>And this is the external network equation, we could �nd form the Kirchhoff</p><p>laws.</p><p>If we add (15.13) and (15.15), we �nd</p><p>−ε0 +</p><p>Q2</p><p>C</p><p>+ (R2 +R3) Q̇2 = 0 (15.17)</p><p>And this is the other external network equation, we could �nd form the Kirch-</p><p>hoff laws.</p><p>If we subtract (15.14) and (15.15), we �nd</p><p>L1Q̈1 + (R1 +R4) Q̇1 −</p><p>Q2</p><p>C</p><p>− (R2 +R3) Q̇2 = 0 (15.18)</p><p>And this is the internal network equation.</p><p>Exercise 16 - Geovane Naysing er</p><p>FIGURE 16.1</p><p>We will just applying the Euler-Lagrange equation and do some especi�cations.</p><p>The Euler-Lagrange equation is</p><p>∂L</p><p>∂q</p><p>− d</p><p>dt</p><p>(</p><p>∂L</p><p>∂q̇</p><p>)</p><p>= 0, (16.1)</p><p>and using the lagrangian of the problem, we can write</p><p>71</p><p>72 CHAPTER 16. EXERCISE 16 - GEOVANE NAYSINGER</p><p>∂L</p><p>∂q</p><p>= −kqeγt</p><p>∂L</p><p>∂q̇</p><p>= eγtmq̇</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂q̇</p><p>)</p><p>= γeγtmq̇ + eγtmq̈</p><p>and putting in (16.1) and doing some simple algebra, we have</p><p>q̈ + γq̇ +</p><p>k</p><p>m</p><p>q = 0 (16.2)</p><p>and it’s a equation of a damped harmonic oscillator. The cases of overdamped, criti-</p><p>cally damped and underdamped is well known. To analyse the constants of motion,</p><p>let see</p><p>P =</p><p>∂L</p><p>∂q̇</p><p>= mq̇eγt</p><p>dP</p><p>dt</p><p>= mγq̇eγt +mq̈eγt 6= 0</p><p>and</p><p>E =</p><p>(</p><p>1</p><p>2</p><p>mq̇2 +</p><p>k</p><p>2</p><p>q2</p><p>)</p><p>eγt</p><p>dE</p><p>dt</p><p>= γeγt</p><p>(</p><p>mq̇2</p><p>2</p><p>+</p><p>kq2</p><p>2</p><p>)</p><p>6= 0.</p><p>Consequently, there is no constants of motion.</p><p>For now, we’ll do the transformation of variable as said in the question. So</p><p>s = eγtq (16.3)</p><p>and we can rewrite as</p><p>q = e−γts</p><p>and the time derivative of q</p><p>q̇ = −γe−γts+ e−γtṡ.</p><p>Then, substituting q and q̇ in the lagrangian of the problem, we have</p><p>73</p><p>L = eγt</p><p>[</p><p>m</p><p>2</p><p>(</p><p>−γe−γts+ e−γtṡ</p><p>)2 − k</p><p>2</p><p>(</p><p>e−γts</p><p>)2]</p><p>with some algebra, we can write the lagrangian as</p><p>L = e−γt</p><p>[</p><p>mṡ2</p><p>2</p><p>−mγsṡ+</p><p>(</p><p>mγ2</p><p>2</p><p>− k</p><p>2</p><p>)</p><p>s2</p><p>]</p><p>(16.4)</p><p>Now, we can write equation of motion using the Euler-Lagrange equation. Lets do it.</p><p>Write the derivatives</p><p>∂L</p><p>∂s</p><p>=</p><p>[</p><p>−mγṡ+ (mγ2 − k)s</p><p>]</p><p>e−γt</p><p>∂L</p><p>∂ṡ</p><p>= [mṡ−mγs] e−γt</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ṡ</p><p>)</p><p>= −γe−γt[mṡ−mγs] + e−γt[ms̈−mγṡ]</p><p>and substituting in (16.1), we obtain the equation of motion</p><p>s̈− γṡ+</p><p>k</p><p>m</p><p>s = 0. (16.5)</p><p>This equation don’t say anything about constant of motion. Thus,</p><p>s = e</p><p>γ</p><p>2</p><p>tq (16.6)</p><p>is a better transformation and we’ll see why. So, substituting</p><p>q = e−</p><p>γ</p><p>2</p><p>ts</p><p>in the lagrangian given by the problem, we have a new lagrangian in term of s</p><p>L =</p><p>1</p><p>2</p><p>m</p><p>(</p><p>ṡ2 +</p><p>γ2s2</p><p>4</p><p>− γsṡ</p><p>4</p><p>)</p><p>− 1</p><p>2</p><p>ks2. (16.7)</p><p>Using the Euler-Lagrange equation, with the algebra</p><p>d</p><p>dt</p><p>[</p><p>mṡ− 1</p><p>4</p><p>γms</p><p>]</p><p>−mγ</p><p>4</p><p>s+</p><p>1</p><p>4</p><p>mγṡ+ ks = 0</p><p>ms̈− 1</p><p>4</p><p>γmṡ−mγ2</p><p>4</p><p>s+</p><p>1</p><p>4</p><p>mγṡ+ ks = 0,</p><p>then, we have</p><p>s̈+</p><p>[</p><p>k</p><p>m</p><p>−</p><p>(γ</p><p>2</p><p>)2]</p><p>s = 0 (16.8)</p><p>which is the harmonic oscillator equation with a effective frequency ω =</p><p>√</p><p>k</p><p>m</p><p>−</p><p>(</p><p>γ</p><p>2</p><p>)2.</p><p>As we already know, in this case, the energy is a constant of motion. To see it, just write</p><p>74 CHAPTER 16. EXERCISE 16 - GEOVANE NAYSINGER</p><p>the solution s = Acos(ωt)+Bsin(ωt) and put in the equation of energyE = 1</p><p>2</p><p>mṡ2+ 1</p><p>2</p><p>ks2</p><p>and take the time derivative.</p><p>16.1 Rascunho Questao 11 cap 8</p><p>This solution is based on the book Problems and Solution on Mechanics of</p><p>QiangYuan-qi, Gu En-pu, Cheng Jia-fu, ti Ze-hua,Ylang De-tian</p><p>a) We call v as the incident speed, re�ected back with a speed v′ and V is the</p><p>wall speed towards the particle. Then, consider a collision of the particle with one wall</p><p>and, as the collision is inelastic, the kinectic energy is conserved and the relative speed</p><p>before and after collision is equal. So, we can write</p><p>v + V = v′ − V</p><p>that is</p><p>v′ = v + 2V .</p><p>After each collision, the intensity of the particle momentum gains an amount 2mV</p><p>and m is the mass of the particle. When the walls are at a distance x apart, as V is</p><p>much smaller than the speed of the particle, the time interval between two consecutive</p><p>collisions is</p><p>T =</p><p>x</p><p>p/m</p><p>=</p><p>xm</p><p>p</p><p>p is the particle linear momentum. The change momentum in time dt is</p><p>dp = 2mV</p><p>dt</p><p>T</p><p>=</p><p>2V pdt</p><p>x</p><p>.</p><p>The movement of the wall with speed V can be write as</p><p>x = x0 − 2V t,</p><p>then</p><p>dp = −pdx</p><p>x</p><p>the integration of it, with p = p0 and x = x0, is</p><p>p =</p><p>p0x0</p><p>x0 − 2V t</p><p>.</p><p>16.1. RASCUNHO QUESTAO 11 CAP 8 75</p><p>b) The momentum acquired by the particle with one collision is</p><p>p+ 2mV − (−p) = 2p+ 2mV</p><p>and the interval with two consecutive collisions with the wall is</p><p>T ′ =</p><p>2xm</p><p>p</p><p>so the change of momentum is</p><p>dp = 2(p+mV )</p><p>dt</p><p>T ′</p><p>.</p><p>Then,</p><p>dp</p><p>dt</p><p>=</p><p>(p+mV )p</p><p>xm</p><p>≈ p2</p><p>xm</p><p>=</p><p>p20x</p><p>2</p><p>0</p><p>mx3</p><p>due to p</p><p>m</p><p>>> V . This is the force exerted by the wall on the particle. To keep the walls</p><p>moving at constant speed, a force of the same magnitude must be apply to each wall.</p><p>The problem can also be solved using the Hamiltonian formalism. The hamiltonian is</p><p>H =</p><p>1</p><p>2</p><p>m</p><p>( p</p><p>m</p><p>+ V</p><p>)2</p><p>and</p><p>H =</p><p>1</p><p>2m</p><p>(p+mV )2 ≈ p2</p><p>2m</p><p>=</p><p>p20x</p><p>2</p><p>0</p><p>2mx2</p><p>.</p><p>The force applied on the particle is ṗ which is given by the hamiltonian equation</p><p>ṗ = −∂H</p><p>∂x</p><p>=</p><p>p20x</p><p>2</p><p>0</p><p>mx3</p><p>.</p><p>16.1.1 Rascunho Questao 24 cap 8</p><p>16.1.2 Rascunho Questao 34 cap 8</p><p>Exercise 17 - Gustavo Cor tazzi Garcia</p><p>Kessler</p><p>77</p><p>Exercise 18 - Rodrigo Weber Pereira</p><p>A point mass is constrained to move on a massless hoop of radius a �xed in</p><p>a vertical plane that rotates about its vertical symmetry axis with constant angular</p><p>speed ω. Obtain the Lagrange equations of motion assuming the only external forces</p><p>arise from gravity. What are the constants of motion? Show that if ω is greather than a</p><p>critical value ω0, there can be a solution in which the particle remains stationary on the</p><p>hoop at a point other than at the bottom, but that if ω < ω0, the only stationary point</p><p>for the particle is at the bottom of the hoop. What is the value of ω0?</p><p>The angular velocity is constant, so ω = φ̇ = cte, so φ = φ0 + ωt. The potential</p><p>energy of the point mass is:</p><p>V = mgz = mga cos θ (18.1)</p><p>The kinetic energy of the point mass is:</p><p>T =</p><p>ma2</p><p>2</p><p>(ω2 sin2 θ + θ̇2) (18.2)</p><p>So the Lagrangian can be written as:</p><p>L =</p><p>ma2</p><p>2</p><p>(ω2 sin2 θ + θ̇2)−mga cos θ (18.3)</p><p>From the Lagrangian, we obtain the following equation of motion:</p><p>θ̈ − sin θ(cos θ − g</p><p>a</p><p>) = 0 (18.4)</p><p>To identify the constants of motion, we note that the Lagrangian is time independent.</p><p>Also, all the forces that do not come from constraint are derivable from potentials.</p><p>79</p><p>80 CHAPTER 18. EXERCISE 18 - RODRIGO WEBER PEREIRA</p><p>From this we get:</p><p>dh</p><p>dt</p><p>+</p><p>∂L</p><p>∂t︸︷︷︸</p><p>=0</p><p>=</p><p>∑</p><p>j</p><p>Qj︸︷︷︸</p><p>=0</p><p>q̇j ⇒</p><p>dh</p><p>dt</p><p>= 0⇒ h = C, (18.5)</p><p>Where C is a constant. Now, from the de�nition of h:</p><p>h =</p><p>∑</p><p>j</p><p>∂L</p><p>∂q̇j</p><p>q̇j − L (18.6)</p><p>We may write:</p><p>ma2</p><p>2</p><p>(θ̇2 − ω2 sin2 θ) +mga cos θ = C (18.7)</p><p>A boundary condition of this problem is that, for t = 0, θ = π and θ̇ = 0. From this we</p><p>�nd the value of C to be C = −mga. So, the least equation is:</p><p>a2</p><p>2</p><p>(θ̇2 − ω2 sin2 θ) + ga(cos θ − 1) = 0 (18.8)</p><p>The stationary solution (that is, a solution in which the point mass does not move) can</p><p>be obtained setting θ̇ = 0 in the equation, witch simpli�es to:</p><p>cos2</p><p>θ</p><p>2</p><p>(sin2 θ</p><p>2</p><p>− g</p><p>aω2</p><p>) = 0, (18.9)</p><p>Where we used the following trigonometric identities: cos2 θ</p><p>2</p><p>= 1</p><p>2</p><p>(1 + cos θ) and sin θ =</p><p>2 sin θ</p><p>2</p><p>cos θ</p><p>2</p><p>. The last equation is true only if</p><p>θ = π (18.10)</p><p>or</p><p>θ = 2 sin−1(</p><p>ω0</p><p>ω</p><p>) ; ω0 =</p><p>√</p><p>g</p><p>a</p><p>(18.11)</p><p>The second solution is valid only if ω ≥ ω0, as the argument of the inverse sine</p><p>function can take only values between [−1; 1]. So, if ω < ω0, the only solution is θ = π</p><p>(the bottom of the hoop).</p><p>Exercise 19 - Jackson Galvªo</p><p>VIEW SLIDE 8 - AULA 3.</p><p>A particle moves without friction in a conservative �led of produced by various</p><p>mass distributions. In each instance, the force generated by a volume element of the</p><p>distribution is derived from a potential that is proportional to the mass of the volume</p><p>element and is a function only of the scalar distance from the volume element. For the</p><p>following �xed, homogeneous mass distributions, state the conserved quantities in</p><p>the motion of the particle:</p><p>a) The mass is uniformly distributed in the plane z = 0.</p><p>The force and the potential don’t depend of x and y. So, the Lagrangian is in-</p><p>variant under translation in x and y (x,y - cyclic). Thus, the linear momenta, px and py</p><p>are conserved. However, the force and potential don’t change with a rotation around</p><p>z− axis. The setup is symetric under rotation. So we have the conservation of angular</p><p>momentum Lz, such that</p><p>81</p><p>82 CHAPTER 19. EXERCISE 19 - JACKSON GALVˆO</p><p>L̇x = 0 → px</p><p>L̇y = 0 → py</p><p>}</p><p>L̇z 6= 0</p><p>b) The mass is uniformly distributed in the plane z = 0, y > 0.</p><p>The setup has a symmetry in x direction - translational symmetry. Te conserva-</p><p>tion,</p><p>L̇x = 0 → px</p><p>c) The mass is uniformly distributed in a circular cylinder of �nite length, with axis</p><p>along the z axis.</p><p>The setup has translational arsymmetry in z direction. The setup has a rotational sym-</p><p>metry arround</p><p>z direction. The conservation has,</p><p>Lz, pz</p><p>d) The mass is uniformly distributed in a circular cylinder of in�nite length, with axis</p><p>83</p><p>along the z axis.</p><p>The setup has only a rotational symmetry around z direction. The conservation has,</p><p>Lz</p><p>e) The mass is uniformly distributed in a right cylinder of elliptical cross sections and</p><p>in�nite length, with axis along the z axis.</p><p>The setup has only a translational symmetry in z direction. The conservations has,</p><p>pz</p><p>f) The mass is uniformly distributed in a dumbbell whose axis is oriented along the z</p><p>axis.</p><p>The setup has only a rotational symmetry around z axis. The conservations has,</p><p>Lz</p><p>84 CHAPTER 19. EXERCISE 19 - JACKSON GALVˆO</p><p>g) The mass is in the form a uniform wire wound in the geometry of an in�nite helical</p><p>solenoid, with axis along the z axis.</p><p>The setup does not have any translational or rotational symmetry. But a linear combi-</p><p>nation can be conserved combined preserce the Lagragian. (tranlation by a+ rotatinal</p><p>by L). Being a</p><p>apz + Lz</p><p>Exercise 20 - Lucas Bourscheidt</p><p>Solution: The following �gure illustrates the situation described in the problem:</p><p>85</p><p>86 CHAPTER 20. EXERCISE 20 - LUCAS BOURSCHEIDT</p><p>Coordinates: x, y, and s completely specify the con�guration of the system.</p><p>A. Constraints</p><p>x− s = l cos α</p><p>y = l sin α</p><p>Multiplying the �rst equation by sin α and the second by cos α:</p><p>(x− s) sin α = l sin α cos α</p><p>y cos α = l sin α cos α</p><p>Eliminating l:</p><p>f(x, y, s) = y cos α− (x− s) sin α = 0</p><p>Thus, the number of degrees of freedom is 3− 1 = 2.</p><p>B. Lagrangian function, equations of motion, and forces of constraint</p><p>L =</p><p>1</p><p>2</p><p>m(ẋ2 + ẏ2) +</p><p>1</p><p>2</p><p>Mṡ2 −mgy</p><p>87</p><p>For a generalized coordinate qk:</p><p>d</p><p>dt</p><p>∂L</p><p>∂q̇k</p><p>− ∂L</p><p>∂qk</p><p>= −λ ∂f</p><p>∂qk</p><p>= Qk</p><p>For x:</p><p>d</p><p>dt</p><p>mẋ = −λ(−sin α)</p><p>mẍ = λsin α = Qx (A1)</p><p>ẍ =</p><p>λ</p><p>m</p><p>sin α (A2)</p><p>For y:</p><p>d</p><p>dt</p><p>mẏ +mg = −λcos α</p><p>mÿ +mg = −λcos α = Qy (B1)</p><p>ÿ = −λcos α +mg</p><p>m</p><p>(B2)</p><p>For s:</p><p>d</p><p>dt</p><p>Mṡ = −λsin α</p><p>Ms̈ = −λsin α = Qs (C1)</p><p>s̈ = − λ</p><p>M</p><p>sin α (C2)</p><p>Taking the second time derivative of the constraint relation:</p><p>ÿ cos α− (ẍ− s̈) sin α = 0 (D)</p><p>Inserting (A2), (B2), and (C2) in (D):</p><p>−λ cos α +mg</p><p>m</p><p>cos α−</p><p>(</p><p>λ</p><p>m</p><p>sin α +</p><p>λ</p><p>M</p><p>sin α</p><p>)</p><p>sin α = 0</p><p>λ cos2 α +mg cosα + λ</p><p>(</p><p>1 +</p><p>m</p><p>M</p><p>)</p><p>sin2 α = 0</p><p>λ+mg cosα + λ</p><p>m</p><p>M</p><p>sin2 α = 0</p><p>Isolating λ:</p><p>λ = − mg cosα</p><p>1 + m</p><p>M</p><p>sin2α</p><p>(E)</p><p>Returning with (E) in (A1), (B1), and (C1) we found for the forces of constraint:</p><p>88 CHAPTER 20. EXERCISE 20 - LUCAS BOURSCHEIDT</p><p>For Qx:</p><p>Qx = −mg sin α cos α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>(F1)</p><p>For Qy:</p><p>Qy =</p><p>mg cos2α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>(F2)</p><p>For Qs:</p><p>Qs =</p><p>mg sin α cos α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>(F3)</p><p>Inserting equations (F) into equations of motion (A1), (B1), and (C1) we �nd:</p><p>For ẍ:</p><p>ẍ = −g sin α cos α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>(G1)</p><p>For ÿ:</p><p>ÿ =</p><p>g cos2α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>− g (G2)</p><p>For s̈:</p><p>s̈ =</p><p>m</p><p>M</p><p>g sin α cos α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>(G3)</p><p>C. Work done by the forces of constraint</p><p>On the particle:</p><p>Wparticle =</p><p>∫ t</p><p>0</p><p>Q p · d r = Qx(x− x0) +Qy(y − y0)</p><p>= −mg sin α cos α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>(</p><p>ẋ0 +</p><p>1</p><p>2</p><p>ẍ t2</p><p>)</p><p>+</p><p>mg cos2α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>(</p><p>ẏ0 +</p><p>1</p><p>2</p><p>ÿ t2</p><p>)</p><p>89</p><p>Assuming the initial velocities are null and using the results (G):</p><p>Wparticle =</p><p>mg2 t2</p><p>2</p><p>[</p><p>sin α cos α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>sin α cos α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>+</p><p>cos2α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>(</p><p>cos2α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>− 1</p><p>)]</p><p>=</p><p>mg2 t2</p><p>2</p><p>[</p><p>sin2α cos2α(</p><p>1 + m</p><p>M</p><p>sin2α</p><p>)2 +</p><p>cos2α(</p><p>1 + m</p><p>M</p><p>sin2α</p><p>)2 (cos2α− 1− m</p><p>M</p><p>sin2α</p><p>)]</p><p>=</p><p>mg2 t2</p><p>2</p><p>[</p><p>sin2α cos2α(</p><p>1 + m</p><p>M</p><p>sin2α</p><p>)2 +</p><p>cos2α(</p><p>1 + m</p><p>M</p><p>sin2α</p><p>)2 (−sin2α− m</p><p>M</p><p>sin2α</p><p>)]</p><p>=</p><p>mg2 t2</p><p>2</p><p>[</p><p>sin2α cos2α(</p><p>1 + m</p><p>M</p><p>sin2α</p><p>)2 − sin2α cos2α(</p><p>1 + m</p><p>M</p><p>sin2α</p><p>)2 − m</p><p>M</p><p>sin2α cos2α(</p><p>1 + m</p><p>M</p><p>sin2α</p><p>)2</p><p>]</p><p>= −m</p><p>2g2 t2</p><p>2M</p><p>sin2α cos2α(</p><p>1 + m</p><p>M</p><p>sin2α</p><p>)2</p><p>On the wedge:</p><p>Wwedge =</p><p>∫ t</p><p>0</p><p>Q w · d r = Qs(s− s0)</p><p>=</p><p>mg sin α cos α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>(</p><p>ṡ0 +</p><p>1</p><p>2</p><p>s̈ t2</p><p>)</p><p>Assuming ṡ0 = 0 and using the result (G3):</p><p>Wwedge =</p><p>t2</p><p>2</p><p>mg sin α cos α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>(</p><p>m</p><p>M</p><p>g sin α cos α</p><p>1 + m</p><p>M</p><p>sin2α</p><p>)</p><p>=</p><p>m2g2 t2</p><p>2M</p><p>sin2α cos2α(</p><p>1 + m</p><p>M</p><p>sin2α</p><p>)2</p><p>D. Constants of motion</p><p>· Linear momentum in the horizontal direction</p><p>From (A1) and (C1) we have Qs = −Qx, and adding these equations:</p><p>mẍ+Ms̈ = 0</p><p>d</p><p>dt</p><p>(mẋ+Mṡ) = 0</p><p>Alternatively, using the constraint relation to write the Lagrangian in terms of x</p><p>90 CHAPTER 20. EXERCISE 20 - LUCAS BOURSCHEIDT</p><p>and y:</p><p>L =</p><p>1</p><p>2</p><p>m(ẋ2 + ẏ2) +</p><p>1</p><p>2</p><p>M</p><p>(</p><p>ẋ− ẏ</p><p>tg α</p><p>)2</p><p>−mgy</p><p>ẋ− ẏ</p><p>tg α</p><p>= ṡ</p><p>x is a cyclic coordinate and the corresponding equation of motion becomes:</p><p>d</p><p>dt</p><p>[</p><p>mẋ+M</p><p>(</p><p>ẋ− ẏ</p><p>tg α</p><p>)]</p><p>= 0</p><p>that is</p><p>d</p><p>dt</p><p>(mẋ+Mṡ) = 0</p><p>· Energy</p><p>E =</p><p>1</p><p>2</p><p>m(ẋ2 + ẏ2) +</p><p>1</p><p>2</p><p>Mṡ2 +mgy</p><p>The energy value can be found from the solutions of the equations of motion:</p><p>E =</p><p>1</p><p>2</p><p>m</p><p>[</p><p>(ẋ0 + ẍt)2 + (ẏ0 + ÿt)2</p><p>]</p><p>+</p><p>1</p><p>2</p><p>M(ṡ0 + s̈t)2 +mg</p><p>(</p><p>y0 + ẏ0t+</p><p>1</p><p>2</p><p>ÿt2</p><p>)</p><p>Grouping the terms according to the power of t:</p><p>E = E0 + β t+ γ</p><p>t2</p><p>2</p><p>Where:</p><p>E0 =</p><p>1</p><p>2</p><p>m(ẋ0</p><p>2 + ẏ0</p><p>2) +</p><p>1</p><p>2</p><p>Mṡ0</p><p>2 +mgy0</p><p>β = mẋ0ẍ+mẏ0ÿ +Mṡ0s̈+mgẏ0</p><p>γ = mẍ2 +mÿ2 +Ms̈2 +mgÿ</p><p>Now using the expressions (A2), (B2), and (C2) for the accelerations we �nd,</p><p>after some algebra:</p><p>β = λ [(ẋ0 − ṡ0) sin α− ẏ0 cos α] = 0</p><p>γ =</p><p>λ</p><p>m</p><p>[</p><p>λ</p><p>(</p><p>1 +</p><p>m</p><p>M</p><p>sin2α</p><p>)</p><p>+mg cos α</p><p>]</p><p>= 0</p><p>β is null due to the constraint relation and γ because of the expression of the</p><p>91</p><p>Lagrange multiplier λ (formula (E)). So we found:</p><p>E = E0 =</p><p>1</p><p>2</p><p>m(ẋ0</p><p>2 + ẏ0</p><p>2) +</p><p>1</p><p>2</p><p>Mṡ0</p><p>2 +mgy0</p><p>E. Comparison with a �xed wedge system</p><p>Coordinates: x and y completely specify the con�guration of the system.</p><p>Constraints (with s = 0)</p><p>f(x, y) = y cos α− x sin α = 0</p><p>Thus, the number of degrees of freedom is 2− 1 = 1.</p><p>Lagrangian function, equations of motion, and forces of constraint</p><p>L =</p><p>1</p><p>2</p><p>m(ẋ2 + ẏ2)−mgy</p><p>For x:</p><p>mẍ = λsin α = Qx (H)</p><p>For y:</p><p>mÿ +mg = −λcos α = Qy (I)</p><p>Multiplying the constraint relation by m and taking the second time derivative:</p><p>mÿ cos α−mẍ sin α = 0 (J)</p><p>Inserting (H) and (I) in (J) and isolating λ:</p><p>λ = −mg cosα (K)</p><p>Returning with (K) in (H) and (I) we found for the forces of constraint:</p><p>For Qx:</p><p>Qx = −mg sin α cos α (L)</p><p>92 CHAPTER 20. EXERCISE 20 - LUCAS BOURSCHEIDT</p><p>For Qy:</p><p>Qy = mg cos2α (M)</p><p>These forces of constraint precisely correspond to the horizontal and vertical</p><p>components of the normal force. They are obtained from the previous case, with the</p><p>wedge free to move, in the limit m/M → 0.</p><p>Work done by the constraint forces on the particle (assuming the initial veloc-</p><p>ities are null):</p><p>Wparticle =</p><p>∫ t</p><p>0</p><p>Q p · dr = Qx(x− x0) +Qy(y − y0)</p><p>= −mg sin α cos α</p><p>(</p><p>1</p><p>2</p><p>ẍ t2</p><p>)</p><p>+mg cos2α</p><p>(</p><p>1</p><p>2</p><p>ÿ t2</p><p>)</p><p>Using the results H - M :</p><p>Wparticle =</p><p>mg2 t2</p><p>2</p><p>[</p><p>sin2α cos2α + cos2α(cos2α− 1)</p><p>]</p><p>= 0</p><p>Constants of motion</p><p>· Linear momentum in the horizontal direction</p><p>Using the constraint relation to write the Lagrangian in terms of x:</p><p>L =</p><p>1</p><p>2</p><p>m(1 + tg2α)ẋ2 −mgx tg α</p><p>As x is NOT a cyclic coordinate, px is not a constant of motion. The correspond-</p><p>ing equation of motion becomes:</p><p>d</p><p>dt</p><p>[</p><p>m(1 + tg2α)ẋ</p><p>]</p><p>+mg tg α = 0</p><p>that is</p><p>d</p><p>dt</p><p>(mẋ) = −mg sin α cos α</p><p>· Energy</p><p>E =</p><p>1</p><p>2</p><p>m(ẋ2 + ẏ2) +mgy</p><p>93</p><p>The energy value can be found from the solutions of the equations of motion,</p><p>and is determined by the initial conditions:</p><p>E = E0 =</p><p>1</p><p>2</p><p>m(ẋ0</p><p>2 + ẏ0</p><p>2) +mgy0</p><p>Exercise 21 - Edgard Kretschmann</p><p>The potential energy of the given problem is easily written as</p><p>V (R, r) =</p><p>K(R−R0)</p><p>2</p><p>2</p><p>+</p><p>kr2</p><p>2</p><p>(21.1)</p><p>We can obtain the transformations between the coordinates (R,r) to (x,y) and vice-versa</p><p>by the equations: [</p><p>x</p><p>y</p><p>]</p><p>=</p><p>[</p><p>cosθ −sinθ</p><p>sinθ cosθ</p><p>] [</p><p>R</p><p>r</p><p>]</p><p>(21.2)</p><p>[</p><p>R</p><p>r</p><p>]</p><p>=</p><p>[</p><p>cosθ sinθ</p><p>−sinθ cosθ</p><p>] [</p><p>x</p><p>y</p><p>]</p><p>(21.3)</p><p>So, for the lab reference frame (x,y) our Lagrangian becomes:</p><p>L(x, y, ẋ, ẏ, t) =</p><p>m(ẋ2 + ẏ2)</p><p>2</p><p>−K(xcos(ωt) + ysin(ωt)−R0)</p><p>2</p><p>2</p><p>− k(−xsin(ωt) + ycos(ωt)2</p><p>2</p><p>(21.4)</p><p>We now realize that the Lagrangian depends explicitly of t, so the energy is not</p><p>conserved.</p><p>dh</p><p>dt</p><p>= −∂L</p><p>∂t</p><p>(21.5)</p><p>Since the kinetic energy is quadratic independent of ẋ and ẏ, then the energy function</p><p>is the mechanical energy of the system. Using the Jacobi’s Integral we have</p><p>h =</p><p>∑</p><p>q̇j</p><p>∂L</p><p>∂q̇j</p><p>− L (21.6)</p><p>h = ẋ</p><p>∂L</p><p>∂ẋ</p><p>+ ẏ</p><p>∂L</p><p>∂ẏ</p><p>− (T − V ) (21.7)</p><p>h = 2T − (T − V ) = T + V = E (21.8)</p><p>95</p><p>96</p><p>CHAPTER 21. EXERCISE 21 - EDGARD KRETSCHMANN</p><p>Analysing now to a rotating frame with a angular velocity ω. Differentiating (2)[</p><p>ẋ</p><p>ẏ</p><p>]</p><p>= ω</p><p>[</p><p>−sinθ −cosθ</p><p>cosθ −sinθ</p><p>] [</p><p>R</p><p>r</p><p>]</p><p>+</p><p>[</p><p>cosθ −sinθ</p><p>sinθ cosθ</p><p>] [</p><p>Ṙ</p><p>ṙ</p><p>]</p><p>(21.9)</p><p>We can now write the Lagrangian in the rotating frame</p><p>L(R, r, Ṙ, ṙ, t) =</p><p>m</p><p>2</p><p>(Ṙ2 + ṙ2 +ω2(R2 + r2) + 2ω(ṙR− Ṙr))− K(R−R0)</p><p>2</p><p>2</p><p>+</p><p>kr2</p><p>2</p><p>(21.10)</p><p>We can notice that this Lagrangian depends implicitly of t. So the energy function will</p><p>be conserved. But the energy function isn’t the mechanical energy of the system, once</p><p>the kinetic energy has linear terms.</p><p>T (R, r, Ṙ, ṙ, t) =</p><p>m</p><p>2</p><p>(Ṙ2 + ṙ2 + ω2(R2 + r2) + 2ω(ṙR− Ṙr)) (21.11)</p><p>The Jacobi’s Integral will result that</p><p>h(rot) = Ṙ</p><p>∂L</p><p>∂Ṙ</p><p>+ ṙ</p><p>∂L</p><p>∂ṙ</p><p>− (T − V ) (21.12)</p><p>h(rot) = Ṙ</p><p>m</p><p>2</p><p>(2Ṙ− 2ωr) + ṙ</p><p>m</p><p>2</p><p>(2ṙ + 2ωR)− (T − V ) (21.13)</p><p>h(rot) = m(Ṙ2 + ṙ2 + ω(ṙR− Ṙr))− (T − V ) (21.14)</p><p>Rewriting equation (?) we can �nd that</p><p>2T −mω2(R2 + r2)−mω(ṙR− Ṙr) = m(Ṙ2 + ṙ2 + ω(ṙR− Ṙr)) (21.15)</p><p>h(rot) = T + V −mω2(R2 + r2)−mω(ṙR− Ṙr)) (21.16)</p><p>We can see with this relation that the mechanical energy is not conserved. But it</p><p>can be conserved if ω = 0, then we will have the same Jacobi’s integral for both frames.</p><p>h(rot) = h = E.</p><p>Exercise 22 - Jefferson Santana Mar-</p><p>tins</p><p>Suppose a particle moves in space subject to a conservative potential V (r) but</p><p>is constrained to always move on a surface whose equation is σ(r, t) = 0 (Note the</p><p>explicit time-dependence!). The instantaneous force of constraint is perpendicular to</p><p>the surface. Show analytically that the energy of the particle is not conserved if the</p><p>surface moves in time.</p><p>R. In general, the conservation of the total mechanical energy of the system (de-</p><p>�ned asE = T +U ) needs to be established by looking explicitly at the total differential</p><p>of the energy with respect to time, i.e. dE</p><p>dt</p><p>. Under special circumstances, i.e. when the</p><p>relation between Cartesian and generalized coordinates does not depend on time and</p><p>the active forces are conservative, it is true that dE</p><p>dt</p><p>= −∂L</p><p>∂t</p><p>and the conservation of</p><p>energy can be established by simply looking at the explicit time dependence/indepen-</p><p>dence of the Lagrangian. In the case of the problem, since the equation of the constraint</p><p>is time-dependent (σ(r, t) = 0), the relation between Cartesian and generalized coordi-</p><p>nates is in general time dependent and the conservation of energy cannot be deduced</p><p>from the Lagrangian. In general, we can then write that,</p><p>dE</p><p>dt</p><p>=</p><p>dT</p><p>dt</p><p>+</p><p>dU</p><p>dt</p><p>dE</p><p>dt</p><p>=</p><p>∂T</p><p>∂qi</p><p>q̇i +</p><p>∂T</p><p>∂q̇i</p><p>q̈i +</p><p>∂T</p><p>∂t</p><p>+</p><p>∂U</p><p>∂qi</p><p>q̇i +</p><p>dU</p><p>dt</p><p>(22.1)</p><p>where repeated indexes indicate summation. Using that for a system subject to both</p><p>97</p><p>98 CHAPTER 22. EXERCISE 22 - JEFFERSON SANTANA MARTINS</p><p>potential (Qi = − ∂U</p><p>∂qi</p><p>) and non-potential (Q̃i) forces, the Euler-Lagrange equations read,</p><p>d</p><p>dt</p><p>∂T</p><p>∂q̇i</p><p>− ∂T</p><p>∂qi</p><p>= −∂U</p><p>∂qi</p><p>+ Q̃i (22.2)</p><p>We can recast Eq. (22.2) in the following form,</p><p>dE</p><p>dt</p><p>=</p><p>(</p><p>d</p><p>dt</p><p>∂T</p><p>∂q̇i</p><p>+</p><p>∂U</p><p>∂qi</p><p>− Q̃i</p><p>)</p><p>q̇i +</p><p>∂T</p><p>∂q̇i</p><p>q̈i +</p><p>∂T</p><p>∂t</p><p>+</p><p>∂U</p><p>∂qi</p><p>q̇i +</p><p>∂U</p><p>∂t</p><p>(22.3)</p><p>dE</p><p>dt</p><p>=</p><p>d</p><p>dt</p><p>(</p><p>∂T</p><p>∂q̇i</p><p>q̇i</p><p>)</p><p>+</p><p>∂T</p><p>∂t</p><p>+ 2</p><p>∂U</p><p>∂qi</p><p>q̇i +</p><p>∂U</p><p>∂t</p><p>− Q̃iq̇i</p><p>dE</p><p>dt</p><p>= 2</p><p>∂T</p><p>∂t</p><p>− d (T1 + 2T0)</p><p>dt</p><p>+</p><p>∂T</p><p>∂t</p><p>+ 2</p><p>∂U</p><p>∂t</p><p>− ∂U</p><p>∂t</p><p>− Q̃iq̇i (22.4)</p><p>or equivalently,</p><p>dE</p><p>dt</p><p>=</p><p>d (T1 + 2T0)</p><p>dt</p><p>− ∂T</p><p>∂t</p><p>+</p><p>∂U</p><p>∂t</p><p>+ Q̃iq̇i (22.5)</p><p>where we have written the kinetic energy T as a polynomial in the q̇i,</p><p>T = Aij q̇iq̇j +Biq̇i + C = T2 + T1 + T0 (22.6)</p><p>with,</p><p>Aij =</p><p>∂r</p><p>∂qi</p><p>· ∂r</p><p>∂qj</p><p>, Bi =</p><p>∂r</p><p>∂qi</p><p>· ∂r</p><p>∂t</p><p>, and C =</p><p>∂r</p><p>∂qi</p><p>· ∂r</p><p>∂t</p><p>(22.7)</p><p>and we have used that,</p><p>dT2</p><p>dt</p><p>= 2T2,</p><p>dT1</p><p>dt</p><p>= T1, and</p><p>dT0</p><p>dT</p><p>= 0 (22.8)</p><p>The constraints are time dependent (or rheonomic), the relations between Carte-</p><p>sian and generalized coordinates will in general be also time dependent ( r = r(qi, t))</p><p>and one or more terms on the right-hand side of Eq. (22.5) will be non-zero, causing</p><p>the total mechanical energy of the system not to be conserved. Physically, we can in-</p><p>terpret this by observing that, in the case of rheonomic or time-dependent constraints,</p><p>the work of the forces of constraints is non-zero since they do not remain orthogonal</p><p>to the trajectory during the time evolution of the system (given that the constraints</p><p>change with time). The work of the forces of constraints per unit time corresponds</p><p>indeed to the last term in Eq. (22.5), and as we see, it is just one of the reasons why</p><p>the total mechanical energy of the system is not conserved. Occasionally the explicit</p><p>99</p><p>time-dependence of either kinetic or potential energies, or both, can contribute to it as</p><p>well.</p><p>Exercise 23 - Francisco Augusto Fer-</p><p>reira Almeida</p><p>Consider two particles of masses m1 and m2. Let m1 be con�ned to move on a</p><p>circle of radius a in the z=0 plane, centered at x=y=0. Let m2 be con�ned to move on</p><p>a circle of radius b in the z=c plane, centered at x=y=0. A light (massless) spring of</p><p>spring constant k is attached between the two particles.</p><p>(a) Find the Lagrangian for the system.</p><p>(b) Solve the problem using Lagrange multipliers and give a physical interpretation for</p><p>each multiplier.</p><p>Solution</p><p>T = T1 + T2 =</p><p>m1</p><p>2</p><p>(ẋ1</p><p>2 + ẏ1</p><p>2) +</p><p>m2</p><p>2</p><p>(ẋ2</p><p>2 + ẏ2</p><p>2) (23.1)</p><p>U = U1 + U2 = m2gc+</p><p>k</p><p>2</p><p>(</p><p>√</p><p>(x2 − x1)2 + (y2 − y1)2 − b)2 (23.2)</p><p>L = T + U =</p><p>m1</p><p>2</p><p>(ẋ1</p><p>2 + ẏ1</p><p>2) +</p><p>m2</p><p>2</p><p>(ẋ2</p><p>2 + ẏ2</p><p>2) +m2gc+</p><p>k</p><p>2</p><p>(</p><p>√</p><p>(x2 − x1)2 + (y2 − y1)2 − b)2</p><p>(23.3)</p><p>101</p><p>Exercise 24 - Guilherme Shoiti Yoshi-</p><p>matsu Giardini</p><p>The one-dimensional harmonic oscillator has the Lagrangian L = mẋ2/2 −</p><p>k x2/2. Suppose you did not know the solution to the motion, but realized that the</p><p>motion must be periodic and therefore could be described by a Fourier series of the</p><p>form</p><p>x(t) =</p><p>∑</p><p>j=0</p><p>aj cos(j ω t) , (24.1)</p><p>(taking t = 0 at a turning point) where ω is the (unknown) angular frequency of the</p><p>motion. This representation for x(t) de�nes a many-parameter path for the system</p><p>point in con�guration space. Consider the action integral I for two points, t1 and t2</p><p>separated by the period 2π</p><p>ω</p><p>. Show that with this form for the system path, I is an</p><p>extremum for non-vanishing x only if aj = 0, for j 6= 1, and only if ω2 = k</p><p>m</p><p>.</p><p>In spite of the enunciation, there are other possible values for aj for aj 6= 1. The</p><p>correct condition should be there must be only one aj 6= 0 while the others should be</p><p>aj′ = 0 ; j 6= j′. In other words, there should be only one mode of vibration.</p><p>We begin by noting the Lagrangian</p><p>L =</p><p>mẋ2</p><p>2</p><p>− k x2</p><p>2</p><p>and (24.2)</p><p>103</p><p>104 CHAPTER 24. EXERCISE 24 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>the supposed solution to the position</p><p>x(t) =</p><p>∑</p><p>j=0</p><p>aj cos(j ω t) also (24.3)</p><p>that the action integral is de�ned as</p><p>I =</p><p>∫ t0+</p><p>2π</p><p>ω</p><p>t0</p><p>L(x, ẋ, t)dt assuming (24.4)</p><p>that x(t) = x(0) + α g(t), where g(t0) = g(t0 + 2π/ω) = 0, we can minimize the action</p><p>via the derivative of I with respect to α, by making this derivative zero, we will �nd</p><p>an extrema to the action. The derivative is</p><p>dI</p><p>dα</p><p>=</p><p>∫ t0</p><p>t0</p><p>[</p><p>∂L</p><p>∂x</p><p>∂x</p><p>∂α</p><p>+</p><p>∂L</p><p>∂ẋ</p><p>∂ẋ</p><p>∂α</p><p>]</p><p>dt .</p><p>Now we rewrite the second term of the integral as ∂L</p><p>∂ẋ</p><p>∂ẋ</p><p>∂α</p><p>= ∂L</p><p>∂ẋ</p><p>∂2x</p><p>∂t∂α</p><p>and through</p><p>the partial fractions method, we �nd that</p><p>∫ t0+</p><p>2π</p><p>ω</p><p>t0</p><p>∂L</p><p>∂ẋ</p><p>∂2x</p><p>∂t∂α</p><p>dt =</p><p>∂L</p><p>∂ẋ �</p><p>�</p><p>�∂x</p><p>∂α</p><p>∣∣∣∣∣</p><p>t0+</p><p>2π</p><p>ω</p><p>t0</p><p>−</p><p>∫ t0+</p><p>2π</p><p>ω</p><p>t0</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ẋ</p><p>)</p><p>∂L</p><p>∂ẋ</p><p>dt , (24.5)</p><p>where</p><p>u =</p><p>∂L</p><p>∂ẋ</p><p>, (24.6)</p><p>du =</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ẋ</p><p>)</p><p>dt , (24.7)</p><p>dv =</p><p>∂2x</p><p>∂α∂t</p><p>dt and (24.8)</p><p>105</p><p>v =</p><p>∂x</p><p>∂α</p><p>. (24.9)</p><p>Then the derivative of the action becomes</p><p>dI</p><p>dα</p><p>=</p><p>∫ t0+</p><p>2π</p><p>ω</p><p>t0</p><p>[</p><p>∂L</p><p>∂x</p><p>− d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ẋ</p><p>)]</p><p>∂x</p><p>∂α</p><p>dt = 0 . (24.10)</p><p>Because ∂x</p><p>∂α</p><p>= g(x) that is an arbitrary function that vanishes at the extrema of</p><p>the integral, the differential equation of L that minimizes the action is</p><p>∂L</p><p>∂x</p><p>− d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ẋ</p><p>)</p><p>= 0 . (24.11)</p><p>Now, if we suppose that x(t) =</p><p>∑</p><p>j=0 ajcos(j ω t) is indeed a solution to the</p><p>oscillator problem, it must obey the condition that the action in minimal. To �nd it</p><p>is true or not, we substitute this solution into the Lagrangian and test if it obeys the</p><p>Euler-Lagrange equation condition.</p><p>L =</p><p>m</p><p>2</p><p>(</p><p>−</p><p>∑</p><p>j=0</p><p>j ω aj sin(j ω t)</p><p>)2</p><p>− k</p><p>2</p><p>(∑</p><p>j=0</p><p>ajcos(j ω t)</p><p>)2</p><p>then (24.12)</p><p>we proceed to the Euler-Lagrange equation. Calculating each term separately, we have</p><p>∂L</p><p>∂x</p><p>= −kx</p><p>= −k</p><p>∑</p><p>j=0</p><p>ajcos(j ω t) , (24.13)</p><p>we also have</p><p>∂L</p><p>∂ẋ</p><p>= mẋ</p><p>= −m</p><p>∑</p><p>j=0</p><p>j ω aj sin(j ω t) then (24.14)</p><p>106 CHAPTER 24. EXERCISE 24 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>d</p><p>dt</p><p>(</p><p>L</p><p>∂ẋ</p><p>)</p><p>= −m</p><p>∑</p><p>j=0</p><p>j2 ω2 aj cos(j ω t) . (24.15)</p><p>To minimize the action we consider equation (24.15) equal to equation (24.13)</p><p>k</p><p>∑</p><p>j=0</p><p>ajcos(j ω t) = m</p><p>∑</p><p>j=0</p><p>j2 ω2 aj cos(j ω t)</p><p>0 =</p><p>∑</p><p>j=0</p><p>(</p><p>k −m(jω)2</p><p>)</p><p>aj cos(j ω t) . (24.16)</p><p>Because the solution is non trivial, cos(j ω t) 6= 0, it means that</p><p>k = m(jω)2</p><p>ω =</p><p>√</p><p>k</p><p>m j2</p><p>, (24.17)</p><p>because ω, m and k are constants, j can only assume one value. In other words, there</p><p>is only one mode of vibration into the one-dimensional harmonic oscillator.</p><p>Going back to equation 24.16 j = 0, we have a0 k = 0, considering that k 6= 0,</p><p>then a0 = 0. The other coef�cients aj are all zero, with the exception of an arbitrary</p><p>one. Mathematically</p><p>aj = 0 ∀j ∈ ([1,∞]− j0) where (24.18)</p><p>j0∈ [1,∞] is an arbitrary integer. To comply with what was asked, we can assume that</p><p>a1 6= 0 and ai = 0 ∀i 6= 1 without loss of generality, such that ω becomes</p><p>ω =</p><p>√</p><p>k</p><p>m</p><p>. (24.19)</p><p>Exercise 25 - Jackson Galvão</p><p>A disk of radius R rolls without slipping inside the stationary parabola y = ax2 . Find</p><p>the equations of constraint. What condition allows the disk to roll so that it touches</p><p>the parabola at one and only one point independent of its position?</p><p>The disk roll by through length s of parabola y = ax2. Then, the length s of</p><p>parabola is given by</p><p>s =</p><p>Z x</p><p>0</p><p>ds =</p><p>Z x</p><p>0</p><p>p</p><p>dx2 + dy2 =</p><p>p</p><p>1 + [y0(x)]2 dx</p><p>y0(x) = 2 ax ! [y0(x)]2 = 4a2x2</p><p>Then</p><p>s =</p><p>Z x</p><p>0</p><p>p</p><p>1 + 4a2x2 dx (25.1)</p><p>The disk have a continuum step through of length s from rolling your bow r� in contact</p><p>with the parabola. For not slipping, the relative velocity vs;� between the tangent point</p><p>of disck in contact of parabola and s way should be null. Hence</p><p>vs;� = _s � r _� = 0</p><p>107</p><p>108 CHAPTER 25. EXERCISE 25 - JACKSON GALVÃO</p><p>Then</p><p>ṡ = rφ̇</p><p>where we have to ṡ</p><p>ṡ =</p><p>ds</p><p>dt</p><p>=</p><p>ds</p><p>dx</p><p>dx</p><p>dt</p><p>=</p><p>ds</p><p>dx</p><p>ẋ</p><p>Hence,</p><p>ṡ =</p><p>d</p><p>dx</p><p>∫ x</p><p>0</p><p>√</p><p>1 + 4a2x2 dx ẋ =</p><p>√</p><p>1 + 4a2x2 ẋ</p><p>Therefore, we must be have unholonomic conditions</p><p>rφ̇ =</p><p>√</p><p>1 + 4a2x2 ẋ (25.2)</p><p>being x the coordinate of contact with parabola.</p><p>For the holonomic constraint, the contact of disk with parabola’s point must be</p><p>have the coordinates that satisfy y = x2. The center of disk (x0, y0) about �gure</p><p>109</p><p>has be</p><p>x = x0 + r sinα</p><p>y = y0 − r cosα</p><p>From the relation function and angle, we have</p><p>sinα =</p><p>tanα√</p><p>1 + tan2 α</p><p>and</p><p>cosα =</p><p>1√</p><p>1 + tan2 α</p><p>Where tanα is</p><p>tanα = y0 = 2ax</p><p>thereby that,</p><p>sinα =</p><p>2ax√</p><p>1 + 4a2x2</p><p>and</p><p>cosα =</p><p>1√</p><p>1 + 4a2x2</p><p>Therefore, our holonomic constraint with the center coordinates of disk (x0, y0) have</p><p>must be</p><p>x0 = x−</p><p>(</p><p>2ax√</p><p>1 + 4a2x2</p><p>)</p><p>r (25.3)</p><p>and</p><p>y0 = ax2 +</p><p>(</p><p>1√</p><p>1 + 4a2x2</p><p>)</p><p>r (25.4)</p><p>To the disk touches in one only point of parabola, r have must be limited. And x0 = 0</p><p>like as in the �gure</p><p>110 CHAPTER 25. EXERCISE 25 - JACKSON GALVˆO</p><p>Thereby, the limit of r is given by</p><p>0 = x−</p><p>(</p><p>2ax√</p><p>1 + 4a2x2</p><p>)</p><p>r</p><p>what takes us</p><p>r2 =</p><p>1</p><p>4a2</p><p>+ x2</p><p>So, we can �nd r can touche x coordinate about y = y(x) in two points, we have</p><p>x = ±</p><p>√</p><p>r2 − 1</p><p>4a2</p><p>such that r have must be</p><p>r ></p><p>1</p><p>2a</p><p>(25.5)</p><p>111</p><p>Hence, for touches one only point of parabola, we have</p><p>r 6</p><p>1</p><p>2a</p><p>(25.6)</p><p>Exercise 26 - Jhordan Silveira de</p><p>Borba</p><p>A particle of mass m is suspended by a massless spring of length L. It hangs,</p><p>without initial motion, in a gravitational �eld of strength g. It is struck by an impulsive</p><p>horizontal blow, which introduces an angular velocity ! . If ! is suf�ciently small, it</p><p>is obvious that the mass moves as a simple pendulum. If ! is suf�ciently large, the</p><p>mass will rotate about support. Use a Lagrange multiplier to determine the conditions</p><p>under which the string becomes slack at some point in the motion.</p><p>Answer : The �rst observation we must make is that at one point is mentioned</p><p>'spring' and at another 'string', for the context we will assume that it is a string.</p><p>So working in polar coordinates, we have x = r cos� and y = r sin� so the kinetic</p><p>energy can be written as:</p><p>T =</p><p>1</p><p>2</p><p>m ( _r � _r )</p><p>=</p><p>1</p><p>2</p><p>m</p><p>� �</p><p>_r cos� � r sin� _�</p><p>� 2</p><p>+</p><p>�</p><p>_r sin� + r cos� _�</p><p>� 2</p><p>�</p><p>=</p><p>1</p><p>2</p><p>m</p><p>��</p><p>_r 2 cos2 � + r 2 sin2 � _� 2 � 2 _rr _� cos� sin�</p><p>�</p><p>+</p><p>�</p><p>_r 2 sin2 � + r 2 cos2 � _� 2 + 2 _rr _� cos� sin�</p><p>��</p><p>=</p><p>1</p><p>2</p><p>m</p><p>�</p><p>_r 2</p><p>�</p><p>cos2 � + sin2 �</p><p>�</p><p>+ r 2 _� 2</p><p>�</p><p>sin2 � + cos2 �</p><p>� �</p><p>=</p><p>1</p><p>2</p><p>m</p><p>�</p><p>_r 2 + r 2 _� 2</p><p>�</p><p>And for the potential energy, putting V = 0 at the lowest level the particle can reach</p><p>113</p><p>114 CHAPTER 26. EXERCISE 26 - JHORDAN SILVEIRA DE BORBA</p><p>trapped in a string of size L, that is V = 0 when sin θ = −1 and r = L. So we have:</p><p>V = mgy</p><p>= mg (L+ r sin θ)</p><p>And the system’s Lagrangian is written as:</p><p>L = T − V</p><p>=</p><p>1</p><p>2</p><p>m</p><p>(</p><p>ṙ2 + r2θ̇2</p><p>)</p><p>−mg (L+ r sin θ)</p><p>We have a condition that the position of the particle obeys the following inequality</p><p>r ≤ L where L is the size of the string. So while the string is stretched r = L and we</p><p>have the following constraint:</p><p>f1 = r − L = 0</p><p>And the generalized force is given by:</p><p>Qr = −λ1r</p><p>[</p><p>∂f1</p><p>∂r</p><p>]</p><p>= −λ1r</p><p>Again let’s denote just Qr = λ because as discussed in question 13, the sign is arbitrary.</p><p>We also have to:</p><p>Q� = −λ1�</p><p>[</p><p>∂f1</p><p>∂θ</p><p>]</p><p>= 0</p><p>So solving the equation for r, we have:</p><p>Qr =</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ṙ</p><p>)</p><p>− ∂L</p><p>∂r</p><p>λ =</p><p>d</p><p>dt</p><p>(mṙ)−</p><p>(</p><p>mrθ̇2 −mg sin θ</p><p>)</p><p>= mr̈ −mrθ̇2 +mg sin θ</p><p>Solving for θ:</p><p>Q� =</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂θ̇</p><p>)</p><p>− ∂L</p><p>∂θ</p><p>0 =</p><p>d</p><p>dt</p><p>(</p><p>mr2θ̇</p><p>)</p><p>− (−mgr cos θ)</p><p>= 2mrṙθ̇ +mr2θ̈ +mgr cos θ</p><p>Focusing on the situation where the string is stretched, we have r = L and ṙ = 0,</p><p>and the string slackens when Qr = 0, that is, when there is no more force generalized</p><p>capable of maintaining the constraint. So in this situation the equations of motion</p><p>115</p><p>reduce to:</p><p>0 = −mLθ̇2 +mg sin θ (26.1)</p><p>0 = mL2θ̈ +mgL cos θ (26.2)</p><p>From the 26.1 equation, we have:</p><p>θ̇2 =</p><p>g</p><p>L</p><p>sin θ (26.3)</p><p>This is only possible if sin θ > 0, then θ ∈ [0, π], which makes sense as it must be</p><p>in this angle range where gravity can affect the movement so that the rope slackens.</p><p>Considering that the system is conservative, the total energy is given by:</p><p>E = L+ V</p><p>0 =</p><p>1</p><p>2</p><p>m</p><p>(</p><p>ṙ2 + r2θ̇2</p><p>)</p><p>+mg (L+ r sin θ)</p><p>With the string stretched then, for the initial instant being r = L, ṙ = 0, θ̇ = ω and</p><p>θ = 3</p><p>2</p><p>π:</p><p>E0 =</p><p>1</p><p>2</p><p>mL2ω2</p><p>Since energy is conserved, it must be the same for any value of θ. With the rope still</p><p>stretched, then we have to:</p><p>E = E0</p><p>1</p><p>2</p><p>mL2θ̇2 +mgL (1 + sin θ) =</p><p>1</p><p>2</p><p>mL2ω2</p><p>1</p><p>2</p><p>Lθ̇2 + g (1 + sin θ) =</p><p>1</p><p>2</p><p>Lω2</p><p>1</p><p>2</p><p>Lθ̇2 + g + g sin θ =</p><p>1</p><p>2</p><p>Lω2</p><p>Replacing the obtained in 26.3 to θ̇2:</p><p>1</p><p>2</p><p>L</p><p>( g</p><p>L</p><p>sin θ</p><p>)</p><p>+ g sin θ =</p><p>1</p><p>2</p><p>Lω2 − g</p><p>3</p><p>2</p><p>g sin θ =</p><p>1</p><p>2</p><p>Lω2 − g</p><p>sin θ =</p><p>Lω2</p><p>3g</p><p>− 2</p><p>3</p><p>So as sin θ ∈ [0, 1] we have that Lω2</p><p>3g</p><p>− 2</p><p>3</p><p>∈ [0,1], and:</p><p>0 ≤ Lω2</p><p>3g</p><p>− 2</p><p>3</p><p>≤ 1</p><p>2</p><p>3</p><p>≤ Lω2</p><p>3g</p><p>≤ 5</p><p>3</p><p>So we have the condition that if</p><p>2g</p><p>L</p><p>≤ ω2 ≤ 5g</p><p>L</p><p>116 CHAPTER 26. EXERCISE 26 - JHORDAN SILVEIRA DE BORBA</p><p>The rope will becomes slack at some point in the movement. We can assume that if</p><p>! 2 � 2g</p><p>L corresponding to the case where the mass moves like a pendulum, and if</p><p>! 2 � 5g</p><p>L has enough initial angular velocity to rotate around the support.</p><p>Bibliography</p><p>GOLDSTEIN, H.; POOLE, C.; SAFKO, J. Classical Mechanics. Addison Wesley, 2002.</p><p>ISBN 9780201657029. Disponível em: <https://books.google.com.br/books?id=</p><p>tJCuQgAACAAJ>.</p><p>RAY, J. R. Erratum: Nonholonomic constraints. The American Journal of</p><p>Physics, v. 34, n. 12, p. 1202�1203, 1966. ISSN 0002-9505. Disponível em:</p><p><https://www.deepdyve.com/lp/american-association-of-physics-teachers/</p><p>erratum-nonholonomic-constraints-am-j-phys-34-406-1966-ekgJcxFgHm>.</p><p>RAY, J. R. Nonholonomic constraints. The American Journal of Physics, v. 34, n. 7, p. 406,</p><p>1966. ISSN 0002-9505. Disponível em: <https://aapt.scitation.org/doi/10.1119/1.</p><p>1973007>.</p><p>117</p><p>https://books.google.com.br/books?id=tJCuQgAACAAJ</p><p>https://books.google.com.br/books?id=tJCuQgAACAAJ</p><p>https://www.deepdyve.com/lp/american-association-of-physics-teachers/erratum-nonholonomic-constraints-am-j-phys-34-406-1966-ekgJcxFgHm</p><p>UNIVERSIDADE FEDERAL DO RIO GRANDE DO SUL</p><p>INSTITUTO DE FÍSICA</p><p>SOLUTIONS MANUAL HERBERT B. GOLDSTEIN 3RD ED.</p><p>L IST 3 OF CLASSICAL MECHANICS</p><p>CLASS : FIP00001</p><p>TEACHER : MARIA BEATRIZ DE LEONE GAY DUCATI</p><p>PORTO ALEGRE , RS</p><p>2021/01</p><p>Contents</p><p>1 Exercise 1 - JENIFER ANDRADE DE MATOS 3</p><p>2 Exercise 2 - EDGARD KRETSCHMANN 7</p><p>3 Exercise 3 - LUCAS BOURSCHEIDT 9</p><p>4 Exercise 4 - GUSTAVO CORTAZZI GARCIA KESSLER 11</p><p>5 Exercise 5 - FRANCISCO AUGUSTO FERREIRA ALMEIDA 13</p><p>6 Exercise 6 - DOUGLAS OLIVEIRA NOVAES 15</p><p>7 Exercise 7 - JACKSON GALVÃO 19</p><p>8 Exercise 8 - RODRIGO WEBER PEREIRA 23</p><p>9 Exercise 9 - JEFFERSON SANTANA MARTINS 25</p><p>10 Exercise 10 - GUILHERME SHOITI YOSHIMATSU GIARDINI 29</p><p>11 Exercise 11 - GEOVANE NAYSINGER 37</p><p>12 Exercise 12 - JHORDAN SILVEIRA DE BORBA 41</p><p>13 Exercise 13 - EDGARD KRETSCHMANN 45</p><p>14 Exercise 14 - JENIFER ANDRADE DE MATOS 47</p><p>15 Exercise 15 - GUILHERME SHOITI YOSHIMATSU GIARDINI 49</p><p>16 Exercise 16 - JACKSON GALVÃO 57</p><p>17 Exercise 17 - GUSTAVO CORTAZZI GARCIA KESSLER 59</p><p>18 Exercise 18 - RODRIGO WEBER PEREIRA 61</p><p>19 Exercise 19 - JEFFERSON SANTANA MARTINS 65</p><p>20 Exercise 20 - JHORDAN SILVEIRA DE BORBA 69</p><p>iii</p><p>CONTENTS 1</p><p>21 Exercise 21 - DOUGLAS OLIVEIRA NOVAES 75</p><p>22 Exercise 22 - FRANCISCO AUGUSTO FERREIRA ALMEIDA 81</p><p>23 Exercise 23 - LUCAS BOURSCHEIDT 85</p><p>24 Exercise 24 - GEOVANE NAYSINGER 91</p><p>25 Exercise 25 - LUCAS BOURSCHEIDT 95</p><p>26 Exercise 26 - RODRIGO WEBER PEREIRA 101</p><p>27 Exercise 27 - JENIFER ANDRADE DE MATOS 103</p><p>28 Exercise 28 - JACKSON GALVÃO 107</p><p>29 Exercise 29 - EDGARD KRETSCHMANN 113</p><p>30 Exercise 30 - JEFFERSON SANTANA MARTINS 117</p><p>31 Exercise 31 - GUILHERME SHOITI YOSHIMATSU GIARDINI 119</p><p>32 Exercise 32 - DOUGLAS OLIVEIRA NOVAES 121</p><p>33 Exercise 33 - FRANCISCO AUGUSTO FERREIRA ALMEIDA 127</p><p>34 Exercise 34 - GEOVANE NAYSINGER 133</p><p>35 Exercise 35 - GUSTAVO CORTAZZI GARCIA KESSLER 135</p><p>Bibliography 137</p><p>Exercise 1 - JENIFER ANDRADE DE</p><p>MATOS</p><p>(a) Reverse the Legendre transformation to derive the properties of L(qi,q̇i, t)</p><p>from H(qi, pi, t), treating the q̇i as independent quantities, and show that it leads to</p><p>the Lagrangian equations of motion. (b) By the same procedure find the equations of</p><p>motion in terms of the function</p><p>L′(p, ṗ, t) = −ṗiqi −H(q, p, t)</p><p>To solve this problem, we start by writing the Lagrangian as a Legendre trans-</p><p>formation from the Hamiltonian.</p><p>L(qi, q̇i, t) = q̇ipi −H(q, p, t) (1.1)</p><p>Using the Hamilton’s equations:</p><p>q̇i =</p><p>∂H</p><p>∂pi</p><p>(1.2)</p><p>ṗi = −</p><p>∂H</p><p>∂qi</p><p>(1.3)</p><p>Let´s diferenciate (1.1):</p><p>dL = q̇dp+ pq̇ − ∂H</p><p>∂q</p><p>dq − ∂H</p><p>∂p</p><p>dp−−∂H</p><p>∂t</p><p>dt (1.4)</p><p>3</p><p>4 CHAPTER 1. EXERCISE 1 - JENIFER ANDRADE DE MATOS</p><p>Using the relation (1.2) and (1.3) and rearranging the terms, we find:</p><p>dL = pdq̇ + ṗdq − ∂H</p><p>∂t</p><p>dt (1.5)</p><p>But, we can also write dL as:</p><p>dL =</p><p>∂L</p><p>∂q</p><p>dq +</p><p>∂L</p><p>∂q̇</p><p>dq̇ +</p><p>∂L</p><p>∂t</p><p>dt (1.6)</p><p>Comparing the terms of (1.5) and (1.6), we must have:</p><p>pi =</p><p>∂L</p><p>∂q̇i</p><p>(1.7)</p><p>ṗi =</p><p>∂L</p><p>∂qi</p><p>(1.8)</p><p>∂L</p><p>∂t</p><p>= −∂H</p><p>∂t</p><p>(1.9)</p><p>Deriving (1.7), we find:</p><p>ṗi =</p><p>d</p><p>dt</p><p>∂L</p><p>∂q̇i</p><p>(1.10)</p><p>Substituting (1.10) in (1.8), we find:</p><p>d</p><p>dt</p><p>∂L</p><p>∂q̇i</p><p>=</p><p>∂L</p><p>∂qi</p><p>(1.11)</p><p>We can write this expression as:</p><p>d</p><p>dt</p><p>∂L</p><p>∂q̇i</p><p>− ∂L</p><p>∂qi</p><p>= 0 (1.12)</p><p>5</p><p>Which are the Euler-Lagrange Equations, as we wanted to show.</p><p>b) To solve this item, let́s use the same procedures we used in the preceding</p><p>item, but to L’(p, ṗ, t)</p><p>L′(p, ṗ, t) = −ṗiqi −H(q, p, t) (1.13)</p><p>We also have that:</p><p>q̇i =</p><p>∂H</p><p>∂pi</p><p>(1.14)</p><p>ṗi = −</p><p>∂H</p><p>∂qi</p><p>(1.15)</p><p>Differentiating:</p><p>dL′ = −ṗdq − qdṗ− ∂H</p><p>∂q</p><p>dq − ∂H</p><p>∂p</p><p>dp− ∂H</p><p>∂t</p><p>dt (1.16)</p><p>Rearranging the terms in (1.16) and using (1.14) and (1.15), we find:</p><p>dL′ = −qdṗ− q̇dp− ∂H</p><p>∂t</p><p>dt (1.17)</p><p>We can write the dL’ as:</p><p>dL′ =</p><p>∂L′</p><p>∂p</p><p>dp+</p><p>∂L′</p><p>∂ṗ</p><p>dṗ+</p><p>∂L′</p><p>∂t</p><p>dt (1.18)</p><p>Comparing (1.17) and (1.18), we find:</p><p>−q̇i =</p><p>∂L′</p><p>∂pi</p><p>(1.19)</p><p>6 CHAPTER 1. EXERCISE 1 - JENIFER ANDRADE DE MATOS</p><p>−qi =</p><p>∂L′</p><p>∂ṗ</p><p>(1.20)</p><p>−∂H</p><p>∂t</p><p>=</p><p>∂L′</p><p>∂t</p><p>(1.21)</p><p>Deriving (1.20), we have:</p><p>q̇i =</p><p>d</p><p>dt</p><p>∂L′</p><p>∂ṗi</p><p>(1.22)</p><p>Substituting (1.22) in (1.19) and rearranging the terms, we find the equations of</p><p>motion:</p><p>∂L′</p><p>∂pi</p><p>− d</p><p>dt</p><p>∂L′</p><p>∂ṗi</p><p>= 0 (1.23)</p><p>Exercise 2 - EDGARD KRETSCHMANN</p><p>We have already proven that we can write a new Lagrangian adding the total</p><p>time derivative of a functionf = f(qj, t) to the original Lagrangian will give the same</p><p>equations of motion. Now we will analyze what happens to the Hamiltonian and to</p><p>the generalized momenta. So, the new Lagrangian is written as</p><p>L′(qj; q̇j; t) = L(qj; q̇j; t) +</p><p>df(qj; t)</p><p>dt</p><p>(2.1)</p><p>The generalized momenta is computed by</p><p>pj =</p><p>∂L</p><p>∂q̇j</p><p>(2.2)</p><p>Using the new Lagrangian to obtain the generalized momenta, we now have that</p><p>p′j =</p><p>∂L′</p><p>∂q̇j</p><p>=</p><p>∂L</p><p>∂q̇j</p><p>+</p><p>∂</p><p>∂q̇j</p><p>df</p><p>dt</p><p>(2.3)</p><p>Expanding df</p><p>dt</p><p>into</p><p>∑ ∂f</p><p>∂qi</p><p>q̇i +</p><p>∂f</p><p>∂t</p><p>then the last term of the equation (3) is now equals to</p><p>∂f</p><p>∂qj</p><p>. Then, the new generalized momenta is given by</p><p>p′j = pj +</p><p>∂f</p><p>∂qj</p><p>(2.4)</p><p>Now, analyzing what happens to the new Hamiltonian we will make a Legendre trans-</p><p>formation</p><p>H ′(qj; p</p><p>′</p><p>j; t) =</p><p>∑</p><p>p′kq̇k − L′(qj; q̇j; t) (2.5)</p><p>Using the equations (1) and (4) we can compare the n ew Hamiltonian to the original</p><p>one.</p><p>H ′(qj; pj; t) =</p><p>∑</p><p>(pk +</p><p>∂f</p><p>∂qk</p><p>)q̇k − L(qj; q̇j; t)−</p><p>df(qj; t)</p><p>dt</p><p>7</p><p>8 CHAPTER 2. EXERCISE 2 - EDGARD KRETSCHMANN</p><p>H ′(qj; pj; t) =</p><p>∑</p><p>pkq̇k − L(qj; q̇j; t) +</p><p>∑ ∂f</p><p>∂qk</p><p>q̇k −</p><p>df(qj; t)</p><p>dt</p><p>The two first terms of the right hand of the equation is the original Hamiltonian and</p><p>the difference between the two last terms is−∂f</p><p>∂t</p><p>. Therefore,</p><p>H ′(qj; pj; t) = H(qj; pj; t)−</p><p>∂f</p><p>∂t</p><p>(2.6)</p><p>So, for the new Hamiltonian the equations of motion are given by</p><p>q̇j =</p><p>∂H ′</p><p>∂p′j</p><p>; ṗ′j = −</p><p>∂H ′</p><p>∂qj</p><p>(2.7)</p><p>Applying equations (4) and (6) on (7) we will have that</p><p>q̇j =</p><p>∂H</p><p>∂pj</p><p>; ṗj = −</p><p>∂H</p><p>∂qj</p><p>(2.8)</p><p>Exercise 3 - LUCAS BOURSCHEIDT</p><p>Solution:</p><p>Differentiating the LagrangianL(qi, q̇i, t):</p><p>dL =</p><p>∂L</p><p>∂qi</p><p>dqi +</p><p>∂L</p><p>∂q̇i</p><p>dq̇i +</p><p>∂L</p><p>∂t</p><p>dt</p><p>Then:</p><p>dL = d</p><p>(</p><p>∂L</p><p>∂qi</p><p>qi</p><p>)</p><p>− qid</p><p>(</p><p>∂L</p><p>∂qi</p><p>)</p><p>+</p><p>∂L</p><p>∂q̇i</p><p>dq̇i +</p><p>∂L</p><p>∂t</p><p>dt</p><p>d</p><p>(</p><p>∂L</p><p>∂qi</p><p>qi − L</p><p>)</p><p>= qid</p><p>(</p><p>∂L</p><p>∂qi</p><p>)</p><p>− ∂L</p><p>∂q̇i</p><p>dq̇i −</p><p>∂L</p><p>∂t</p><p>dt</p><p>From the Lagrange equations of motion:</p><p>ṗi =</p><p>∂L</p><p>∂qi</p><p>pi =</p><p>∂L</p><p>∂q̇i</p><p>Thus:</p><p>d (ṗiqi − L) = qidṗi − pidq̇i −</p><p>∂L</p><p>∂t</p><p>dt (A)</p><p>9</p><p>10 CHAPTER 3. EXERCISE 3 - LUCAS BOURSCHEIDT</p><p>Defining:</p><p>G(ṗi, q̇i, t) = ṗiqi − L</p><p>DifferentiatingG:</p><p>dG =</p><p>∂G</p><p>∂ṗi</p><p>dṗi +</p><p>∂G</p><p>∂q̇i</p><p>dq̇i +</p><p>∂G</p><p>∂t</p><p>dt (B)</p><p>By comparing (A) with (B), we find that</p><p>∂G</p><p>∂t</p><p>= −∂L</p><p>∂t</p><p>and the equations of motion:</p><p>qi =</p><p>∂G</p><p>∂ṗi</p><p>pi = −</p><p>∂G</p><p>∂q̇i</p><p>Exercise 4 - GUSTAVO CORTAZZI GAR-</p><p>CIA KESSLER</p><p>11</p><p>Exercise 5 - FRANCISCO AUGUSTO</p><p>FERREIRA ALMEIDA</p><p>Verify that the matrix J has the properties given in Eqs. (8.38c) and (8.38e) and</p><p>that its determinant has the value +1</p><p>Solution:</p><p>Let J be the 2nx2n matrix composed of four nxn zero and unit matrices according</p><p>to the scheme:</p><p>J =</p><p>[</p><p>0 1</p><p>−1 0</p><p>]</p><p>with the following transpose matrix, which is its inverse:</p><p>JT =</p><p>[</p><p>0 −1</p><p>1 0</p><p>]</p><p>Thus</p><p>JTJ =</p><p>[</p><p>0 −1</p><p>1 0</p><p>] [</p><p>0 −1</p><p>1 0</p><p>]</p><p>JTJ =</p><p>[</p><p>0 · 0 + (−1) · (−1) 0 · 1 + (−1) · 0</p><p>1 · 0 + 0 · (−1) 1 · 1 + 0 · 0</p><p>]</p><p>13</p><p>14 CHAPTER 5. EXERCISE 5 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>JTJ =</p><p>[</p><p>1 0</p><p>0 1</p><p>]</p><p>JTJ = ⊮ (5.1)</p><p>The property given in 8.38c is proved.</p><p>J2 = J · J =</p><p>[</p><p>0 1</p><p>−1 0</p><p>] [</p><p>0 1</p><p>−1 0</p><p>]</p><p>J2 =</p><p>[</p><p>0 · 0 + 1 · (−1) 0 · 1 + 1 · 0</p><p>−1 · 0 + 0 · (−1) −1 · 1 + 0 · 0</p><p>]</p><p>J2 =</p><p>[</p><p>−1 0</p><p>0 −1</p><p>]</p><p>J2 = −</p><p>[</p><p>1 0</p><p>0 1</p><p>]</p><p>JTJ = −⊮ (5.2)</p><p>The property given in 8.38e is proved.</p><p>the determinant of matrix J is:</p><p>|J | = 0− (−1) = 1 (5.3)</p><p>Exercise 6 - DOUGLAS OLIVEIRA NO-</p><p>VAES</p><p>Question: Show that Hamilton’s principle can be written as</p><p>δ</p><p>∫ 2</p><p>1</p><p>[</p><p>2H(η, t) + ηJη̇</p><p>]</p><p>dt = 0</p><p>Response: The Hamilton’s principle is</p><p>δ</p><p>∫ t2</p><p>t1</p><p>L dt = 0</p><p>or equivalently</p><p>δ</p><p>∫ t2</p><p>t1</p><p>2L dt = 0</p><p>One can subtract the total time derivative of a function whose variation vanishes</p><p>at the endpoints of the path, from the integrand, without invalidating the variational</p><p>principle. This is because such a function will only contribute to boundary terms in-</p><p>volving the variation ofqi and pi at the end points of the path, which vanish</p><p>equations of constraint of the form</p><p>n∑</p><p>i=1</p><p>gi(x1, . . . , xn)dxi = 0</p><p>A constraint of this type is holonomic only if an integrating function</p><p>f(x1, . . . , xn) can be found that turns it into an exact differential. Clearly the function</p><p>must be such that</p><p>∂(fgi)</p><p>∂xj</p><p>=</p><p>∂(fgj)</p><p>∂xi</p><p>for all i 6= j. Show that no such integrating factor can be found for either of Eqs.</p><p>(1.39).</p><p>Solution: For the disk (Eq. 1.39), we have:</p><p>dy + a cos θdφ = 0</p><p>dx− a sin θdφ = 0</p><p>Expanding the given relation, we have:</p><p>∂(fgi)</p><p>∂xj</p><p>= f</p><p>∂gi</p><p>∂xj</p><p>+ gi</p><p>∂f</p><p>∂xj</p><p>9</p><p>10 CHAPTER 4. EXERCISE 4 - RODRIGO WEBER P.</p><p>where, gi = ∂f</p><p>∂xi</p><p>. Now, we can compare both equations. For the first, taking</p><p>xj = θ and gi = a cos θ, we have:</p><p>∂</p><p>∂θ</p><p>(f(a cos θ)) = f(−a sin θ) + a cos θ × 0 = −af sin θ</p><p>Conversely, using, xi = φ and gj = 0</p><p>∂</p><p>∂θ</p><p>(f × 0) = 0</p><p>We see that −af sin θ = 0 ⇐⇒ f = 0, so the first equation can’t be integrated.</p><p>For the second equation, taking xj = θ and gi = −a sin θ, we have:</p><p>∂</p><p>∂θ</p><p>(f(−a sin θ)) = f(−a cos θ) + (−a sin θ × 0 = −a cos θ</p><p>Conversely, using, xi = φ and gj = 0</p><p>∂</p><p>∂θ</p><p>(f × 0) = 0</p><p>We see, again, that −af cos θ = 0 ⇐⇒ f = 0, so the second equation also can’t</p><p>be integrated, wich concludes the exercise.</p><p>Exercise 5 - Jenifer A. de Matos</p><p>Two wheel of radius a are mounted on the ends of a common axle of length</p><p>b such that the wheels rotate independently. The whole combination rolls without</p><p>slipping on a plane. Show that there are 2 nonholonomic equations of constraint,</p><p>cosθdx+ sinθdy = 0</p><p>sinθdx− cosθdy =</p><p>1</p><p>2</p><p>a(dφ+ dφ′),</p><p>(where θ, φ and φ′ have meanings similar to those in the problem of a single vertical</p><p>disk, and (x, y), are the coordinates of a point on the axle midway between the two</p><p>wheels) and one holonomic equation of constraint,</p><p>θ = C − a</p><p>b</p><p>(φ− φ′),</p><p>where c is a constant.</p><p>To solve this problem, we are initially going to draw the situation using a frame</p><p>of reference:</p><p>11</p><p>12 CHAPTER 5. EXERCISE 5 - JENIFER A. DE MATOS</p><p>From the figure and the elements of the question, we have:</p><p>x =</p><p>x1 + x2</p><p>2</p><p>(5.1)</p><p>y =</p><p>y1 + y2</p><p>2</p><p>(5.2)</p><p>From that:</p><p>dx =</p><p>dx1 + dx2</p><p>2</p><p>(5.3)</p><p>dy =</p><p>dy1 + dy2</p><p>2</p><p>(5.4)</p><p>13</p><p>Using the equations of constraint to both wheels, we have:</p><p>dx1 = a,sin(θ)dφ′ (5.5)</p><p>dx2 = a,sin(θ)dφ (5.6)</p><p>Substituting (5.5) and (5.6) in (5.3), we have:</p><p>dx =</p><p>a,sin(θ)</p><p>2</p><p>(dφ′ + dφ) (5.7)</p><p>For y:</p><p>dy1 = −a,cos(θ)dφ′ (5.8)</p><p>dy2 = −a,cos(θ)dφ (5.9)</p><p>Substituting (5.8) and (5.9) in (5.4), we have:</p><p>dy =</p><p>−a,cos(θ)</p><p>2</p><p>(dφ′ + dφ) (5.10)</p><p>From that, we multiply (5.7) by sin(θ) and (5.10) by -cos(θ). So we have:</p><p>sin(θ)dx =</p><p>a,sin2(θ)</p><p>2</p><p>(dφ′ + dφ) (5.11)</p><p>−cos(θ)dy =</p><p>a,cos2(θ)</p><p>2</p><p>(dφ′ + dφ) (5.12)</p><p>14 CHAPTER 5. EXERCISE 5 - JENIFER A. DE MATOS</p><p>Adding (5.11) and (5.12), we find the second nonholonomic equation asked in</p><p>the question.</p><p>sin(θ)dx− cos(θ)dy =</p><p>a</p><p>2</p><p>(dφ′ + dφ) (5.13)</p><p>Back to the expression (5.7) and (5.10), we multiply (5.7) by cos(θ) and (5.10) by</p><p>sin(θ):</p><p>cos(θ)dx =</p><p>a,sin(θ),cos(θ)</p><p>2</p><p>(dφ′ + dφ) (5.14)</p><p>sin(θ)dy =</p><p>−a,sin(θ),cos(θ)</p><p>2</p><p>(dφ′ + dφ) (5.15)</p><p>Adding (5.14) and (5.15), we find the first nonholonomic equation asked in the</p><p>question:</p><p>cosθdx+ sinθdy = 0 (5.16)</p><p>According to the situation, we have the following relation:</p><p>cos(θ) =</p><p>x2 − x1</p><p>b</p><p>(5.17)</p><p>Deriving (5.17), we have:</p><p>−sin(θ)dθ = dx2 − dx1</p><p>b</p><p>(5.18)</p><p>Substituting (5.5) and (5.6) in (5.18), we find:</p><p>−sin(θ)dθ = a</p><p>b</p><p>sin(θ)(dφ− dφ′) (5.19)</p><p>Then,</p><p>dθ = − a</p><p>b</p><p>(dφ− dφ′) (5.20)</p><p>15</p><p>Integrating both sides:∫</p><p>dθ =</p><p>∫</p><p>− a</p><p>b</p><p>(dφ− dφ′) (5.21)</p><p>We find the desired holonomic equation:</p><p>θ = C − a</p><p>b</p><p>(φ− φ′) (5.22)</p><p>Exercise 6 - Jackson Galvão</p><p>A particle moves in the xy plane under the constraint that its velocitys vector</p><p>is always direct towards a point on the x axis whose abcissa is some given function</p><p>of time f(t). Show that for f(t) differentiable, but otherwise arbitrary, the constraint is</p><p>nonholonomic.</p><p>dx = x(t)− f(t)</p><p>dy = y(t)</p><p>vx =</p><p>dx</p><p>dt</p><p>vy =</p><p>dy</p><p>dt</p><p>For the same direction on the xy plane, we have</p><p>vx</p><p>vy</p><p>=</p><p>dx/dy</p><p>dy/dt</p><p>=</p><p>dx</p><p>dy</p><p>=</p><p>x− f(t)</p><p>y</p><p>then,</p><p>dy</p><p>y</p><p>=</p><p>dx</p><p>x− f(t)</p><p>For a holonomic constraint, f(t) can be dependents only of spatial coordinates</p><p>and time.</p><p>Therefore, for exemple:</p><p>17</p><p>18 CHAPTER 6. EXERCISE 6 - JACKSON GALVÃO</p><p>f(t) = βt,</p><p>where β is a constant, we have</p><p>dy</p><p>y</p><p>=</p><p>dx</p><p>x− βt</p><p>ln y = ln(x− βt) + c1</p><p>y = eln(x−βt)ec1</p><p>y = C0(x− βt)</p><p>In tohe cases, where f(t) depend of variables that not be spatial and times coor-</p><p>dinates, then we have nonholonomic function, and so not integrables.</p><p>Exercise 7 - Jackson Galvão</p><p>Show that Lagrange’s equations in the form of Eqs. (1.53) can also be written as</p><p>∂Ṫ</p><p>∂q̇j</p><p>− 2</p><p>∂T</p><p>∂qj</p><p>= Qj (7.1)</p><p>This are sometimes known as the Nielsen form of the Lagrange equations.</p><p>The Lagrange’s equation, by (1.53) of Goldstein book, to generalized force Qj is</p><p>given by</p><p>d</p><p>dt</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>− ∂T</p><p>∂qj</p><p>= Qj (7.2)</p><p>Firstly, we observe that the Kinetic Energy T is a function of the generalized</p><p>coordinates and velocities and the time, such that</p><p>T = T (q, q̇, t)</p><p>and that the derivative of the time of T is can written as</p><p>Ṫ =</p><p>dT</p><p>dt</p><p>where the exact differential of T is the sum of yours partial differentials.</p><p>19</p><p>20 CHAPTER 7. EXERCISE 7 - JACKSON GALVÃO</p><p>dT</p><p>dt</p><p>=</p><p>∑</p><p>i</p><p>∂T</p><p>∂qi</p><p>qi +</p><p>∑</p><p>i</p><p>∂T</p><p>∂q̇i</p><p>q̈i +</p><p>∂T</p><p>∂t</p><p>Obtaining the partial derivative about q̇i to Ṫ , we have</p><p>∂Ṫ</p><p>∂q̇i</p><p>=</p><p>∂</p><p>∂q̇i</p><p>(∑</p><p>i</p><p>∂T</p><p>∂qi</p><p>q̇i +</p><p>∑</p><p>i</p><p>∂T</p><p>∂q̇i</p><p>q̈i +</p><p>∂T</p><p>∂t</p><p>)</p><p>(7.3)</p><p>Applying the partial ∂/∂q̇i at the first term of (7.3), we obtaing</p><p>∂</p><p>∂q̇j</p><p>∑</p><p>i</p><p>∂T</p><p>∂qi</p><p>q̇i =</p><p>[∑</p><p>i</p><p>∂</p><p>∂q̇j</p><p>(</p><p>∂T</p><p>∂qi</p><p>)</p><p>q̇i +</p><p>∂T</p><p>∂qj</p><p>]</p><p>where ∂T</p><p>∂qj</p><p>arise from derivative of the product between</p><p>(</p><p>∂T</p><p>∂qi</p><p>q̇i</p><p>)</p><p>.</p><p>To second term of (7.3), applying ∂/∂q̇i over the second term, we have</p><p>∂</p><p>∂q̇j</p><p>∑</p><p>i</p><p>∂T</p><p>∂q̇i</p><p>q̈i =</p><p>∑</p><p>i</p><p>∂</p><p>∂q̇j</p><p>(</p><p>∂T</p><p>∂q̇i</p><p>)</p><p>q̈i</p><p>And to third term, of it naturally follows that (7.3) is</p><p>∂Ṫ</p><p>∂q̇j</p><p>=</p><p>∑</p><p>i</p><p>∂</p><p>∂q̇j</p><p>(</p><p>∂T</p><p>∂qi</p><p>)</p><p>q̇i +</p><p>∂T</p><p>∂qj</p><p>+</p><p>∑</p><p>i</p><p>∂</p><p>∂q̇j</p><p>(</p><p>∂T</p><p>∂q̇i</p><p>)</p><p>q̈i +</p><p>∂</p><p>∂q̇j</p><p>(</p><p>∂T</p><p>∂t</p><p>)</p><p>that we can rewrite in form of (7.4), highlighting ∂T/∂q̇j in the terms</p><p>∂Ṫ</p><p>∂q̇j</p><p>=</p><p>∑</p><p>i</p><p>∂</p><p>∂qi</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>q̇i +</p><p>∑</p><p>i</p><p>∂</p><p>∂q̇i</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>q̈i +</p><p>∂</p><p>∂t</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>+</p><p>∂T</p><p>∂qj</p><p>(7.4)</p><p>The first three highlighted terms of (7.4) are the exact differential of ∂T/∂q̇i, such</p><p>that</p><p>21</p><p>d</p><p>dt</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>=</p><p>∑</p><p>i</p><p>∂</p><p>∂qi</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>q̇i +</p><p>∑</p><p>i</p><p>∂</p><p>∂q̇i</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>q̈i +</p><p>∂</p><p>∂t</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>Therefore, (7.3) can be rewritten as</p><p>∂Ṫ</p><p>∂q̇j</p><p>=</p><p>d</p><p>dt</p><p>(</p><p>∂T</p><p>∂qj</p><p>)</p><p>+</p><p>∂T</p><p>∂qj</p><p>that given us,</p><p>d</p><p>dt</p><p>(</p><p>∂T</p><p>∂qj</p><p>)</p><p>=</p><p>∂Ṫ</p><p>∂q̇j</p><p>− ∂T</p><p>∂qj</p><p>(7.5)</p><p>Substituting (7.5) in Euler-Lagrange form, in the (1.53) equation given for (7.2),</p><p>we have</p><p>(</p><p>∂Ṫ</p><p>∂q̇j</p><p>− ∂T</p><p>∂qj</p><p>)</p><p>− ∂T</p><p>∂qj</p><p>= Qj</p><p>which gives us the Nielsen form for the Lagrange equation of (7.1) equation:</p><p>∂Ṫ</p><p>∂q̇j</p><p>− 2</p><p>∂T</p><p>∂qj</p><p>= Qj</p><p>Exercise 8 - Gustavo C. G. Kessler</p><p>23</p><p>Exercise 9 - Rodrigo Weber P.</p><p>The eletromagnetic field is invariant under a gauge transformation of the scalar</p><p>and vector potential given by</p><p>A→ A+∇ψ(r, t)</p><p>φ→ φ− 1</p><p>c</p><p>∂ψ</p><p>∂t</p><p>Where ψ is arbitrary (but differentiable). What effect does this gauge transfor-</p><p>mation have on the Lagrangian of a particle moving in the electromagnetic field? Is</p><p>the motion affected?</p><p>Solution: The Lagrangian of a charge moving in the electromagnetic field is</p><p>given by, in Gauss units (Eq. 1.62):</p><p>L = T − q</p><p>(</p><p>φ− A · v</p><p>c</p><p>)</p><p>Substituting the proposed gauge transformation in the Lagrangian:</p><p>L′ = T − q</p><p>(</p><p>φ− 1</p><p>c</p><p>∂ψ</p><p>∂t</p><p>− A · v</p><p>c</p><p>− ∇ψ · v</p><p>c</p><p>)</p><p>Rearanginf terms, we get:</p><p>25</p><p>26 CHAPTER 9. EXERCISE 9 - RODRIGO WEBER P.</p><p>L′ = T − q</p><p>(</p><p>φ− A · v</p><p>c</p><p>)</p><p>+</p><p>q</p><p>c</p><p>(</p><p>∇ψ · v +</p><p>∂ψ</p><p>∂t</p><p>)</p><p>The first term can be identified as the Lagrangian L of the system. The second,</p><p>given that the ψ function is a scalar, can be written as the derivative, with respect to</p><p>time, of the function ψ, because</p><p>dψ(r, t)</p><p>dt</p><p>=</p><p>n∑</p><p>i=1</p><p>∂ψ</p><p>∂xi</p><p>ẋi +</p><p>∂ψ</p><p>∂t</p><p>= ∇ψ · v +</p><p>∂ψ</p><p>∂t</p><p>So, we can write:</p><p>L′ = L+</p><p>q</p><p>c</p><p>(</p><p>dψ(r, t)</p><p>dt</p><p>)</p><p>So that:</p><p>L′ = L+</p><p>d</p><p>dt</p><p>F (r, t)</p><p>Where, F (r, t) = q</p><p>c</p><p>ψ(r, t). As seen in the derivation 8, the above equation means that</p><p>the Lagrangian L′ produce the same equations of motion as the Lagrangian L.</p><p>Exercise</p><p>by as-</p><p>sumption. Such a function ispiqi. So, the ‘modified’ Hamilton’s principle is</p><p>δ</p><p>∫ t2</p><p>t1</p><p>[</p><p>2L − d</p><p>dt</p><p>(piqi)</p><p>]</p><p>dt = 0 (6.1)</p><p>Since</p><p>H (q, p, t) = q̇ipi −L (q, q̇, t)</p><p>15</p><p>16 CHAPTER 6. EXERCISE 6 - DOUGLAS OLIVEIRA NOVAES</p><p>then</p><p>L (q; _q; t) = _qi pi � H (q; p; t)</p><p>that replacing in (6.1) come</p><p>�</p><p>Z t2</p><p>t1</p><p>�</p><p>�2 _qi pi � 2H (q; p; t)� � �� pi _qi � _pi qi</p><p>�</p><p>dt = 0 (6.2)</p><p>Since</p><p>_qi =</p><p>@H</p><p>@pi</p><p>and � _pi =</p><p>@H</p><p>@qi</p><p>then (6.2) come</p><p>�</p><p>Z t2</p><p>t1</p><p>�</p><p>� 2H (q; p; t) + pi</p><p>@H</p><p>@pi</p><p>+ qi</p><p>@H</p><p>@qi</p><p>�</p><p>dt = 0 (6.3)</p><p>Let the column matrix � with 2n elements for a system of n degrees of freedom</p><p>such that</p><p>qi = � i and pi = � i + n</p><p>and the elements of the matrix @H =@�as</p><p>@H</p><p>@pi</p><p>=</p><p>�</p><p>@H</p><p>@�</p><p>�</p><p>i + n</p><p>and</p><p>@H</p><p>@qi</p><p>=</p><p>�</p><p>@H</p><p>@�</p><p>�</p><p>i</p><p>with i � n</p><p>Then, replacing them in (6.3)</p><p>�</p><p>Z t2</p><p>t1</p><p>�</p><p>� 2H (�; t ) + � i</p><p>�</p><p>@H</p><p>@�</p><p>�</p><p>i</p><p>+ � i + n</p><p>�</p><p>@H</p><p>@�</p><p>�</p><p>i + n</p><p>�</p><p>dt = 0 (6.4)</p><p>Since</p><p>_� = J</p><p>@H</p><p>@�</p><p>and J =</p><p>�</p><p>0 1</p><p>� 1 0</p><p>�</p><p>one can write</p><p>�</p><p>@H</p><p>@�</p><p>= �</p><p>_�</p><p>J</p><p>= �</p><p>J</p><p>J</p><p>_�</p><p>J</p><p>= �</p><p>J</p><p>J 2</p><p>_� = � �J _�; J 2 = � 1</p><p>17</p><p>Replacing it in (6.4)</p><p>�</p><p>Z t2</p><p>t1</p><p>�</p><p>� 2H (�; t ) � �J _�</p><p>�</p><p>dt = 0</p><p>and finally, the Hamilton’s principle can be written as</p><p>�</p><p>Z 2</p><p>1</p><p>�</p><p>2H (�; t ) + �J _�</p><p>�</p><p>dt = 0 (6.5)</p><p>Exercise 7 - JACKSON GALVÃO</p><p>(a) Show that angle of recoil of the target particle relative to the incident direction of</p><p>the scartterd particle is simplyΦ = 1</p><p>2</p><p>(π −Θ).</p><p>The center of mass frame both the particles momentump′1 and p′2 is</p><p>m1v</p><p>′</p><p>1 = m2v</p><p>′</p><p>2</p><p>by eq.(3.2) by Golsdtein</p><p>r⃗1</p><p>′ = −</p><p>(</p><p>m2</p><p>m1 +m2</p><p>)</p><p>r⃗2</p><p>′ = −</p><p>(</p><p>m1</p><p>m1 +m2</p><p>)</p><p>Where prime refer to CM frame. With the eq.(3.2) and eq.(3.110), we defined relative</p><p>speed v after collision relative the CM of system.</p><p>v′1 =</p><p>µ</p><p>m1</p><p>v =</p><p>m2</p><p>m1 +m2</p><p>v0</p><p>And v is a relative speed after collision, being this collision must be have an elastic,</p><p>such that the kinetic energy it is conserved.</p><p>v′1 =</p><p>m2</p><p>m1 +m2</p><p>v0</p><p>and so letv2</p><p>v′2 =</p><p>m1</p><p>m1 +m2</p><p>v0</p><p>19</p><p>20 CHAPTER 7. EXERCISE 7 - JACKSON GALVˆO</p><p>For the total momentum, we know has be it conserved. Then</p><p>(m1 +m2)vcm = m1v0</p><p>So, we must be have</p><p>vcm =</p><p>m1</p><p>m1 +m2</p><p>v0</p><p>and the same tov′2, such that v′2 = vcm.</p><p>This forms an isosceles triangle where the two sides are equal. Therefore,</p><p>Φ + Φ +Θ = π</p><p>so,</p><p>Φ =</p><p>1</p><p>2</p><p>(π − Θ)</p><p>(b) It is observed that in elastic scattering the scattering cross section is isotropic in</p><p>terms ofΘ. What are the corresponding probability distributions for the scattered en-</p><p>ergy of the incident particle,E1, and for the recoil energy of the target particle,E2?</p><p>For a central potential V(r), the angle of scattering is determined by impact parameter,</p><p>that calledb(θ).</p><p>For a single or two particles, the scattered can be rated betweenθ and θ + dθ bounded</p><p>by b at b+ db. Let see under figure.</p><p>Thereby, we can written in terms of solid angle, like as</p><p>dΘ = 2π sin θdθ</p><p>21</p><p>So, we must be have</p><p>dσ(θ)</p><p>dΘ</p><p>=</p><p>b</p><p>sin θ</p><p>∣∣∣∣dbdθ</p><p>∣∣∣∣</p><p>For elastic scattering from collision, we have then</p><p>b(θ) = R sinα = R sin</p><p>(</p><p>π − θ</p><p>2</p><p>)</p><p>= −R cos</p><p>(</p><p>θ</p><p>2</p><p>)</p><p>The solution has be</p><p>dσ(θ)</p><p>dΘ</p><p>=</p><p>R2</p><p>4</p><p>Considering thatdσ and dΘ just a bounded by velocitiesv′1 and v′2 so it is reasonable to</p><p>say that dσ and dΘ are suffer variation withv′1 and v′2, such that</p><p>dσ =</p><p>dr′1</p><p>dt</p><p>sin θdθ</p><p>R2dΘ =</p><p>dr′2</p><p>dt</p><p>cos θdθ</p><p>Therefore, taking into account the kinetic energy relationships ofm1 and m2,the</p><p>probability have must be</p><p>v′1</p><p>v′2</p><p>=</p><p>m2</p><p>8m1</p><p>tan</p><p>(</p><p>π − θ</p><p>2</p><p>)</p><p>Exercise 8 - RODRIGO WEBER</p><p>PEREIRA</p><p>Show that the modified Hamilton‘s principle, in the form of Eq. (8.71), leads to</p><p>Hamilton‘s equations of motion.</p><p>The modified hamilton‘s principle:</p><p>δ</p><p>∫ t2</p><p>t1</p><p>[ n∑</p><p>i</p><p>q̇ipi −H −</p><p>d</p><p>dt</p><p>( n∑</p><p>i</p><p>qipi</p><p>)]</p><p>dt = 0 (8.1)</p><p>leads to the following equation:</p><p>δ</p><p>∫ t2</p><p>t1</p><p>[ n∑</p><p>i</p><p>−ṗiqi −H</p><p>]</p><p>dt = δ</p><p>∫ t2</p><p>t1</p><p>f(q⃗, p⃗, ˙⃗q, ˙⃗p, t)dt = 0, (8.2)</p><p>where</p><p>f(q⃗, p⃗, ˙⃗q, ˙⃗p, t) =</p><p>n∑</p><p>i</p><p>−ṗiqi −H (8.3)</p><p>By the Euler-Lagrange equations, we have (for the q‘s):</p><p>d</p><p>dt</p><p>( ∂f</p><p>∂q̇j</p><p>)</p><p>− ∂f</p><p>∂qj</p><p>= 0 ⇒ ṗj +</p><p>∂H</p><p>∂qj</p><p>= 0 (8.4)</p><p>For the p‘s, we have:</p><p>d</p><p>dt</p><p>( ∂f</p><p>∂ṗj</p><p>)</p><p>− ∂f</p><p>∂pj</p><p>= 0 ⇒ q̇j −</p><p>∂H</p><p>∂pj</p><p>= 0 (8.5)</p><p>Wich are the Hamilton‘s equations of motion, as we wished to show.</p><p>23</p><p>Exercise 9 - JEFFERSON SANTANA</p><p>MARTINS</p><p>If the canonical variables are not all independent; but are connected by auxiliary</p><p>conditions of the form:</p><p>ψk(qi, P i, t) = 0,</p><p>show that the canonical equations of motion can be written</p><p>∂H</p><p>∂pi</p><p>+</p><p>∑</p><p>k</p><p>λk</p><p>∂ψk</p><p>∂pi</p><p>= q̇i ,</p><p>∂H</p><p>∂qi</p><p>+</p><p>∑</p><p>k</p><p>λk</p><p>∂ψk</p><p>∂qi</p><p>= −ṗi</p><p>where the λk are the undetermined Lagrange multipliers. The formulation of</p><p>the Hamiltonian equations in whicht is a canonical variable is a case in point; since a</p><p>relation exists betweenpn+1 and the other canonical variables:</p><p>H(q1, · · · , qn+1; p1, · · · , pn) + pn+1 = 0</p><p>Show that as a result of these circumstances the2n + 2 Hamilton’s equations of this</p><p>formulation can be reduced to the2n ordinary Hamilton’s equations plus Eq. (8.41)</p><p>and the relation</p><p>λ =</p><p>dt</p><p>dθ</p><p>Note that while these results are reminiscent of the relativistic covariant Hamiltonian</p><p>formulation, they have been arrived at entirely within the framework of nonrelativistic</p><p>mechanics.</p><p>25</p><p>26 CHAPTER 9. EXERCISE 9 - JEFFERSON SANTANA MARTINS</p><p>Solution</p><p>The constraints can be incorporated into the LagrangianL by defining a “con-</p><p>strained Lagrangian” Lc, as</p><p>Lc (q, q̇, t) = L (q, q̇, t)−</p><p>∑</p><p>k</p><p>λkψk (q, p, t) (9.1)</p><p>Applying Hamilton’s principle, and using the Legendre transformation forL, we get</p><p>δ</p><p>∫ (</p><p>piq̇i −H(q, p, t)−</p><p>∑</p><p>k</p><p>λkψk (q, p, t)</p><p>)</p><p>dt = 0 (9.2)</p><p>By analogy with the constrained Lagrangian, we can define a “constrained Hamilto-</p><p>nian” Hc as</p><p>Hc (q, q, t) = H (q, q, t)−</p><p>∑</p><p>k</p><p>λkψk (q, p, t) (9.3)</p><p>Since both the terms are functions ofqi, pi and t, this is a “good” Hamiltonian. Equation</p><p>(9.2) can then be written as</p><p>δ</p><p>∫</p><p>(piq̇i −Hc(q, p, t)) dt = 0 (9.4)</p><p>This bears a resemblance to the usual variational principle in Hamiltonian mechanics,</p><p>for a HamiltonianHc. So the Hamilton equations are</p><p>q̇i =</p><p>∂Hc</p><p>∂pi</p><p>ṗi = −</p><p>∂Hc</p><p>∂qi</p><p>which become</p><p>q̇i =</p><p>∂H</p><p>∂pi</p><p>+</p><p>∑</p><p>k</p><p>λk</p><p>∂ψk</p><p>∂pi</p><p>(9.5)</p><p>−ṗi =</p><p>∂H</p><p>∂qi</p><p>+</p><p>∑</p><p>k</p><p>λk</p><p>∂ψk</p><p>∂qi</p><p>(9.6)</p><p>If timet is treated as a canonical variable, we defineqn+1 = t. By Hamilton’s equations</p><p>ṗn+1 =</p><p>−∂H</p><p>∂qn+1</p><p>(9.7)</p><p>= −∂H</p><p>∂t</p><p>(9.8)</p><p>27</p><p>= −dH</p><p>dt</p><p>(9.9)</p><p>and</p><p>q̇n+1 =</p><p>∂H</p><p>∂pn+1</p><p>(9.10)</p><p>= 1 (since qn+1 = t) (9.11)</p><p>As the Hamiltonian contains terms of the formpi = q̇i for each coordinate and its</p><p>canonical momentum, in order to incorporate the constraint imposed by the inclusion</p><p>of time as the(n+1)th canonical variable, we include a term of the formpn+1q̇n+1 = pn+1</p><p>to the Hamiltonian to set up the constraint. Equivalently, the constraint can be obtained</p><p>by integrating equation (9.9) above, and is given by</p><p>H(q1, · · · , qn, qn+1; p1, · · · , pn) + pn+1 = 0 (9.12)</p><p>Hamilton’s principle,</p><p>δ</p><p>∫</p><p>(piq̇i −H) dt = 0 (9.13)</p><p>can be written as</p><p>δ</p><p>∫</p><p>(piq̇i −H) t′dθ = 0 (9.14)</p><p>where t′ = dt</p><p>dθ</p><p>and θ is some parameter.Using the constrained form of Hamilton’s equa-</p><p>tions, we get</p><p>q̇i = (1 + λ)</p><p>∂H</p><p>∂pi</p><p>, i = 1, 2, · · ·n (9.15)</p><p>ṗi = − (1 + λ)</p><p>∂H</p><p>∂qi</p><p>, i = 1, 2, · · ·n (9.16)</p><p>q̇n+1 = λ (9.17)</p><p>ṗn+1 = − (1 + λ)</p><p>∂H</p><p>∂t</p><p>= −∂L</p><p>∂t</p><p>(9.18)</p><p>By regarding H ′ = (1 + λ)H as an equivalent Hamiltonian, these equations are the</p><p>required (2n+ 2) equations of motion. Alsoλ = q̇n+1 =</p><p>dt</p><p>dθ</p><p>.</p><p>Exercise 10 - GUILHERME SHOITI</p><p>YOSHIMATSU GIARDINI</p><p>Assume that the Lagrangian is a polynomial inq̇ of no higher order than</p><p>quadratic. Convert the2n equations (8.2) and (8.14)</p><p>pi =</p><p>∂L</p><p>∂q̇i</p><p>ṗi =</p><p>∂L</p><p>∂qi</p><p>, (10.1)</p><p>into 2n equations forq̇i and ṗi in terms ofq and p, using the matrix form of the</p><p>Lagrangian. Show that these are the same equations as would be obtained from the</p><p>Hamilton’s equations of motion.</p><p>First and foremost, we must note that the dependencies of the Lagrangian and</p><p>the Hamiltonian are different, such that we have to use the same treatment of ther-</p><p>modynamic potentials in the classical mechanics subject. Otherwise, we would have a</p><p>paradox where</p><p>H = T + V (10.2)</p><p>L = T − V and (10.3)</p><p>∇⃗q⃗H = − ˙⃗p (10.4)</p><p>∇⃗q⃗L = ˙⃗p . (10.5)</p><p>29</p><p>30 CHAPTER 10. EXERCISE 10 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>The equation</p><p>above (10.5) is only correct when we consider the dependencies of</p><p>the Lagrangian and the Hamiltonian. The Lagrangian has dependencies on</p><p>L (~q;_~q; t) (10.6)</p><p>and the Hamiltonian</p><p>H (~q; ~p(~q;_~q); t) (10.7)</p><p>thus if we want</p><p>dH = dL (10.8)</p><p>we must write</p><p>dL =</p><p>�</p><p>@L</p><p>@q</p><p>�</p><p>dq+</p><p>�</p><p>@L</p><p>@_q</p><p>�</p><p>d_q (10.9)</p><p>and</p><p>dH =</p><p>�</p><p>@H</p><p>@q</p><p>�</p><p>dq+</p><p>�</p><p>@H</p><p>@p</p><p>�</p><p>dp , (10.10)</p><p>now considering explicitly that ~p = ~p(~q;_~q), the variation of the Hamiltonian with re-</p><p>spect only to ~q( _~qis held constant) may be written as</p><p>dH =</p><p>��</p><p>@H</p><p>@q</p><p>�</p><p>+</p><p>�</p><p>@H</p><p>@p</p><p>@p</p><p>@q</p><p>��</p><p>dq . (10.11)</p><p>31</p><p>In other words,</p><p>∇⃗q⃗L = −∇⃗q⃗H is true (10.12)</p><p>only if</p><p>∇⃗q⃗</p><p>∣∣∣∣∣</p><p>˙⃗q</p><p>L = −∇⃗q⃗</p><p>∣∣∣∣∣</p><p>p⃗</p><p>H meaning that (10.13)</p><p>˙⃗q is held constant when taking the derivative of the Lagrangian and⃗p is held</p><p>constant when taking the derivative of the Hamiltonian.</p><p>With these ideas in mind, we may solve the exercise. Assuming a Lagrangian as</p><p>a polynomial with no higher order inq̇</p><p>L(q⃗, ˙⃗q, t) = a(q⃗, t) + ˙⃗q † · b⃗(q⃗, t) + 1</p><p>2</p><p>˙⃗q †←→C (q⃗, t) ˙⃗q , (10.14)</p><p>where a(q⃗, t) is an arbitrary function with dependence on⃗q and t, b⃗(q⃗, t) an arbitrary</p><p>vector that also depends onq⃗ and t, lastly</p><p>←→</p><p>C (q⃗, t) a tensor with the same dependencies</p><p>as the aforementioned mathematical entities. In Einstein notation we can find⃗p by</p><p>taking the derivative of the Lagrangian with respect tȯ⃗q (holding q⃗) constant</p><p>pj =</p><p>∂L</p><p>∂q̇j</p><p>=</p><p>�</p><p>�</p><p>�</p><p>��∂a(q⃗, t)</p><p>∂q̇j</p><p>+</p><p>∂q̇i</p><p>∂q̇j</p><p>bi(q⃗, t) +</p><p>1</p><p>2</p><p>(</p><p>∂q̇i</p><p>∂q̇j</p><p>Cikq̇k + q̇iCik</p><p>∂q̇k</p><p>∂q̇j</p><p>)</p><p>.</p><p>We know that</p><p>∂q̇i</p><p>∂q̇j</p><p>= δij thus (10.15)</p><p>32 CHAPTER 10. EXERCISE 10 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>we have</p><p>∂L</p><p>∂q̇j</p><p>= bj(q⃗, t) +</p><p>1</p><p>2</p><p>(Cjkq̇k + q̇iCij) now</p><p>because the Hamiltonian is Hermitian and the Lagrangian is a Legendre trans-</p><p>form away from the Hamiltonian, which implies it is also Hermitian, thenCij = Cji,</p><p>meaning that</p><p>∂L</p><p>∂q̇j</p><p>= bj(q⃗, t) + Cjkq̇k = pj .</p><p>Now, we must isolate theq̇ variable as a function ofp, then we must reverse the</p><p>equation obtained above</p><p>bj(q⃗, t) + Cjkq̇k = pj</p><p>Cjkq̇k = pj − bj(q⃗, t)</p><p>q̇k = C−1</p><p>jk (pj − bj(q⃗, t))</p><p>or in matricial notation</p><p>˙⃗q =</p><p>←→</p><p>C −1(p⃗− b⃗(q⃗, t)) . (10.16)</p><p>Now we must find an equation of ˙⃗p in terms of q⃗ and p⃗, for this we use the</p><p>equation suggested in the exercise</p><p>˙⃗p = ∇⃗q⃗</p><p>∣∣∣∣∣</p><p>˙⃗q</p><p>L .</p><p>33</p><p>This calculation should be straightforward, but⃗̇p has dependencies on ˙⃗q, so be-</p><p>fore taking the derivative, we must replacė⃗q by using the relation (10.16), also, we will</p><p>omit the dependencies of the functions i.e.f(x, y)→ f to simplify the calculations and</p><p>we will use standard matrix notation for the calculations below (in the derivative with</p><p>respect to q⃗ we keep ˙⃗q constant)</p><p>˙⃗p = ∇⃗q⃗a+ ˙⃗q y</p><p>(</p><p>∇⃗q⃗ · b⃗</p><p>)</p><p>+</p><p>1</p><p>2</p><p>(</p><p>˙⃗q y</p><p>(</p><p>∇⃗q⃗</p><p>←→</p><p>C</p><p>)</p><p>˙⃗q</p><p>)</p><p>= ∇⃗q⃗a+ [p⃗ y − b⃗ y]</p><p>←→</p><p>C � 1 y</p><p>(</p><p>∇⃗q⃗⃗b</p><p>)</p><p>+</p><p>1</p><p>2</p><p>(</p><p>[p⃗ y − b⃗ y]</p><p>←→</p><p>C � 1 y</p><p>(</p><p>∇⃗q⃗</p><p>←→</p><p>C</p><p>)←→</p><p>C � 1[p⃗− b⃗]</p><p>)</p><p>.</p><p>(10.17)</p><p>These are the movement equations of ˙⃗q and ˙⃗p obtained in terms of q⃗ and p⃗</p><p>through the Lagrange equations, now, we must compare these results with the Hamil-</p><p>ton equations ones. To do this, we must first find said Hamiltonian through the Leg-</p><p>endre transform</p><p>H (q⃗, p⃗, t) = ˙⃗q · p⃗− L</p><p>= ˙⃗q · p⃗− a− ˙⃗q y · b⃗− 1</p><p>2</p><p>˙⃗q y←→C ˙⃗q</p><p>= [p⃗ y − b⃗ y]</p><p>←→</p><p>C � 1 yp⃗− a− [p⃗ y − b⃗ y]</p><p>←→</p><p>C � 1 y⃗b− 1</p><p>2</p><p>[p⃗ y − b⃗ y]</p><p>←→</p><p>C � 1 y←→C</p><p>←→</p><p>C � 1[p⃗− b⃗]</p><p>= −a+ [p⃗ y − b⃗ y]</p><p>←→</p><p>C � 1 y[p⃗− b⃗]− 1</p><p>2</p><p>[p⃗ y − b⃗ y]</p><p>←→</p><p>C � 1 y←→C</p><p>←→</p><p>C � 1[p⃗− b⃗] . (10.18)</p><p>to proceed we note that</p><p>←→</p><p>C</p><p>←→</p><p>C 1 = I , then</p><p>H (q⃗, p⃗, t) = −a+ [p⃗ y − b⃗ y]</p><p>←→</p><p>C � 1 y[p⃗− b⃗]− 1</p><p>2</p><p>[p⃗ y − b⃗ y]</p><p>←→</p><p>C � 1 y[p⃗− b⃗]</p><p>= −a+ 1</p><p>2</p><p>[p⃗ y − b⃗ y]</p><p>←→</p><p>C � 1 y[p⃗− b⃗] or (10.19)</p><p>= −a+ 1</p><p>2</p><p>[</p><p>p⃗ y←→C � 1 yp⃗− b⃗ y←→C � 1 yp⃗− p⃗ y←→C � 1 y⃗b+ b⃗ y←→C � 1 y⃗b</p><p>]</p><p>.(10.20)</p><p>Now in possession of the Hamiltonian, we use Hamilton’s equations to verify if</p><p>the movement equations are equal to the ones obtained from the Lagrangian method.</p><p>The first Hamilton equation is</p><p>34 CHAPTER 10. EXERCISE 10 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>˙⃗q = ∇⃗p⃗</p><p>∣∣∣∣∣</p><p>q⃗</p><p>H , (10.21)</p><p>where we note that∇⃗p⃗ · p⃗ = 1, (∇⃗p⃗ · p⃗)† = 1, ∇⃗p⃗ a = 0, ∇⃗p⃗ · b⃗ = 0 and ∇⃗p⃗</p><p>←→</p><p>C = 0.</p><p>We may make these assumptions because in a closed physical system momentum is</p><p>conserved, which implies in an independence between⃗p and q⃗ such that ∇⃗p⃗ · f(q⃗) = 0,</p><p>with these remarks, we may proceed with the calculations to obtain</p><p>∇⃗p⃗ H(q⃗, p⃗, t) =</p><p>∇⃗p⃗</p><p>2</p><p>[</p><p>p⃗ †←→C −1 †p⃗− b⃗ †←→C −1 †p⃗− p⃗ †←→C −1 †⃗b+ b⃗ †←→C −1 †⃗b</p><p>]</p><p>− ∇⃗p⃗ a</p><p>=</p><p>1</p><p>2</p><p>[←→</p><p>C −1 †p⃗+ p⃗ †←→C −1 † − b⃗ †←→C −1 † −</p><p>←→</p><p>C −1 †⃗b</p><p>]</p><p>=</p><p>1</p><p>2</p><p>←→</p><p>C −1 †</p><p>(</p><p>p⃗− b⃗</p><p>)</p><p>+</p><p>1</p><p>2</p><p>(</p><p>p⃗ † − b⃗ †</p><p>)←→</p><p>C −1 . (10.22)</p><p>Because of our previous assumption that the Lagrangian must be Hermitian</p><p>(real and equal to its transpose), then the matrix</p><p>←→</p><p>C should also be Hermitian, then</p><p>we may write the result from above as</p><p>∇⃗p⃗ H(q⃗, p⃗, t) =</p><p>←→</p><p>C −1 †</p><p>(</p><p>p⃗− b⃗</p><p>)</p><p>,</p><p>which is the same result we obtained by calculating</p><p>∇⃗ ˙⃗q</p><p>∣∣∣∣∣</p><p>q⃗</p><p>L = p⃗ and (10.23)</p><p>isolatingp⃗ in terms of ˙⃗q. Now we must verify the second result from Hamilton’s</p><p>equations</p><p>˙⃗p = −∇⃗q⃗</p><p>∣∣∣∣∣</p><p>p⃗</p><p>H</p><p>(10.24)</p><p>35</p><p>The Hamiltonian has a dependence on</p><p>←→</p><p>C −1, meaning that we first have to find</p><p>the derivatives of</p><p>←→</p><p>C −1 to obtain a reasonable result. We start by noting that</p><p>∇⃗q⃗I = 0</p><p>= ∇⃗q⃗</p><p>(←→</p><p>C</p><p>←→</p><p>C −1</p><p>)</p><p>= ∇⃗q⃗</p><p>(←→</p><p>C</p><p>)←→</p><p>C −1 +</p><p>←→</p><p>C ∇⃗q⃗</p><p>(←→</p><p>C −1</p><p>)</p><p>,</p><p>now we solve for∇⃗q⃗</p><p>←→</p><p>C −1</p><p>←→</p><p>C ∇⃗q⃗</p><p>(←→</p><p>C −1</p><p>)</p><p>= −∇⃗q⃗</p><p>(←→</p><p>C</p><p>)←→</p><p>C −1 , (10.25)</p><p>now we multiply both sides by</p><p>←→</p><p>C −1, getting</p><p>∇⃗q⃗</p><p>(←→</p><p>C −1</p><p>)</p><p>= −</p><p>←→</p><p>C −1∇⃗q⃗</p><p>(←→</p><p>C</p><p>)←→</p><p>C −1 , (10.26)</p><p>which is the derivative of the inverse of a matrix. Now we may take the deriva-</p><p>tive of the Hamiltonian with respect to⃗q while maintaining⃗p constant.</p><p>− ˙⃗p = ∇⃗q⃗H</p><p>= −∇⃗q⃗a+</p><p>[p⃗ † − b⃗ †]∇⃗q⃗</p><p>(←→</p><p>C −1 †</p><p>)</p><p>[p⃗− b⃗]−</p><p>(</p><p>∇⃗q⃗⃗b</p><p>†</p><p>)←→</p><p>C −1 †[p⃗− b⃗]− [p⃗ † − b⃗ †]</p><p>←→</p><p>C −1 †</p><p>(</p><p>∇⃗q⃗⃗b</p><p>)</p><p>2</p><p>,</p><p>now we substitute the derivative of matrix</p><p>←→</p><p>C −1 found earlier into the calcula-</p><p>tion</p><p>− ˙⃗p = −∇⃗q⃗a+</p><p>[p⃗ † − b⃗ †]</p><p>←→</p><p>C −1†∇⃗q⃗</p><p>(←→</p><p>C</p><p>)←→</p><p>C −1[p⃗− b⃗]</p><p>2</p><p>−</p><p>(</p><p>∇⃗q⃗⃗b</p><p>†</p><p>)←→</p><p>C −1 †[p⃗− b⃗] + [p⃗ † − b⃗ †]</p><p>←→</p><p>C −1 †</p><p>(</p><p>∇⃗q⃗⃗b</p><p>)</p><p>2</p><p>, (10.27)</p><p>36 CHAPTER 10. EXERCISE 10 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>finally, because</p><p>←→</p><p>C is Hermitian andp⃗ and b⃗ are real, we may write</p><p>(</p><p>∇⃗q⃗⃗b</p><p>†</p><p>)←→</p><p>C −1 †[p⃗− b⃗] = [p⃗ † − b⃗ †]</p><p>←→</p><p>C −1 †</p><p>(</p><p>∇⃗q⃗⃗b</p><p>)</p><p>, (10.28)</p><p>which brings us to</p><p>− ˙⃗p = −∇⃗q⃗a+</p><p>−[p⃗ † − b⃗ †]</p><p>←→</p><p>C −1†∇⃗q⃗</p><p>(←→</p><p>C</p><p>)←→</p><p>C −1[p⃗− b⃗]− 2[p⃗ † − b⃗ †]</p><p>←→</p><p>C −1 †</p><p>(</p><p>∇⃗q⃗⃗b</p><p>)</p><p>2</p><p>= −∇⃗q⃗L = ∇⃗q⃗H (10.29)</p><p>which proves the result that the Lagrange equations and the Hamilton equations</p><p>produce the same dynamics.</p><p>Exercise 11 - GEOVANE NAYSINGER</p><p>FIGURE 11.1</p><p>This solution is based on the book Problems and Solution on Mechanics of</p><p>QiangYuan-qi, Gu En-pu, Cheng Jia-fu, ti Ze-hua,Ylang De-tian</p><p>a) We call v as the incident speed, reflected back with a speedv′ and V is the</p><p>wall speed towards the particle. Then, consider a collision of the particle with one wall</p><p>and, as the collision is inelastic, the kinectic energy is conserved and the relative speed</p><p>before and after collision is equal. So, we can write</p><p>v + V = v′ − V</p><p>that is</p><p>v′ = v + 2V .</p><p>After each collision, the intensity of the particle momentum gains an amount 2mV</p><p>and m is the mass of the particle. When the walls are at a distance x apart, as V is</p><p>37</p><p>38 CHAPTER 11. EXERCISE 11 - GEOVANE NAYSINGER</p><p>much smaller than the speed of the particle, the time interval between two consecutive</p><p>collisions is</p><p>T =</p><p>x</p><p>p/m</p><p>=</p><p>xm</p><p>p</p><p>p is the particle linear momentum. The change momentum in time dt is</p><p>dp = 2mV</p><p>dt</p><p>T</p><p>=</p><p>2V pdt</p><p>x</p><p>.</p><p>The movement of the wall with speed V can be write as</p><p>x = x0 − 2V t,</p><p>then</p><p>dp = −pdx</p><p>x</p><p>the integration of it, withp = p0 and x = x0, is</p><p>p =</p><p>p0x0</p><p>x0 − 2V t</p><p>.</p><p>b) The momentum acquired by the particle with one collision is</p><p>p+ 2mV − (−p) = 2p+ 2mV</p><p>and the interval with two consecutive collisions with the wall is</p><p>T ′ =</p><p>2xm</p><p>p</p><p>so the change of momentum is</p><p>dp = 2(p+mV )</p><p>dt</p><p>T ′ .</p><p>Then,</p><p>dp</p><p>dt</p><p>=</p><p>(p+mV )p</p><p>xm</p><p>≈ p2</p><p>xm</p><p>=</p><p>p20x</p><p>2</p><p>0</p><p>mx3</p><p>due to p</p><p>m</p><p>>> V . This is the</p><p>force exerted by the wall on the particle. To keep the walls</p><p>moving at constant speed, a force of the same magnitude must be apply to each wall.</p><p>The problem can also be solved using the Hamiltonian formalism. The hamiltonian is</p><p>H =</p><p>1</p><p>2</p><p>m</p><p>( p</p><p>m</p><p>+ V</p><p>)2</p><p>39</p><p>and</p><p>H =</p><p>1</p><p>2m</p><p>(p+mV )2 ≈ p2</p><p>2m</p><p>=</p><p>p20x</p><p>2</p><p>0</p><p>2mx2</p><p>.</p><p>The force applied on the particle iṡp which is given by the hamiltonian equation</p><p>ṗ = −∂H</p><p>∂x</p><p>=</p><p>p20x</p><p>2</p><p>0</p><p>mx3</p><p>.</p><p>Exercise 12 - JHORDAN SILVEIRA DE</p><p>BORBA</p><p>Write the problem o central force motion of two mass points in Hamiltonian</p><p>formulation, eliminating the cyclic variables, and reducing the problem to quadratures</p><p>Resposta:</p><p>The reduced mass of the system being given byµ = m1m2</p><p>m1+m2</p><p>, the center of mass</p><p>is given byr and the Lagrangian of the system is:</p><p>L = T − V =</p><p>1</p><p>2</p><p>µ</p><p>(</p><p>ṙ2 + r2θ̇2</p><p>)</p><p>− V (r)</p><p>Where polar coordinate was used. For a two-dimensional system, we haver =</p><p>(r sin θ, r cos θ), then ṙ =</p><p>(</p><p>ṙ sin θ + r cos θθ̇, ṙ cos θ − r sin θθ̇</p><p>)</p><p>, and consequently:</p><p>ṙ2 =</p><p>(</p><p>ṙ sin θ + r cos θθ̇</p><p>)2</p><p>+</p><p>(</p><p>ṙ cos θ − r sin θθ̇</p><p>)2</p><p>=ṙ2</p><p>(</p><p>cos2 θ + sin2 θ</p><p>)</p><p>+ r2</p><p>(</p><p>cos2 θ + sin2 θ</p><p>)</p><p>θ̇2 + 2rṙ cos θ sin θθ̇ (1− 1)</p><p>=ṙ2 + r2θ̇2</p><p>The conjugate moment can then be given by the equations:</p><p>pr =</p><p>∂L</p><p>∂ṙ</p><p>= µṙ</p><p>pθ =</p><p>∂L</p><p>∂θ̇</p><p>= µr2θ̇</p><p>41</p><p>42 CHAPTER 12. EXERCISE 12 - JHORDAN SILVEIRA DE BORBA</p><p>And the Hamiltoanin is:</p><p>H =</p><p>∑</p><p>i</p><p>piq̇i − L</p><p>= prṙ + pθθ̇ − T + V</p><p>=</p><p>p2r</p><p>µ</p><p>+</p><p>(</p><p>pθ</p><p>µr</p><p>)2</p><p>− 1</p><p>2</p><p>µ</p><p>((</p><p>pr</p><p>µ</p><p>)2</p><p>+ r2</p><p>(</p><p>pθ</p><p>µr2</p><p>)2</p><p>)</p><p>+ V (r)</p><p>=</p><p>p2r</p><p>µ</p><p>+</p><p>pθ</p><p>µr2</p><p>2</p><p>− 1</p><p>2</p><p>p2r</p><p>µ</p><p>− 1</p><p>2</p><p>p2θ</p><p>µr2</p><p>+ V (r)</p><p>=</p><p>p2r</p><p>2µ</p><p>+</p><p>p2θ</p><p>2µr2</p><p>+ V (r)</p><p>We can also notice thatH = T + V = E. In addition we get the relationships:</p><p>ṙ =</p><p>∂H</p><p>∂pr</p><p>=</p><p>pr</p><p>µ</p><p>θ̇ =</p><p>∂H</p><p>∂pθ</p><p>=</p><p>pθ</p><p>µr2</p><p>ṗr = −</p><p>∂H</p><p>∂r</p><p>=</p><p>p2θ</p><p>µr3</p><p>− ∂V</p><p>∂r</p><p>ṗθ = −</p><p>∂H</p><p>∂θ</p><p>= 0</p><p>∂H</p><p>∂t</p><p>= −∂L</p><p>∂t</p><p>= 0</p><p>We have two motion constants:pθ and H = E. Let’s use them to eliminate the cyclic</p><p>variables and reduce the problem to quadratures. In differential equations, solving</p><p>an equation by quadrature means expressing the solution in terms of integrals. The</p><p>motion constants are:</p><p>pθ = µr2θ̇ → θ̇ =</p><p>pθ</p><p>µr2</p><p>E =</p><p>1</p><p>2</p><p>µ</p><p>(</p><p>ṙ2 +</p><p>p2θ</p><p>µ2r2</p><p>)</p><p>+ V (r) → ṙ =</p><p>√</p><p>2</p><p>E</p><p>µ</p><p>− 2</p><p>V</p><p>µ</p><p>− p2θ</p><p>µ2r2</p><p>Thus, the problem is then reduced to solving two differential equations forθ̇ and ṙ. To</p><p>r:</p><p>dr</p><p>dt</p><p>=</p><p>√</p><p>2</p><p>E</p><p>µ</p><p>− 2</p><p>V</p><p>µ</p><p>− p2θ</p><p>µ2r2</p><p>dt =</p><p>dr√</p><p>2E</p><p>µ</p><p>− 2V</p><p>µ</p><p>− p2θ</p><p>µ2r2</p><p>43</p><p>Then: ∫</p><p>rdr√</p><p>r2 (E − V )− p2θ/ (2µ)</p><p>=</p><p>√</p><p>2</p><p>µ</p><p>∫</p><p>dt</p><p>And for θ:</p><p>dθ</p><p>dt</p><p>dr</p><p>dt</p><p>=</p><p>pθ</p><p>µr2√</p><p>2E</p><p>µ</p><p>− 2V</p><p>µ</p><p>− p2θ</p><p>µ2r2√</p><p>2µ</p><p>pθ</p><p>dθ</p><p>dr</p><p>=</p><p>1</p><p>r2</p><p>r√</p><p>r2 (E − V )− p2θ/ (2µ)√</p><p>2µ</p><p>pθ</p><p>dθ =</p><p>dr</p><p>r</p><p>√</p><p>r2 (E − V )− p2θ/ (2µ)</p><p>So we have: √</p><p>2µ</p><p>pθ</p><p>∫</p><p>dθ =</p><p>∫</p><p>dr</p><p>r</p><p>√</p><p>r2 (E − V )− p2θ/ (2µ)</p><p>That is, with the following integrals written in terms of potential energy and constants</p><p>of motion, we were able to find the equations of motion by eliminating the cyclic vari-</p><p>ables:</p><p>√</p><p>2µ</p><p>pθ</p><p>∫</p><p>dθ =</p><p>∫</p><p>dr</p><p>r</p><p>√</p><p>r2 (E − V (r))− p2θ/ (2µ)</p><p>, e</p><p>∫</p><p>rdr√</p><p>r2 (E − V (r))− p2θ/ (2µ)</p><p>=</p><p>√</p><p>2</p><p>µ</p><p>∫</p><p>dt</p><p>Exercise 13 - EDGARD</p><p>KRETSCHMANN</p><p>It is already known that the Lagrangian of the double pendulum is given by</p><p>L =</p><p>(m1 +m2)l</p><p>2</p><p>1θ̇1</p><p>2</p><p>2</p><p>+</p><p>m2l</p><p>2</p><p>2θ̇2</p><p>2</p><p>2</p><p>+m2l1l2θ̇1θ̇2cos(θ2−θ1)−[(m1+m2)gl1cos(θ1)+m2gl2cos(θ2)]</p><p>(13.1)</p><p>Therefore, we can obtain the generalized momenta with the definition</p><p>p1 =</p><p>∂L</p><p>∂θ̇1</p><p>; p2 =</p><p>∂L</p><p>∂θ̇2</p><p>(13.2)</p><p>So</p><p>p1 = (m1 +m2)l</p><p>2</p><p>1θ̇1 +m2l1l2θ̇2cos(θ2 − θ1) (13.3)</p><p>p2 = m2l</p><p>2</p><p>2θ̇2 +m2l1l2θ̇1cos(θ2 − θ1) (13.4)</p><p>Now we can obtain the generalized velocities in terms of the generalized momenta and</p><p>coordinates</p><p>θ̇1 =</p><p>1</p><p>m2l21l</p><p>2</p><p>2</p><p>m2l</p><p>2</p><p>2p1 −m2l1l2cos(θ2 − θ1)p2</p><p>m1 +m2sin2(θ2 − θ1)</p><p>(13.5)</p><p>θ̇2 =</p><p>1</p><p>m2l21l</p><p>2</p><p>2</p><p>(m1 +m2)l</p><p>2</p><p>1p2 −m2l1l2cos(θ2 − θ1)p1</p><p>m1 +m2sin2(θ2 − θ1)</p><p>(13.6)</p><p>So, aplying the Legendre transformation to obtain the Hamiltonian we will find that</p><p>H =</p><p>1</p><p>m1 +m2sin2(θ2 − θ1)</p><p>(</p><p>p21</p><p>2l21</p><p>+</p><p>p22</p><p>2l22</p><p>m1 +m2</p><p>m2</p><p>−p1p2</p><p>l1l2</p><p>cos(θ2−θ1))+(m1+m2)gl1cos(θ1)+m2gl2cos(θ2)</p><p>(13.7)</p><p>45</p><p>46 CHAPTER 13. EXERCISE 13 - EDGARD KRETSCHMANN</p><p>So the equations of motion in the Hamiltonian formulation are</p><p>q̇j =</p><p>∂H</p><p>∂pj</p><p>; ṗj = −</p><p>∂H</p><p>∂qj</p><p>(13.8)</p><p>Therefore</p><p>θ̇1 =</p><p>1</p><p>m1 +m2sin2(θ2 − θ1)</p><p>[</p><p>p1</p><p>l21</p><p>− p2</p><p>l1l2</p><p>cos(θ2 − θ1)] (13.9)</p><p>θ̇2 =</p><p>1</p><p>m1 +m2sin2(θ2 − θ1)</p><p>[</p><p>p2</p><p>l22</p><p>(</p><p>m1 +m2</p><p>m2</p><p>)− p1</p><p>l1l2</p><p>cos(θ2 − θ1)] (13.10)</p><p>ṗ1 = −(m1 +m2)gl1sinθ1 −</p><p>∂T</p><p>∂θ1</p><p>(13.11)</p><p>ṗ2 = −m2gl2sinθ2 −</p><p>∂T</p><p>∂θ2</p><p>(13.12)</p><p>Where T is the kinetic energy. Then we have that</p><p>∂T</p><p>∂θ2</p><p>= A−B (13.13)</p><p>A =</p><p>(m1 +m2(1 + cos2(θ2 − θ1)))p1p2sin(θ2 − θ1)</p><p>l1l2(m1 +m2sin2(θ2 − θ1))2</p><p>(13.14)</p><p>B =</p><p>p21</p><p>2l21</p><p>+</p><p>p22</p><p>2l22</p><p>m1+m2</p><p>m2</p><p>(m1 +m2sin2(θ2 − θ1))2</p><p>(13.15)</p><p>Since we are differentiating a functionf(θ2 − θ1) so</p><p>∂T</p><p>∂θ2</p><p>= − ∂T</p><p>∂θ1</p><p>(13.16)</p><p>We have obtained the equations of motion</p><p>Exercise 14 - JENIFER ANDRADE DE</p><p>MATOS</p><p>The Lagrangian for a system can be written as</p><p>L = aẋ2 + b</p><p>ẏ</p><p>x</p><p>+ cẋẏ + fy2ẋż + gẏ2 − k</p><p>√</p><p>x2 + y2,</p><p>where a, b, c, f, g and k are constant. What is the Hamiltonian? What quantities</p><p>are conserved?</p><p>From the Lagrangian we can obtain the conjugated momenta:</p><p>px =</p><p>∂L</p><p>∂ẋ</p><p>, py =</p><p>∂L</p><p>∂ẏ</p><p>, pz =</p><p>∂L</p><p>∂ż</p><p>(14.1)</p><p>px = 2aẋ+ cẏ + fy2ż (14.2)</p><p>py =</p><p>b</p><p>x</p><p>+ cẋ+ 2gẏ (14.3)</p><p>pz = fy2ẋ (14.4)</p><p>From that we write the Hamiltonian:</p><p>47</p><p>48 CHAPTER 14. EXERCISE 14 - JENIFER ANDRADE DE MATOS</p><p>H = pxẋ+ pyẏ + pz ż − L (14.5)</p><p>Now, let́ s write ẋ, ẏ, ż in terms ofpx, py and pz, from the equations (14.2), (14.3)</p><p>and (14.4):</p><p>ẋ =</p><p>pz</p><p>fy2</p><p>(14.6)</p><p>ẏ =</p><p>py</p><p>2g</p><p>− b</p><p>2gx</p><p>− cpz</p><p>2fgy2</p><p>(14.7)</p><p>ż =</p><p>px</p><p>fy2</p><p>− cpy</p><p>2fgy2</p><p>+</p><p>c2pz</p><p>2f 2gy4</p><p>− 2apz</p><p>f 2y4</p><p>+</p><p>bc</p><p>2fgxy2</p><p>(14.8)</p><p>Rewriting the Hamiltonian:</p><p>H = pxẋ+ pyẏ + pz ż − aẋ2 − b</p><p>ẏ</p><p>x</p><p>− cẋẏ − fy2ẋż − gẏ2 + k</p><p>√</p><p>x2 + y2 (14.9)</p><p>Substituting (14.6), (14.7) and (14.8) in (14.9) we can show that the Hamiltonian</p><p>can be written as:</p><p>H =</p><p>pxpz</p><p>fy2</p><p>− bpy</p><p>2gx</p><p>+</p><p>p2y</p><p>4g</p><p>− cpypz</p><p>2fgy2</p><p>+</p><p>bcpz</p><p>2fgxy2</p><p>+</p><p>p2z</p><p>f 2y4</p><p>(</p><p>c2</p><p>4g</p><p>− a</p><p>)</p><p>+</p><p>b2</p><p>4gx2</p><p>−k</p><p>√</p><p>x2 + y2 (14.10)</p><p>In this Halmitonian, the H is conserved, since it’s not explicitly dependent of the</p><p>time.</p><p>Exercise 15 - GUILHERME SHOITI</p><p>YOSHIMATSU GIARDINI</p><p>A dynamical systema has the Lagrangian</p><p>L = q̇21 +</p><p>q̇22</p><p>a+ b q21</p><p>+ k1q</p><p>2</p><p>1 + k2q̇1q̇2 , (15.1)</p><p>where a,b,k1 and k2 are constants. Find the equations of motion in the Hamilto-</p><p>nian formulation.</p><p>To solve this problem, first we note that the Hamiltonian is related to the La-</p><p>grangian via a Legendre transform</p><p>H(q⃗, p⃗, t) = ˙⃗q · p⃗−L( ˙⃗q, ˙⃗p, t) (15.2)</p><p>or in Einstein’s index notation</p><p>H(q⃗, p⃗, t) = q̇ipi −L( ˙⃗q, ˙⃗p, t) . (15.3)</p><p>Notice that while the Lagrangian depends oṅ⃗q and ˙⃗p, the Hamiltonian depends</p><p>on q⃗ and p⃗ producing 2n first order differential problems oḟq and ṗ when the Hamil-</p><p>ton equations are solved. In opposition to then second order differential equations</p><p>49</p><p>50 CHAPTER 15. EXERCISE 15 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>produced in the Lagrangian case. This procedure can sometimes facilitate the systems’</p><p>solutions.</p><p>Due to this dependence in⃗q and p⃗, we need to replace ˙⃗q and ˙⃗p for the aforemen-</p><p>tioned variables while we make the transform. In other words we wantL ( ˙⃗q, ˙⃗p, t) →</p><p>L (q⃗, p⃗, t), it can be done through the relations</p><p>ṗi =</p><p>∂L</p><p>∂qi</p><p>(15.4)</p><p>pi =</p><p>L</p><p>∂q̇i</p><p>, (15.5)</p><p>then we proceed to find the differential equations of⃗̇q and ˙⃗p in terms ofp⃗ and q⃗.</p><p>p1 =</p><p>∂L</p><p>∂q̇1</p><p>=</p><p>∂</p><p>∂q̇1</p><p>(</p><p>q̇21 +</p><p>q̇22</p><p>a+ b q21</p><p>+ k1q</p><p>2</p><p>1 + k2q̇1q̇2</p><p>)</p><p>= 2q̇1 + k2q̇2 and (15.6)</p><p>p2 =</p><p>∂L</p><p>∂q̇2</p><p>=</p><p>∂</p><p>∂q̇2</p><p>(</p><p>q̇21 +</p><p>q̇22</p><p>a+ b q21</p><p>+ k1q</p><p>2</p><p>1 + k2q̇1q̇2</p><p>)</p><p>=</p><p>2q̇2</p><p>a+ b q21</p><p>+ k2q̇1 . (15.7)</p><p>With p1 and p2 equations in hands, we must remove the dependencies of the</p><p>Lagrangian of ˙⃗q and ˙⃗p. To do this, we only have to solve the system of equations</p><p>of (15.6) and (15.7) to obtain q̇1 and q̇2, the canonical position coordinates in terms</p><p>of the canonical momenta. It is feasible by takingp1 − p2 and doing some algebra</p><p>manipulation</p><p>p2 − k2</p><p>p1</p><p>2</p><p>=</p><p>2q̇2</p><p>a+ b q21</p><p>− k22 q̇2</p><p>2</p><p>+ k2q̇1 − k2q̇1</p><p>=</p><p>2q̇2</p><p>a+ b q21</p><p>− k22 q̇2</p><p>2</p><p>, (15.8)</p><p>51</p><p>then</p><p>q̇2 =</p><p>(</p><p>p2 �</p><p>k2p1</p><p>2</p><p>)(</p><p>2</p><p>a+ b q2</p><p>1</p><p>�</p><p>k2</p><p>2</p><p>2</p><p>)� 1</p><p>=</p><p>(</p><p>p2 �</p><p>k2p1</p><p>2</p><p>)(</p><p>4 � k2</p><p>2(a+ b q2</p><p>1)</p><p>2(a+ b q2</p><p>1)</p><p>)� 1</p><p>=</p><p>(</p><p>p2 �</p><p>k2p1</p><p>2</p><p>)(</p><p>2(a+ b q2</p><p>1)</p><p>4 � k2</p><p>2(a+ b q2</p><p>1)</p><p>)</p><p>= (2p2 � k2p1)</p><p>(</p><p>a+ b q2</p><p>1</p><p>4 � k2</p><p>2(a+ b q2</p><p>1)</p><p>)</p><p>. (15.9)</p><p>For q̇1</p><p>we do a similar manipulation</p><p>p1 �</p><p>k2p2</p><p>2</p><p>(a+ b q2</p><p>1) = 2q̇1 �</p><p>k2</p><p>2</p><p>2</p><p>(a+ b q2</p><p>1)q̇1 + k2q̇2 � k2q̇2</p><p>=</p><p>(</p><p>2 �</p><p>k2</p><p>2</p><p>2</p><p>(a+ b q2</p><p>1)</p><p>)</p><p>q̇1 , (15.10)</p><p>then</p><p>q̇1 =</p><p>(</p><p>p1 �</p><p>k2p2</p><p>2</p><p>(a+ b q2</p><p>1)</p><p>)(</p><p>2 � k2</p><p>2(a+ b q2</p><p>1)</p><p>)� 1</p><p>=</p><p>(</p><p>p1 �</p><p>k2p2</p><p>2</p><p>(a+ b q2</p><p>1)</p><p>)(</p><p>2</p><p>4 � k2</p><p>2(a+ b q2</p><p>1)</p><p>)</p><p>=</p><p>(</p><p>2p1 � k2p2(a+ b q2</p><p>1)</p><p>)( 1</p><p>4 � k2</p><p>2(a+ b q2</p><p>1)</p><p>)</p><p>=</p><p>(</p><p>2p1 � k2p2(a+ b q2</p><p>1)</p><p>4 � k2</p><p>2(a+ b q2</p><p>1)</p><p>)</p><p>. (15.11)</p><p>Now we substitute equations (15.10) and15,11 into the Lagrangian, effectively</p><p>makingL( ˙⃗q, ˙⃗p, t) ! L(q⃗, p⃗, t).</p><p>52 CHAPTER 15. EXERCISE 15 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>L = q̇21 +</p><p>q̇22</p><p>a+ b q21</p><p>+ k1q</p><p>2</p><p>1 + k2q̇1q̇2</p><p>=</p><p>(</p><p>2p1 − k2p2(a+ b q21)</p><p>4− k22(a+ b q21)</p><p>)2</p><p>+ (2p2 − k2p1)2</p><p>(</p><p>a+ b q21</p><p>4− k22(a+ b q21)</p><p>)2</p><p>1</p><p>a+ b q21</p><p>+ k1q</p><p>2</p><p>1</p><p>+ k2 (2p2 − k2p1)</p><p>(</p><p>a+ b q21</p><p>4− k22(a+ b q21)</p><p>)(</p><p>2p1 − k2p2(a+ b q21)</p><p>4− k22(a+ b q21)</p><p>)</p><p>. (15.12)</p><p>Now we obtain the Hamiltonian via the Legendre transform</p><p>H(q⃗, p⃗, t) = ˙⃗q · p⃗−L(q⃗, p⃗, t)</p><p>= p1q̇1 + p2q̇2 −L(q⃗, p⃗, t)</p><p>= p1</p><p>(</p><p>2p1 − k2p2(a+ b q21)</p><p>4− k22(a+ b q21)</p><p>)</p><p>+ p2 (2p2 − k2p1)</p><p>(</p><p>a+ b q21</p><p>4− k22(a+ b q21)</p><p>)</p><p>−</p><p>(</p><p>2p1 − k2p2(a+ b q21)</p><p>4− k22(a+ b q21)</p><p>)2</p><p>− (2p2 − k2p1)2</p><p>(</p><p>a+ b q21</p><p>4− k22(a+ b q21)</p><p>)2</p><p>1</p><p>a+ b q21</p><p>− k1q21</p><p>− k2 (2p2 − k2p1)</p><p>(</p><p>a+ b q21</p><p>4− k22(a+ b q21)</p><p>)(</p><p>2p1 − k2p2(a+ b q21)</p><p>4− k22(a+ b q21)</p><p>)</p><p>.</p><p>Now we have to expand all the terms, which results in</p><p>H(q⃗, p⃗, t) = p1</p><p>(</p><p>2p1 − k2p2(a+ b q21)</p><p>4− k22(a+ b q21)</p><p>)</p><p>+ p2 (2p2 − k2p1)</p><p>(</p><p>a+ b q21</p><p>4− k22(a+ b q21)</p><p>)</p><p>−</p><p>(</p><p>4p2q + k22(a+ b q21)</p><p>2p22 − 4p1p2k2(a+ b q21)</p><p>(4− k22[a+ b q21])</p><p>2</p><p>)</p><p>−</p><p>(</p><p>4p22 + k22p</p><p>2</p><p>1 − 4k2p1p2</p><p>)( a+ b q21</p><p>(4− k22[a+ b q21])</p><p>2</p><p>)</p><p>− k1q21</p><p>− k2(a+ b q21)</p><p>(4− k22[a+ b q21])</p><p>2</p><p>(</p><p>4p1p2 − 2k2p</p><p>2</p><p>1 − k2(a+ b q21)p</p><p>2</p><p>2 + k22(a+ b q21)p1p2</p><p>)</p><p>.(15.13)</p><p>(15.14)</p><p>Now we proceed to multiply and cancel all the separated terms</p><p>53</p><p>H(q⃗, p⃗, t) =</p><p>1</p><p>(4− k22[a+ b q21])</p><p>2</p><p>{(</p><p>2p21 − k2p1p2(a+ b q21)</p><p>) (</p><p>4− k22[a+ b q21]</p><p>)</p><p>+</p><p>(</p><p>2p22(a+ b q21)− k2p1p2(a+ b q21)</p><p>) (</p><p>4− k22[a+ b q21]</p><p>)</p><p>− 4p21 +((((((((</p><p>k22p</p><p>2</p><p>2(a+ b q21)</p><p>2 − 4p1p2k2(a+ b q21)</p><p>−</p><p>(</p><p>4p22 + k22p</p><p>2</p><p>1 −�����4k2p1p2</p><p>)</p><p>(a+ b q21)</p><p>−</p><p>(</p><p>�����4k2p1p2 − 2k22p</p><p>2</p><p>1 − ���</p><p>1</p><p>2 k22p</p><p>2</p><p>2(a+ b q21) + k32(a+ b q21)p1p2</p><p>)</p><p>(a+ b q21)</p><p>}</p><p>− k1q21 ,</p><p>(15.15)</p><p>then</p><p>H(q⃗, p⃗, t) =</p><p>1</p><p>(4− k22[a+ b q21])</p><p>2</p><p>{</p><p>2p21(���</p><p>2</p><p>4 −������</p><p>k22[a+ b q21])− k2p1p2(a+ b q21)(�4−������</p><p>k22[a+ b q21 ])</p><p>+ 2p22(a+ b q21)(���</p><p>2</p><p>4 − ���</p><p>(1/2)</p><p>1 k22[a+ b q21])− k2p1p2(a+ b q21)(4− k22[a+ b q21])</p><p>− �</p><p>�4p21 +(((((((((</p><p>4k2p1p2(a+ b q21) −�������</p><p>4p22(a+ b q21) − k22p21(a+ b q21) +((((((((</p><p>2k22p</p><p>2</p><p>1(a+ b q21)</p><p>− (((((((((</p><p>k32p1p2(a+ b q21)</p><p>2 +((((((((</p><p>k22p</p><p>2</p><p>2(a+ b q21)</p><p>}</p><p>− k1q21 , (15.16)</p><p>after the cancellations we get</p><p>H(q⃗, p⃗, t) =</p><p>1</p><p>(4− k22[a+ b q21])</p><p>2</p><p>{</p><p>4p21 + p22(a+ b q21)(4− k22[a+ b q21]) + k32p1p2(a+ b q21)</p><p>− k2p1p2(a+ b q21)(4− k22[a+ b q21])</p><p>}</p><p>− k1q21 . (15.17)</p><p>Then, simplifying even more, we finally get</p><p>H(q⃗, p⃗, t) =</p><p>1</p><p>(4− k22[a+ b q21])</p><p>{</p><p>p21 + p22(a+ b q21)− p1p2k2(a+ b q21)</p><p>}</p><p>− k1q21</p><p>(15.18)</p><p>54 CHAPTER 15. EXERCISE 15 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>Now, in possession of the Hamiltonian as a function of the right coordinates, we</p><p>find the motion differential equations through the Hamilton equations, written as</p><p>q̇i =</p><p>∂H</p><p>∂pi</p><p>(15.19)</p><p>ṗi = −∂H</p><p>∂qi</p><p>(15.20)</p><p>or in vector notation</p><p>˙⃗q = ∇⃗p⃗ H (15.21)</p><p>˙⃗p = −∇⃗q⃗ H . (15.22)</p><p>Thus, we find</p><p>q̇1 =</p><p>∂H</p><p>∂p1</p><p>=</p><p>∂</p><p>∂p1</p><p>[</p><p>1</p><p>(4− k22[a+ b q21])</p><p>{</p><p>p21 + p22(a+ b q21)− p1p2k2(a+ b q21)</p><p>}</p><p>− k1q21</p><p>]</p><p>=</p><p>1</p><p>(4− k22[a+ b q21])</p><p>(2p1 − p2k2(a+ b q1)) , (15.23)</p><p>q̇2 =</p><p>∂H</p><p>∂p2</p><p>=</p><p>∂</p><p>∂p2</p><p>[</p><p>1</p><p>(4− k22[a+ b q21])</p><p>{</p><p>p21 + p22(a+ b q21)− p1p2k2(a+ b q21)</p><p>}</p><p>− k1q21</p><p>]</p><p>=</p><p>(a+ b q21)</p><p>(4− k22[a+ b q21])</p><p>(2p2 − p1k2) , (15.24)</p><p>55</p><p>ṗ1 = −∂H</p><p>∂q1</p><p>= −</p><p>{</p><p>2b q1k</p><p>2</p><p>2(</p><p>4− k22[a+ b q21]</p><p>)2 (p21 + p22(a+ b q21)− p1p2k2(a+ b q21)</p><p>)</p><p>+</p><p>1(</p><p>4− k22[a+ b q21]</p><p>) (2b q1 p22 − 2b q1 p1p2k2</p><p>)</p><p>− 2q1k1</p><p>}</p><p>and (15.25)</p><p>ṗ2 = −∂H</p><p>∂q2</p><p>= 0 (15.26)</p><p>Exercise 16 - JACKSON GALVÃO</p><p>16. Prove that in a Kepler elliptic orbit with small eccentricitye the angular motion</p><p>of a particle as viewed from theempty focus of the ellipse is uniform (the empty focus</p><p>is the focus that isnot the center of attraction) to first order ine. It is this theorem</p><p>that enables the Ptolemaic picture of planetary motion to be reasonably on a circle</p><p>approximation. On this picture the Sun is assumed to move uniformly on a circle</p><p>whose center is shifted form Earth by a distance called theequant. If the equant is taken</p><p>as the distance between the tow foci of the correct elliptical orbit, then the angular</p><p>motion is thus described by the Ptolemaic picture accurately to first order ine.</p><p>57</p><p>Exercise 17 - GUSTAVO CORTAZZI</p><p>GARCIA KESSLER</p><p>59</p><p>Exercise 18 - RODRIGO WEBER</p><p>PEREIRA</p><p>Repeat the preceding exercise except this time allow thependulum to move in</p><p>three dimensions, that is, a spring-loaded spherical pendulum. Either the direct or the</p><p>matrix approach may be used.</p><p>My conceptualization of the problem is like follows</p><p>The spring is constrained to move only in the vertical direction, while the rigid</p><p>61</p><p>62 CHAPTER 18. EXERCISE 18 - RODRIGO WEBER PEREIRA</p><p>bar of leght2l is free to move in three dimensions. The position of the i-th particle in</p><p>the rigid bar is:</p><p>r⃗i = zk̂ + r⃗′i (18.1)</p><p>Where r⃗′i is the position of the i-th particle relative to the point that joints the spring to</p><p>the bar. So the kinetic energy of the bar is:</p><p>T =</p><p>N∑</p><p>i</p><p>miv⃗</p><p>2</p><p>i</p><p>2</p><p>=</p><p>N∑</p><p>i</p><p>mi</p><p>2</p><p>(żk̂ +</p><p>⃗̇</p><p>r′i)</p><p>2 (18.2)</p><p>For all particles of the bar,ṙ′i = 0, ϕi = ϕj = ϕ and θi = θj = θ ∀i, j (rigid). This way,</p><p>the velocity of the i-th particle can be written as:</p><p>⃗̇</p><p>r′i = v⃗i = żk̂ + ṙ′ie⃗r︸︷︷︸</p><p>=0</p><p>+r′iϕ̇ sin e⃗ϕ + r′iθ̇e⃗θ (18.3)</p><p>Taking the square of this equation, remembering that̂k · e⃗ϕ = 0 and k̂ · e⃗θ = − sin θ, we</p><p>have:</p><p>v2i = ż2 − 2riżθ̇ sin θ + r′2i (θ̇</p><p>2 + ϕ̇2 sin2 θ) (18.4)</p><p>So, the total kinetic energy is:</p><p>T =</p><p>N∑</p><p>i</p><p>mi</p><p>2</p><p>[ż2 − 2riżθ̇ sin θ + r′2i (θ̇</p><p>2 + ϕ̇2 sin2 θ)]</p><p>T =</p><p>ż2</p><p>2</p><p>( N∑</p><p>i</p><p>mi</p><p>)</p><p>︸ ︷︷ ︸</p><p>M</p><p>−żθ̇ sin θ</p><p>( N∑</p><p>i</p><p>miri</p><p>)</p><p>︸ ︷︷ ︸∫M</p><p>0 r′dm=Ml</p><p>+</p><p>1</p><p>2</p><p>(θ̇2 + ϕ̇2 sin2 θ)</p><p>( N∑</p><p>i</p><p>mir</p><p>2</p><p>i</p><p>)</p><p>︸ ︷︷ ︸∫M</p><p>0 r′2dm= 4</p><p>3</p><p>Ml2=I</p><p>T =</p><p>M</p><p>2</p><p>[</p><p>ż2 − 2lżθ̇ sin θ +</p><p>4</p><p>3</p><p>l2(θ̇2 + ϕ̇2 sin2 θ)</p><p>]</p><p>(18.5)</p><p>The potential energy of the system is the sum of the potentials coming from gravity</p><p>−Mg(z + z′) and from the springk</p><p>2</p><p>z2, so:</p><p>V =</p><p>k</p><p>2</p><p>z2 −Mg(z + l cos θ) (18.6)</p><p>From T and V we construct the Lagrangian:</p><p>L = T − V =</p><p>M</p><p>2</p><p>[</p><p>ż2 − 2lżθ̇ sin θ +</p><p>4</p><p>3</p><p>l2(θ̇2 + ϕ̇2 sin2 θ)</p><p>]</p><p>− k</p><p>2</p><p>z2 +Mg(z + l cos θ),</p><p>witch can be arranged using the inertia momentI = 4</p><p>3</p><p>Ml2:</p><p>L =Mg(z + l cos θ)− kz2</p><p>2</p><p>+</p><p>M ż2</p><p>2</p><p>−Ml sin θżθ̇ + Iθ̇2 + I sin2 θϕ̇2 (18.7)</p><p>63</p><p>Using the definition of the conjugate moments, we can obtain the expressions relating</p><p>them with the generalized velocities:</p><p>ż =</p><p>pz</p><p>M</p><p>ϕ̇ =</p><p>pϕ</p><p>2I sin2 θ</p><p>θ̇ =</p><p>pθ + l sin θpz</p><p>2I</p><p>Now we can construct the Hamiltonian of the system:</p><p>H(q⃗, p⃗, t) = żpz + ϕ̇pϕ + θ̇pθ − L(q⃗, p⃗, t)</p><p>Which, after some algebra turns into:</p><p>H(q⃗, p⃗, t) =</p><p>p2z</p><p>2M</p><p>(</p><p>1+</p><p>3</p><p>2</p><p>sin2 θ</p><p>)</p><p>+</p><p>p2ϕ</p><p>4I sin2 θ</p><p>+</p><p>p2θ</p><p>4I</p><p>+</p><p>l</p><p>2I</p><p>sin θpθpz−Mg(z+l cos θ)+</p><p>kz2</p><p>2</p><p>(18.8)</p><p>The Hamilton equations of motion are</p><p>ż =</p><p>∂H</p><p>∂pz</p><p>⇒ ż =</p><p>pz</p><p>M</p><p>(</p><p>1 +</p><p>3</p><p>2</p><p>sin2 θ</p><p>)</p><p>+</p><p>l</p><p>2I</p><p>sin θpθ (18.9)</p><p>ϕ̇ =</p><p>∂H</p><p>∂pϕ</p><p>⇒ ϕ̇ =</p><p>pϕ</p><p>2I sin2 θ</p><p>(18.10)</p><p>θ̇ =</p><p>∂H</p><p>∂pθ</p><p>⇒ θ̇ =</p><p>pθ</p><p>2I</p><p>+</p><p>l</p><p>2I</p><p>sin θpz (18.11)</p><p>ṗz = −</p><p>∂H</p><p>∂z</p><p>⇒ ṗz =Mg − kz (18.12)</p><p>ṗϕ = −</p><p>∂H</p><p>∂ϕ</p><p>⇒ ṗϕ = 0 (18.13)</p><p>ṗθ = −</p><p>∂H</p><p>∂θ</p><p>⇒ ṗθ = −</p><p>3</p><p>2M</p><p>sin θ cos θp2z +</p><p>cos θp2ϕ</p><p>2I sin3 θ</p><p>− l cos θpθpz</p><p>2I</p><p>−Mgl sin θ (18.14)</p><p>Exercise 19 - JEFFERSON SANTANA</p><p>MARTINS</p><p>The point of suspension of a simple pendulum of lengthl and mass m is con-</p><p>strained to move on a parabolaz = ax2 in the vertical plane. Derive a hamiltonian</p><p>governing the motion of the pendulum and its point of suspension. Obtain the hamil-</p><p>ton’s equations of motion.</p><p>FIGURE 19.1</p><p>Solution</p><p>The point of suspension of a simple pendulum of lengthl and mass m is con-</p><p>strained to move on a parabola defined byz = ax2 in the vertical plane, in a uniform</p><p>65</p><p>66 CHAPTER 19. EXERCISE 19 - JEFFERSON SANTANA MARTINS</p><p>downwards gravitational acceleration</p><p>g. The position of the pivot point in thexz plane</p><p>is given by(x, ax2). Now, let a coordinateθ be the angle the pendulum makes with the</p><p>vertical. The position of the mass is then(x+ lsin(θ), ax2− lcos(θ)). So the generalized</p><p>coordinates for this system arex and θ, and the kinetic energy is</p><p>T =</p><p>1</p><p>2</p><p>m</p><p>(</p><p>ẋ2 + ż2</p><p>)</p><p>=</p><p>m</p><p>2</p><p>[(</p><p>ẋ+ lθ̇cos(θ)</p><p>)2</p><p>+</p><p>(</p><p>2axẋ+ lθ̇sin(θ)</p><p>)2]</p><p>(19.1)</p><p>=</p><p>m</p><p>2</p><p>[(</p><p>1 + 4a2x2</p><p>)</p><p>ẋ2 + 2l (cos(θ) + 2axsin(θ)) ẋθ̇ + l2θ̇2c</p><p>]</p><p>· (19.2)</p><p>Let the zero potential be defined such that whenx = θ = 0, U = 0. Therefore, the zero</p><p>potential surface is the plane defined byz = l, so the potential energy is</p><p>U = mg(l + ax2 − lcos(θ) (19.3)</p><p>With this information, the Lagrangian is</p><p>L =</p><p>1</p><p>2</p><p>mẋ2 + 2ma2x2ẋ2 +mlasin(θ)xẋ+</p><p>1</p><p>2</p><p>ml2θ̇2 −mgl −mgax2 +mglcos(θ) (19.4)</p><p>so the conjugate momenta are</p><p>px = mẋ+ 4ma2x2ẋ+mlcos(θ)θ̇ + 2mlasin(θ)xθ̇ (19.5)</p><p>pθ = mlcos(θ)ẋ+ 2mlasin(θ)xẋ+ml2θ̇ (19.6)</p><p>With these the Hamiltonian can be written as</p><p>H = ẋpx + θ̇pθ −L (19.7)</p><p>so Equations 19.5 and 19.6 must be solved simultaneously foṙx and θ̇. The solutions</p><p>are</p><p>ẋ = −2apθxsin(θ)− lpxsin2(θ) + pθcos(θ)</p><p>lm(sin(θ)− 2axcos(θ))2</p><p>= − lpx − pθ(2axsin(θ) + cos(θ))</p><p>lm(sin(θ)− 2axcos(θ))2</p><p>(19.8)</p><p>θ̇ = −−4a</p><p>2pθx</p><p>2 + 2alpxxsin(θ) + lpxcos(θ)− pθ</p><p>l2m(sin(θ)− 2axcos(θ))2</p><p>= −pθ(1 + 4a2x2)− lpx(2axsin(θ)− cos(θ))</p><p>l2m(sin(θ)− 2axcos(θ))2</p><p>(19.9)</p><p>67</p><p>and the Hamiltonian is given by</p><p>H = pθ</p><p>[</p><p>pθ(1 + 4a2x2)− lpx(2axsin(θ)− cos(θ))</p><p>l2m(sin(θ)− 2axcos(θ))2</p><p>]</p><p>+px</p><p>[</p><p>lpx − pθ(2axsin(θ) + cos(θ))</p><p>lm(sin(θ)− 2axcos(θ))2</p><p>]</p><p>−L</p><p>(19.10)</p><p>From this the final two of Hamilton’s equations (thėx and θ̇ Hamilton’s equations are</p><p>shown above),</p><p>H = pθ</p><p>[</p><p>pθ(1 + 4a2x2)− lpx(2axsin(θ)− cos(θ))</p><p>l2m(sin(θ)− 2axcos(θ))2</p><p>]</p><p>+px</p><p>[</p><p>lpx − pθ(2axsin(θ) + cos(θ))</p><p>lm(sin(θ)− 2axcos(θ))2</p><p>]</p><p>−L</p><p>(19.11)</p><p>can be found</p><p>ṗθ = −</p><p>∂H</p><p>∂θ</p><p>ṗx =</p><p>∂H</p><p>∂x</p><p>· (19.12)</p><p>The derivatives of the Hamiltonian are</p><p>ṗθ =</p><p>[p2θ(a</p><p>2x2 + 1) + l2p2x] (2axsin(θ) + cos(θ))</p><p>l2m(sin(θ)− 2axcos(θ))3</p><p>+ glm+</p><p>lpθpx(4ax(ax) [cos(2θ)− 3]− sin(2θ)− cos(2θ)− 3)</p><p>l2m(sin(θ)− 2axcos(θ))3</p><p>(19.13)</p><p>ṗx = 2agmx+</p><p>2a</p><p>l2m(sin(θ − 2axcos(θ)))3</p><p>{</p><p>1</p><p>2</p><p>lpθpx [2axsin(2θ) + cos(2θ) + 3]−</p><p>p2θ(2axsin(θ) + cos(θ)) + l2p2xcos(θ) (19.14)</p><p>The equations of motion can then be found by differentiating Equations 19.8 and</p><p>19.9 and plugging in the above results, then substituting in the expressions forpθ and</p><p>px.</p><p>Exercise 20 - JHORDAN SILVEIRA DE</p><p>BORBA</p><p>Obtain Hamilton’s equations of motion for a plane pendulum of lengthl with</p><p>maas point m whose radius of suspension rotates uniformly on the circumference of a</p><p>vertical circle of radiusa. Describe physically the nature of the canonical momentum</p><p>and the Hamiltonian.</p><p>Resposta: Considering then a pendulum of lengthl in which its suspension</p><p>point moves with constant angular velocityω in a circle of radiusa, we have:</p><p>The holding position vector can be written asR = (a cos (ωt) , a sin (ωt)), where</p><p>69</p><p>70 CHAPTER 20. EXERCISE 20 - JHORDAN SILVEIRA DE BORBA</p><p>that α = ωt. So we can write the cartesian components of the massm as:</p><p>r = R+ l (sin β,− cos β)</p><p>And the speed:</p><p>ṙ = Ṙ+ lβ̇ (cos β, sin β)</p><p>Where Ṙ = (−aω sin (ωt) , aω cos (ωt)). We then have a single generalized coordinate</p><p>β. Kinetic energy can then be written as:</p><p>T =</p><p>1</p><p>2</p><p>mv2</p><p>=</p><p>1</p><p>2</p><p>m</p><p>[(</p><p>−aω sin (ωt) + lβ̇ cos β</p><p>)2</p><p>+</p><p>(</p><p>aω cos (ωt) + lβ̇ sin β</p><p>)2]</p><p>=</p><p>1</p><p>2</p><p>m</p><p>[</p><p>a2ω2 sin2 (ωt) + l2β̇2 cos2 β − 2alβ̇ω sin (ωt) cos β + a2ω2 cos2 (ωt) + l2β̇2 sin2 β + 2aωlβ̇ cos (ωt) sin β</p><p>]</p><p>=</p><p>1</p><p>2</p><p>m</p><p>[</p><p>a2ω2</p><p>(</p><p>sin2 (ωt) + cos2 (ωt)</p><p>)</p><p>+ l2β̇2</p><p>(</p><p>cos2 β</p><p>)</p><p>+ 2alβ̇ω (cos (ωt) sin β − sin (ωt) cos β)</p><p>]</p><p>=</p><p>1</p><p>2</p><p>m</p><p>[</p><p>a2ω2 + l2β̇2 + 2alβ̇ω sin (β − ωt)</p><p>]</p><p>And puttingV = 0 in y = 0 we have:</p><p>V = mgy = mg (a sin (ωt)− l cos β)</p><p>So the lagrangian is:</p><p>L = T − V</p><p>=</p><p>m</p><p>2</p><p>[</p><p>a2ω2 + l2β̇2 + 2alβ̇ω sin (β − ωt)</p><p>]</p><p>−mg (a sin (ωt)− b cos β)</p><p>And the conjugated moment is:</p><p>pβ =</p><p>∂L</p><p>∂β̇</p><p>= ml2β̇ + ωalm sin (β − ωt) (20.1)</p><p>The hamiltonian can then be written as:</p><p>H =</p><p>∑</p><p>i</p><p>piq̇i − L</p><p>= β̇pβ − L</p><p>= β̇</p><p>[</p><p>ml2β̇ + alωm sin (β − ωt)</p><p>]</p><p>− m</p><p>2</p><p>[</p><p>a2ω2 + l2β̇2 + 2alβ̇ω sin (β − ωt)</p><p>]</p><p>+mg (a sin (ωt)− l cos β)</p><p>= ml2β̇2 + alωmβ̇ sin (β − ωt)− ma2ω2</p><p>2</p><p>− ml2β̇2</p><p>2</p><p>− alβ̇ω sin (β − ωt) +mga sin (ωt)−mgl cos β</p><p>=</p><p>m</p><p>2</p><p>l2β̇2 −mgl cos β − ma2ω2</p><p>2</p><p>+mga sin (ωt)</p><p>71</p><p>Let’s take a break to review the first two terms, rewriting in terms ofR = (Rx, Ry) =</p><p>(a cos (ωt) , a sin (ωt)):</p><p>H =</p><p>[m</p><p>2</p><p>l2β̇2 −mgl cos β</p><p>]</p><p>−</p><p>[</p><p>mṘ2</p><p>2</p><p>−mgRy</p><p>]</p><p>The first 2 terms are simply the energy for a pendulum on a fixed support. If we had</p><p>R = 0, we would have this case. Or yet if the support were stopped witḣR = 0, we</p><p>would only have an increase in energy due to the position of the support. Hamiltonian</p><p>must be written in terms ofβ and pβ , so we need to replaceβ̇. From the 20.1 equation</p><p>we have:</p><p>β̇ =</p><p>pβ</p><p>ml2</p><p>− aω</p><p>l</p><p>sin (β − ωt)</p><p>Then:</p><p>H =</p><p>m</p><p>2</p><p>l2</p><p>( pβ</p><p>ml2</p><p>− aω</p><p>l</p><p>sin (β − ωt)</p><p>)2</p><p>−mgl cos β − ma2ω2</p><p>2</p><p>+mga sin (ωt)</p><p>=</p><p>m</p><p>2</p><p>l2</p><p>(( pβ</p><p>ml2</p><p>)2</p><p>+</p><p>a2ω2</p><p>l2</p><p>sin2 (β − ωt)− 2</p><p>aωpβ</p><p>ml3</p><p>sin (β − ωt)</p><p>)</p><p>−mgl cos β − ma2ω2</p><p>2</p><p>+mga sin (ωt)</p><p>Therefore the Hamiltonian of the system is given by:</p><p>H =</p><p>p2β</p><p>2ml2</p><p>− pβaω</p><p>l</p><p>sin (β − ωt)+1</p><p>2</p><p>ma2ω2 sin2 (β − ωt)− 1</p><p>2</p><p>ma2ω2+mga sin (ωt)−mgl cos β</p><p>(20.2)</p><p>In addition, we have to:</p><p>β̇ =</p><p>∂H</p><p>∂pβ</p><p>=</p><p>pβ</p><p>ml2</p><p>− aω</p><p>l</p><p>sin (β − ωt) (20.3)</p><p>We can notice that 20.3 retrieves the result obtained in 20.1. Another relationship is</p><p>given by:</p><p>∂H</p><p>∂t</p><p>= −∂L</p><p>∂t</p><p>So we have:</p><p>∂H</p><p>∂t</p><p>=</p><p>pβaω</p><p>2</p><p>l</p><p>cos (β − ωt)−ma2ω3 sin (β − ωt) cos (β − ωt) +mgωa cos (ωt)</p><p>=</p><p>pβaω</p><p>2</p><p>l</p><p>−ma2ω3 sin (β − ωt)</p><p>=</p><p>pβ</p><p>ml2</p><p>− aω</p><p>l</p><p>sin (β − ωt)</p><p>72 CHAPTER 20. EXERCISE 20 - JHORDAN SILVEIRA DE BORBA</p><p>And:</p><p>−∂L</p><p>∂t</p><p>= −</p><p>(</p><p>−malβ̇ω2 cos (β − ωt)−mωga cos (ωt)</p><p>)</p><p>= malβ̇ω2</p><p>= β̇</p><p>Then:</p><p>pβ</p><p>ml2</p><p>− aω</p><p>l</p><p>sin (β − ωt) = β̇ (20.4)</p><p>Which is the same result obtained in 20.3 and 20.1. We can have one more equation:</p><p>ṗβ = −∂H</p><p>∂β</p><p>= −</p><p>(</p><p>−pβaω</p><p>l</p><p>cos (β − ωt) +ma2ω2 sin (β − ωt) cos (β − ωt) +mgl sin β</p><p>)</p><p>=</p><p>pβaω</p><p>l</p><p>cos (β − ωt)−ma2ω2 sin (β − ωt) cos (β − ωt)−mgl sin β</p><p>Then differentiating the equation 20.4:</p><p>β̈ =</p><p>ṗβ</p><p>ml2</p><p>− aω</p><p>l</p><p>cos (β − ωt)</p><p>(</p><p>β̇ − ω</p><p>)</p><p>=</p><p>1</p><p>ml2</p><p>(pβaω</p><p>l</p><p>cos (β − ωt)−ma2ω2 sin (β − ωt) cos (β − ωt)−mgl sin β</p><p>)</p><p>− aω</p><p>l</p><p>cos (β − ωt)</p><p>(</p><p>β̇ − ω</p><p>)</p><p>=</p><p>pβaω</p><p>ml3</p><p>cos (β − ωt)− a2ω2</p><p>l2</p><p>sin (β − ωt) cos (β − ωt)− g</p><p>l</p><p>sin β − aωβ̇</p><p>l</p><p>cos (β − ωt) + aω2</p><p>l</p><p>cos (β − ωt)</p><p>Substituting the first term, using 20.1 we have:</p><p>pβaω</p><p>ml3</p><p>cos (β − ωt) =</p><p>(</p><p>ml2β̇ + lmωa sin (β − ωt)</p><p>) aω</p><p>ml3</p><p>cos (β − ωt)</p><p>= ml2β̇</p><p>aω</p><p>ml3</p><p>cos (β − ωt) + lm</p><p>aω</p><p>ml3</p><p>cos (β − ωt)ωa sin (β − ωt)</p><p>=</p><p>aωβ̇</p><p>l</p><p>cos (β − ωt) + a2ω2</p><p>l2</p><p>cos (β − ωt)ωa sin (β − ωt)</p><p>We are then left with the following equation of motion:</p><p>β̈ =</p><p>aω2</p><p>l</p><p>cos (β − ωt)− g</p><p>l</p><p>sin β</p><p>Again if we haveω = 0 it reduces to an equation of motion for a simple pendulum.</p><p>Rewriting as:</p><p>β̈ +</p><p>g</p><p>l</p><p>sin β =</p><p>aω2</p><p>l</p><p>cos (β − ωt)</p><p>73</p><p>We then have an equation of motion for a pendulum with an external torque as ev-</p><p>idenced by the term on the right. This is the same result obtained using lagrangena</p><p>in Classical Dynamics of Particles and Systems (THORNTON; MARION, 2004). In</p><p>addition, an interesting discussion about other moves towards support was made by</p><p>Professor González (GONZáLEZ, 2016).</p><p>Exercise 21 - DOUGLAS OLIVEIRA NO-</p><p>VAES</p><p>Question: a) The point of suspension of a plane simple pendulum of massm and</p><p>length l is constrained to move along a horizontal track and is connected to a point on</p><p>the circumference of a uniform flywheel of massM and radius a through a massless</p><p>connecting rod also of length0, as shown in the figure. The flywheel rotates about a</p><p>center fixed on the track. Find a Hamiltonian for the combined system and determine</p><p>Hamilton’s equations of motion.</p><p>Response: The combined system involves the disc and the mass suspended. The</p><p>Lagrangian is</p><p>L(q, q̇, t) = T − U</p><p>75</p><p>76 CHAPTER 21. EXERCISE 21 - DOUGLAS OLIVEIRA NOVAES</p><p>The kinetic energy is</p><p>T =</p><p>m</p><p>2</p><p>v2 +</p><p>I</p><p>2</p><p>θ̇2</p><p>where θ is the angle described by radius of the flywheel andI = 1</p><p>2</p><p>Ma2 is its moment</p><p>of inertia.</p><p>Since</p><p>x = 2a cos θ + l sinϕ</p><p>y = −l cosϕ</p><p>where ϕ is the angle described by length of the pendulum, the kinetic energy</p><p>can be</p><p>written as</p><p>T =</p><p>m</p><p>2</p><p>(</p><p>4a2θ̇2 sin2 θ + l2ϕ̇2 cos2 ϕ− 4alθ̇ϕ̇ sin θ cosϕ+ l2ϕ̇2 sin2 ϕ</p><p>)</p><p>+</p><p>1</p><p>4</p><p>Ma2θ̇2</p><p>or T =</p><p>1</p><p>2</p><p>(</p><p>4m sin2 θ +</p><p>1</p><p>2</p><p>M</p><p>)</p><p>a2θ̇2 +</p><p>m</p><p>2</p><p>l2ϕ̇2 − 2malθ̇ϕ̇ sin θ cosϕ</p><p>and the potential energy as</p><p>U = −mgl cosϕ</p><p>Replacing the kinetic and potential energy in the Lagrangian</p><p>L(q, q̇, t) =</p><p>1</p><p>2</p><p>(</p><p>4m sin2 θ +</p><p>1</p><p>2</p><p>M</p><p>)</p><p>a2θ̇2 +</p><p>m</p><p>2</p><p>l2ϕ̇2 − 2mal sin θ cosϕ θ̇ϕ̇+mgl cosϕ</p><p>since</p><p>L(q, q̇, t) = L0(q, t) + q̇iai(q, t) + q̇2i Ti(q, t)</p><p>or L(q, q̇, t) = L0(q, t) + ˜̇qa +</p><p>1</p><p>2</p><p>˜̇qTq̇</p><p>where one recognizes each term as</p><p>L0(ϕ) = mgl cosϕ</p><p>L1(q̇) = q̇iai(q, t) = 0</p><p>L2(θ̇</p><p>2, θ̇ϕ̇, ϕ̇θ̇, ϕ̇2) =</p><p>1</p><p>2</p><p>(</p><p>4m sin2 θ +</p><p>1</p><p>2</p><p>M</p><p>)</p><p>a2 − 2mal sin θ cosϕ+</p><p>m</p><p>2</p><p>l2ϕ̇2</p><p>Being the column matrix and its transpose</p><p>q̇ =</p><p>[</p><p>θ̇</p><p>ϕ̇</p><p>]</p><p>and ˜̇q =</p><p>[</p><p>θ̇ ϕ̇</p><p>]</p><p>77</p><p>and the matrixT</p><p>T =</p><p>[</p><p>T θ̇2 T θ̇ϕ̇</p><p>Tϕ̇θ̇ Tϕ̇2</p><p>]</p><p>=</p><p>(4m sin2 θ + 1</p><p>2</p><p>M</p><p>)</p><p>a2 −2mal sin θ cosϕ</p><p>−2mal sin θ cosϕ ml2ϕ̇2</p><p></p><p>Since the Hamiltonian is</p><p>H(q, p, t) =</p><p>1</p><p>2</p><p>(</p><p>p̃ − ���</p><p>0</p><p>ã</p><p>)</p><p>T−1</p><p>(</p><p>p −��</p><p>0</p><p>a</p><p>)</p><p>−L0(q, q̇, t)</p><p>the inverse matrixT−1</p><p>T−1 =</p><p>T̃c</p><p>detT</p><p>where Tc is the cofactor matrix whose elements(Tc)jk are (−1)j+k times the determi-</p><p>nant of the matrix obtained by striking out thejth row and thekth column ofT. (Gold-</p><p>stein, 3rd ed.)</p><p>Hence</p><p>T−1 = (detT)−1</p><p> ml2ϕ̇2 +2mal sin θ cosϕ</p><p>+2mal sin θ cosϕ</p><p>(</p><p>4m sin2 θ + 1</p><p>2</p><p>M</p><p>)</p><p>a2</p><p></p><p>T−1 =</p><p>1</p><p>4m2a2l2 sin2 θ sin2 ϕ+ m</p><p>2</p><p>Ma2l2</p><p> ml2ϕ̇2 +2mal sin θ cosϕ</p><p>+2mal sin θ cosϕ</p><p>(</p><p>4m sin2 θ + 1</p><p>2</p><p>M</p><p>)</p><p>a2</p><p></p><p>(21.1)</p><p>and</p><p>p̃T−1p = (detT)−1</p><p>[</p><p>pθ pϕ</p><p>]  ml2ϕ̇2 +2mal sin θ cosϕ</p><p>+2mal sin θ cosϕ</p><p>(</p><p>4m sin2 θ + 1</p><p>2</p><p>M</p><p>)</p><p>a2</p><p>[pθ</p><p>pϕ</p><p>]</p><p>= (detT)−1</p><p>[</p><p>ml2pθ + 2mal sin θ cosϕpϕ 2mal sin θ cosϕpθ +</p><p>(</p><p>4m sin2 θ + 1</p><p>2</p><p>M</p><p>)</p><p>a2pϕ</p><p>] [</p><p>pθ</p><p>pϕ</p><p>]</p><p>= (detT)−1</p><p>[</p><p>ml2p2θ +</p><p>(</p><p>4m sin2 θ +</p><p>1</p><p>2</p><p>M</p><p>)</p><p>a2p2ϕ + 4mal sin θ cosϕpθpϕ</p><p>]</p><p>Then the Hamiltonian is</p><p>H(q, p, t) =</p><p>1</p><p>2</p><p>[</p><p>ml2p2θ +</p><p>(</p><p>4m sin2 θ + 1</p><p>2</p><p>M</p><p>)</p><p>a2p2ϕ + 4mal sin θ cosϕpθpϕ</p><p>]</p><p>4m2a2l2 sin2 θ sin2 ϕ+ m</p><p>2</p><p>Ma2l2</p><p>−mgl cosϕ</p><p>78 CHAPTER 21. EXERCISE 21 - DOUGLAS OLIVEIRA NOVAES</p><p>or</p><p>H(q, p, t) =</p><p>m</p><p>2</p><p>l2p2θ +</p><p>(</p><p>4m sin2 θ + 1</p><p>2</p><p>M</p><p>)</p><p>a2</p><p>2</p><p>p2ϕ + 2mal sin θ cosϕpθpϕ</p><p>4m2a2l2 sin2 θ sin2 ϕ+ m</p><p>2</p><p>Ma2l2</p><p>−mgl cosϕ</p><p>The equations of motion are given by</p><p>q̇i =</p><p>∂H</p><p>∂pi</p><p>and − ṗi =</p><p>∂H</p><p>∂qi</p><p>Hence, directly, for the velocities</p><p>θ̇ =</p><p>ml2p2θ + 2mal sin θ cosϕpϕ</p><p>4m2a2l2 sin2 θ sin2 ϕ+ m</p><p>2</p><p>Ma2l2</p><p>and</p><p>ϕ̇ =</p><p>(</p><p>4m sin2 θ + 1</p><p>2</p><p>M</p><p>)</p><p>a2</p><p>2</p><p>pϕ + 2mal sin θ cosϕpθ</p><p>4m2a2l2 sin2 θ sin2 ϕ+ m</p><p>2</p><p>Ma2l2</p><p>For canonical momenta,</p><p>−pθ =</p><p>∂H</p><p>∂θ</p><p>and − pϕ =</p><p>∂H</p><p>∂ϕ</p><p>resulting, after a lot of algebra, in</p><p>pθ = 8m2a2l2 sin2 ϕ sin (2θ)</p><p>[</p><p>m</p><p>2</p><p>l2p2θ +</p><p>(</p><p>4m sin2 θ + 1</p><p>2</p><p>M</p><p>)</p><p>a2</p><p>2</p><p>p2ϕ + 2mal sin θ cosϕpθpϕ(</p><p>4m2a2l2 sin2 θ sin2 ϕ+ m</p><p>2</p><p>Ma2l2</p><p>)2</p><p>]</p><p>−</p><p>ap2ϕ sin</p><p>2 θ sin (2θ)− l cos θ cosϕpϕpθ</p><p>mal2 sin2 θ sin2 ϕ+ M</p><p>4</p><p>al2</p><p>and</p><p>pϕ =</p><p>2m2al sin2 θ cosϕ</p><p>[</p><p>l</p><p>a</p><p>p2θ +</p><p>(</p><p>4m sin2 θ + 1</p><p>2</p><p>M</p><p>)</p><p>a</p><p>ml</p><p>p2ϕ + 4 sin θ cosϕpθpϕ</p><p>]</p><p>2mal sin2 θ sin2 ϕ+ M</p><p>4</p><p>al</p><p>+</p><p>sin θ cosϕ pθpϕ</p><p>2mal sin2 θ sin2 ϕ+ M</p><p>4</p><p>al</p><p>b) Suppose the point of suspension were moved along the track according to</p><p>some function of timex = f(t), where x reverses at x = ±2a (relative to the center of</p><p>the fly wheel). Again, find a Hamiltonian and Hamilton’s equations of motion.</p><p>79</p><p>Response: The coordinates of the particle in polar coordinates are:</p><p>x = f(t) + l sinϕ</p><p>y = −l cosϕ</p><p>and the velocities</p><p>ẋ = ḟ(t) + lϕ̇ cosϕ</p><p>ẏ = −lϕ̇ sinϕ</p><p>As θ = ωt = ω(t) there is a single generalized coordinate ϕ and the Hamiltonian</p><p>is</p><p>H(ϕ, pϕ, t) = ϕ̇pϕ−L′(ϕ, ϕ̇, t)</p><p>The Lagrangian is</p><p>L′(ϕ, ϕ̇, t) = T − U</p><p>Then, the kinetic and potential energy is given by</p><p>T =</p><p>1</p><p>2</p><p>m</p><p>(</p><p>ḟ 2 + l2ϕ2 + 2ḟ lϕ̇ cosϕ</p><p>)</p><p>U = −mgl cosϕ</p><p>So, the Lagrangian come</p><p>L′ =</p><p>1</p><p>2</p><p>m</p><p>(</p><p>l2ϕ2 + ḟ 2 + 2ḟ lϕ̇ cosϕ</p><p>)</p><p>+mgl cosϕ</p><p>Since the Lagrangian is invariant up to a full time derivative</p><p>L′ = L +</p><p>dF</p><p>dt</p><p>one can recognize two of them in the Lagrangian L′ obtained above, that are</p><p>ḟ ϕ̇ cosϕ =</p><p>d</p><p>dt</p><p>(ḟ sinϕ)− f̈ sinϕ and ḟ 2 =</p><p>dF</p><p>dt</p><p>Hence, the Lagrangian results in</p><p>L =</p><p>1</p><p>2</p><p>m</p><p>(</p><p>l2ϕ2 − lf̈ sinϕ) +mgl cosϕ</p><p>80 CHAPTER 21. EXERCISE 21 - DOUGLAS OLIVEIRA NOVAES</p><p>Since the conjugate momentum is</p><p>pϕ =</p><p>∂L</p><p>∂ϕ̇</p><p>= ml2ϕ̇</p><p>and the Hamiltonian given by</p><p>H = ϕ̇pϕ −L</p><p>to replace ϕ̇ and ϕ̇2, from conjugate momentum, and the Lagrangian, the Hamiltonian</p><p>is</p><p>H(ϕ, pϕ) =</p><p>p2ϕ</p><p>ml2</p><p>− 1</p><p>2</p><p>mlf̈ sinϕ)−mgl cosϕ</p><p>and, since ϕ̇ = ∂H</p><p>∂pϕ</p><p>and ϕ̇ = ∂H</p><p>∂pϕ</p><p>, the Hamilton’s equations of motion are</p><p>ϕ̇ =</p><p>2</p><p>ml2</p><p>pϕ</p><p>ṗϕ =</p><p>1</p><p>2</p><p>mlf cosϕ−mgl sinϕ</p><p>Exercise 22 - FRANCISCO AUGUSTO</p><p>FERREIRA ALMEIDA</p><p>For the arrangement described in Exercise 21 of Chapter 2, find the Hamiltonian</p><p>of the system, first in terms of coordinates in the laboratory system and then in terms</p><p>of coordinates in the rotating systems. What are the conservation properties of the</p><p>Hamiltonians, and how are they related to the energy of the system?</p><p>Solution:</p><p>We start by placing the laboratory coordinate system (x,y) at the point of attach-</p><p>ment of the first spring. Let the rotating coordinate system be (x’,y’) with x’ being along</p><p>the first spring. two coordinate systems are related by[</p><p>x′</p><p>y′</p><p>]</p><p>=</p><p>[</p><p>cosωt sinωt</p><p>−sinωt cosωt</p><p>]</p><p>(22.1)</p><p>Kinetic energy of the system is only duo to particle of mass m since rest of the</p><p>components are assumed massless.</p><p>T =</p><p>1</p><p>2</p><p>mṙ2 =</p><p>1</p><p>2</p><p>m(ẋ2 + ẏ2) (22.2)</p><p>The potential energy due to two springs is given by</p><p>V =</p><p>1</p><p>2</p><p>K(x′ − ro)2 +</p><p>1</p><p>2</p><p>ky′2 (22.3)</p><p>The Lagrangian of the system is given by L=T-V. Once we substitute x’ and y’,</p><p>81</p><p>82 CHAPTER 22. EXERCISE 22 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>we get</p><p>L =</p><p>1</p><p>2</p><p>m(ẋ2 + ẏ2)− 1</p><p>2</p><p>K(xcosωt+ ysinωt− ro)2 −</p><p>1</p><p>2</p><p>k(−xsinωt+ ycosωt)2 (22.4)</p><p>Sincepx = mẋ and py = mẏ, the Hamiltonian of the system é given by</p><p>H = ẋpx + ẏpy − L = 2T − L = T + V (22.5)</p><p>H =</p><p>p2x</p><p>2m</p><p>+</p><p>p2y</p><p>2m</p><p>+</p><p>1</p><p>2</p><p>K(xcosωt+ ysinωt− ro)2 +</p><p>1</p><p>2</p><p>k(ycosωt− xsinωt)2 (22.6)</p><p>Next, we relate positions and velocities in the two reference frames</p><p>˙⃗r =</p><p>˙⃗</p><p>r′ + ω⃗ × r⃗ (22.7)</p><p>r⃗ = r⃗′ (22.8)</p><p>Note that time derivative ofr⃗′ refers to derivative with respect to rotating coor-</p><p>dinate system . Kynetic energy in terms of coordinates of the rotating system is given</p><p>by</p><p>T =</p><p>1</p><p>2</p><p>m⃗̇r2 =</p><p>1</p><p>2</p><p>m</p><p>⃗̇</p><p>r′2 +mω · (r⃗′ × ⃗̇r′) + 1</p><p>2</p><p>mω2r⃗′</p><p>2</p><p>(22.9)</p><p>Here, we extended our two coordinates systems to include z and z’ which coin-</p><p>cide. Sinceω⃗ = ωẑ = ωẑ′, we have</p><p>T =</p><p>1</p><p>2</p><p>m(ẋ′</p><p>2</p><p>+ ẏ′</p><p>2</p><p>) +</p><p>1</p><p>2</p><p>mω2(x′</p><p>2</p><p>+ y′</p><p>2</p><p>) +mω(x′ẏ′ − y′ẋ′) (22.10)</p><p>The Lagrangian of the system is given by</p><p>83</p><p>L =</p><p>1</p><p>2</p><p>m(ẋ′2 + ẏ′2) +</p><p>1</p><p>2</p><p>mω2(x′2 + y′2) +mω(x′ẏ′ − y′ẋ′)− 1</p><p>2</p><p>K(x′ − ro)2 −</p><p>1</p><p>2</p><p>ky′2 (22.11)</p><p>To find the Hamiltonian of the system, we make use of equations (8.23) and</p><p>(8.27) from textbook. Comparing term by term our Lagrangian with equation (8.23)</p><p>we see that our Hamiltonian is given by</p><p>H̃ =</p><p>1</p><p>2</p><p>[</p><p>px′ +mωy′ py′ +mωx′</p><p>] [1/m 0</p><p>0 1/m</p><p>] [</p><p>px′ +mωy′</p><p>py′ +mωx′</p><p>]</p><p>− 1</p><p>2</p><p>mω2(x′2 + y′2) + V</p><p>(22.12)</p><p>After we simplify, the Hamiltonian of the system is given by</p><p>H̃ =</p><p>(px′ +mωy′)2</p><p>2m</p><p>+</p><p>(py′ +mωx′)2</p><p>2m</p><p>− 1</p><p>2</p><p>mω2(x′2 + y′2) +</p><p>1</p><p>2</p><p>K(x′ − ro)2 +</p><p>1</p><p>2</p><p>ky′2 (22.13)</p><p>Notice that the Hamiltonian in terms of rotating coordinate system doesn’t ex-</p><p>plicity depend on time and is a conserved quantity i.e.0 =</p><p>∂H̃</p><p>∂t</p><p>=</p><p>dH̃</p><p>dt</p><p>Next, we need to relate the two Hamiltonians, we do so by writing out H and̃H</p><p>in terms of(ẋ, ẏ, ẋ′, ẏ′, x′, y′)</p><p>H =</p><p>1</p><p>2</p><p>m(ẋ2 + ẏ2) + V (x′, y′) (22.14)</p><p>H̃ =</p><p>1</p><p>2</p><p>m(ẋ′</p><p>2</p><p>+ ẏ′</p><p>2</p><p>) +</p><p>1</p><p>2</p><p>mω2(x′2 + y′2) + V (x′, y′) (22.15)</p><p>We notice that</p><p>ẋ2 + ẏ2 = ⃗̇r2 =</p><p>⃗̇</p><p>r′2 + 2ω · (r⃗′ × ·(r⃗′) + ω2r⃗′</p><p>2</p><p>(22.16)</p><p>ẋ2 + ẏ2 = ẋ′2 + ẏ′2 + ω2(x′2 + y′2) + 2ω(x′ẏ′ − y′ẋ′) (22.17)</p><p>84 CHAPTER 22. EXERCISE 22 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>by comparing equations (22.14) and (22.15) we see that</p><p>H = H̃ +mω(x′ẏ′ − y′ẋ′) (22.18)</p><p>Exercise 23 - LUCAS BOURSCHEIDT</p><p>Solution:</p><p>(a) Employing cylindrical coordinates</p><p>Position:</p><p>r = rr̂</p><p>Velocity:</p><p>85</p><p>86 CHAPTER 23. EXERCISE 23 - LUCAS BOURSCHEIDT</p><p>v = ṙr̂+ rθ̇θ̂</p><p>v2 = ṙ2 + r2θ̇2</p><p>Magnetic �eld:</p><p>B = Bẑ</p><p>Vector potential:</p><p>A =</p><p>1</p><p>2</p><p>B× r =</p><p>1</p><p>2</p><p>(Bẑ)× (rr̂) =</p><p>B</p><p>2</p><p>rθ̂</p><p>Lagrangian:</p><p>L =</p><p>1</p><p>2</p><p>mv2 + eA · v − V (r)</p><p>=</p><p>1</p><p>2</p><p>m(ṙ2 + r2θ̇2) + e</p><p>(</p><p>B</p><p>2</p><p>rθ̂</p><p>)</p><p>· (ṙr̂+ rθ̇θ̂)− V (r)</p><p>So:</p><p>L =</p><p>1</p><p>2</p><p>m(ṙ2 + r2θ̇2)</p><p>+</p><p>eB</p><p>2</p><p>r2θ̇ − V (r) (A)</p><p>Momenta:</p><p>From (A):</p><p>pr =</p><p>∂L</p><p>∂ṙ</p><p>= mṙ (B)</p><p>87</p><p>pθ =</p><p>∂L</p><p>∂θ̇</p><p>= mr2θ̇ +</p><p>eB</p><p>2</p><p>r2 = mr2</p><p>(</p><p>θ̇ +</p><p>eB</p><p>2m</p><p>)</p><p>(C)</p><p>Now, from (B) and (C) we find the time derivatives:</p><p>ṙ =</p><p>pr</p><p>m</p><p>(D)</p><p>θ̇ =</p><p>pθ</p><p>mr2</p><p>− eB</p><p>2m</p><p>(E)</p><p>Hamiltonian:</p><p>H = piq̇i − L</p><p>Employing (A), (B), and (C):</p><p>H = mṙṙ +mr2</p><p>(</p><p>θ̇ +</p><p>eB</p><p>2m</p><p>)</p><p>θ̇ − 1</p><p>2</p><p>m(ṙ2 + r2θ̇2)− eB</p><p>2</p><p>r2θ̇ + V (r)</p><p>= mṙ2 +mr2θ̇2 +</p><p>eB</p><p>2</p><p>r2θ̇ − 1</p><p>2</p><p>mṙ2 − 1</p><p>2</p><p>mr2θ̇2 − eB</p><p>2</p><p>r2θ̇ + V (r)</p><p>Canceling the third term on the right with the sixth and grouping similar terms to-</p><p>gether:</p><p>H =</p><p>1</p><p>2</p><p>mṙ2 +</p><p>1</p><p>2</p><p>mr2θ̇2 + V (r) (F )</p><p>The Hamiltonian corresponds to the total mechanical energy of the system. Replacing</p><p>(D) and (E) in (F):</p><p>H =</p><p>1</p><p>2</p><p>m</p><p>(pr</p><p>m</p><p>)2</p><p>+</p><p>1</p><p>2</p><p>mr2</p><p>(</p><p>pθ</p><p>mr2</p><p>− eB</p><p>2m</p><p>)2</p><p>+ V (r)</p><p>=</p><p>p2r</p><p>2m</p><p>+</p><p>1</p><p>2</p><p>mr2</p><p>(</p><p>pθ</p><p>mr2</p><p>− eB</p><p>2m</p><p>)2</p><p>+ V (r)</p><p>88 CHAPTER 23. EXERCISE 23 - LUCAS BOURSCHEIDT</p><p>Finally:</p><p>H =</p><p>p2r</p><p>2m</p><p>+</p><p>1</p><p>2m</p><p>(</p><p>pθ</p><p>r</p><p>− eB</p><p>2</p><p>r</p><p>)2</p><p>+ V (r)</p><p>which is a particular case of the more general expression:</p><p>H =</p><p>1</p><p>2m</p><p>(p− eA)2 + V (r, t)</p><p>(b) Assuming the coordinate axes coincide att = 0:</p><p>[</p><p>x′</p><p>y′</p><p>]</p><p>=</p><p>[</p><p>cos ωt sin ωt</p><p>−sin ωt cos ωt</p><p>] [</p><p>x</p><p>y</p><p>]</p><p>Employing cylindrical coordinates in both reference frames,x = r cos θ and y = r sin θ</p><p>in the inertial frame andx′ = r′ cos θ′ and y′ = r′ sin θ′ in the rotating frame:</p><p>{</p><p>r′ cos θ′ = r (cos θ cos ωt+ sin θ sin ωt) = r cos(θ − ωt)</p><p>r′ sin θ′ = r (−cos θ sin ωt+ sin θ cos ωt) = r sin(θ − ωt)</p><p>From this we conclude thatr′ = r and θ′ = θ− ωt. So, insertingr = r′ and θ̇ = θ̇′ + ω =</p><p>θ̇′ − eB/2m in (A) we find:</p><p>L =</p><p>1</p><p>2</p><p>m</p><p>[</p><p>ṙ′</p><p>2</p><p>+ r′2</p><p>(</p><p>θ̇′ − eB</p><p>2m</p><p>)2</p><p>]</p><p>+</p><p>eB</p><p>2</p><p>r′2</p><p>(</p><p>θ̇′ − eB</p><p>2m</p><p>)</p><p>− V (r′)</p><p>=</p><p>1</p><p>2</p><p>m</p><p>[</p><p>ṙ′</p><p>2</p><p>+ r′2(θ̇′ + ω)2</p><p>]</p><p>−mωr′2(θ̇′ + ω)− V (r′)</p><p>After a little algebra we get:</p><p>L =</p><p>1</p><p>2</p><p>m(ṙ′</p><p>2</p><p>+ r′2θ̇′2)− 1</p><p>2</p><p>mω2r′2 − V (r′) (G)</p><p>89</p><p>That is, the velocity-dependent potential in the Lagrangian of equation (A) is replaced</p><p>by the term 1</p><p>2</p><p>mω2r′2 in the rotating frame. From (G), replacingr′ → r and θ′ → θ, we</p><p>find the canonical momenta:</p><p>pr =</p><p>∂L</p><p>∂ṙ</p><p>= mṙ (H)</p><p>pθ =</p><p>∂L</p><p>∂θ̇</p><p>= mr2θ̇ +</p><p>eB</p><p>2</p><p>r2 = mr2θ̇ (I)</p><p>Thus, in the rotating reference frame, the Hamiltonian is given by:</p><p>H =</p><p>1</p><p>2</p><p>m(ṙ2 + r2θ̇2) +</p><p>1</p><p>2</p><p>mω2r2 + V (r) (J)</p><p>Finally, inserting (H) and (I) into (J):</p><p>H =</p><p>1</p><p>2m</p><p>(</p><p>ṗr</p><p>2 +</p><p>ṗθ</p><p>2</p><p>r2</p><p>)</p><p>+</p><p>1</p><p>2</p><p>mω2r2 + V (r)</p><p>Exercise 24 - GEOVANE NAYSINGER</p><p>FIGURE 24.1</p><p>Firstly, we need to find the lagrangian of the sistem. So, to start, define the</p><p>displacement vector in cillindrical coordinates as</p><p>r⃗ = (a.senθ, a.cosθ, z) (24.1)</p><p>and the velocity is</p><p>˙⃗r = (aθ̇senθ, aθ̇cosθ, ż), (24.2)</p><p>where a is radius of cylinder andθ is the polar angular displacement. Then, the dot</p><p>91</p><p>92 CHAPTER 24. EXERCISE 24 - GEOVANE NAYSINGER</p><p>product is</p><p>˙⃗r. ˙⃗r = ṙ2 = a2θ̇2 + ż2 (24.3)</p><p>and the kinectic energy relative to particle is</p><p>Tp =</p><p>m</p><p>2</p><p>(a2θ̇2 + ż2). (24.4)</p><p>The rotation kinect energy relative to cylinder can be write as</p><p>Tc =</p><p>Iω2</p><p>2</p><p>=</p><p>M(z)R2ω2</p><p>4</p><p>=</p><p>πa4z0ρω</p><p>2</p><p>4</p><p>, (24.5)</p><p>where ρ is the density andω is angular velocity relative to cylinder. The mass M of</p><p>cylinder is fixed, so we usez0 as the lenght of cylinder. We callα̇ = ω, so α is an-</p><p>gular displacement of cylinder. The interactive potential in this problem is only the</p><p>gravitation potentialV = −mgz. So the Lagrangian can be write as</p><p>L =</p><p>m</p><p>2</p><p>(a2θ̇2 + ż2) +</p><p>πa4z0ρα̇</p><p>2</p><p>4</p><p>+mgz. (24.6)</p><p>Note that we chose as generalized coordinate the variables z, θ, α.</p><p>For now, let’s work with hamiltonian. The hamiltonian of the system is like</p><p>H = H(z, θ, α, Pz, Pθ, Pα, t). We write the canonical momenta of the system</p><p>Pz =</p><p>∂L</p><p>∂ż</p><p>= mż</p><p>Pθ =</p><p>∂L</p><p>∂θ̇</p><p>= ma2θ̇</p><p>Pα =</p><p>∂L</p><p>∂α̇</p><p>= Bα̇</p><p>whereB ≡ πa4ρz0</p><p>2</p><p>. The Hamiltonian is</p><p>H = żPz + θ̇Pθ + α̇Pα − L</p><p>and substituting the valuesż, θ̇ and α̇, we find</p><p>H =</p><p>1</p><p>2</p><p>P 2</p><p>z</p><p>m</p><p>+</p><p>1</p><p>2</p><p>P 2</p><p>θ</p><p>ma2</p><p>+</p><p>1</p><p>2</p><p>P 2</p><p>α</p><p>B</p><p>−mgz. (24.7)</p><p>Now we can write the Hamilton equations of motion. Then</p><p>93</p><p>ż =</p><p>∂H</p><p>∂Pz</p><p>=</p><p>Pz</p><p>m</p><p>Ṗz = −</p><p>∂H</p><p>∂z</p><p>= −mg</p><p>θ̇ =</p><p>∂H</p><p>∂Pθ</p><p>=</p><p>Pθ</p><p>ma2</p><p>Ṗθ = −</p><p>∂H</p><p>∂θ</p><p>= 0</p><p>α̇ =</p><p>∂H</p><p>∂Pα</p><p>=</p><p>Pα</p><p>B</p><p>Ṗα = −∂H</p><p>∂α</p><p>= 0</p><p>that describe the system of the problem.</p><p>Exercise 25 - LUCAS BOURSCHEIDT</p><p>Solution:</p><p>In the rotating frame, in which the helix is at rest, the helix equation parametrized by</p><p>the cylindrical coordinateθ is given by:</p><p>95</p><p>96 CHAPTER 25. EXERCISE 25 - LUCAS BOURSCHEIDT</p><p>r′(θ) = a cos θ x̂′ + a sin θ ŷ′ + (z0 − bθ) ẑ′ (A)</p><p>with r′(0) = ax̂′ + z0 ẑ′ and b = h/2π, where h is the circular helix step. The transfor-</p><p>mation from the coordinates of the rotating referential (S’) to the laboratory referential</p><p>(S) is given by:</p><p>xy</p><p>z</p><p> =</p><p>cos ωt −sin ωt 0</p><p>sin ωt cos ωt 0</p><p>0 0 1</p><p>x′y′</p><p>z′</p><p></p><p>With this transformation, the position vectorr in the laboratory reference is obtained</p><p>as follows: </p><p>x = a (cos θ cos ωt− sin θ sin ωt) = a cos(θ + ωt)</p><p>y = a (cos θ sin ωt+ sin θ cos ωt) = a sin(θ + ωt)</p><p>z = z0 − bθ</p><p>So:</p><p>r(θ) = a cos (θ + ωt) x̂+ a sin (θ + ωt) ŷ + (z0 − bθ) ẑ (A′)</p><p>Settingϕ = θ + ωt as the angle of rotation of the particle in the laboratory frame, we</p><p>have for r:</p><p>r(ϕ) = a cos ϕ x̂+ a sin ϕ ŷ + [z0 − b(ϕ− ωt)] ẑ (B)</p><p>(a) Hamiltonian in the laboratory system</p><p>Velocity:</p><p>97</p><p>From (B):</p><p>v =</p><p>dr</p><p>dt</p><p>= −a ϕ̇ sin ϕ x̂+ a ϕ̇ cos ϕ ŷ − b (ϕ̇− ω) ẑ</p><p>Kinetic energy:</p><p>T =</p><p>1</p><p>2</p><p>mv2 =</p><p>1</p><p>2</p><p>ma2 ϕ̇2 +</p><p>1</p><p>2</p><p>mb2 (ϕ̇− ω)2</p><p>Potential energy:</p><p>V = mgz = mg [z0 − b(ϕ− ωt)]</p><p>Lagrangian:</p><p>L = T − V =</p><p>1</p><p>2</p><p>ma2 ϕ̇2 +</p><p>1</p><p>2</p><p>mb2 (ϕ̇− ω)2 −mg [z0 − b(ϕ− ωt)] (C)</p><p>Canonical momentum:</p><p>pϕ =</p><p>∂L</p><p>∂ϕ̇</p><p>= m(a2 + b2)ϕ̇−mb2ω (D)</p><p>Isolatingϕ̇:</p><p>ϕ̇ =</p><p>pϕ +mb2ω</p><p>m(a2 + b2)</p><p>(E)</p><p>Hamiltonian:</p><p>H = pϕ ϕ̇− L</p><p>98 CHAPTER 25. EXERCISE 25 - LUCAS BOURSCHEIDT</p><p>Using (C) and (D):</p><p>H =</p><p>[</p><p>m(a2 + b2)ϕ̇−mb2ω</p><p>]</p><p>ϕ̇− 1</p><p>2</p><p>ma2 ϕ̇2 − 1</p><p>2</p><p>mb2 (ϕ̇− ω)2 +mg [z0 − b(ϕ− ωt)]</p><p>Performing the necessary algebra we find:</p><p>H =</p><p>1</p><p>2</p><p>m(a2 + b2) ϕ̇2 − 1</p><p>2</p><p>mb2 ω2 +mg [z0 − b(ϕ− ωt)] (F )</p><p>Inserting (E) into (F):</p><p>H =</p><p>(pϕ +mb2 ω)2</p><p>2m(a2 + b2)</p><p>− 1</p><p>2</p><p>mb2 ω2 +mg [z0 − b(ϕ− ωt)]</p><p>From (F), since H is NOT the sum of kinetic energy and potential energy, H is NOT</p><p>the total energy. Since H is explicitly dependent on time, it is also NOT a constant of</p><p>motion.</p><p>(b) Hamiltonian in the rotating system</p><p>Velocity:</p><p>From (A’), which naturally includes the inertial effects due to dependence onω,</p><p>without having to consider the time dependence of the basis vectors (which would be</p><p>necessary if we used (A)):</p><p>v′ =</p><p>dr</p><p>dt</p><p>= −a (θ̇ + ω) sin (θ + ωt) x̂+ a (θ̇ + ω) cos (θ + ωt) ŷ − b θ̇ ẑ</p><p>Kinetic energy:</p><p>T =</p><p>1</p><p>2</p><p>mv′2 =</p><p>1</p><p>2</p><p>ma2(θ̇ + ω)2 +</p><p>1</p><p>2</p><p>mb2θ̇2</p><p>99</p><p>Potential energy:</p><p>V = mgz = mg (z0 − bθ)</p><p>Lagrangian:</p><p>L = T − V =</p><p>1</p><p>2</p><p>ma2(θ̇ + ω)2 +</p><p>1</p><p>2</p><p>mb2θ̇2 −mg (z0 − bθ) (G)</p><p>Canonical momentum:</p><p>pθ =</p><p>∂L</p><p>∂θ̇</p><p>= m(a2 + b2) θ̇ +ma2 ω (H)</p><p>Isolatingθ̇:</p><p>θ̇ =</p><p>pθ −ma2 ω</p><p>m(a2 + b2)</p><p>(I)</p><p>Hamiltonian:</p><p>H = pθ θ̇ − L</p><p>Using (G) and (H):</p><p>H =</p><p>[</p><p>m(a2 + b2) θ̇ +ma2 ω</p><p>]</p><p>θ̇ − 1</p><p>2</p><p>ma2(θ̇ + ω)2 − 1</p><p>2</p><p>mb2θ̇2 +mg (z0 − bθ)</p><p>That is:</p><p>H =</p><p>1</p><p>2</p><p>m(a2 + b2) θ̇2 − 1</p><p>2</p><p>ma2 ω2 +mg (z0 − bθ) (J)</p><p>100 CHAPTER 25. EXERCISE 25 - LUCAS BOURSCHEIDT</p><p>Inserting (I) into (J):</p><p>H =</p><p>(pθ −ma2 ω)2</p><p>2m(a2 + b2)</p><p>− 1</p><p>2</p><p>ma2 ω2 +mg (z0 − bθ)</p><p>From (J), again, since H is NOT the sum of kinetic energy and potential energy, H is</p><p>NOT the total energy. However, since now H is NOT explicitly dependent on time, it</p><p>is a constant of motion.</p><p>Exercise 26 - RODRIGO WEBER</p><p>PEREIRA</p><p>A particle of massm can move in one dimension under the influence of two</p><p>springs connected to fixed points a distancea apart (see figure). The springs obey</p><p>Hooke’s law and have zero unstretched lengths and force constantsk1 and k2 respec-</p><p>tively.</p><p>a) Using the position of the particle from one fixed point as the generalized coordinate,</p><p>find the Lagrangian and the corresponding Hamiltonian. Is the energy conserved? is</p><p>the Hamiltonian conserved?</p><p>b) Introduce a new coordinateQ defined by</p><p>Q = q − b sinωt, b =</p><p>k2a</p><p>k1 + k2</p><p>,</p><p>What is the Lagrangian in terms ofQ? What is the corresponding Hamiltonian? is the</p><p>energy conserved? Is the Hamiltonian conserved?</p><p>Resposta: Taking∆x1 = q and ∆x2 = a − q as the deformations of the strings,</p><p>101</p><p>102 CHAPTER 26. EXERCISE 26 - RODRIGO WEBER PEREIRA</p><p>we may write</p><p>the kinetic and potential energy as:</p><p>T =</p><p>mq̇2</p><p>2</p><p>V =</p><p>k1q</p><p>2</p><p>2</p><p>+</p><p>k2(a− q)2</p><p>2</p><p>,</p><p>the Lagrangian of the system is:</p><p>L = T − V =</p><p>mq̇2</p><p>2</p><p>− k1q</p><p>2</p><p>2</p><p>− k2(a− q)2</p><p>2</p><p>, (26.1)</p><p>Since ∂L</p><p>∂t</p><p>= 0, H (the Hamiltonian) is constant. Also, since the components of the</p><p>position vector don’t depend explicitly of the time (∂r⃗</p><p>∂t</p><p>= 0), the Hamiltonian is equal</p><p>to the total energy of the system:H = E = T + V . Taking the conjugate momentp</p><p>associated with the coordinateq to be</p><p>p =</p><p>∂L</p><p>∂q̇</p><p>= mq̇,</p><p>we may write the Hamiltonian of the system as:</p><p>H = T + V =</p><p>p2</p><p>2m</p><p>+</p><p>k1q</p><p>2</p><p>2</p><p>+</p><p>k2(a− q)2</p><p>2</p><p>(26.2)</p><p>Now, using the coordinateQ = q−b sinωt, we have: q = Q+b sinω. So q̇ = Q̇+bω cosωt.</p><p>Substituting that in the already constructed Lagrangian:</p><p>L =</p><p>m</p><p>2</p><p>(Q̇+ bω cosωt)2 − 1</p><p>2</p><p>k1(Q+ b sinωt)2 − 1</p><p>2</p><p>k2(a−Q− b sinωt)2</p><p>Rearranging terms to match the matrix form:</p><p>L = L0 + Q̇︸︷︷︸</p><p>⃗̇qt</p><p>bω cosωt︸ ︷︷ ︸</p><p>a⃗(Q,t)</p><p>+</p><p>1</p><p>2</p><p>Q̇︸︷︷︸</p><p>⃗̇qt</p><p>m︸︷︷︸</p><p>T (Q,t)</p><p>Q̇︸︷︷︸</p><p>⃗̇q</p><p>(26.3)</p><p>WhereL0(Q, t) =</p><p>m</p><p>2</p><p>(bω cosωt)2 − 1</p><p>2</p><p>k1(Q + b sinω)2 − 1</p><p>2</p><p>k2(a−Q− b sinω)2. SinceT is a</p><p>symmetric (1× 1) matrix, the Hamiltonian can be written as:</p><p>H =</p><p>1</p><p>2</p><p>(p− a)T−1(p− a)− L0 =</p><p>(p− a)2</p><p>2</p><p>T−1 − L0</p><p>SinceT = m is a 1× 1 matrix, it’s inverse is justT−1 = m−1, the Hamiltonian is:</p><p>H =</p><p>(p− bω cosωt)2</p><p>2m</p><p>− m</p><p>2</p><p>(bω cosωt)2 +</p><p>1</p><p>2</p><p>k1(Q+ b sinω)2 +</p><p>1</p><p>2</p><p>k2(a−Q− b sinω)2 (26.4)</p><p>Wherep = ∂L</p><p>∂Q̇</p><p>. This time, as the Lagrangian is an explicit function oft, the Hamiltonian</p><p>is not conserved. Also, since⃗r(Q, t) =< Q + b sinωt, 0, 0 >, ∂r⃗</p><p>∂t</p><p≯= 0, which means that</p><p>the Hamiltonian is not equal to the total energy of the system.</p><p>Exercise 27 - JENIFER ANDRADE DE</p><p>MATOS</p><p>(a) The Lagrangian for a system of 1 degree of freedom can be written as</p><p>L =</p><p>m</p><p>2</p><p>(q̇2sin2ωt+ q̇qωsin2ωt+ q2ω2),</p><p>What is the corresponding Hamiltonian? Is it conserved?</p><p>(b) Introduce a new coordinate defined by</p><p>Q = qsinωt,</p><p>Find the Lagrangian in terms of the new coordinate and the corresponding</p><p>Hamiltonian. Is H conserved?</p><p>Resolution:</p><p>Using the conjugated momentum:</p><p>p =</p><p>∂L</p><p>∂q̇</p><p>= mq̇sin2(ωt) +</p><p>1</p><p>2</p><p>mqωsin(2ωt) (27.1)</p><p>Let´s isolate q̇ from (27.1)</p><p>103</p><p>104 CHAPTER 27. EXERCISE 27 - JENIFER ANDRADE DE MATOS</p><p>q̇ =</p><p>p</p><p>msin2(ωt)</p><p>− qωcot(ωt) (27.2)</p><p>To determine the Hamiltonian, let’s use:</p><p>H = pq̇ − L (27.3)</p><p>H = pq̇ − m</p><p>2</p><p>(q̇2sin2ωt+ q̇qωsin2ωt+ q2ω2) (27.4)</p><p>Substituting (27.2) in (27.4), we find:</p><p>H =</p><p>p2</p><p>2msin2(ωt)</p><p>− qpωcot(ωt)− 1</p><p>2</p><p>mq2ω2sin2(ωt) (27.5)</p><p>This Hamiltonian depends explicitly of the time and therefore it doesn’t con-</p><p>serve.</p><p>b) Resolution:</p><p>Now, we have to substitute q for Q andq̇ for Q̇</p><p>Since,</p><p>Q = qsinωt, (27.6)</p><p>We can derive both sides and isolateq, and q̇:</p><p>q =</p><p>Q</p><p>sin(ωt)</p><p>(27.7)</p><p>q̇ =</p><p>Q̇</p><p>sin(ωt)</p><p>− Qωcos(ωt)</p><p>sin2(ωt)</p><p>(27.8)</p><p>105</p><p>Substituting 27.7 and 27.8 in the original Lagrangian, we find:</p><p>L =</p><p>m</p><p>2</p><p>(</p><p>Q̇2 + ω2Q2</p><p>)</p><p>(27.9)</p><p>To determine the Hamiltonian, we have:</p><p>P =</p><p>∂L</p><p>∂Q̇</p><p>= mQ̇ (27.10)</p><p>The Hamiltonian can be written as:</p><p>H = PQ̇− L (27.11)</p><p>Substituting (27.9) and (27.10) in (27.11), we find the Hamiltonian:</p><p>H =</p><p>P 2</p><p>2m</p><p>− mω2Q2</p><p>2</p><p>(27.12)</p><p>Since, using this set of coordinates the Hamiltonian is not explicitly dependent</p><p>of the time, it is conserved.</p><p>Exercise 28 - JACKSON GALVÃO</p><p>28. A magnetic monopole is defined (if one exist) by a magnetic field singularity of the</p><p>form B = br/r3, where b is a constant (a measure of the magnetic charge, as it were).</p><p>Suppose a particle os massm moves in the field of a magnetic monopole and a central</p><p>force derived from the potentialV (r) = −k/r.</p><p>Review the derivation of the Runge-Lenz vector:</p><p>˙⃗p = f(r)e⃗r</p><p>f(r) = −∂V</p><p>∂V</p><p>d</p><p>dt</p><p>(p⃗xL⃗) = ˙⃗pxL⃗ = f(r)e⃗× L⃗</p><p>this brings us to</p><p>e⃗r × (v⃗ × p⃗) = mrr⃗r × (r⃗r × ˙⃗r) = −mr2dr⃗r</p><p>dt</p><p>e⃗rṙ − ˙⃗r = −rde⃗r</p><p>dt</p><p>Thereby, for T whether or not, the⃗L is conserved.</p><p>e⃗r × L⃗) = −mr2</p><p>de⃗r</p><p>dt</p><p>Therefore, we assumed thatL⃗ is conserved.</p><p>d</p><p>dt</p><p>(p⃗× L⃗) = −mr2f(r)de⃗r</p><p>dt</p><p>107</p><p>108 CHAPTER 28. EXERCISE 28 - JACKSON GALVˆO</p><p>d</p><p>dt</p><p>(p⃗× L⃗) = −mr2f(r)de⃗r</p><p>dt</p><p>(a) Find the form of Newton’s equations of motion, using the Lorentz force given by</p><p>Eq.(1.60). By looking at the productr × show that while the mechanical angular mo-</p><p>mentum is not conserved (the filed of force is noncentral) there is a conserved vector</p><p>D = L − qb</p><p>c</p><p>r</p><p>r</p><p>Then, after observed, we doing</p><p>˙⃗p = f(r)e⃗r +</p><p>q</p><p>c</p><p>˙⃗v × B⃗</p><p>for</p><p>B⃗ =</p><p>b</p><p>r3</p><p>r⃗ =</p><p>b</p><p>r2</p><p>e⃗r</p><p>where</p><p>b</p><p>r3</p><p>˙⃗r × r⃗ = − b</p><p>mr3</p><p>r⃗ × p⃗ = − b</p><p>mr3</p><p>L⃗</p><p>That bring us to the equation of motions, such that</p><p>˙⃗p = f(r)e⃗r −</p><p>qb</p><p>mcr3</p><p>L⃗</p><p>With this, we have</p><p>dL⃗</p><p>dt</p><p>=</p><p>d</p><p>dt</p><p>(r⃗ × p⃗) = ˙r⃗ × p⃗ + r⃗ × ˙⃗p = −qb</p><p>c</p><p>de⃗r</p><p>dt</p><p>where</p><p>r⃗ × ˙⃗p = − qb</p><p>mcr3</p><p>r⃗ × L⃗</p><p>= − qb</p><p>mcr</p><p>e⃗r × L⃗</p><p>= −qb</p><p>c</p><p>de⃗r</p><p>dt</p><p>109</p><p>Therefore, we have the solutions to anyf(r), in the form of 28.1</p><p>d</p><p>dt</p><p>(L⃗ − qb</p><p>c</p><p>e⃗r) = (28.1)</p><p>where</p><p>L⃗ − qb</p><p>c</p><p>e⃗r ≡ D⃗</p><p>(b) By paralleling the steps leading from Eq.(3.79) to Eq.(3.82), show that for somef(r)</p><p>there is a conserved vector analogues to the Laplace-Runge-Lenz vector in whichD</p><p>plays the same role asL in the pure Kepler force problem.</p><p>So, firstly we can do</p><p>d</p><p>dt</p><p>(p⃗× D⃗) = ˙⃗p× D⃗</p><p>=</p><p>(</p><p>f(r)e⃗r −</p><p>qb</p><p>mcr3</p><p>L⃗</p><p>)</p><p>×</p><p>(</p><p>L⃗ − qb</p><p>c</p><p>e⃗r</p><p>)</p><p>=</p><p>(</p><p>f(r) − 1</p><p>mr3</p><p>(</p><p>qb</p><p>c</p><p>)2</p><p>)</p><p>e⃗r × L⃗</p><p>=</p><p>(</p><p>−mr2f(r) +</p><p>1</p><p>r</p><p>(</p><p>qb</p><p>c</p><p>)2</p><p>)</p><p>de⃗r</p><p>dt</p><p>If the terms inner parentheses, we can have a conserved quantity, such that</p><p>−mr2f(r) +</p><p>1</p><p>r</p><p>(</p><p>qb</p><p>c</p><p>)2</p><p>then</p><p>f(r) =</p><p>−k</p><p>r2</p><p>+</p><p>(qb/c)2</p><p>mr3</p><p>(28.2)</p><p>Therefore, we must be have</p><p>d</p><p>dt</p><p>(p⃗× D⃗ − mke⃗r) = 0 (28.3)</p><p>You might have wedered why you’re not ashed to do part (a) by finding the conserved</p><p>momentum associated with an angular coordinate.</p><p>110 CHAPTER 28. EXERCISE 28 - JACKSON GALVÃO</p><p>The veason is that a magnetic monopole is not a solution of the Maxweell equations</p><p>((r � ~B 6= 0) at v = 0, so tecnically, thers is no vector potential ~A such that r � ~A = ~B.</p><p>Nonetheless, one can introduce a vector potential that works everywhers except along</p><p>a Ime form r = 0 to r ! 1 . Pick as arbitrary z-direction, an let this line be the negative</p><p>z-axis.</p><p>Aplly Stroke's theorem to the "cap" on the sohere de�ned by � � � . Then,</p><p>A � 2�r sin � =</p><p>Z</p><p>C</p><p>~A � ~ds =</p><p>Z</p><p>S</p><p>~B � ~ds</p><p>= b</p><p>Z</p><p>1</p><p>� �x2 � �x2 sin�d�d�</p><p>= 2�b</p><p>Z �</p><p>0</p><p>d� sin�</p><p>= 2�b (1 � cos �)</p><p>Then we have the singular on negative z-axis at (� = � ), such that</p><p>A � =</p><p>b</p><p>r</p><p>1 � cos�</p><p>sin�</p><p>Thereby, we veri�ed that</p><p>~A =</p><p>b</p><p>r</p><p>1 � cos�</p><p>sin�</p><p>~er</p><p>111</p><p>satisfying</p><p>∇× A⃗ =</p><p>b</p><p>r2</p><p>e⃗r = B⃗</p><p>Thus the particle is</p><p>L =</p><p>1</p><p>2</p><p>m ˙⃗v2 − V (r) +</p><p>q</p><p>c</p><p>A⃗ · ˙⃗r</p><p>In spherical coordinates, we must be have</p><p>A⃗ · r⃗ = AϕVϕ =</p><p>b</p><p>�r</p><p>1 − cos θ</p><p>���sin θ �r�</p><p>��sin θϕ̇ = b(1 − cos θ)ϕ̇</p><p>So, the Lagrangian is</p><p>L =</p><p>1</p><p>2</p><p>mṙ2 +</p><p>1</p><p>2</p><p>mr2θ̇2 +</p><p>1</p><p>2</p><p>mr2 sin2 θϕ̇2</p><p>Now, the azimutal coordinate ϕ is cyclic. Then,</p><p>pϕ =</p><p>∂L</p><p>∂ϕ̇</p><p>= mr2 sin θϕ̇ +</p><p>qb</p><p>c</p><p>(1 − cos θ</p><p>where</p><p>mr2 sin θϕ̇ = Lz = L⃗ · e⃗z</p><p>is conserved. Then, we can writepϕ as</p><p>pϕ = L⃗ · e⃗z +</p><p>qb</p><p>c</p><p>− qb</p><p>c</p><p>e⃗r · e⃗z</p><p>=</p><p>(</p><p>L − qb</p><p>c</p><p>e⃗r</p><p>)</p><p>e⃗z +</p><p>qb</p><p>c</p><p>where qb/c is a constant.</p><p>Therefore, the z-direction was arbitrary, so</p><p>L⃗ − qb</p><p>c</p><p>e⃗r</p><p>is conserved.</p><p>Exercise 29 - EDGARD</p><p>KRETSCHMANN</p><p>The lagrangian of a heavy symmetric top with one point �xed, as seen in the</p><p>previous chapters, is given by</p><p>L =</p><p>I 1</p><p>2</p><p>( _� 2 + _� 2sin2� +</p><p>I 3</p><p>2</p><p>( _ + _�cos� )2 � Mglcos� (29.1)</p><p>So, now we can obtain the generalized momenta for each generalized coordinate. We</p><p>can observe thatp� and p are cyclic coordinates, since the lagrangian doesn't depend</p><p>explicitly of these coordinates. Therefore, they are constants of motion in our problem</p><p>p =</p><p>@L</p><p>@_</p><p>= I 3( _ + _�cos� ) (29.2)</p><p>p� =</p><p>@L</p><p>@_�</p><p>= _� (I 1sin2� + I 3cos2� ) + I 3</p><p>_ cos� (29.3)</p><p>p� =</p><p>@L</p><p>@_�</p><p>= I 1</p><p>_� (29.4)</p><p>Now, we can obtain the generalized velocities in terms of the generalized mo-</p><p>menta</p><p>_ = p (</p><p>1</p><p>I 3</p><p>+</p><p>1</p><p>I 1tan2�</p><p>) � p�</p><p>1</p><p>I 1tan�sin�</p><p>(29.5)</p><p>_� =</p><p>p� � p cos�</p><p>I 1sin2�</p><p>(29.6)</p><p>_� =</p><p>p�</p><p>I 1</p><p>(29.7)</p><p>113</p><p>114 CHAPTER 29. EXERCISE 29 - EDGARD KRETSCHMANN</p><p>Now we are ready to obtain the Hamiltonian of this system using a Legendre transfor-</p><p>mation</p><p>H = pϕϕ̇+ pψψ̇ + pθθ̇ − L (29.8)</p><p>So, using equations (1), (5), (6) and (7) on (8), we will obtain that</p><p>H =</p><p>p2ψ</p><p>2</p><p>(</p><p>1</p><p>I3</p><p>+</p><p>1</p><p>I1tan2θ</p><p>) +</p><p>p2ϕ</p><p>2</p><p>1</p><p>I1sin2θ</p><p>− pϕpψcosθ</p><p>I1sin2θ</p><p>+</p><p>p2θ</p><p>2I1</p><p>+Mglcosθ (29.9)</p><p>So, three of the six equations of motions are equations (5), (6) and (7), when we com-</p><p>pute q̇i = ∂H</p><p>∂pi</p><p>. The other three equations are</p><p>ṗψ = −∂H</p><p>∂ψ</p><p>= 0 (29.10)</p><p>ṗϕ = −</p><p>∂H</p><p>∂ϕ</p><p>= 0 (29.11)</p><p>ṗθ = −</p><p>∂H</p><p>∂θ</p><p>=</p><p>(p2ψ + p2ϕ)cosθ − pψpϕ(1 + cos2θ)</p><p>I1sin3θ</p><p>+Mglsinθ (29.12)</p><p>The generalized momenta pϕ and pψ are constants as we had already concluded</p><p>before. And, since the Hamiltonian is a quantity that conserves as well, we can write</p><p>the generalized momentum pθ in terms ofθ using equation (9), as it follows</p><p>pθ =</p><p>r</p><p>2I1E − p2ψ(</p><p>I1</p><p>I3</p><p>+</p><p>1</p><p>tan2θ</p><p>)− p2ϕ</p><p>1</p><p>sin2θ</p><p>+ 2</p><p>pϕpψcosθ</p><p>sin2θ</p><p>− 2I1Mglcosθ (29.13)</p><p>So using equation (7) we can reduce this problem to a quadrature</p><p>t =</p><p>Z θ(t)</p><p>θ(0)</p><p>I1</p><p>dθ</p><p>q</p><p>2I1E − p2ψ( I1</p><p>I3</p><p>+ 1</p><p>tan2θ</p><p>)− p2ϕ 1</p><p>sin2θ</p><p>+ 2</p><p>p� p cosθ</p><p>sin2θ</p><p>− 2I1Mglcosθ</p><p>(29.14)</p><p>We can write as well the Routhian of this system. The definition of the Routhian is</p><p>R = Hcycl − Lnoncycl (29.15)</p><p>Therefore, remembering that the generalized coordinatesϕ andψ are cyclic coordinates</p><p>we write the Routhian as</p><p>R =</p><p>p2ψ</p><p>2</p><p>(</p><p>1</p><p>I3</p><p>+</p><p>1</p><p>I1tan2θ</p><p>) +</p><p>p2ϕ</p><p>2</p><p>1</p><p>I1sin2θ</p><p>− pϕpψcosθ</p><p>I1sin2θ</p><p>− I1</p><p>2</p><p>(θ̇2 +Mglcosθ (29.16)</p><p>115</p><p>We have just one nonignorable coordinate and so we will have only one equation of</p><p>motion</p><p>d</p><p>dt</p><p>∂R</p><p>∂θ̇</p><p>− ∂R</p><p>∂θ</p><p>= 0 (29.17)</p><p>This equation will result in an equivalent equation when we input equation (3)</p><p>on equation (12)</p><p>Exercise 30 - JEFFERSON SANTANA</p><p>MARTINS</p><p>In Exercise 16 of chapter 1, there is given the velocity-dependent potential as-</p><p>sumed in Weber's electrodynamics. What is the Hamiltonian for a simple particle</p><p>moving under the in�uence of such a potential?</p><p>Solution Starting from the de�nition of generalized force:</p><p>Qr = �</p><p>@U</p><p>@r</p><p>+</p><p>d</p><p>dt</p><p>@U</p><p>@_r</p><p>=</p><p>1</p><p>r 2</p><p>�</p><p>1</p><p>c2</p><p>�</p><p>_r 2</p><p>r 2</p><p>�</p><p>2•r</p><p>r</p><p>�</p><p>=</p><p>1</p><p>r 2</p><p>�</p><p>1</p><p>c2</p><p>_r 2</p><p>r 2</p><p>+</p><p>2</p><p>c2</p><p>d_r</p><p>dt</p><p>1</p><p>r</p><p>=</p><p>1</p><p>r 2</p><p>�</p><p>1</p><p>c2</p><p>_r 2</p><p>r 2</p><p>+</p><p>2</p><p>c2</p><p>�</p><p>d</p><p>dt</p><p>�</p><p>_r</p><p>r</p><p>�</p><p>� _r</p><p>d</p><p>dt</p><p>�</p><p>1</p><p>r</p><p>��</p><p>=</p><p>1</p><p>r 2</p><p>�</p><p>1</p><p>c2</p><p>_r 2</p><p>r 2</p><p>+</p><p>d</p><p>dt</p><p>�</p><p>2 _r</p><p>c2r</p><p>�</p><p>�</p><p>2 _r</p><p>c2</p><p>dr</p><p>dt</p><p>d</p><p>dr</p><p>�</p><p>1</p><p>r</p><p>�</p><p>=</p><p>1</p><p>r 2</p><p>�</p><p>1</p><p>c2</p><p>_r 2</p><p>r 2</p><p>+</p><p>d</p><p>dt</p><p>�</p><p>2 _r</p><p>c2r</p><p>�</p><p>+</p><p>2</p><p>c2</p><p>_r 2</p><p>r 2</p><p>=</p><p>1</p><p>r 2</p><p>+</p><p>1</p><p>c2</p><p>_r 2</p><p>r 2</p><p>+</p><p>d</p><p>dt</p><p>�</p><p>1</p><p>c2r</p><p>@_r 2</p><p>@_r</p><p>�</p><p>=</p><p>1</p><p>r 2</p><p>+</p><p>1</p><p>r 2</p><p>_r 2</p><p>c2</p><p>+</p><p>d</p><p>dt</p><p>@</p><p>@_r</p><p>�</p><p>1</p><p>r</p><p>_r 2</p><p>c2</p><p>�</p><p>Then</p><p>�</p><p>@U</p><p>@r</p><p>=</p><p>1</p><p>r 2</p><p>+</p><p>1</p><p>r 2</p><p>_r 2</p><p>c2</p><p>= �</p><p>@</p><p>@r</p><p>�</p><p>1</p><p>r</p><p>+</p><p>1</p><p>r</p><p>_r 2</p><p>c2</p><p>�</p><p>(A)</p><p>and</p><p>@U</p><p>@_r</p><p>=</p><p>@</p><p>@_r</p><p>�</p><p>1</p><p>r</p><p>_r 2</p><p>c2</p><p>�</p><p>=</p><p>@</p><p>@_r</p><p>�</p><p>1</p><p>r</p><p>+</p><p>1</p><p>r</p><p>_r 2</p><p>c2</p><p>�</p><p>(B )</p><p>117</p><p>118 CHAPTER 30. EXERCISE 30 - JEFFERSON SANTANA MARTINS</p><p>From equations (A) and (B) we conclude that</p><p>U (r, ṙ) =</p><p>1</p><p>r</p><p>+</p><p>1</p><p>r</p><p>ṙ2</p><p>c2</p><p>=</p><p>1</p><p>r</p><p>(</p><p>1 +</p><p>ṙ2</p><p>c2</p><p>)</p><p>Thus, the Lagrangian function is given by</p><p>L = T − U =</p><p>1</p><p>2</p><p>m</p><p>(</p><p>ṙ2 + r2θ̇2 + r2sin2θ ϕ̇2</p><p>)</p><p>− 1</p><p>r</p><p>(</p><p>1 +</p><p>ṙ2</p><p>c2</p><p>)</p><p>The generalized coordinates are r, θ, and ϕ. The generalized momentums being</p><p>pθ =</p><p>∂L</p><p>∂θ̇</p><p>= mr2θ̇</p><p>pr =</p><p>∂L</p><p>∂ṙ</p><p>=</p><p>(</p><p>m− 2</p><p>rc2</p><p>)</p><p>ṙ</p><p>pr =</p><p>∂L</p><p>∂ϕ̇</p><p>= mr2sin2(θ)ϕ̇·</p><p>We can get the Hamiltonian by the expression</p><p>H(r, θ, pθ, pr, pϕ) = pθθ̇ + prṙ + pϕϕ̇− L (30.1)</p><p>isolating</p><p>θ̇ =</p><p>pθ</p><p>mr2</p><p>ϕ̇ =</p><p>pϕ</p><p>mr2sen2(θ)</p><p>ṙ =</p><p>prc</p><p>2r</p><p>mrc2 − 2</p><p>·</p><p>Substituting these expressions in equation 30.1, we obtain the Hamiltonian</p><p>H(r, θ, pθ, pr, pϕ) =</p><p>p2θ</p><p>mr2</p><p>+</p><p>p2ϕ</p><p>2mr2sin2(θ)</p><p>+</p><p>p2rc</p><p>2r</p><p>mrc2 − 2</p><p>[</p><p>1− 1</p><p>2</p><p>mc2r</p><p>(mrc2 − 2)</p><p>]</p><p>+</p><p>1</p><p>r</p><p>[</p><p>1 +</p><p>p2rc</p><p>4r2</p><p>c2(mrc2 − 2)2</p><p>]</p><p>·</p><p>Exercise 31 - GUILHERME SHOITI</p><p>YOSHIMATSU GIARDINI</p><p>119</p><p>Exercise 32 - DOUGLAS OLIVEIRA NO-</p><p>VAES</p><p>A symmetrical top is mounted so that it pivots about its center of mass.The pivot</p><p>in turn is �xed a distance r from the center of a horizontal disk free to rotate about a</p><p>vertical axis. The top is started with an initial rotation about its �gure axis, which is</p><p>initially at an angle � 0 to the vertical. Analyze the possible nutation of the top as a case</p><p>of small oscillations about steady motion.</p><p>Response: One set the horizontal disk in xy plane and let the vertical direction</p><p>be z. It will be assumed that pivot is a distance r from disk's center of mass and is �xed</p><p>to the disk. This implies that symmetrical top is free to rotate about it's center of mass</p><p>but is �xed to the disk.</p><p>One has two angular velocities: 1) one describes rotation of the disk</p><p>= _� = ẑ</p><p>and 2) ! describes rotation of the symmetrical top.</p><p>Kinetic energy of the disk is simply T d = 1</p><p>2 I _� 2. Next, one considers the rigid</p><p>top. To de�ne Euler angles one simply translates our coordinate system by r (center of</p><p>top's mass)</p><p>(x0; y0; z0) = r + ( x; y; z)</p><p>One can de�ne Euler angles and relate top's principal axes to our inertial coor-</p><p>dinate system. For convenience, one decomposes the position vector of an arbitrary</p><p>particle of the top r i = r + x i . The velocity of an arbitrary particle of the top is given</p><p>121</p><p>122 CHAPTER 32. EXERCISE 32 - DOUGLAS OLIVEIRA NOVAES</p><p>by</p><p>ṙi = Ω � r+ ω � xi</p><p>Since the top is symmetric, the elements of the inertia tensorI1 = I2 6= I3. Kinetic</p><p>energy of the top has the following form</p><p>Tt =</p><p>1</p><p>2</p><p>X</p><p>i</p><p>miṙ</p><p>2</p><p>i</p><p>=</p><p>1</p><p>2</p><p>r2ξ̇2</p><p>X</p><p>i</p><p>mi + (Ω � r),</p><p>�</p><p>ω �</p><p>X</p><p>i</p><p>mixi</p><p>�</p><p>+</p><p>1</p><p>2</p><p>X</p><p>i</p><p>mi(ω � xi)</p><p>2</p><p>Since</p><p>P</p><p>mixi = 0 and the last term in the equation above corresponds to mo-</p><p>ment of inertia, one has</p><p>Tt =</p><p>1</p><p>2</p><p>Mr2ξ̇2 +</p><p>1</p><p>2</p><p>eω � I � ω</p><p>One can express angular velocity in terms of Euler angles as,</p><p>ωx = ϕ̇ sin θ sinψ + θ̇ cosψ</p><p>ωy = ϕ̇ sin θ cosψ � θ̇ sinψ</p><p>ωz = ϕ̇ cos θ + ψ̇</p><p>solve the products of matrices as</p><p>eω � I � ω =</p><p>�</p><p>ωx ωy ωz</p><p>�</p><p>2</p><p>4</p><p>I1 0 0</p><p>0 I2 0</p><p>0 0 I3</p><p>3</p><p>5</p><p>2</p><p>4</p><p>ωx</p><p>ωy</p><p>ωz</p><p>3</p><p>5</p><p>and write the total kinetic energy of the system as</p><p>T =</p><p>1</p><p>2</p><p>Iξ̇2 +</p><p>1</p><p>2</p><p>Mr2ξ̇2 +</p><p>1</p><p>2</p><p>I1</p><p>�</p><p>θ̇2 + ϕ̇2 sin2 θ</p><p>�</p><p>+</p><p>1</p><p>2</p><p>I3</p><p>�</p><p>ψ̇ + ϕ̇ cos θ</p><p>� 2 (32.1)</p><p>Notice thatT = E = L , the Lagrangian. Sinceξ, ϕ and ψ are cyclic coordinates,</p><p>one concludes that it is necessary to remove them. Then, from energyE = E(θ̇, θ), one</p><p>can find equation of motion forθ. Hence</p><p>E =</p><p>1</p><p>2</p><p>I1θ̇</p><p>2 +</p><p>1</p><p>2</p><p>˜̇qTq̇</p><p>=</p><p>1</p><p>2</p><p>I1θ̇</p><p>2 +</p><p>1</p><p>2</p><p>p̃T−1p</p><p>(32.2)</p><p>123</p><p>Where ˜̇q =</p><p>[</p><p>ξ̇, ϕ̇, ψ̇</p><p>]</p><p>and p̃ =</p><p>[</p><p>pξ, pϕ, pψ</p><p>]</p><p>. Inverting matrixT will eliminate cyclic coor-</p><p>dinates. pξpϕ</p><p>pψ</p><p> =</p><p>I +Mr2 0 0</p><p>0 I1 sin</p><p>2 θ + I3 cos</p><p>2 θ I3 cos θ</p><p>0 I3 cos θ I3</p><p> ξ̇ϕ̇</p><p>ψ̇</p><p></p><p>After inversion of matrixT , as done in equation12 of Exercise 21, one has</p><p>T � 1 =</p><p></p><p>1</p><p>I+Mr2 0 0</p><p>0 1</p><p>I1 sin2 θ</p><p>− cos θ</p><p>I1 sin2 θ</p><p>0 − cos θ</p><p>I1 sin2 θ</p><p>1</p><p>I3</p><p>+ cos2 θ</p><p>I1 sin2 θ</p><p></p><p>Hence,</p><p>E =</p><p>1</p><p>2</p><p>I1θ̇</p><p>2 +</p><p>(pϕ − pψ cos θ)2</p><p>2I1 sin</p><p>2 θ</p><p>+ const, (32.3)</p><p>Entering the following initial conditions</p><p>θ(0) = θ0</p><p>θ̇(0) = 0</p><p>ξ̇(0) =?</p><p>ϕ̇(0) =?</p><p>ψ̇(0) = ψ̇0</p><p>(32.4)</p><p>The question marks are assigned to initial conditions that are not given. Sinceϕ is</p><p>defined relative to our inertial system we will assume thaṫξ = ξ̇(0) = ϕ̇(0) = ξ̇0.</p><p>Additionally, one will assume thatψ̇ ≫ ξ̇0.</p><p>These two assumptions will model a gyroscope on earth. To see this, if you have</p><p>a gyroscope at rest on Earth’s surface in rotating reference frame, then this leads to</p><p>ϕ̇(0) = ξ̇0 (due to friction on bearings). Now, one rotates the gyroscope, as described</p><p>by equations (32.1), (32.2) and first from (32.4).</p><p>Inserting the initial conditions forθ into equation (32.3)</p><p>0 =</p><p>1</p><p>2</p><p>I1θ̇</p><p>2 +</p><p>(pϕ − pψ cos θ)2</p><p>2I1 sin</p><p>2 θ</p><p>− (pϕ − pψ cos θ0)2</p><p>2I1 sin</p><p>2 θ0</p><p>,</p><p>124 CHAPTER 32. EXERCISE 32 - DOUGLAS OLIVEIRA NOVAES</p><p>And, inserting initial conditions for the rest</p><p>p� = ( I 1 sin2 � 0 + I 3 cos2 � 0) _� 0 + I 3 cos� 0</p><p>_ 0</p><p>p = I 3 cos� 0</p><p>_� 0 + I 3</p><p>_ 0</p><p>Using substitution u = cos� , shorthand notation I I b= p� and I I a = p ,</p><p>0 =</p><p>1</p><p>2</p><p>_u2 +</p><p>1</p><p>2</p><p>(b� au)2 �</p><p>1</p><p>2</p><p>_� 2</p><p>0 sin2 � 0(1 � u2)</p><p>Taking the time derivative of equation above</p><p>•u +</p><p>�</p><p>a2 + _� 2</p><p>0 sin2 � 0</p><p>�</p><p>u = ab+ O</p><p>� _� 2</p><p>0</p><p>�</p><p>Since,u = cos� , one �nds the frequency of nutation to be</p><p>! =</p><p>q</p><p>a2 + _� 2</p><p>0 sin2 � 0</p><p>= _ 0 +</p><p>I 3</p><p>I 1</p><p>cos� 0</p><p>_� 2</p><p>0</p><p>Next, one can also shows that amplitude of nutation is small. From initial con-</p><p>ditions on u</p><p>u =</p><p>ab</p><p>a2 + _� 2</p><p>0 sin2 � 0</p><p>+ A cos!t</p><p>=</p><p>b</p><p>a</p><p>+ A cos!t + O</p><p>� _� 2</p><p>0</p><p>�</p><p>Sinceu(0) = cos � 0, the amplitude A come</p><p>A = cos� 0 �</p><p>b</p><p>a</p><p>= �</p><p>I 1 sin2 � 0</p><p>I 3</p><p>_� 0</p><p>_ 0</p><p>+ O</p><p>� _� 2</p><p>0</p><p>_ 0</p><p>� 2</p><p>!</p><p>So it can be seen thatA is indeed small. The gyroscope will process unless it</p><p>is aligned with an axis of</p><p>a rotating body. In general one will have small oscillations</p><p>whenever A � 1.</p><p>125</p><p>From above equations, one can �nd such condition in terms of initial conditions</p><p>and frequency of nutation</p><p>A = cos� 0 �</p><p>p� p</p><p>p2</p><p>� + I 2</p><p>1</p><p>_� 2(0) sin2 � 0</p><p>� 1</p><p>! =</p><p>1</p><p>I 1</p><p>q</p><p>p2</p><p>� + I 2</p><p>1</p><p>_� 2(0) sin2 � 0</p><p>Physically this problem would be a model for say a gyroscope rotating on the</p><p>surface of Earth if one neglects nutation of Earth.</p><p>Exercise 33 - FRANCISCO AUGUSTO</p><p>FERREIRA ALMEIDA</p><p>Two mass points,m1 andm2. are connected by a string that acts as a Hooke’s-law</p><p>spring with force constant k. One particle is free to move without friction on a smooth</p><p>horizontal plane surface, the other hangs vertically down from the string through a</p><p>hole in the surface. Find the condition for steady motion in which the mass point on</p><p>the plane rotates uniformly at constant distance from the hole. Investigate the small</p><p>oscillations in die radial distance from the hole, and in the vertical height of the second</p><p>particle.</p><p>Solution:</p><p>FIGURE 33.1</p><p>127</p><p>128CHAPTER 33. EXERCISE 33 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>Notice that we can't apply small oscillation approximation since � for small os-</p><p>cillations will be almost linear function of time. We start by writing the Lagrangian</p><p>in cylindrical coordinate system (r, � ,z). Let the massm2 which hangs vertically down</p><p>from the string through a hole in the surface be along positive z axis. Kinetic energy of</p><p>the system is given by</p><p>T =</p><p>1</p><p>2</p><p>m1( _r1</p><p>2 + r 2</p><p>1</p><p>_� 2) +</p><p>1</p><p>2</p><p>m2 _z2</p><p>2 (33.1)</p><p>We will let gravity point upwards ~F = m2gẑ. Potential energy of the system is</p><p>due to string and gravity</p><p>V =</p><p>1</p><p>2</p><p>k(r1 + z2)2 � mgz2 (33.2)</p><p>Lagrangian of the system is given by</p><p>L =</p><p>1</p><p>2</p><p>m1( _r1</p><p>2 + r 2</p><p>1</p><p>_� 1</p><p>2</p><p>) +</p><p>1</p><p>2</p><p>m2 _z2</p><p>2 � V (33.3)</p><p>Notice that � is a cyclic coordinate, we can eliminate it by using Routhian pro-</p><p>cedure. Sincep� = m1r 2</p><p>1</p><p>_� 1 = constant, we �nd Routhian to be</p><p>R = p�</p><p>_� � L (33.4)</p><p>R = V +</p><p>p2</p><p>�</p><p>2m1r 2</p><p>1</p><p>�</p><p>�</p><p>1</p><p>2</p><p>m1 _r1</p><p>2 +</p><p>1</p><p>2</p><p>m2 _z2</p><p>2</p><p>�</p><p>(33.5)</p><p>Notice that our Routhian is of form R = Vef f � T is given by</p><p>Vef f =</p><p>1</p><p>2</p><p>k(r1 + z2)2 � m2gz2 +</p><p>p2</p><p>�</p><p>2m1r 2</p><p>1</p><p>(33.6)</p><p>129</p><p>For non-cyclic variable equations of motion are given by equation (8.50) from</p><p>the textbook.</p><p>d</p><p>dt</p><p>�</p><p>∂(� R)</p><p>∂ _qj</p><p>�</p><p>=</p><p>∂(� R)</p><p>∂qj</p><p>= �</p><p>∂Vef f</p><p>∂qj</p><p>(33.7)</p><p>Hence we see that∂Vef f /∂qj = 0 will give us equilibrium position ofr1 and z2.</p><p>Furthermore, we do not care about the cyclic part of Routhian and will therefore set</p><p>p� = m1r</p><p>2</p><p>10ωo. Conditions for equilibrium atr1 = r10andz2 = z20 will give</p><p>0 = k(r1 + z2) �</p><p>m2</p><p>1ω</p><p>2</p><p>or</p><p>4</p><p>10</p><p>m1r31</p><p>(33.8)</p><p>0 = k(r1 + z2) � m2g (33.9)</p><p>We see that equations (33.7) and (33.8) of the first paragraph correspond to bal-</p><p>ance of centrifugal force with gravitational force. Solving forr10 and z20 we get</p><p>r10 =</p><p>m2g</p><p>m1ω2</p><p>o</p><p>(33.10)</p><p>z20 =</p><p>m2g</p><p>k</p><p>�</p><p>m2g</p><p>m1ω2</p><p>o</p><p>(33.11)</p><p>To get small oscillations in radial and vertical direction we approximateVef f</p><p>about equilibrium position</p><p>Vef f = constant +</p><p>1</p><p>2</p><p>∂2Vef f</p><p>∂r21</p><p>(r1 � r10)2 +</p><p>∂2Vef f</p><p>∂r1∂z1</p><p>(r1 � r10)(z1 � z10) +</p><p>1</p><p>2</p><p>∂2Vef f</p><p>∂z22</p><p>(z2 � z20)2</p><p>(33.12)</p><p>Letting r1 = r10 + η1 and z2 = z20 + η2, we find Vef f to be</p><p>130CHAPTER 33. EXERCISE 33 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>Veff =</p><p>1</p><p>2</p><p>(k + 3m1ω</p><p>2</p><p>o)η</p><p>2</p><p>1 + kη1η2 +</p><p>1</p><p>2</p><p>kη22 (33.13)</p><p>Again, using equation (8.50) from textbook and our approximated potential, we</p><p>have</p><p>m1η̈1 + V11η1 + V12η2 = 0 (33.14)</p><p>m2η̈2 + V21η1 + V22η2 = 0 (33.15)</p><p>Where Vij = ∂2Veff/∂qi∂qj evaluated at equilibrium position. Here we assume</p><p>oscillatory solutionηi = aie</p><p>iωt where i in the exponential is the imaginary unit. Insert-</p><p>ing oscillatory solution into equations (33.4) and (33.5) we get</p><p>[</p><p>V11 −m1ω</p><p>2 V12</p><p>V21 V22 −m2ω</p><p>2</p><p>] [</p><p>a1</p><p>a2</p><p>]</p><p>= 0 (33.16)</p><p>The determinant of above matrix equation must vanish, hence we find frequency</p><p>of oscillations</p><p>ω1,2 =</p><p>√√√√ V11</p><p>2m1</p><p>+</p><p>V22</p><p>2m2</p><p>+</p><p>√(</p><p>V11</p><p>2m1</p><p>+</p><p>V22</p><p>2m2</p><p>)2</p><p>− (V11V22 − V 2</p><p>12)</p><p>m1m2</p><p>(33.17)</p><p>We will use shorthand notation</p><p>√</p><p>k/m = ωio. InsertingVij and simplifying, we</p><p>find the frequencies to be</p><p>ω1,2 =</p><p>√√√√1</p><p>2</p><p>(ω2</p><p>10 + ω202) +</p><p>3</p><p>2</p><p>ω2</p><p>o +</p><p>√(</p><p>1</p><p>2</p><p>(ω2</p><p>10 + ω2</p><p>20) +</p><p>3</p><p>2</p><p>ω2</p><p>o</p><p>)2</p><p>− 3ω2</p><p>20ω</p><p>2</p><p>o (33.18)</p><p>Since the nested square root must be positive we find additional conditional for</p><p>small oscillation</p><p>131</p><p>1</p><p>2</p><p>(ω2</p><p>10 + ω2</p><p>20) +</p><p>3</p><p>2</p><p>ω2</p><p>o ></p><p>√</p><p>3ω20ωo (33.19)</p><p>Considering that k is given and doesn’t vary, we must find allowed values ofωo.</p><p>Discriminant∆ of inequality 2 is</p><p>∆ = −3ω2</p><p>10 (33.20)</p><p>Since∆ is not positive we see that all values of k andωo are allowed.</p><p>Exercise 34 - GEOVANE NAYSINGER</p><p>FIGURE 34.1</p><p>Let’s start rewriting the lagrangian as</p><p>Λ =</p><p>1</p><p>2</p><p>muλuλ +Dλνm(xλuν − xνuλ),</p><p>so the lagrangian has the quadri-formΛ = Λ(xµ, ∂µxµ) and we will let implicit</p><p>Dλν(xµ) = Dλν . The Hamiltonian is</p><p>H = πλuλ + πνuν − Λ (34.1)</p><p>where π is the canonical quadri-momentum. Calculating the quadri-momenta</p><p>πλ =</p><p>∂Λ</p><p>∂uλ</p><p>= muλ −Dλνmxν</p><p>πν =</p><p>∂Λ</p><p>∂uν</p><p>= Dλνmxλ.</p><p>Then we can rewrite the equation (34.1) substituting the momenta and the lagragian.</p><p>Thus</p><p>133</p><p>134 CHAPTER 34. EXERCISE 34 - GEOVANE NAYSINGER</p><p>H = ( mu� � D �� mx � ) u� + D �� mx � u� �</p><p>1</p><p>2</p><p>mu� u� � D �� m(x � u� � x � u� )</p><p>=</p><p>1</p><p>2</p><p>mu� u� + D �� m (� x � u� + x � u� + x � u� � x � u� )</p><p>=</p><p>1</p><p>2</p><p>mu� u� + D �� m [(x � u� � x � u� ) � (x � u� � x � u� )]</p><p>and the anti-symmetric tensor</p><p>m[�� ] =</p><p>1</p><p>2</p><p>(m�� � m�� ) ;</p><p>the Hamiltonian is then</p><p>H =</p><p>1</p><p>2</p><p>mu� u� + 2D �� m[�� ] (34.2)</p><p>Exercise 35 - GUSTAVO CORTAZZI</p><p>GARCIA KESSLER</p><p>teste</p><p>135</p><p>Bibliography</p><p>GONZáLEZ, G. A pendulum with a moving support point. 2016. Disponível</p><p>em: <http://www.phys.lsu.edu/faculty/gonzalez/Teaching/Phys7221/</p><p>PendulumWithMovingSupport.pdf>.</p><p>THORNTON, S.; MARION, J.Classical Dynamics of Particles and Systems. Brooks/Cole,</p><p>2004. ISBN 9780534408961. Disponível em: <https://books.google.com.br/books?</p><p>id=HOqLQgAACAAJ>.</p><p>137</p><p>http://www.phys.lsu.edu/faculty/gonzalez/Teaching/Phys7221/PendulumWithMovingSupport.pdf</p><p>UNIVERSIDADE FEDERAL DO RIO GRANDE DO SUL</p><p>INSTITUTO DE FÍSICA</p><p>SOLUTIONS MANUAL HERBERT B. GOLDSTEIN 3RD ED.</p><p>LIST 4 OF CLASSICAL MECHANICS</p><p>CLASS: FIP00001</p><p>TEACHER: MARIA BEATRIZ DE LEONE GAY DUCATI</p><p>PORTO ALEGRE, RS</p><p>2021/01</p><p>Contents</p><p>1 Exercise 1 - LUCAS BOURSCHEIDT 3</p><p>2 Exercise 6 - JENIFER ANDRADE DE MATOS 7</p><p>3 Exercise 7 - FRANCISCO AUGUSTO FERREIRA ALMEIDA 11</p><p>4 Exercise 9 - JACKSON GALVÃO 19</p><p>5 Exercise 10 - RODRIGO WEBER PEREIRA 23</p><p>6 Exercise 13 - DOUGLAS OLIVEIRA NOVAES 27</p><p>7 Exercise 14 - JHORDAN SILVEIRA DE BORBA 31</p><p>8 Exercise 15 - JEFFERSON SANTANA MARTINS 35</p><p>9 Exercise 17 - EDGARD KRETSCHMANN 39</p><p>10 Exercise 19 - GEOVANE NAYSINGER 41</p><p>11 Exercise 20 - GUILHERME SHOITI YOSHIMATSU GIARDINI 43</p><p>11.1 (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43</p><p>11.1.1 Solution (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43</p><p>11.2 (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46</p><p>11.2.1 Solution (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46</p><p>12 Exercise 22 - LUCAS BOURSCHEIDT 51</p><p>13 Exercise 23 -JENIFER ANDRADE DE MATOS 59</p><p>14 Exercise 26 - FRANCISCO AUGUSTO FERREIRA ALMEIDA 67</p><p>15 Exercise 28 - JACKSON GALVÃO 71</p><p>16 Exercise 30 - RODRIGO WEBER PEREIRA 75</p><p>17 Exercise 32 - DOUGLAS OLIVEIRA NOVAES 79</p><p>18 Exercise 33 - JHORDAN SILVEIRA DE BORBA 81</p><p>iii</p><p>CONTENTS 1</p><p>19 Exercise 35 - JEFFERSON SANTANA MARTINS 87</p><p>20 Exercise 36 - EDGARD KRETSCHMANN 89</p><p>21 Exercise 37 - GEOVANE NAYSINGER 91</p><p>22 Exercise 41 - GUILHERME SHOITI YOSHIMATSU GIARDINI 93</p><p>22.1 (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93</p><p>22.1.1 Solution (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93</p><p>22.2 (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95</p><p>22.2.1 Solution (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95</p><p>22.3 (c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97</p><p>22.3.1 solution (c) . . . . . . . . . . . . . .</p><p>10 - Jhordan S. de Borba</p><p>Let q1, . . . , qn be a set of independent generalized coordinates for a system of n</p><p>degrees of freedom, with a Lagrangian L (q, q̇, t). Suppose we transform to another set</p><p>of independent coordinates s1, . . . , sn by means of transformation equations</p><p>qi = qi (s1, . . . , sn, t) , i = 1, . . . , n, (10.1)</p><p>(such a transformation is called a point transformation.) Show that if the Lagrangian is</p><p>expressed as a function of sj, ṡj and t through the equations of transformation, then L</p><p>satisfies Lagrange’s equations with respect to the s coordinates:</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ṡj</p><p>)</p><p>− ∂L</p><p>∂sj</p><p>= 0 (10.2)</p><p>In other words, the form of Lagrange’s equations is invariant under a point transfor-</p><p>mation.</p><p>Solução:</p><p>Partindo então da equação de Lagrange:</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂q̇i</p><p>)</p><p>− ∂L</p><p>∂q̇i</p><p>= 0</p><p>Uma vez que qi = qi (s1, . . . , sn, t) conforme foi apresentado no enunciado, então pela</p><p>regra da cadeia, uma vez que L = L (q1, . . . , qn, q̇1 . . . , q̇n, t), temos:</p><p>∂L</p><p>∂sj</p><p>=</p><p>∑</p><p>i</p><p>[</p><p>∂L</p><p>∂qi</p><p>∂qi</p><p>∂sj</p><p>+</p><p>∂L</p><p>∂q̇i</p><p>∂q̇i</p><p>∂sj</p><p>+</p><p>∂L</p><p>∂t</p><p>∂t</p><p>∂sj</p><p>]</p><p>27</p><p>28 CHAPTER 10. EXERCISE 10 - JHORDAN S. DE BORBA</p><p>Evidentemente como t é uma variável independente, podemos reduzir a:</p><p>@L</p><p>@sj</p><p>=</p><p>X</p><p>i</p><p>�</p><p>@L</p><p>@qi</p><p>@qi</p><p>@sj</p><p>+</p><p>@L</p><p>@_qi</p><p>@_qi</p><p>@sj</p><p>�</p><p>(10.3)</p><p>Para a derivada de qi temos então que:</p><p>_qi =</p><p>dq</p><p>dt</p><p>=</p><p>X</p><p>j</p><p>@qi</p><p>@sj</p><p>dsj</p><p>dt</p><p>+</p><p>@qj</p><p>@t</p><p>=</p><p>X</p><p>j</p><p>@qi</p><p>@sj</p><p>_sj +</p><p>@qi</p><p>@t</p><p>Derivando novamente:</p><p>@_qi</p><p>@_sk</p><p>=</p><p>X</p><p>j</p><p>@</p><p>@_sk</p><p>�</p><p>@qi</p><p>@sj</p><p>_sj</p><p>�</p><p>+</p><p>@</p><p>@_sk</p><p>�</p><p>@qj</p><p>@t</p><p>�</p><p>=</p><p>X</p><p>j</p><p>�</p><p>@</p><p>@_sk</p><p>�</p><p>@qi</p><p>@sj</p><p>�</p><p>_sj +</p><p>@qi</p><p>@sj</p><p>@_sj</p><p>@_sk</p><p>�</p><p>+</p><p>@</p><p>@_sk</p><p>�</p><p>@qj</p><p>@t</p><p>�</p><p>Uma vez que podemos trocar a ordem da derivação se a função possui segundas</p><p>derivadas contínuas, e como @qi</p><p>@_s = 0, então</p><p>@_qi</p><p>@_sk</p><p>=</p><p>X</p><p>j</p><p>�</p><p>@</p><p>@sj</p><p>�</p><p>@qi</p><p>@_sk</p><p>�</p><p>_sj +</p><p>@qi</p><p>@sj</p><p>� jk</p><p>�</p><p>+</p><p>@</p><p>@t</p><p>�</p><p>@qj</p><p>@_sk</p><p>�</p><p>(10.4)</p><p>@_qi</p><p>@_sk</p><p>=</p><p>@qi</p><p>@sj</p><p>(10.5)</p><p>Derivando L novamente em relação a _sj : análogo ao que foi feito anteriormente:</p><p>@L</p><p>@_sj</p><p>=</p><p>X</p><p>i</p><p>�</p><p>@L</p><p>@qi</p><p>@qi</p><p>@_sj</p><p>+</p><p>@L</p><p>@_qi</p><p>@_qi</p><p>@_sj</p><p>+</p><p>@L</p><p>@t</p><p>@t</p><p>@_sj</p><p>�</p><p>Usando novamente o fato de que t é uma variável independente, @qi</p><p>@_s = 0 e o resultado</p><p>obtido em 10.5, �camos apenas com:</p><p>@L</p><p>@_sj</p><p>=</p><p>X</p><p>i</p><p>�</p><p>@L</p><p>@_qi</p><p>@qi</p><p>@sj</p><p>�</p><p>Derivando então no tempo:</p><p>d</p><p>dt</p><p>�</p><p>@L</p><p>@_sj</p><p>�</p><p>=</p><p>X</p><p>i</p><p>�</p><p>@qi</p><p>@sj</p><p>d</p><p>dt</p><p>�</p><p>@L</p><p>@_qi</p><p>�</p><p>+</p><p>@L</p><p>@_qi</p><p>d</p><p>dt</p><p>�</p><p>@qi</p><p>@sj</p><p>��</p><p>(10.6)</p><p>=</p><p>X</p><p>i</p><p>�</p><p>@qi</p><p>@sj</p><p>d</p><p>dt</p><p>�</p><p>@L</p><p>@_qi</p><p>�</p><p>+</p><p>@L</p><p>@_qi</p><p>@_qi</p><p>@sj</p><p>�</p><p>(10.7)</p><p>29</p><p>Onde usamos novamente a mudança na ordem de derivação no segundo termo. Sub-</p><p>traindo então a equação 10.7 da 10.3, temos:</p><p>@L</p><p>@sj</p><p>�</p><p>d</p><p>dt</p><p>�</p><p>@L</p><p>@_sj</p><p>�</p><p>=</p><p>X</p><p>i</p><p>�</p><p>@L</p><p>@qi</p><p>@qi</p><p>@sj</p><p>+</p><p>@L</p><p>@_qi</p><p>@_qi</p><p>@sj</p><p>�</p><p>@qi</p><p>@sj</p><p>d</p><p>dt</p><p>�</p><p>@L</p><p>@_qi</p><p>�</p><p>�</p><p>@L</p><p>@_qi</p><p>@_qi</p><p>@sj</p><p>�</p><p>=</p><p>X</p><p>i</p><p>�</p><p>@qi</p><p>@sj</p><p>�</p><p>@L</p><p>@qi</p><p>�</p><p>d</p><p>dt</p><p>�</p><p>@L</p><p>@_qi</p><p>��</p><p>+</p><p>�</p><p>@L</p><p>@_qi</p><p>@_qi</p><p>@sj</p><p>�</p><p>@L</p><p>@_qi</p><p>@_qi</p><p>@sj</p><p>��</p><p>Como o primeiro termo entre parêntese é a própria equação de Legendre antes da</p><p>transformação, ficamos com:</p><p>@L</p><p>@sj</p><p>�</p><p>d</p><p>dt</p><p>�</p><p>@L</p><p>@_sj</p><p>�</p><p>= 0</p><p>Que era o resultado desejado.</p><p>Exercise 11 - Jhordan S. de Borba</p><p>Consider a uniform thin disk that rolls without slipping on a horizontal plane.</p><p>A horizontal force is applied to the center of the disk in a direction parallel to the plane</p><p>of disk.</p><p>(a) Derive Lagrange’s equations and find the generalized force.</p><p>(b) Discuss the motion if the force i not applied parallel to the plane o the disk.</p><p>Solução:</p><p>Se a força é aplicada no centro do disco e paralelo ao plano do mesmo, então</p><p>o movimento ocorre apenas em uma dimensão. Colocando por exemplo um plano</p><p>x×y paralelo ao disco, então as coordenadas generalizadas do centro de massa podem</p><p>serem escritas com (µ, η). Porém a altura é sempre constante, portanto η = y = R,</p><p>sendo R o raio do disco. Temos a seguinte coordenada generalizada dada simples-</p><p>mente por µ = x. Uma vez que não desliza, também podemos introduzir um ângulo</p><p>de rotação do disco de forma que:</p><p>µ = Rθ → µ̇ = Rθ̇</p><p>31</p><p>32 CHAPTER 11. EXERCISE 11 - JHORDAN S. DE BORBA</p><p>Considerando que a posição inicial do disco seja de forma que θ = µ = 0 em</p><p>t = 0. A energia cinética é então:</p><p>T =</p><p>mv2cm</p><p>2</p><p>+</p><p>Iω2</p><p>2</p><p>Sendo a velocidade do centro de massa dado por v2cm = µ̇2 + η̇2 = µ̇2, a velocidade</p><p>angular é simplesmente ω = θ̇ e o momento de inércia é dado por I = mR2</p><p>2</p><p>. A energia</p><p>cinética pode ser então escrita como:</p><p>T =</p><p>mv2cm</p><p>2</p><p>+</p><p>mR2θ̇2</p><p>4</p><p>=</p><p>mµ̇2</p><p>2</p><p>+</p><p>mR2µ̇2</p><p>R24</p><p>=</p><p>mµ̇2</p><p>2</p><p>+</p><p>mµ̇2</p><p>4</p><p>=</p><p>3</p><p>4</p><p>mµ̇2</p><p>Colocando o zero do potencial em η = R, ficamos então com o seguinte lagrangeano:</p><p>L = T − U</p><p>=</p><p>3</p><p>4</p><p>mµ̇2</p><p>Resolvendo a equação de Lagrange, temos</p><p>d</p><p>dt</p><p>[</p><p>∂L</p><p>∂µ̇</p><p>]</p><p>− ∂L</p><p>∂µ</p><p>= Q</p><p>d</p><p>dt</p><p>[</p><p>3</p><p>2</p><p>mµ̇</p><p>]</p><p>= Q</p><p>3</p><p>2</p><p>mµ̈ = Q</p><p>E ainda, da equação 1.49 do livro, temos que a força generalizada é definida como:</p><p>Qj =</p><p>∑</p><p>F i ·</p><p>∂ri</p><p>∂qj</p><p>Das nossas coordenadas do centro de massa r (µ, η) = (µ,R), temos que:</p><p>∂r</p><p>∂µ</p><p>=</p><p>(</p><p>∂µ</p><p>∂µ</p><p>,</p><p>∂R</p><p>∂µ</p><p>)</p><p>= (1, 0)</p><p>∂r</p><p>∂η</p><p>=</p><p>(</p><p>∂µ</p><p>∂η</p><p>,</p><p>∂R</p><p>∂µ</p><p>)</p><p>= (0, 0)</p><p>Então nossa força generalizada reduz-se a apenas:</p><p>Qj = F</p><p>33</p><p>b)</p><p>Aqui, há alguns cenários possíveis. Considerando que a foça continue a ser apli-</p><p>cado no centro do disco, então precisamos decompor a força pelo ângulo � formado</p><p>entre o eixo que atravessa o centro do disco e a própria força. Enquanto na componente</p><p>paralela ao disco apenas precisamos substituir F pela nova força decomposta, F cos�</p><p>por exemplo, na componente perpendicular, se mantiverrmos a condição de que o</p><p>disco não deslize, então vai ter uma força de atrito de reação de mesma magnitiude</p><p>e sentido oposto para balancea-la. Ainda se se não for aplicado no centro do disco,</p><p>precisaríamos novas coordenadas para permitir que o disco se desloce em um plano</p><p>bidimensional, assim como considerar a rotação do disco em torno do seu diâmetro</p><p>e/ou mesmo considerar uma variação no ângulo entre o eixo do disco e a normal do</p><p>chão, isto é, a possibilidade de ’balançar’. O artigo de Paris e Zhang (2002) tem uma</p><p>discussão bastante interessante sobre um disco rolando sobre uma superfície horizon-</p><p>tal sem deslizar.</p><p>Exercise 12 - Jefferson S. Martins</p><p>The escape velocity of a particle on Earth is the minimum velocity required at</p><p>Earth’s surface in order that the particle can escape from Earth’s gravitational field.</p><p>Neglecting the resistance of the atmosphere, the system is conservative. From the con-</p><p>servation theorem for potential plus kinetic energy show that the escape velocity for</p><p>Earth, ignoring the presence of the Moon, is 11,2km/s.</p><p>Solution: The escape velocity, it’s defined as the minimum velocity in order that</p><p>the particle escape Earth’s gravitational field, which means the least amount of kinetic</p><p>energy to overcome the particle potential due to the gravity.</p><p>The gravitational force is a conservative force, so we can apply the mechanical</p><p>energy conservation theorem:</p><p>EM = U +K ⇒ Esurface</p><p>M = lim</p><p>r→∞</p><p>EM = 0 (12.1)</p><p>where K is the kinetic energy, U the gravitational potential energy and r represents the</p><p>distance from a position to the center of the earth.</p><p>Replacing kinetic energy and gravitational potential energy by their respective</p><p>expressions in equation (12.1), we obtain the following equation:</p><p>−GMm</p><p>R</p><p>+</p><p>mv2e</p><p>2</p><p>= 0⇒ ve =</p><p>√</p><p>2GM</p><p>R</p><p>(12.2)</p><p>Substituting the values of the Earth’s mass (M = 5,97 × 1024 kg), of the gravitational</p><p>constant (G = 6,67×10−11 m3</p><p>kg,s2</p><p>) and of the Earth’s radius (R = 6,371×106 m) in equation</p><p>35</p><p>36 CHAPTER 12. EXERCISE 12 - JEFFERSON S. MARTINS</p><p>(12.3), we get the Earth’s escape velocity</p><p>ve =</p><p>√</p><p>2 · 6,67× 10−11 · 5,97× 1024</p><p>6,371× 106</p><p>= 11,2× 103</p><p>m</p><p>s</p><p>.</p><p>Exercise 13 - Geovane Naysinger</p><p>FIGURE 13.1</p><p>For the second Newton’s Law, we have</p><p>mr̈ =</p><p>d</p><p>dt</p><p>(−mv′)−mg</p><p>that is resulting force is equal to force due to gas escaping minus the weight of the</p><p>rocket. Keeping the calculation, we have</p><p>37</p><p>38 CHAPTER 13. EXERCISE 13 - GEOVANE NAYSINGER</p><p>m</p><p>dv</p><p>dt</p><p>= −v′dm</p><p>dt</p><p>−mg,</p><p>using the chain rule and replacing to the equation, we write</p><p>dv</p><p>dm</p><p>dm</p><p>dt</p><p>=</p><p>dv</p><p>dt</p><p>m</p><p>dv</p><p>dm</p><p>dm</p><p>dt</p><p>= −v′dm</p><p>dt</p><p>−mg</p><p>dv</p><p>dm</p><p>dm</p><p>dt</p><p>= − v</p><p>′</p><p>m</p><p>dm</p><p>dt</p><p>− g,</p><p>The problem say to us the mass loss per second is equal 1/60th of the initial mass, tha</p><p>is</p><p>dm</p><p>dt</p><p>= −m0</p><p>60</p><p>and substituting, we have</p><p>dv</p><p>dm</p><p>(</p><p>−m0</p><p>60</p><p>)</p><p>= − v</p><p>′</p><p>m</p><p>(</p><p>−m0</p><p>60</p><p>)</p><p>−</p><p>. . . . . . . . . . . . . . . . . . 97</p><p>Bibliography 99</p><p>Exercise 1 - LUCAS BOURSCHEIDT</p><p>Solution:</p><p>It is given that:</p><p></p><p>Q = q + ip</p><p>P = q − ip</p><p>Adding and subtracting the above equations we can express q and p as:</p><p></p><p>q = 1</p><p>2</p><p>(Q+ P )</p><p>p = 1</p><p>2i</p><p>(Q− P )</p><p>From Hamilton’s equation for q, q̇ = ∂H</p><p>∂p</p><p>:</p><p>3</p><p>4 CHAPTER 1. EXERCISE 1 - LUCAS BOURSCHEIDT</p><p>1</p><p>2</p><p>(Q̇+ Ṗ ) =</p><p>∂H</p><p>∂Q</p><p>∂Q</p><p>∂p</p><p>+</p><p>∂H</p><p>∂P</p><p>∂P</p><p>∂p</p><p>1</p><p>2</p><p>(Q̇+ Ṗ ) = i</p><p>∂H</p><p>∂Q</p><p>− i∂H</p><p>∂P</p><p>(A)</p><p>From Hamilton’s equation for p, ṗ = −∂H</p><p>∂q</p><p>:</p><p>1</p><p>2i</p><p>(Q̇− Ṗ ) = −∂H</p><p>∂Q</p><p>∂Q</p><p>∂q</p><p>− ∂H</p><p>∂P</p><p>∂P</p><p>∂q</p><p>1</p><p>2i</p><p>(Q̇− Ṗ ) = −∂H</p><p>∂Q</p><p>− ∂H</p><p>∂P</p><p>1</p><p>2</p><p>(Q̇− Ṗ ) = −i∂H</p><p>∂Q</p><p>− i∂H</p><p>∂P</p><p>(B)</p><p>Taking (A) + (B) and (A) - (B), we find, respectively:</p><p></p><p>Q̇ = −2i∂H</p><p>∂P</p><p>Ṗ = 2i∂H</p><p>∂Q</p><p>which are clearly not Hamiltonian equations of motion. However, these equations can</p><p>be expressed suggestively:</p><p></p><p>Q̇√</p><p>2</p><p>= ∂H</p><p>∂</p><p>(</p><p>ip√</p><p>2</p><p>)</p><p>i Ṗ√</p><p>2</p><p>= − ∂H</p><p>∂</p><p>(</p><p>Q√</p><p>2</p><p>)</p><p>This last pair of equations is presented in Hamilton’s form for the canonical variables:</p><p></p><p>Q′ = 1√</p><p>2</p><p>Q = 1√</p><p>2</p><p>(q + ip)</p><p>P ′ = i√</p><p>2</p><p>P = i√</p><p>2</p><p>(q − ip) = 1√</p><p>2</p><p>(iq + p)</p><p>Thus, for Q′ and P ′ we have Hamilton’s equations:</p><p>5</p><p></p><p>Q̇′ = ∂H</p><p>∂P ′</p><p>Ṗ ′ = − ∂H</p><p>∂Q′</p><p>Exercise 6 - JENIFER ANDRADE DE</p><p>MATOS</p><p>The transformation equations between two sets of coordinates are</p><p>Q = log(1 + q1/2cos(p))</p><p>P = 2(1 + q1/2cos(p))q1/2sin(p)</p><p>(a) Show directly from these transformation equations that Q, P are canonical</p><p>variables if q and p are.</p><p>(b) Show that the function that generates this transformation is</p><p>F3 = −(eQ − 1)2tan(p))</p><p>To determine if Q and P are canonical variables, we will evaluate the value of the</p><p>Poisson Brackets. If the transformation is canonical, then the following relation must</p><p>be true.</p><p>[Q,P ]q,p = 1 (2.1)</p><p>7</p><p>8 CHAPTER 2. EXERCISE 6 - JENIFER ANDRADE DE MATOS</p><p>Let´s open the Poisson brackets:</p><p>[Q,P ]q,p =</p><p>∂Q</p><p>∂q</p><p>∂P</p><p>∂p</p><p>− ∂P</p><p>∂q</p><p>∂Q</p><p>∂p</p><p>(2.2)</p><p>Now, let´s determine the value of each partial derivative:</p><p>∂Q</p><p>∂q</p><p>=</p><p>q−1/2cos(p)</p><p>2(1 + q1/2cos(p))</p><p>(2.3)</p><p>∂P</p><p>∂p</p><p>= 2q1/2cos(p) + 2q[cos2(p)− sin2(p)] (2.4)</p><p>∂P</p><p>∂q</p><p>= q−1/2sin(p) + 2cos(p)sin(p) (2.5)</p><p>∂Q</p><p>∂p</p><p>= − sin(p)q1/2</p><p>1 + q1/2cos(p)</p><p>(2.6)</p><p>Now we will substitute (2.3), (2.4), (2.5) and (2.6) in (2.2):</p><p>[Q,P ]q,p =</p><p>(</p><p>q−1/2cos(p)</p><p>2(1+q1/2cos(p))</p><p>) (</p><p>2q1/2cos(p) + 2q[cos2(p)− sin2(p)]</p><p>)</p><p>−</p><p>(</p><p>q−1/2sin(p) + 2cos(p)sin(p)</p><p>) (</p><p>− sin(p)q1/2</p><p>1+q1/2cos(p)</p><p>)</p><p>It can be writen as:</p><p>[Q,P ]q,p =</p><p>(</p><p>cos2(p) + q1/2cos3(p)− q1/2cos(p)sin2(p)</p><p>1 + q1/2cos(p)</p><p>)</p><p>+</p><p>(</p><p>sin2(p) + 2q1/2cos(p)sin2(p)</p><p>1 + q1/2cos(p)</p><p>)</p><p>(2.7)</p><p>Since:</p><p>cos2(p) + sin2(p) = 1 (2.8)</p><p>Then:</p><p>[Q,P ]q,p =</p><p>(</p><p>1 + q1/2cos(p)</p><p>1 + q1/2cos(p)</p><p>)</p><p>= 1 (2.9)</p><p>As we wanted to demonstrate. Therefore this is a canonical transformation.</p><p>9</p><p>Now let´s solve the second part of the question. The generating function pro-</p><p>posed is:</p><p>F3 = −(eQ − 1)2tan(p))</p><p>This is a function type 3. And therefore we can write:</p><p>q = −∂F</p><p>∂p</p><p>and P = −∂F</p><p>∂Q</p><p>(2.10)</p><p>Now let´s determine q:</p><p>q = −(−(eQ − 1)2sec2(p)) (2.11)</p><p>q = (eQ − 1)2sec2(p) (2.12)</p><p>From this, let´s determine Q:√</p><p>q</p><p>sec2(p)</p><p>=</p><p>√</p><p>(eQ − 1)2 (2.13)</p><p>eQ = 1 + q1/2cos(p) (2.14)</p><p>Q = ln(1 + q1/2cos(p)) (2.15)</p><p>Now let´s determine P:</p><p>P = 2eQ(eQ − 1)tan(p) (2.16)</p><p>Now we substitute 2.14 in 2.16 to determine P as a function of p and q. Now</p><p>let´s determine P:</p><p>P = 2tan(p)(q1/2cos(p) + 1)(q1/2cos(p) + 1− 1) (2.17)</p><p>Therefore:</p><p>10 CHAPTER 2. EXERCISE 6 - JENIFER ANDRADE DE MATOS</p><p>P = 2q1/2sin(p)(q1/2cos(p) + 1) (2.18)</p><p>The relations (2.15) and (2.18) are the same transformation equations proposed</p><p>in the question, showing that F3 is the generating function of the transformation equa-</p><p>tions.</p><p>Exercise 7 - FRANCISCO AUGUSTO</p><p>FERREIRA ALMEIDA</p><p>(a) If each of the four types of generating functions exist for a given canonical</p><p>transformation, use the Legendre transformation to derive relations between them. (b)</p><p>Find a generating function of the F4 type for the identify transformation and of the F3</p><p>type for the exchange transformation. (c) For an orthogonal point transformation of q</p><p>in a system of n degrees of freedom, show that the new momenta are likewise given by</p><p>the orthogonal transformation of an n-dimensional vector whose components are the</p><p>old momenta plus a gradient in configuration space.</p><p>Solution:</p><p>(a)</p><p>F = F1(q,Q, t) (3.1)</p><p>dF1 =</p><p>∂F1</p><p>∂q</p><p>dq +</p><p>∂F1</p><p>∂Q</p><p>dQ (3.2)</p><p>dF1 = pdq − PdQ (3.3)</p><p>With generating function derivatives:</p><p>11</p><p>12 CHAPTER 3. EXERCISE 7 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>p =</p><p>∂F1</p><p>∂q</p><p>(3.4)</p><p>and</p><p>P = −∂F1</p><p>∂Q</p><p>(3.5)</p><p>Let’s do a Legendre transformation:</p><p>F2 = F1 +QP (3.6)</p><p>dF2 = d(F1 +QP ) (3.7)</p><p>dF2 = dF1 + d(QP ) (3.8)</p><p>dF2 = dF1 +QdP + PdQ (3.9)</p><p>dF2 = pdq − PdQ+QdP + PdQ (3.10)</p><p>dF2 = pdq +QdP (3.11)</p><p>But</p><p>dF2 =</p><p>∂F2</p><p>∂q</p><p>dq +</p><p>∂F2</p><p>∂P</p><p>dP (3.12)</p><p>So, we obtain the generating function derivatives:</p><p>13</p><p>p =</p><p>∂F2</p><p>∂q</p><p>(3.13)</p><p>and</p><p>Q = −∂F2</p><p>∂P</p><p>(3.14)</p><p>Let’s do another Legendre transformation:</p><p>F3 = F1 − qp (3.15)</p><p>dF3 = d(F1 − qp) (3.16)</p><p>dF3 = dF1 − d(qp) (3.17)</p><p>dF3 = dF1 − (pdq + qdp) (3.18)</p><p>dF3 = dF1 − pdq − qdp (3.19)</p><p>dF3 = pdq − PdQ− pdq − qdp (3.20)</p><p>dF3 = −PdQ− qdp (3.21)</p><p>But</p><p>dF3 =</p><p>∂F3</p><p>∂Q</p><p>dQ+</p><p>∂F1</p><p>∂p</p><p>dp (3.22)</p><p>14 CHAPTER 3. EXERCISE 7 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>So, we obtain the generating function derivatives:</p><p>P = −∂F3</p><p>∂Q</p><p>(3.23)</p><p>and</p><p>q = −∂F3</p><p>∂p</p><p>(3.24)</p><p>Let’s do another Legendre transformation:</p><p>F4 = F1 − qp+QP (3.25)</p><p>dF4 = d(F1 − qp+QP ) (3.26)</p><p>dF4 = dF1 − d(qp) + d(QP ) (3.27)</p><p>dF4 = dF1 − (qdp+ pdq) + (QdP + PdQ) (3.28)</p><p>dF4 = dF1 − qdp− pdq +QdP + PdQ (3.29)</p><p>dF4 = pdq − PdQ− qdp− pdq +QdP + PdQ (3.30)</p><p>dF4 = −− qdp+QdP (3.31)</p><p>But</p><p>15</p><p>dF4 =</p><p>∂F4</p><p>∂p</p><p>dp+</p><p>∂F4</p><p>∂P</p><p>dP (3.32)</p><p>So, we obtain the generating function derivatives:</p><p>q = −∂F4</p><p>∂p</p><p>(3.33)</p><p>and</p><p>Q =</p><p>∂F4</p><p>∂P</p><p>(3.34)</p><p>(b)</p><p>F2 = qp is a identify transformation.</p><p>p =</p><p>∂F2</p><p>∂q</p><p>= P (3.35)</p><p>and</p><p>Q =</p><p>∂F2</p><p>∂P</p><p>= q (3.36)</p><p>So, we have:</p><p>F4 = F2 − pq (3.37)</p><p>F4 = qp− pq (3.38)</p><p>F4 = 0 (3.39)</p><p>16 CHAPTER 3. EXERCISE 7 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>F1 = qQ is a exchange transformation.</p><p>p =</p><p>∂F1</p><p>∂q</p><p>= Q (3.40)</p><p>and</p><p>P =</p><p>∂F1</p><p>∂Q</p><p>= q (3.41)</p><p>So, we have:</p><p>F3 = F1 − pq (3.42)</p><p>F3 = qQ− pq = qp− pq (3.43)</p><p>F3 = 0 (3.44)</p><p>(c)</p><p>Consider a type 2 generating function F2(q; P; t) of the old coordinates and the</p><p>new momenta, of the form</p><p>F2(q;P ; t) = fi(q1; . . . ; qn; t)Pi − g(q1; . . . ; qn; t) (3.45)</p><p>where f ′</p><p>is are a set of independent functions, and g′is are diferentiable functions</p><p>of the old coordinates and time. The new coordinates Qi are given by</p><p>Qi =</p><p>∂F2</p><p>∂Pi</p><p>= fi(q1, . . . , qn; t) (3.46)</p><p>In particular, the function</p><p>17</p><p>fi(q1, . . . , qn; t) = Rijqj (3.47)</p><p>where Rij is the (i; j)-th element of a N × N orthogonal matrix, generates an</p><p>orthogonal transformation of the coordinates. Now,</p><p>pj =</p><p>∂F2</p><p>∂qj</p><p>=</p><p>∂fi</p><p>∂qj</p><p>Pi −</p><p>∂g</p><p>∂qj</p><p>= RijPi −</p><p>∂g</p><p>∂qj</p><p>(3.48)</p><p>This equation can be written in matrix form, as</p><p>p =</p><p>∂f</p><p>∂q</p><p>P − ∂g</p><p>∂q</p><p>(3.49)</p><p>where p denotes the N × 1 column vector (p1; . . . ; pN)T , ∂g/∂q denotes the N ×</p><p>1 column vector (∂g/∂q1; . . . ; ∂g/∂qn)T , and ∂f</p><p>∂q</p><p>denotes the N × N matrix with entries</p><p>(</p><p>∂f</p><p>∂q</p><p>)</p><p>ij</p><p>=</p><p>∂fi</p><p>∂qi</p><p>= Rij (3.50)</p><p>From (3.49), the new momenta are given by</p><p>P =</p><p>(</p><p>∂f</p><p>∂q</p><p>)−1(</p><p>p+</p><p>∂g</p><p>∂q</p><p>)</p><p>(3.51)</p><p>P = R−1</p><p>(</p><p>p+</p><p>∂g</p><p>∂q</p><p>)</p><p>(3.52)</p><p>P = R−1(p+∇qg) (3.53)</p><p>As R is an orthogonal matrix, RRT = RTR = I , so R−1 = RT is also an orthogo-</p><p>nal transformation.</p><p>This gives the required result: the new momenta are given by the orthogonal</p><p>transformation (R−1) of an n-dimensional vector (p+∇qg), whose components are the</p><p>old momenta (p) plus a gradient in configuration space (∇qg).</p><p>Exercise 9 - JACKSON GALVÃO</p><p>(a) For a single show directly (that is, by direct evaluation of Poisson bracket), that if u</p><p>is scalar function only of r2, p2 and r · p, then</p><p>[u,L] = 0</p><p>let u, v, such that</p><p>[u, v]q,p =</p><p>∂u</p><p>∂qi</p><p>∂ν</p><p>∂pi</p><p>− ∂u</p><p>∂pi</p><p>∂ν</p><p>∂qi</p><p>Doing u = u(r2, p2, r× p), and v = L, so</p><p>[u,L] = 0</p><p>for n→ k, Lk is, using the convention summation with Levi-Civita,</p><p>Lk = ϵijkqipj (4.1)</p><p>where j, i = (1, 2, 3,,), limited at 3-D i, j = 1, 2, 3. So, by Poisson bracket, we must be</p><p>have</p><p>[u, Lk] = [u, ϵijkqipj] = ϵ[u, qipj]</p><p>so, by Poisson properties,</p><p>[u, Lk] = ϵijk[u, qi]pj + ϵijk[u, pj]qi (4.2)</p><p>Knowing that</p><p>[u, pj] =</p><p>∂u</p><p>∂qj</p><p>(4.3)</p><p>19</p><p>20 CHAPTER 4. EXERCISE 9 - JACKSON GALVÃO</p><p>and</p><p>[u, qi] = −</p><p>∂u</p><p>∂pj</p><p>(4.4)</p><p>to r2, p2, we have</p><p>r2 =</p><p>3∑</p><p>i</p><p>q2i (4.5)</p><p>and</p><p>p2 =</p><p>3∑</p><p>i</p><p>p2j (4.6)</p><p>Let do it, to partial derivatives in the form</p><p>∂u</p><p>∂fi</p><p>=</p><p>∂u</p><p>∂α</p><p>dαi +</p><p>∂u</p><p>∂β</p><p>dβi</p><p>∂u</p><p>∂gj</p><p>=</p><p>∂u</p><p>∂γ</p><p>dγj +</p><p>∂u</p><p>∂β</p><p>dβj</p><p>(4.7)</p><p>like α = p2, β = r · p, α = r2, thereby</p><p>∂u</p><p>∂fi</p><p>=</p><p>∂u</p><p>∂pi</p><p>=</p><p>∂u</p><p>∂p2</p><p>d(p2i ) +</p><p>∂u</p><p>∂(r · p)</p><p>d(r · pi)</p><p>= 2pi</p><p>∂u</p><p>∂p2</p><p>+ qi</p><p>∂u</p><p>∂(r · p)</p><p>(4.8)</p><p>and</p><p>∂u</p><p>∂gj</p><p>=</p><p>∂u</p><p>∂qj</p><p>=</p><p>∂u</p><p>∂r2</p><p>d(r2j ) +</p><p>∂u</p><p>∂(r · p)</p><p>d(rj · p)</p><p>= 2qj</p><p>∂u</p><p>∂r2</p><p>+ pj</p><p>∂u</p><p>∂(r · p)</p><p>(4.9)</p><p>after an arrangement by Eqs.(4.8) and (4.9), we must be have</p><p>∂u</p><p>∂qj</p><p>qi −</p><p>∂u</p><p>∂pi</p><p>pj = 2qiqj</p><p>∂u</p><p>∂r2</p><p>− 2pipj</p><p>∂u</p><p>∂p2</p><p>what makes us able to write the Eq.(4.10).</p><p>[u, qipj] = 2qiqj</p><p>∂u</p><p>∂r2</p><p>− 2pipj</p><p>∂u</p><p>∂p2</p><p>(4.10)</p><p>21</p><p>By symmetric for the indices i and j, the summation has vanishes. Therefore, from</p><p>equivalence with Eq.(4.2), we cn be write the Eq. (4.11)</p><p>[u, Lk] = ϵijk[u, qipj] = 0 (4.11)</p><p>Then, knowing that for the tensor Tij = Tji, let do we have</p><p>ϵijkTij = ϵjikTji = −ϵijkTij (4.12)</p><p>Therefore,</p><p>2ϵjikTji = 0. (4.13)</p><p>(b) Similarly show directly that if F is a vector function,</p><p>F = ur + vp + w(r× p)</p><p>where u, v and w are scalar function of the same From properties we applied as item</p><p>"a", we have</p><p>Fi = uqi + vpi + wLi (4.14)</p><p>where we can be written</p><p>[Fi, Lj] = u[qi, Lj] + qi[u, Lj] + v[u, Lj] + v[pi, Lj] + pi[v, Lj] +w[Li, Lj] +Li[w,Lj] (4.15)</p><p>Using the result of Eqs.(4.11) and (4.12), we must be have</p><p>[Fi, Lj] = u[qi, Lj] + v[pi, Lj] + w[Li, Lj] (4.16)</p><p>And, therefore, knowing that [Li, Lj] = ϵijkLk by Eq.(9.128 - Goldstein, 3th ed.), we</p><p>have</p><p>[qi, Lj] =</p><p>∂Lj</p><p>∂pi</p><p>= ϵijkqk (4.17)</p><p>22 CHAPTER 4. EXERCISE 9 - JACKSON GALVÃO</p><p>[pi, Lj] = −</p><p>∂Lj</p><p>∂qi</p><p>= ϵijkpk (4.18)</p><p>And, finnaly, the joint the results, we must be have,</p><p>[Fi, Lj] = ϵijk(uqk + vpk + wLk) = ϵijkFk (4.19)</p><p>Exercise 10 - RODRIGO WEBER</p><p>PEREIRA</p><p>Find under what conditions</p><p>Q =</p><p>αp</p><p>x</p><p>P = βx2</p><p>Where α and β are constants, represents a canonical transformation for system of one</p><p>degree of freedom, and obtain a suitable generating function. Apply the transforma-</p><p>tion to the solution of the linear harmonic oscillator.</p><p>Solution: A transformation η⃗ → ζ⃗ , where</p><p>η⃗ =</p><p>(</p><p>x</p><p>p</p><p>)</p><p>ζ⃗ =</p><p>(</p><p>Q</p><p>P</p><p>)</p><p>is canonical if it satisfys the following relation: [ζi, ζj] = Jij , Where J is the symplectic</p><p>matrix. Since the system has only one degree of freedom, the Poisson parenthesis is</p><p>given by:</p><p>[ζi, ζj] =</p><p>∂ζi</p><p>∂x</p><p>∂ζj</p><p>∂p</p><p>− ∂ζi</p><p>∂p</p><p>∂ζj</p><p>∂x</p><p>(5.1)</p><p>We first calculate the derivatives:</p><p>∂Q</p><p>∂x</p><p>= −αp</p><p>x2</p><p>(5.2)</p><p>∂Q</p><p>∂p</p><p>=</p><p>α</p><p>x</p><p>(5.3)</p><p>∂P</p><p>∂x</p><p>= 2βx (5.4)</p><p>23</p><p>24 CHAPTER 5. EXERCISE 10 - RODRIGO WEBER PEREIRA</p><p>@P</p><p>@p</p><p>= 0; (5.5)</p><p>Testing the canonical condition, it is clear that [Q; Q] = [ � 1; � 1] = 0 = J11 and [P; P] =</p><p>[� 2; � 2] = 0 = J22. Now, Using the derivatives:</p><p>[Q; P] = [ � 1; � 2] =</p><p>�</p><p>�</p><p>�p</p><p>x2</p><p>�</p><p>0 �</p><p>� �</p><p>x</p><p>�</p><p>2�x = � 2��</p><p>If the transformation is canonical, the above result should be one. So, the condition for</p><p>it to happen is, clearly:</p><p>�� = �</p><p>1</p><p>2</p><p>(5.6)</p><p>Now we proceed to obtain the generating function. Isolating x in the transformation</p><p>relation, we see that it can be written as a function of p and Q, that is, x(p; Q; t) = �p</p><p>Q .</p><p>So, there must be a generating function of the type F3(p; Q; t) such that</p><p>x = �</p><p>@F3</p><p>@p</p><p>P = �</p><p>@F3</p><p>@Q</p><p>K = H +</p><p>@F3</p><p>@t</p><p>; (5.7)</p><p>where K is the "Kamiltonian", and H the Hamiltonian of the system. By the �rst</p><p>equation we may write (replacing x):</p><p>F3(p; Q; t) = �</p><p>�</p><p>Q</p><p>Z</p><p>pdp+ A(Q; t)</p><p>| {z }</p><p>=0</p><p>;</p><p>where we set A(Q; t) - the the "constant" of integration - to zero for simplicity. So, the</p><p>generating function is</p><p>F3(p; Q; t) = �</p><p>�p 2</p><p>2Q</p><p>(5.8)</p><p>Using Eqs. 5.7, We can easily verify that the above function generates the transforma-</p><p>tion:</p><p>x = �</p><p>@F3</p><p>@p</p><p>=</p><p>�p</p><p>Q</p><p>) Q =</p><p>�p</p><p>x</p><p>P = �</p><p>@F3</p><p>@Q</p><p>= �</p><p>�p 2</p><p>2Q2</p><p>=|{z}</p><p>Q= �p</p><p>x</p><p>�</p><p>x2</p><p>2�</p><p>=|{z}</p><p>�� = � 1</p><p>2</p><p>�x 2</p><p>To obtain the "Kamiltonian" we �rst take the partial derivative of F3 with respect to</p><p>time:</p><p>@F3</p><p>@t</p><p>= 0;</p><p>25</p><p>and then write the Hamiltonian of harmonic oscillator, which is given by (Goldstein,</p><p>3Ed, p.378):</p><p>H =</p><p>1</p><p>2m</p><p>�</p><p>p2 + m2! 2x2</p><p>�</p><p>(5.9)</p><p>Where m is the mass and ! =</p><p>q</p><p>k</p><p>m , with k the spring constant. We also note that, since</p><p>@~r</p><p>@t = 0 , the Hamiltonian is constant. Also, since @H</p><p>@t = 0 , the Hamiltonian is the total</p><p>energy of the system (H = E ). By the third Eq. 5.7, we have:</p><p>K = H = E</p><p>Isolating the old variables (x; p) in terms of the new ones (Q; P) and substituting in the</p><p>Hamiltonian, we can write the "Kamiltonian" as follows:</p><p>K (Q; P; t) =</p><p>P</p><p>2m�</p><p>hQ2</p><p>� 2</p><p>+ m2! 2</p><p>i</p><p>(5.10)</p><p>Finally, we may apply the equations of motion:</p><p>_Q =</p><p>@K</p><p>@P</p><p>_P = �</p><p>@K</p><p>@Q</p><p>(5.11)</p><p>For _Q, we have:</p><p>_Q =</p><p>1</p><p>2m�</p><p>hQ2</p><p>� 2</p><p>+ m2! 2</p><p>i</p><p>=</p><p>E</p><p>P</p><p>;</p><p>where we used the fact that the "Kamiltonian" (H ) equals the total energy (E ) of the</p><p>system, which remains constant. So,</p><p>_Q =</p><p>E</p><p>P</p><p>(5.12)</p><p>For _P , we have:</p><p>_P =</p><p>4�</p><p>m</p><p>PQ (5.13)</p><p>where we used Eq. 5.6. To solve this set of equations, we can use the following trick:</p><p>_Q =</p><p>dQ</p><p>dt</p><p>=</p><p>dQ</p><p>dP</p><p>dP</p><p>dt</p><p>) _Q =</p><p>dQ</p><p>dP</p><p>_P</p><p>Using Eqs. 5.12 e 5.13, we get to the following equation:</p><p>4�</p><p>Em</p><p>Z</p><p>QdQ =</p><p>Z</p><p>dP</p><p>P2</p><p>+ C</p><p>which can be easily solved to:</p><p>2�</p><p>Em</p><p>Q2 = �</p><p>1</p><p>P</p><p>+ C</p><p>26 CHAPTER 5. EXERCISE 10 - RODRIGO WEBER PEREIRA</p><p>Substituting Q and P by the original variables, we get</p><p>p2</p><p>2m</p><p>+</p><p>kx2</p><p>2</p><p>= E,</p><p>with the constant of integration being given by C = −kα</p><p>E</p><p>. The last equation is the</p><p>conservation of energy for the harmonic oscillator, witch concludes the problem.</p><p>Exercise 13 - DOUGLAS OLIVEIRA NO-</p><p>VAES</p><p>Question: The set of restricted canonical transformations forms a group (Ap-</p><p>pendix B). Verify this statement once using the invariance of Hamilton’s principle un-</p><p>der canonical transformation (cf. Eq. (9.11)), and again using the symplectic condition.</p><p>Response: One has to show that restricted canonical transformations form a</p><p>group. For this, the canonical transformations will form a group if (Goldstein, p. 387):</p><p>1. The identity transformation is canonical.</p><p>2. If a transformation is canonical, so is its inverse.</p><p>3. Two successive canonical transformations (the group "product" operation)</p><p>define a transformation that is also canonical.</p><p>4. The product operation is associative.</p><p>For a restricted (time independent) canonical transformation</p><p>pi _qi � H = Pi</p><p>_Qi � K +</p><p>dF</p><p>dt</p><p>(6.1)</p><p>equation (6.1) must hold. This can easily be checked for identity transformation (q; p) =</p><p>(Q; P). In that case, we have</p><p>dF</p><p>dt</p><p>= 0 (6.2)</p><p>27</p><p>28 CHAPTER 6. EXERCISE 13 - DOUGLAS OLIVEIRA NOVAES</p><p>which clearly holds. For second property, one starts with a canonical transformation</p><p>for which equation (6.1) holds. Then, one considers its inverse (Q,P ) −→ (q, p)</p><p>PiQ̇i −K = piq̇i −H +</p><p>dF ′</p><p>dt</p><p>(6.3)</p><p>Since equation (6.1) holds, one sees that equation (6.3) will hold if</p><p>dF ′</p><p>dt</p><p>= −dF</p><p>dt</p><p>(6.4)</p><p>For the third property, consider two successive canonical transformations</p><p>piq̇i −H = PiQ̇i −K +</p><p>dF</p><p>dt</p><p>(6.5)</p><p>PiQ̇i −K = P ′</p><p>i Q̇</p><p>′</p><p>i −K ′ +</p><p>dF ′</p><p>dt</p><p>(6.6)</p><p>Equations (6.4) and (6.5) hold, now one considers the "product" (q, p) −→</p><p>(Q′, P ′) and ask, does the following equation hold</p><p>piq̇i −H = P ′</p><p>i Q̇</p><p>′</p><p>i −K ′ +</p><p>dF ′′</p><p>dt</p><p>(6.7)</p><p>Inserting equations (6.4) and (6.5) into (6.6) will give</p><p>dF ′′</p><p>dt</p><p>=</p><p>dF</p><p>dt</p><p>+</p><p>dF ′</p><p>dt</p><p>(6.8)</p><p>For verify the associative property, one has to consider 3 transformations. Say,</p><p>(q, p) −→ (Q,P ) −→ (Q′, P ′) −→ (Q′′, P ′′) (6.9)</p><p>Again, to skip some unnecessary algebra, one will have something similar to</p><p>equation (6.8)</p><p>dF ′′′</p><p>dt</p><p>=</p><p>dF</p><p>dt</p><p>+</p><p>dF ′</p><p>dt</p><p>+</p><p>dF ′′</p><p>dt</p><p>(6.10)</p><p>29</p><p>Clearly, this operation is associative. Notice that all these properties listed are</p><p>inherent from the behaviour of real numbers under addition. The symplectic approach</p><p>is much nicer. A transformation is canonical if the following equation holds</p><p>MJM̃ = J (6.11)</p><p>For an identity transformation, Jacobian matrix M=1 , clearly equation (6.3) in</p><p>that case holds. Now, for second property: if some transformation is canonical, so is its</p><p>inverse. One can easily check this</p><p>MJM̃ = J (6.12)</p><p>M � 1JM̃</p><p>� 1</p><p>= M � 1MJM̃ M̃</p><p>� 1</p><p>= J (6.13)</p><p>For the third property, consider two successive transformations</p><p>MJM̃ = J (6.14)</p><p>M 0JM̃</p><p>0</p><p>= J (6.15)</p><p>Product of these two successive will also be a canonical transformation</p><p>MM 0JM̃</p><p>0</p><p>M̃ = MJM̃ = J (6.16)</p><p>Fourth property of associativity is inherent from matrix multiplication. Consider</p><p>3 successive transformations:</p><p>MM 0M 00JM̃</p><p>00</p><p>M̃</p><p>0</p><p>M̃ = J (6.17)</p><p>Clearly, this is associative since A(BC) = (AB)C.</p><p>Exercise 14 - JHORDAN SILVEIRA DE</p><p>BORBA</p><p>Prove that the transformation</p><p>Q1 = q21 Q2 = q2 sec p2</p><p>P1 =</p><p>p1 cos p2 − 2q2</p><p>2q1 cos p2</p><p>P2 = sin p2 − 2q1</p><p>is canonical, by any method you choose. Find a suitable generating function that</p><p>will lead to this transformation.</p><p>Answer: Poisson brackets are invariant under canonical transformations, so we</p><p>want to prove that:</p><p>[Qj, Pk]q,p = δj,k = − [Pj, Qk]q,p</p><p>And</p><p>[Qj, Qk]q,p = 0 = [Pj, Pk]q,p</p><p>The Poisson bracket of two functions u, v with respect to the canonical variables (q, p)</p><p>is defined as:</p><p>[u, v]q,p =</p><p>N∑</p><p>i=1</p><p>(</p><p>∂u</p><p>∂qi</p><p>∂v</p><p>∂pi</p><p>− ∂u</p><p>∂pi</p><p>∂v</p><p>∂qi</p><p>)</p><p>So once we have:</p><p>Q1 = q21 Q2 = q2 sec p2</p><p>P1 =</p><p>p1 cos p2 − 2q2</p><p>2q1 cos p2</p><p>=</p><p>1</p><p>2</p><p>p1</p><p>q1</p><p>− q2</p><p>q1</p><p>sec p2 P2 = sin p2 − 2q1</p><p>31</p><p>32 CHAPTER 7. EXERCISE 14 - JHORDAN SILVEIRA DE BORBA</p><p>For [Q1, P1]q,p:</p><p>[Q1, P1]q,p =</p><p>∂Q1</p><p>∂q1</p><p>∂P1</p><p>∂p1</p><p>− ∂Q1</p><p>∂p1</p><p>∂P1</p><p>∂q1</p><p>+</p><p>∂Q1</p><p>∂q2</p><p>∂P1</p><p>∂p2</p><p>− ∂Q1</p><p>∂p2</p><p>∂P1</p><p>∂q2</p><p>= 2q1</p><p>cos p2</p><p>2q1 cos p2</p><p>− 0 + 0− 0</p><p>= 1</p><p>Similarly for the other cases:</p><p>[Q2, P2]q,p = 0− 0 + sec p2 cos p2 − 0 =</p><p>cos p2</p><p>cos p2</p><p>= 1</p><p>[Q1, P2]q,p = 0− 0 + 0− 0 = 0</p><p>[Q2, P1]q,p = 0− 0 + sec p2</p><p>(</p><p>−q2</p><p>q1</p><p>sec p2 tan p2</p><p>)</p><p>+ (q2 sec p2 tan p2)</p><p>1</p><p>q1</p><p>sec p2 = 0</p><p>[Q1, Q2]q,p = 0− 0 + 0− 0 = 0</p><p>[P1, P2]q,p = 0 +</p><p>(</p><p>cos p2</p><p>2q1 cos p2</p><p>)</p><p>(2) +</p><p>(</p><p>−2</p><p>2q1 cos p2</p><p>)</p><p>(cos p2)− 0 = 0</p><p>In this way we were able to demonstrate that the transformation is canonical.</p><p>The next step then is to find a generator function. Looking then for a generating equa-</p><p>tion of the third type, that is, one that depends only on the old generalized moment and</p><p>the new generalized coordinates. Transformation equations can be found by writing F</p><p>as:</p><p>F = F3 (p,Q, t) + q · p</p><p>First we then want to write Pi and qi as a function of pi and Qi. For qi, we can just</p><p>invert:</p><p>q1 =</p><p>√</p><p>Q1 q2 = Q2 cos p2</p><p>And then substituting in Pi to eliminate the qi, we’re left with:</p><p>P1 =</p><p>1</p><p>2</p><p>p1√</p><p>Q1</p><p>− Q2√</p><p>Q1</p><p>P2 = sin p2 − 2</p><p>√</p><p>Q1</p><p>Once we have:</p><p>qi = −</p><p>∂F3</p><p>∂pi</p><p>Pi = −</p><p>∂F3</p><p>∂Qi</p><p>We can calculate for q1:</p><p>33</p><p>q1 =−</p><p>∂F3</p><p>∂p1</p><p>−q1dp1 =dF3</p><p>−q1p1 + c0 =F3</p><p>Where we take into account that in the Hamiltonian formulation, the generalized mo-</p><p>ment is also an independent variable at the same level as the generalized coordinates.</p><p>We want to find F3 just in terms of pi and Qi. Eliminating q1 =</p><p>√</p><p>Q1, then:</p><p>F3 = −</p><p>√</p><p>Q1p1 + c0</p><p>Since we have c0 = c0 (p2, Q1, Q2, t), differentiating from p2, we have:</p><p>q2 = −</p><p>∂F3</p><p>∂p2</p><p>= −∂c0</p><p>∂p2</p><p>So c0 = −q2p2 + c1 (Q1, Q2, t). We stay with:</p><p>F3 = −</p><p>√</p><p>Q1p1 − q2p2 + c1</p><p>Substituting q2 :</p><p>F3 = −</p><p>√</p><p>Q1p1 −Q2p2 cos (p2) + c1</p><p>Differentiating now in Q2:</p><p>P2 = −</p><p>∂F3</p><p>∂Q2</p><p>sin p2 − 2</p><p>√</p><p>Q1 = p2 cos (p2)−</p><p>∂c1</p><p>∂Q2[</p><p>− sin p2 + 2</p><p>√</p><p>Q1 + p2 cos (p2)</p><p>]</p><p>Q2 + c2 (Q1, t) = c1</p><p>Then:</p><p>F3 = −</p><p>√</p><p>Q1p1 −Q2p2 cos (p2)−Q2 sin p2 + 2</p><p>√</p><p>Q1Q2 +Q2p2 cos (p2) + c2 (Q1, t)</p><p>Or:</p><p>F3 = −</p><p>√</p><p>Q1p1 −Q2 sin p2 + 2</p><p>√</p><p>Q1Q2 + c2</p><p>34 CHAPTER 7. EXERCISE 14 - JHORDAN SILVEIRA DE BORBA</p><p>Remembering that c2 = c2 (Q, t). And finally we calculate P1:</p><p>P1 = − ∂F3</p><p>∂Q1</p><p>1</p><p>2</p><p>p1√</p><p>Q1</p><p>− Q2√</p><p>Q1</p><p>=</p><p>1</p><p>2</p><p>p1√</p><p>Q1</p><p>− Q2√</p><p>Q1</p><p>− ∂c2</p><p>∂Q1</p><p>c (t) =</p><p>∂c2</p><p>∂Q1</p><p>So, we have:</p><p>F3 = −</p><p>p</p><p>Q1p1 −Q2 sin p2 + 2</p><p>p</p><p>Q1Q2 + c (t)</p><p>As stated in the book, F3 is uncertain about an arbitrary additive function of t alone,</p><p>but this does not affect the transformation equations. Retrieving the transformation</p><p>equations:</p><p>q1 = −∂F3</p><p>∂p1</p><p>=</p><p>p</p><p>Q1 (7.1)</p><p>q2 = −∂F3</p><p>∂p2</p><p>= Q2 cos p2 (7.2)</p><p>P1 = − ∂F3</p><p>∂Q1</p><p>=</p><p>1</p><p>2</p><p>p1√</p><p>Q1</p><p>− Q2√</p><p>Q1</p><p>=</p><p>1</p><p>2</p><p>p1</p><p>q1</p><p>− q2 sec (p2)</p><p>q1</p><p>(7.3)</p><p>P2 = − ∂F3</p><p>∂Q2</p><p>= sin p2 − 2</p><p>p</p><p>Q1 = sin p2 − 2q1 (7.4)</p><p>Where we use the results of the equations 7.1 and 7.2 in the equations 7.3 and</p><p>7.4.</p><p>Exercise 15 - JEFFERSON SANTANA</p><p>MARTINS</p><p>(a) Using the fundamental Poisson Brackets, find the values of α and β for which</p><p>the equation</p><p>Q = qα cos βp, P = qα sin βp</p><p>represents a canonical transformation.</p><p>(b) For what values of α and β do these equations represent an extended canon-</p><p>ical transformation? Find a generating function of the F3 form for the transformation.</p><p>(c) On the basis of part (b), can the transformation equations be modified so that</p><p>they describe a canonical transformation for all values of β?</p><p>Solution</p><p>(a) We are given a transformation as follows,</p><p>Q = qα cos βp, P = qα sin βp</p><p>We know that the fundamental Poisson Brackets of the transformed variables</p><p>have the same value when evaluated with respect to any canonical coordinate set. In</p><p>other words, the fundamental Poisson Brackets are invariant under canonical transfor-</p><p>mation.</p><p>35</p><p>36 CHAPTER 8. EXERCISE 15 - JEFFERSON SANTANA MARTINS</p><p>Therefore, in order for the given transformation to be canonical, the Poisson</p><p>Bracket of Q,P with respect to q & p should be equal to 1.</p><p>Using the formula for Poisson Bracket,</p><p>[u, v]q,p =</p><p>∂u</p><p>∂qi</p><p>∂v</p><p>∂pi</p><p>− ∂v</p><p>∂qi</p><p>∂u</p><p>∂pi</p><p>∴ [Q,P ]q,p =</p><p>∂Q</p><p>∂q</p><p>∂P</p><p>∂p</p><p>− ∂P</p><p>∂q</p><p>∂Q</p><p>∂p</p><p>[Q,P ]q,p =</p><p>(</p><p>αqα−1 cos βp</p><p>)</p><p>(qαβ cos βp)− αqα−1 sin βpqαβ(− sin βp)</p><p>[Q,P ]q,p = αq2α−1β cos2 βp+ αq2α−1β sin2 βp =⇒ [Q,P ]q,p = αβq2α−1 = 1</p><p>This equation is satisfied if 2α− 1 = 0→ α = 1/2 and β = 1/α = 2.</p><p>(b)When α = 1/2 and β is taken to be an arbitrary constant, we have</p><p>[Q,P ]q,p =</p><p>β</p><p>2</p><p>,</p><p>which represents an extended canonical transformation for any value of β ̸= 2. Now,</p><p>our transformation equations are</p><p>Q = q1/2 cos βp, P = q1/2 sin βp</p><p>For extended canonical transformation for a system with one degree of freedom, we</p><p>have</p><p>λ (pq̇ −H) = PQ̇−K +</p><p>dF</p><p>dt</p><p>·</p><p>when F = F3(p,Q, t) + λpq, we obtain</p><p>λ (pq̇ −H) = PQ̇−K + λṗq +</p><p>∂F3</p><p>∂p</p><p>ṗ+</p><p>∂F3</p><p>∂Q</p><p>Q̇+</p><p>∂F3</p><p>∂t</p><p>,</p><p>which leads to</p><p>∂F3</p><p>∂p</p><p>= λq (8.1)</p><p>∂F3</p><p>∂Q</p><p>= −P (8.2)</p><p>K = λH +</p><p>∂F3</p><p>∂t</p><p>(8.3)</p><p>We first express P and q in terms of p and Q, as below</p><p>q = Q2sec2(βp) (8.4)</p><p>37</p><p>P = Qsec (βp) sin (βp) = Qtan (βp) (8.5)</p><p>Combining Eqs. (8.1) and (8.4), we have</p><p>∂F3</p><p>∂p</p><p>= λQ2sec2(βp) → F3 =</p><p>λ</p><p>β</p><p>Q2tan (βp) + f(Q)</p><p>Using this in combination with Eqs. (8.2) and (8.5), we have</p><p>2</p><p>λ</p><p>β</p><p>Qtan (βp) +</p><p>df</p><p>dQ</p><p>= Qtan (βQ)</p><p>df</p><p>dQ</p><p>= (1− 2λ</p><p>β</p><p>)Qtan (βp)</p><p>f(Q) =</p><p>1</p><p>2</p><p>(</p><p>1− 2λ</p><p>β</p><p>)</p><p>Q2 tan (βp) ,</p><p>leading to the final expression for the generating function</p><p>F3 =</p><p>1</p><p>2</p><p>Q2 tan (βp)</p><p>(c) For the transformation to be canonical, λ must be 1. So from equation 8.1 we</p><p>have</p><p>∂F3</p><p>∂p</p><p>= λq → ∂F3</p><p>∂p</p><p>= q</p><p>q =</p><p>∂F3</p><p>∂p</p><p>=</p><p>1</p><p>2</p><p>Q2βsec2 (βp)</p><p>From this equation, we have Q</p><p>Q =</p><p>√</p><p>2q</p><p>β</p><p>cos(βp)· (8.6)</p><p>From equation 8.5,</p><p>P = Qtg(βp) =</p><p>√</p><p>2q</p><p>β</p><p>cos(βp)tg(βp) (8.7)</p><p>P =</p><p>√</p><p>2q</p><p>β</p><p>sen(βp) (8.8)</p><p>The equations 8.6 and 8.8 describe a canonical transformation for all values of β.</p><p>Exercise 17 - EDGARD</p><p>KRETSCHMANN</p><p>For the commutator</p><p>[A,B] = AB −BA (9.1)</p><p>We can find out the equivalent Jacobi’s identity. Expanding, firstly, each term of the</p><p>sum, we find that</p><p>[A, [B,C]] = A[B,C]− [B,C]A = ABC − ACB −BCA+ CBA (9.2)</p><p>[B, [C,A]] = BCA−BAC − CAB + ACB (9.3)</p><p>[C, [A,B]] = CAB − CBA− ABC +BAC (9.4)</p><p>It’s easy to see that the sum of these three equation will canceal term by term, and the</p><p>sum will be null. So we return the Jacobi’s identity for the commutator</p><p>[A, [B,C]] + [B, [C,A]] + [C, [A,B]] = 0 (9.5)</p><p>We can show as well this other property</p><p>[A,BC] = [A,B]C +B[A,C] (9.6)</p><p>For this, we can expand the left-hand side of the equation</p><p>[A,BC] = ABC −BCA (9.7)</p><p>We have just to add and subtract the term BAC, so we can return the property</p><p>[A,BC] = ABC −BAC +BAC −BCA = [A,B]C +B[A,C] (9.8)</p><p>39</p><p>Exercise 19 - GEOVANE NAYSINGER</p><p>FIGURE 10.1</p><p>Firstly, the Lagrange brackets is</p><p>{u, v}q,p =</p><p>∂q</p><p>∂u</p><p>∂p</p><p>∂v</p><p>− ∂p</p><p>∂u</p><p>∂q</p><p>∂v</p><p>(10.1)</p><p>to any analytic function u,v and similarly to w. Remindiring that both functions u,v</p><p>and w are write in term of (q,p). So, just rewrite putting the lagrande brackets in Jacobi</p><p>identity write in question.</p><p>∂</p><p>∂w</p><p>(</p><p>∂q</p><p>∂u</p><p>∂p</p><p>∂v</p><p>− ∂p</p><p>∂u</p><p>∂q</p><p>∂v</p><p>)</p><p>+</p><p>∂</p><p>∂u</p><p>(</p><p>∂q</p><p>∂v</p><p>∂p</p><p>∂w</p><p>− ∂p</p><p>∂v</p><p>∂q</p><p>∂w</p><p>)</p><p>+</p><p>∂</p><p>∂v</p><p>(</p><p>∂q</p><p>∂w</p><p>∂p</p><p>∂u</p><p>− ∂p</p><p>∂w</p><p>∂q</p><p>∂u</p><p>)</p><p>and now we write in terms of second derivatives and regrouping the term as is follow</p><p>∂p</p><p>∂v</p><p>∂2q</p><p>∂w∂u</p><p>− ∂p</p><p>∂v</p><p>∂2q</p><p>∂u∂w</p><p>+</p><p>∂q</p><p>∂u</p><p>∂2p</p><p>∂w∂v</p><p>− ∂q</p><p>∂u</p><p>∂2p</p><p>∂v∂w</p><p>+</p><p>∂q</p><p>∂v</p><p>∂2p</p><p>∂w∂u</p><p>− ∂q</p><p>∂v</p><p>∂2p</p><p>∂u∂w</p><p>+</p><p>∂p</p><p>∂u</p><p>∂2q</p><p>∂w∂v</p><p>− ∂p</p><p>∂u</p><p>∂2q</p><p>∂v∂w</p><p>+</p><p>∂q</p><p>∂w</p><p>∂2p</p><p>∂u∂v</p><p>− ∂q</p><p>∂w</p><p>∂2p</p><p>∂v∂u</p><p>+</p><p>∂p</p><p>∂w</p><p>∂2q</p><p>∂u∂v</p><p>− ∂p</p><p>∂w</p><p>∂2q</p><p>∂v∂u</p><p>= 0.</p><p>Then, the Jacobi identity is verified.</p><p>41</p><p>Exercise 20 - GUILHERME SHOITI</p><p>YOSHIMATSU GIARDINI</p><p>11.1 (a)</p><p>Verify that the components of the two-dimensional matrix</p><p>←→</p><p>A , defined by Eq.</p><p>(9.141), are constants of the motion for the two-dimensional isotropic harmonic oscil-</p><p>lator problem.</p><p>The matrix can be written as</p><p>Aij =</p><p>1</p><p>2m</p><p>(</p><p>pipj +m2ωexixj</p><p>)</p><p>or (11.1)</p><p>←→</p><p>A =</p><p>1</p><p>2m</p><p>[</p><p>p2x +m2ω2x2 pxpy +m2ω2x y</p><p>pypx +m2ω2y x p2y +m2ω2y2</p><p>]</p><p>. (11.2)</p><p>11.1.1 Solution (a)</p><p>To demonstrate what is asked, we may use first an arbitrary function F (qi, pi)</p><p>that does not depend explicitly on time.</p><p>Then, it is a constant of motion if</p><p>dF (qi, pi)</p><p>dt</p><p>= 0 but (11.3)</p><p>43</p><p>44 CHAPTER 11. EXERCISE 20 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>we know that</p><p>dF (q, p)</p><p>dt</p><p>=</p><p>∂F</p><p>∂qi</p><p>dqi</p><p>dt</p><p>+</p><p>∂F</p><p>∂pi</p><p>dpi</p><p>dt</p><p>+</p><p>�</p><p>�</p><p>��7</p><p>0</p><p>∂F</p><p>∂t</p><p>= 0 where (11.4)</p><p>we know that ∂F</p><p>∂t</p><p>= 0 because it does not depend explicitly on time, also, it is im-</p><p>portant to note that we are using Einstein’s index notation in equation (11.4), where</p><p>the multiplication of two variables with the same index counts as a summation, i.e.</p><p>xipi =</p><p>P</p><p>i xipi.</p><p>To proceed, we may use Hamilton’s equations, which are written as</p><p>dq⃗</p><p>dt</p><p>= ∇⃗p⃗H and (11.5)</p><p>dp⃗</p><p>dt</p><p>= −∇⃗q⃗H or (11.6)</p><p>in Einstein’s index notation</p><p>d qi</p><p>dt</p><p>=</p><p>∂H</p><p>∂pi</p><p>and (11.7)</p><p>d pi</p><p>dt</p><p>= −∂H</p><p>∂qi</p><p>. (11.8)</p><p>Now, we may substitute Hamilton’s equations into the time derivative of the</p><p>arbitrary function F (q⃗, p⃗), which in turn becomes</p><p>dF (q, p)</p><p>dt</p><p>=</p><p>∂F</p><p>∂qi</p><p>∂H</p><p>∂pi</p><p>− ∂F</p><p>∂pi</p><p>∂H</p><p>∂qi</p><p>= 0 . (11.9)</p><p>For an arbitrary function and Hamiltonian, it may look like we have reached at</p><p>a standpoint, however, on closer inspection, the result above is the definition of the</p><p>Poisson brackets. In other words, a function is a constant of motion if the Hamiltonian</p><p>of the system does not depend explicitly on time and the function’s Poisson bracket</p><p>with the Hamiltonian results in zero.</p><p>11.1. (A) 45</p><p>Now we may verify that</p><p>←→</p><p>A is a constant of motion for the two dimensional</p><p>harmonic oscillator whose Hamiltonian is written as</p><p>H =</p><p>1</p><p>2m</p><p>(</p><p>pipi +m2ω2qi qi</p><p>)</p><p>where (11.10)</p><p>we are also using Einstein’s index notation.</p><p>Now we can take the Poisson bracket for</p><p>←→</p><p>A and H as</p><p>{Aij,H} =</p><p>(</p><p>∂Aij</p><p>∂qk</p><p>∂H</p><p>∂pk</p><p>− ∂Aij</p><p>∂pk</p><p>∂H</p><p>∂qk</p><p>)</p><p>. (11.11)</p><p>To simplify our calculations, we obtain each partial derivative separately</p><p>∂Aij</p><p>∂qk</p><p>=</p><p>1</p><p>2m</p><p>(</p><p>m2ω2 [δikqj + δjkqi]</p><p>)</p><p>, (11.12)</p><p>∂Aij</p><p>∂pk</p><p>=</p><p>1</p><p>2m</p><p>(δik qj + δjk qi) , (11.13)</p><p>∂H</p><p>∂qk</p><p>=</p><p>1</p><p>2m</p><p>(</p><p>2m2ω2 δik qi</p><p>)</p><p>and (11.14)</p><p>finally</p><p>∂H</p><p>∂pk</p><p>=</p><p>1</p><p>2m</p><p>(2δik pi) . (11.15)</p><p>Now we substitute all of the partial derivatives into Poisson’s bracket equation</p><p>{Aij,H} =</p><p>(</p><p>∂Aij</p><p>∂qk</p><p>∂H</p><p>∂pk</p><p>− ∂Aij</p><p>∂pk</p><p>∂H</p><p>∂qk</p><p>)</p><p>=</p><p>�</p><p>�m2ω2</p><p>2��m</p><p>(δikqj + δjkqi)</p><p>pk</p><p>��m</p><p>− 1</p><p>2��m</p><p>(δikpj + δjkpi)��mω</p><p>2qk</p><p>=</p><p>1</p><p>2</p><p>[</p><p>ω2qjpi + qipj − pjqi + piqj</p><p>]</p><p>= 0 . (11.16)</p><p>We can see that Poisson’s brackets between</p><p>←→</p><p>A and H is indeed 0, proving that</p><p>←→</p><p>A is a constant of motion.</p><p>46 CHAPTER 11. EXERCISE 20 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>11.2 (b)</p><p>Verify that the quantities Si, i = 1, 2, 3, defined by Eqs. (9.144), (9.145), (9.146),</p><p>have the properties stated in Eqs. (9.147) and (9.148).</p><p>(9,146) = S1</p><p>=</p><p>A12 + A21</p><p>2ω</p><p>=</p><p>1</p><p>2mω</p><p>(</p><p>pxpy +m2ω2x y</p><p>)</p><p>, (11.17)</p><p>(9,145) = S2</p><p>=</p><p>A22 − A11</p><p>2ω</p><p>=</p><p>1</p><p>4mω</p><p>[</p><p>p2y − p2x +m2ω2(y2 − x2)</p><p>]</p><p>and (11.18)</p><p>(9,146) = S3</p><p>=</p><p>L</p><p>2</p><p>=</p><p>1</p><p>2</p><p>(x py − y px) . (11.19)</p><p>The conditions are</p><p>S2</p><p>1 + S2</p><p>2 + S2</p><p>3 =</p><p>H2</p><p>4ω2</p><p>and (11.20)</p><p>[Si, Sj] = ϵijkSk . (11.21)</p><p>11.2.1 Solution (b)</p><p>The solution to the problem is straightforward. We first have to find S2</p><p>1 , S2</p><p>2 , S2</p><p>3</p><p>and H2</p><p>4ω</p><p>.</p><p>S2</p><p>1 =</p><p>1</p><p>4m2ω2</p><p>(</p><p>p2xp</p><p>2</p><p>y +m4ω4x2y2 + 2m2ω2xypxpy</p><p>)</p><p>, (11.22)</p><p>11.2. (B) 47</p><p>S2</p><p>2 =</p><p>(</p><p>p4</p><p>y + p4</p><p>x + m4! 4(y2 � x2)2 � 2p2</p><p>yp2</p><p>x � 2p2</p><p>xm2! 2(y2 � x2) + 2p2</p><p>ym2! 2(y2 � x2)</p><p>)</p><p>16m2! 2</p><p>,</p><p>(11.23)</p><p>S2</p><p>3 =</p><p>1</p><p>4</p><p>(</p><p>x2p2</p><p>y + y2p2</p><p>x � 2pxpyxy</p><p>)</p><p>and (11.24)</p><p>H 2</p><p>4! 2</p><p>=</p><p>1</p><p>16m2! 2</p><p>(</p><p>p2</p><p>x + p2</p><p>y + m2! 2</p><p>(</p><p>x2 + y2</p><p>))2</p><p>=</p><p>(</p><p>p4</p><p>x + p4</p><p>y + m4! 4 (x4 + y4 + 2x2y2) + 2p2</p><p>xp2</p><p>y + 2pxm2! 2 (x2 + y2) + 2pym2! 2 (x2 + y2)</p><p>)</p><p>16m2! 2</p><p>(11.25)</p><p>then the sum of the three S terms become∑</p><p>i</p><p>S2</p><p>i =</p><p>1</p><p>16m2! 2</p><p>[</p><p>� � ��*</p><p>2p2</p><p>xp2</p><p>y</p><p>4p2</p><p>xp2</p><p>y + 4m4! 4x2y2 + � � � � � � � �: 0</p><p>8m2! 2xypxpy</p><p>+ p4</p><p>y + p4</p><p>x + m4! 4(y2 � x2)2 � � � ��* 0</p><p>2p2</p><p>yp2</p><p>x � 2p2</p><p>xm2! 2(y2 � x2) + 2p2</p><p>ym2! 2(y2 � x2)</p><p>+ 4m2! 2x2p2</p><p>y + 4m2! 2y2p2</p><p>x � � � � � � � � �: 0</p><p>8m2! 2pxpyxy</p><p>]</p><p>=</p><p>1</p><p>16m2! 2</p><p>[</p><p>2p2</p><p>xp2</p><p>y + 4m4! 4x2y2 + � � � � � ��: 2m2! 2x2p2</p><p>y</p><p>4m2! 2x2p2</p><p>y + � � � � � ��: 2m2! 2y2p2</p><p>x</p><p>4m2! 2y2p2</p><p>x</p><p>+ p4</p><p>y + p4</p><p>x + m4! 4(y2 � x2)2 � 2p2</p><p>xm2! 2(�</p><p>�7</p><p>0</p><p>y2 � x2) + 2p2</p><p>ym2! 2(y2 � � ��</p><p>0</p><p>x2 )</p><p>]</p><p>=</p><p>1</p><p>16m2! 2</p><p>[</p><p>2p2</p><p>xp2</p><p>y + 4m4! 4x2y2 + 2m2! 2x2p2</p><p>y + 2m2! 2y2p2</p><p>x</p><p>+ p4</p><p>y + p4</p><p>x + m4! 4</p><p>� � � � � �: (y4 + x4 � 2x2y2)</p><p>(y2 � x2)2 + 2p2</p><p>xx2m2! 2 + 2p2</p><p>yy2m2! 2</p><p>]</p><p>=</p><p>1</p><p>16m2! 2</p><p>[</p><p>2p2</p><p>xp2</p><p>y + � � � � � ��: 2m4! 4x2y2</p><p>4m4! 4x2y2 + 2m2! 2x2p2</p><p>y + 2m2! 2y2p2</p><p>x</p><p>+ p4</p><p>y + p4</p><p>x + m4! 4(y4 + x4 � � � ��* 0</p><p>2x2y2 ) + 2p2</p><p>xx2m2! 2 + 2p2</p><p>yy2m2! 2</p><p>]</p><p>=</p><p>[</p><p>p4</p><p>x + p4</p><p>x + m4! 4 (x4 + y4 + 2x2y2) + 2p2</p><p>xp2</p><p>y + 2pxm2! 2 (x2 + y2) + 2pym2! 2 (x2 + y2)</p><p>]</p><p>16m2! 2</p><p>,</p><p>(11.26)</p><p>if we compare this result to (11.25), we notice that they are equivalent, thus, proving</p><p>that S2</p><p>1 + S2</p><p>2 + S2</p><p>3 = H 2=(4! 2).</p><p>With this proof in hands, now we may proceed to prove that</p><p>[Si ; Sj ] = � ijk Sk and (11.27)</p><p>48 CHAPTER 11. EXERCISE 20 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>as stated in Herbert Goldstein Classical Mechanics (3rd ed.) page 417, "By straightfor-</p><p>ward manipulation of the Poission brackets, we can verify that the three Si quantities</p><p>satisfy the relations". To do this verification, we may use the previous proof</p><p>S2 = S2</p><p>1 + S2</p><p>2 + S2</p><p>3 =</p><p>H2</p><p>4ω2</p><p>. (11.28)</p><p>Because S2 is proportional to H2 to a constant and H is a constant of motion,</p><p>we also know that H is a constant of motion. Also, it was stated in Herbert Goldstein</p><p>Classical Mechanics (3rd ed.) page 417, Si, for i = 1,2 or 3 are also constants of motion,</p><p>thus they all result in zero Poisson brackets with the Hamiltonian, in other words</p><p>{Si,H2} = 4ω2{Si, S2} = 0 . (11.29)</p><p>Now we obtain the proof through manipulation</p><p>{Si, S2} = {Si, S2</p><p>1}+ {Si, S2</p><p>2}+ {Si, S2</p><p>3} . (11.30)</p><p>The next step can be done via Leibnitz’s rule, where</p><p>{h, fg} = {h, f}g + f{h, g} where (11.31)</p><p>h = Si and f = g = Sj , for j = 2 or j = 3, then we have</p><p>{Si, S2} = 2S1{Si, S1}+ 2S2{Si, S2}+ 2S3{Si, S3} = 0 . (11.32)</p><p>Considering now, that the index i can only assume the values i = 1, 2 or 3, the</p><p>commutation relation found above, can be expressed in terms of all the possibilities of</p><p>the index values, in other words,</p><p>������:0</p><p>S1{S1, S1} + S2{S1, S2}+ S3{S1, S3} = 0 (11.33)</p><p>S1{S2, S1}+������:0</p><p>S2{S2, S2} + S3{S2, S3} = 0 (11.34)</p><p>S1{S3, S1}+ S2{S3, S2}+������:0</p><p>S3{S3, S3} = 0 where (11.35)</p><p>the terms that go to zero are the Poisson brackets of Sj with Sj , in other words, any</p><p>Poisson bracket of a function with itself is zero. Which leaves us with a system of</p><p>11.2. (B) 49</p><p>equations</p><p>S2{S1, S2} = −S3{S1, S3} (11.36)</p><p>S1{S2, S1} = −S3{S2, S3} (11.37)</p><p>S1{S3, S1} = −S2{S3, S2} where (11.38)</p><p>this system of equation can be represented as</p><p>{Si, Sj} = ϵijkSk . (11.39)</p><p>It may be as is, because the only other option is the trivial solution, in other</p><p>words</p><p>{S1, S2} = 0 (11.40)</p><p>{S1, S3} = 0 (11.41)</p><p>{S2, S3} = 0 , (11.42)</p><p>which is not true. The only non trivial possibility is</p><p>{S1, S3} = −S2 (11.43)</p><p>{S2, S1} = −S3 (11.44)</p><p>{S3, S2} = −S1 (11.45)</p><p>{S1, S2} = S3 (11.46)</p><p>{S3, S1} = S2 (11.47)</p><p>{S2, S3} = S1 . (11.48)</p><p>In reality, there is another possibility with all the signals exchanged i.e. everything</p><p>multiplied by −1, but it would only</p><p>result in an "altered" Levi-Civita Symbol which</p><p>would bring the same results, but in a non-left-handed referential.</p><p>Exercise 22 - LUCAS BOURSCHEIDT</p><p>(a) Pearson version</p><p>Solution:</p><p>It is given that:</p><p></p><p>Q =</p><p>√</p><p>2q eα cos p</p><p>P =</p><p>√</p><p>2q e−α sin p</p><p>Taking the Poisson bracket in relation to the coordinates q and p:</p><p>[Q,P ]q,p =</p><p>∂Q</p><p>∂q</p><p>∂P</p><p>∂p</p><p>− ∂P</p><p>∂q</p><p>∂Q</p><p>∂p</p><p>=</p><p>eα√</p><p>2q</p><p>cos p</p><p>√</p><p>2q e−α cos p− e−α√</p><p>2q</p><p>sin p (−</p><p>√</p><p>2q eα sin p)</p><p>= cos2p+ sin2p = 1</p><p>51</p><p>52 CHAPTER 12. EXERCISE 22 - LUCAS BOURSCHEIDT</p><p>Thus,</p><p>[Q,P ]q,p = [Q,P ]P,Q = 1</p><p>and the proposed transformation is canonical.</p><p>(b) John Wiley version</p><p>Solution:</p><p>This problem will be solved using the invariance of Poisson brackets under canonical</p><p>transformations. We know that for systems with two degrees of freedom:</p><p>[A,B]q,p =</p><p>2∑</p><p>i=1</p><p>[</p><p>∂A</p><p>∂qi</p><p>∂B</p><p>∂pi</p><p>− ∂B</p><p>∂qi</p><p>∂A</p><p>∂pi</p><p>]</p><p>Since Qi’s do not depend on pj’s, we have ∂Qi</p><p>∂pj</p><p>= 0 and thus</p><p>53</p><p>[Q1, Q2]q,p = 0</p><p>as expected. Starting from the Poisson brackets, the general expressions for P1 e P2 are</p><p>determined below.</p><p>For A = Q1 and B = P1:</p><p>[Q1, P1]q,p =</p><p>2∑</p><p>i=1</p><p>[</p><p>∂Q1</p><p>∂qi</p><p>∂P1</p><p>∂pi</p><p>− ∂P1</p><p>∂qi</p><p>∂Q1</p><p>∂pi</p><p>]</p><p>= 2q1</p><p>∂P1</p><p>∂p1</p><p>= 1</p><p>So</p><p>∂P1</p><p>∂p1</p><p>=</p><p>1</p><p>2q1</p><p>(A)</p><p>Integrating (A):</p><p>P1 =</p><p>p1</p><p>2q1</p><p>+ f(q1, q2, p2) (B)</p><p>For A = Q2 and B = P1:</p><p>[Q2, P1]q,p =</p><p>2∑</p><p>i=1</p><p>[</p><p>∂Q2</p><p>∂qi</p><p>∂P1</p><p>∂pi</p><p>− ∂P1</p><p>∂qi</p><p>∂Q2</p><p>∂pi</p><p>]</p><p>=</p><p>∂P1</p><p>∂p1</p><p>+</p><p>∂P1</p><p>∂p2</p><p>= 0</p><p>Inserting (A) in the result above, we find:</p><p>54 CHAPTER 12. EXERCISE 22 - LUCAS BOURSCHEIDT</p><p>∂P1</p><p>∂p2</p><p>= − 1</p><p>2q1</p><p>(C)</p><p>Now inserting (B) into (C):</p><p>∂f</p><p>∂p2</p><p>= − 1</p><p>2q1</p><p>what brings us to</p><p>f(q1, q2, p2) = −</p><p>p2</p><p>2q1</p><p>+ U(q1, q2) (D)</p><p>Finally, inserting (D) into (B) we find the most general expression possible for P1 that</p><p>preserves the canonical character of the transformation:</p><p>P1 =</p><p>p1 − p2</p><p>2q1</p><p>+ U(q1, q2) (E)</p><p>For A = Q1 and B = P2:</p><p>[Q1, P2]q,p =</p><p>2∑</p><p>i=1</p><p>[</p><p>∂Q1</p><p>∂qi</p><p>∂P2</p><p>∂pi</p><p>− ∂P2</p><p>∂qi</p><p>∂Q1</p><p>∂pi</p><p>]</p><p>= 2q1</p><p>∂P2</p><p>∂p1</p><p>= 0</p><p>That is</p><p>∂P2</p><p>∂p1</p><p>= 0 (F )</p><p>55</p><p>Integrating (F):</p><p>P2 = g(q1, q2, p2) (G)</p><p>For A = Q2 and B = P2:</p><p>[Q2, P2]q,p =</p><p>2∑</p><p>i=1</p><p>[</p><p>∂Q2</p><p>∂qi</p><p>∂P2</p><p>∂pi</p><p>− ∂P2</p><p>∂qi</p><p>∂Q2</p><p>∂pi</p><p>]</p><p>=</p><p>∂P2</p><p>∂p2</p><p>= 1</p><p>That is</p><p>∂P2</p><p>∂p2</p><p>= 1 (H)</p><p>Now inserting (G) into (H):</p><p>∂g</p><p>∂p2</p><p>= 1</p><p>what brings us to</p><p>g(q1, q2, p2) = p2 + V (q1, q2) (I)</p><p>Finally, from (G) and (I) we find the most general expression possible for P2 that pre-</p><p>serves the canonical character of the transformation:</p><p>P2 = p2 + V (q1, q2) (J)</p><p>56 CHAPTER 12. EXERCISE 22 - LUCAS BOURSCHEIDT</p><p>Thus, the general expressions we were looking for are:</p><p>P1 =</p><p>p1−p2</p><p>2q1</p><p>+ U(q1, q2) (E)</p><p>P2 = p2 + V (q1, q2) (J)</p><p>where U and V are arbitrary functions of q1 and q2 only.</p><p>Hamiltonian</p><p>From the expression given for H in the problem and using the results (E) and</p><p>(J), we find the Hamiltonian function in terms of the new coordinates and momenta:</p><p>H =</p><p>(</p><p>p1 − p2</p><p>2q1</p><p>)2</p><p>+ p2 + (q1 + q2)</p><p>2</p><p>= (P1 − U)2 + P2 − V +Q2</p><p>2</p><p>= P 2</p><p>1 + P2 + [U(U − 2P1)− (V −Q2</p><p>2)]</p><p>For Q1 and Q2 to be both ignorable coordinates, given that U and V are arbitrary</p><p>functions of q1 and q2 only (and therefore of Q1 and Q2), it is necessary that U = 0 and</p><p>V = Q2</p><p>2. Thereby: </p><p>P1 =</p><p>p1−p2</p><p>2q1</p><p>(E ′)</p><p>P2 = p2 +Q2</p><p>2 = p2 + (q1 + q2)</p><p>2 (J ′)</p><p>and</p><p>H = P 2</p><p>1 + P2 (K)</p><p>57</p><p>Equations of motion and their solutions</p><p>From the Hamiltonian (K) we find the equations of motion and their solutions</p><p>as shown below:</p><p>For Q1 and P1:</p><p>Ṗ1 = −</p><p>∂H</p><p>∂Q1</p><p>= 0</p><p>P1 = a (L)</p><p>Q̇1 =</p><p>∂H</p><p>∂P1</p><p>= 2P1 = 2a</p><p>Q1 = 2at+ b (M)</p><p>For Q2 and P2:</p><p>Ṗ2 = −</p><p>∂H</p><p>∂Q2</p><p>= 0</p><p>P2 = c (N)</p><p>Q̇2 =</p><p>∂H</p><p>∂P2</p><p>= 1</p><p>Q2 = t+ d (O)</p><p>In the solutions obtained, a, b, c and d are constants determined by the initial</p><p>conditions. It is a simple algebraic problem to reverse the canonical transformation to</p><p>58 CHAPTER 12. EXERCISE 22 - LUCAS BOURSCHEIDT</p><p>obtain q1, q2, p1, p2 in terms of Q1, Q2, P1, P2 and, using the solutions (L), (M), (N) and</p><p>(O), show that:</p><p></p><p>q1 =</p><p>√</p><p>2at+ b</p><p>q2 = t+ d−</p><p>√</p><p>2at+ b</p><p>p1 = c− (t+ d)2 + 2a</p><p>√</p><p>2at+ b</p><p>p2 = c− (t+ d)2</p><p>Exercise 23 -JENIFER ANDRADE DE</p><p>MATOS</p><p>By any method you choose, show that the following transformation is canonical:</p><p>x =</p><p>1</p><p>α</p><p>(√</p><p>2P1,sin(Q1) + P2</p><p>)</p><p>, px =</p><p>α</p><p>2</p><p>(√</p><p>2P1,cos(Q1)−Q2</p><p>)</p><p>y =</p><p>1</p><p>α</p><p>(√</p><p>2P1,cos(Q1) +Q2</p><p>)</p><p>, py = −</p><p>α</p><p>2</p><p>(√</p><p>2P1,sin(Q1)− P2</p><p>)</p><p>where α is some fixed parameter.</p><p>Apply this transformation to the problem of a particle of charge q moving in a</p><p>plane that is perpendicular to a constant magnetic field B. Express the Hamiltonian for</p><p>this problem in the (Qi, Pi) coordinates letting the parameter α take the form</p><p>α2 =</p><p>qB</p><p>c</p><p>From this Hamiltonian, obtain the motion of the particle as a function of time.</p><p>Solution: To show that the proposed transformation is canonical, it must respect</p><p>the following relation:</p><p>[ζ, ζ]η = J (13.1)</p><p>Since the variables involved are x, y, px, py,Q1,Q2, P1 and P2, the relation can be written</p><p>as:</p><p>[x, x] = 0− Trivial (13.2)</p><p>59</p><p>60 CHAPTER 13. EXERCISE 23 -JENIFER ANDRADE DE MATOS</p><p>[x, y] = 0 (13.3)</p><p>[x, px] = 1 (13.4)</p><p>[x, py] = 0 (13.5)</p><p>[y, y] = 0− Trivial (13.6)</p><p>[y, px] = 0 (13.7)</p><p>[y, py] = 1 (13.8)</p><p>[px, px] = 0− Trivial (13.9)</p><p>[px, py] = 0 (13.10)</p><p>[py, py] = 0− Trivial (13.11)</p><p>When solving all this poisson brackets, there will appear a lot of partial deriva-</p><p>tives, we will first determine the partial derivatives and then substitute in the Poisson</p><p>brackets.</p><p>∂x</p><p>∂Q1</p><p>=</p><p>√</p><p>2P1</p><p>α</p><p>cos(Q1) (13.12)</p><p>61</p><p>∂x</p><p>∂P1</p><p>=</p><p>√</p><p>2P1</p><p>2α</p><p>sin(Q1) (13.13)</p><p>∂x</p><p>∂Q2</p><p>= 0 (13.14)</p><p>∂x</p><p>∂P2</p><p>=</p><p>1</p><p>α</p><p>(13.15)</p><p>∂y</p><p>∂Q1</p><p>= −</p><p>√</p><p>2P1</p><p>α</p><p>sin(Q1) (13.16)</p><p>∂y</p><p>∂P1</p><p>=</p><p>√</p><p>2P</p><p>(−1/2)</p><p>1</p><p>2α</p><p>cos(Q1) (13.17)</p><p>∂y</p><p>∂Q2</p><p>=</p><p>1</p><p>α</p><p>(13.18)</p><p>∂y</p><p>∂P2</p><p>= 0 (13.19)</p><p>∂px</p><p>∂Q1</p><p>= −α</p><p>√</p><p>2P1</p><p>2</p><p>sin(Q1) (13.20)</p><p>∂px</p><p>∂P1</p><p>=</p><p>√</p><p>2α</p><p>4</p><p>P</p><p>−1/2</p><p>1 cos(Q1) (13.21)</p><p>∂px</p><p>∂Q2</p><p>= −α</p><p>2</p><p>(13.22)</p><p>∂px</p><p>∂P2</p><p>= 0 (13.23)</p><p>62 CHAPTER 13. EXERCISE 23 -JENIFER ANDRADE DE MATOS</p><p>∂py</p><p>∂Q1</p><p>= −α</p><p>√</p><p>2P1cos(Q1)</p><p>2</p><p>(13.24)</p><p>∂py</p><p>∂Q2</p><p>= 0 (13.25)</p><p>∂py</p><p>∂P1</p><p>= −α</p><p>√</p><p>2P</p><p>−1/2</p><p>1 sin(Q1)</p><p>4</p><p>(13.26)</p><p>∂py</p><p>∂P2</p><p>=</p><p>α</p><p>2</p><p>(13.27)</p><p>Now, let´s apply these relations to the poisson brackets:</p><p>[x, y] =</p><p>∂x</p><p>∂Q1</p><p>∂y</p><p>∂P1</p><p>− ∂x</p><p>∂P1</p><p>∂y</p><p>∂Q1</p><p>+</p><p>∂x</p><p>∂Q2</p><p>∂y</p><p>∂P2</p><p>− ∂x</p><p>∂P2</p><p>∂y</p><p>∂Q2</p><p>(13.28)</p><p>Applying 13.12, 13.13, 13.14, 13.15, 13.16, 13.17, 13.18 and 13.19 in 13.28, we find:</p><p>[x, y] =</p><p>1</p><p>α2</p><p>cos2(Q1) +</p><p>1</p><p>α2</p><p>sin2(Q1) + 0− 1</p><p>α2</p><p>= 0 (13.29)</p><p>[x, px] =</p><p>∂x</p><p>∂Q1</p><p>∂px</p><p>∂P1</p><p>− ∂x</p><p>∂P1</p><p>∂px</p><p>∂Q1</p><p>+</p><p>∂x</p><p>∂Q2</p><p>∂px</p><p>∂P2</p><p>− ∂x</p><p>∂P2</p><p>∂px</p><p>∂Q2</p><p>(13.30)</p><p>Applying 13.12, 13.13, 13.14, 13.15, 13.20, 13.21, 13.22 and 13.23 in 13.30, we find:</p><p>[x, px] =</p><p>(√</p><p>2P1</p><p>α</p><p>cos(Q1)</p><p>)(√</p><p>2αP</p><p>−1/2</p><p>1</p><p>4</p><p>cos(Q1)</p><p>)</p><p>−(√</p><p>2P</p><p>−1/2</p><p>1</p><p>2α</p><p>sin(Q1)</p><p>)(</p><p>−α</p><p>√</p><p>2P1</p><p>2</p><p>sin(Q1)</p><p>)</p><p>+ 0−</p><p>(</p><p>1</p><p>α</p><p>) (</p><p>−α</p><p>2</p><p>)</p><p>[x, px] =</p><p>cos2(Q1)</p><p>2</p><p>−</p><p>(</p><p>−sin</p><p>2(Q1)</p><p>2</p><p>)</p><p>+ 0−</p><p>(</p><p>−1</p><p>2</p><p>)</p><p>= 1 (13.31)</p><p>63</p><p>[x, py] =</p><p>∂x</p><p>∂Q1</p><p>∂py</p><p>∂P1</p><p>− ∂x</p><p>∂P1</p><p>∂py</p><p>∂Q1</p><p>+</p><p>∂x</p><p>∂Q2</p><p>∂py</p><p>∂P2</p><p>− ∂x</p><p>∂P2</p><p>∂py</p><p>∂Q2</p><p>(13.32)</p><p>Applying 13.12, 13.13, 13.14, 13.15, 13.24, 13.25, 13.26 and 13.27, we find:</p><p>[x, py] =</p><p>√</p><p>2P1</p><p>α</p><p>cos(Q1)</p><p>(</p><p>−α</p><p>√</p><p>2P</p><p>−1/2</p><p>1</p><p>4</p><p>sin(Q1)</p><p>)</p><p>−</p><p>(√</p><p>2P</p><p>−1/2</p><p>1</p><p>2α</p><p>sin(Q1)</p><p>)(</p><p>−α</p><p>√</p><p>2P1</p><p>2</p><p>cos(Q1)</p><p>)</p><p>+</p><p>0− 0</p><p>[x, py] = −</p><p>1</p><p>2</p><p>cos(Q1)sin(Q1) +</p><p>1</p><p>2</p><p>cos(Q1)sin(Q1) = 0 (13.33)</p><p>[y, px] =</p><p>∂y</p><p>∂Q1</p><p>∂px</p><p>∂P1</p><p>− ∂y</p><p>∂P1</p><p>∂px</p><p>∂Q1</p><p>+</p><p>∂y</p><p>∂Q2</p><p>∂px</p><p>∂P2</p><p>− ∂y</p><p>∂P2</p><p>∂px</p><p>∂Q2</p><p>(13.34)</p><p>Applying 13.16, 13.17, 13.18, 13.19, 13.20, 13.21, 13.22 and 13.23 in the equation,</p><p>we find:</p><p>[y, px] = −</p><p>1</p><p>2</p><p>sin(Q1)cos(Q1)−</p><p>(</p><p>−1</p><p>2</p><p>sin(Q1)cos(Q1)</p><p>)</p><p>+ 0− 0 = 0 (13.35)</p><p>[y, py] =</p><p>∂y</p><p>∂Q1</p><p>∂py</p><p>∂P1</p><p>− ∂y</p><p>∂P1</p><p>∂py</p><p>∂Q1</p><p>+</p><p>∂y</p><p>∂Q2</p><p>∂py</p><p>∂P2</p><p>− ∂y</p><p>∂P2</p><p>∂py</p><p>∂Q2</p><p>(13.36)</p><p>Applying 13.16, 13.17, 13.18, 13.19, 13.24, 13.25, 13.26 and 13.27 in the equation,</p><p>we find:</p><p>[y, py] =</p><p>1</p><p>2</p><p>sin2(Q1)−</p><p>(</p><p>cos2(Q1)</p><p>2</p><p>)</p><p>+</p><p>1</p><p>2</p><p>− 0 = 1 (13.37)</p><p>[px, py] =</p><p>∂px</p><p>∂Q1</p><p>∂py</p><p>∂P1</p><p>− ∂px</p><p>∂P1</p><p>∂py</p><p>∂Q1</p><p>+</p><p>∂px</p><p>∂Q2</p><p>∂py</p><p>∂P2</p><p>− ∂px</p><p>∂P2</p><p>∂py</p><p>∂Q2</p><p>(13.38)</p><p>Applying 13.20, 13.21, 13.22, 13.23, 13.24, 13.25, 13.26 and 13.27 in the equation,</p><p>we find:</p><p>[px, py] =</p><p>α2sin2(Q1)</p><p>4</p><p>−</p><p>(</p><p>−α</p><p>2cos2(Q1)</p><p>4</p><p>)</p><p>+</p><p>(</p><p>−α</p><p>2</p><p>4</p><p>)</p><p>− 0 = 0 (13.39)</p><p>64 CHAPTER 13. EXERCISE 23 -JENIFER ANDRADE DE MATOS</p><p>Since we showed that all the Poisson brackets respect the necessary relations,</p><p>the proposed transformation is canonical.</p><p>Now, let´s find the Hamiltonian of a charged particle in a uniform magnetic field</p><p>B:</p><p>Since the uniform magnetic field points in the z⃗ direction</p><p>and the particle moves</p><p>along the plane xy, we can write this Hamiltonian using the coordinate x, y, px and py:</p><p>The Hamiltonian of a particle immersed in a magnetic field is:</p><p>H =</p><p>1</p><p>2m</p><p>(</p><p>P⃗ − q</p><p>c</p><p>A⃗</p><p>)2</p><p>(13.40)</p><p>Since B is uniform, we can write:</p><p>A⃗ = −1</p><p>2</p><p>r⃗ × B⃗ (13.41)</p><p>Then the Hamiltonian is:</p><p>H =</p><p>1</p><p>2m</p><p>(</p><p>P⃗ +</p><p>q</p><p>2c</p><p>r⃗ × B⃗</p><p>)2</p><p>(13.42)</p><p>B⃗ =</p><p> 00</p><p>B</p><p> , r⃗ =</p><p>xy</p><p>0</p><p> , p⃗ =</p><p>pxpy</p><p>0</p><p> (13.43)</p><p>Then the Hamiltonian can be written as:</p><p>H =</p><p>1</p><p>2m</p><p>[(</p><p>px +</p><p>q</p><p>2c</p><p>By</p><p>)2</p><p>+</p><p>(</p><p>py −</p><p>q</p><p>2c</p><p>Bx</p><p>)2]</p><p>(13.44)</p><p>Now we will rewrite this Hamiltonian using the new coordinates Q1, Q2, P1 and</p><p>P2, and using the relation of α with B, q and c:</p><p>H =</p><p>1</p><p>2m</p><p>[(</p><p>α</p><p>√</p><p>2P1cos(Q1)</p><p>)2</p><p>+</p><p>(</p><p>−α</p><p>√</p><p>2P1sin(Q1)</p><p>)2]</p><p>(13.45)</p><p>65</p><p>H =</p><p>α2</p><p>m</p><p>P1 (13.46)</p><p>H =</p><p>qB</p><p>mc</p><p>P1 (13.47)</p><p>From the Hamiltonian, we can find the movement equations in the Q1, Q2, P1</p><p>and P2 coordinate system:</p><p>Q̇1 =</p><p>∂H</p><p>∂P1</p><p>=</p><p>qB</p><p>mc</p><p>(13.48)</p><p>Q1 =</p><p>(</p><p>qB</p><p>mc</p><p>)</p><p>t+ φ (13.49)</p><p>Q̇2 =</p><p>∂H</p><p>∂P2</p><p>= 0 (13.50)</p><p>Q2 = Q20 (13.51)</p><p>Ṗ1 = −</p><p>∂H</p><p>∂Q1</p><p>= 0 (13.52)</p><p>P1 = P10 (13.53)</p><p>Ṗ2 = −</p><p>∂H</p><p>∂Q2</p><p>= 0 (13.54)</p><p>P2 = P20 (13.55)</p><p>66 CHAPTER 13. EXERCISE 23 -JENIFER ANDRADE DE MATOS</p><p>Equations 13.49, 13.51, 13.53 and 13.55 are the motion equations of the particle</p><p>in the transformed system of coordinates. It´s interesting to see that Q2, P1 and P2 are</p><p>constants of the motion, while Q1 is linearly dependent of the time.</p><p>Now let´s see the equations of motion in the coordinates x, y, px and py system</p><p>of coordinates:</p><p>x =</p><p>√</p><p>2P10</p><p>α</p><p>sin</p><p>(</p><p>qBt</p><p>mc</p><p>+ φ</p><p>)</p><p>+</p><p>P20</p><p>α</p><p>(13.56)</p><p>y =</p><p>√</p><p>2P10</p><p>α</p><p>cos</p><p>(</p><p>qBt</p><p>mc</p><p>+ φ</p><p>)</p><p>+</p><p>Q20</p><p>α</p><p>(13.57)</p><p>px =</p><p>α</p><p>√</p><p>2P10</p><p>2</p><p>cos</p><p>(</p><p>qBt</p><p>mc</p><p>+ φ</p><p>)</p><p>− αQ20</p><p>2</p><p>(13.58)</p><p>py = −</p><p>α</p><p>√</p><p>2P10</p><p>2</p><p>sin</p><p>(</p><p>qBt</p><p>mc</p><p>+ φ</p><p>)</p><p>+</p><p>αP20</p><p>2</p><p>(13.59)</p><p>These equations, as expected, show that the movement of the particle in the x,y</p><p>system is a uniform circular motion.</p><p>Exercise 26 - FRANCISCO AUGUSTO</p><p>FERREIRA ALMEIDA</p><p>A system of n particles moves in a plane under the influence of interaction forces</p><p>derived from potential terms depending only upon the scalar distances between parti-</p><p>cles.</p><p>(a) Using plane polar coordinates for each particle (relative to a common origin), iden-</p><p>tify the form of the Hamiltonian for the system.</p><p>(b) Find a generating function for the canonical transformation that corresponds to a</p><p>transformation to coordinates rotating in the plane counterclockwise with a uniform</p><p>angular rate ω (the same for all particles). What are the transformation equations for</p><p>the momenta?</p><p>(c) What is the new Hamiltonian? What physical significance can you give to the dif-</p><p>ference between the old and the new Hamiltonians?</p><p>Solution:</p><p>(a)</p><p>For a particle, using plane polar coordinates, we have:</p><p>x = rcosθ (14.1)</p><p>and</p><p>y = rsinθ (14.2)</p><p>67</p><p>68 CHAPTER 14. EXERCISE 26 - FRANCISCO AUGUSTO FERREIRA ALMEIDA</p><p>Thus</p><p>ẋ = ṙcosθ + (−rsinθθ̇) (14.3)</p><p>and</p><p>ẏ = ṙsinθ + rcosθθ̇ (14.4)</p><p>Squared, we have:</p><p>ẋ2 = ṙ2cos2θ − 2rṙθ̇sinθcosθ + r2θ̇2sin2θ (14.5)</p><p>and</p><p>ẏ2 = ṙ2sin2θ + 2rṙθ̇sinθcosθ + r2θ̇2cos2θ (14.6)</p><p>Adding</p><p>ẋ2 + ẏ2 = ṙ2 + r2θ̇2 (14.7)</p><p>So, the lagrangiana is:</p><p>L =</p><p>1</p><p>2</p><p>m(ṙ2 + r2θ̇2)− V (r) (14.8)</p><p>Calculating the conjugate moments, we have:</p><p>pr =</p><p>∂L</p><p>∂ṙ</p><p>= mṙ (14.9)</p><p>and</p><p>pθ =</p><p>∂L</p><p>∂θ̇</p><p>= mr2θ̇ (14.10)</p><p>Then using the definition of the classical Hamiltonian</p><p>H = prṙ + pθθ̇ − L (14.11)</p><p>H = mṙ2 +mr2θ̇2 − 1</p><p>2</p><p>mṙ2 − 1</p><p>2</p><p>mr2θ̇2 + V (r) (14.12)</p><p>H =</p><p>1</p><p>2</p><p>mṙ2 +</p><p>1</p><p>2</p><p>mr2θ̇2 + V (r) (14.13)</p><p>H =</p><p>p2r</p><p>2m</p><p>+</p><p>p2θ</p><p>2mr2</p><p>+ V (r) (14.14)</p><p>69</p><p>H =</p><p>1</p><p>2m</p><p>(</p><p>p2r +</p><p>p2θ</p><p>r2</p><p>)</p><p>+ V (r) (14.15)</p><p>For a system of n particles:</p><p>H =</p><p>n∑</p><p>i=1</p><p>1</p><p>2mi</p><p>(</p><p>p2ri +</p><p>pθ</p><p>2</p><p>i</p><p>r2i</p><p>)</p><p>+ V (ri) (14.16)</p><p>Exercise 28 - JACKSON GALVÃO</p><p>A charge particle moves in space with a constant magnetic field B such that the vector</p><p>potential, A, is</p><p>A =</p><p>1</p><p>2</p><p>(B× r)</p><p>(a) If vj are the Cartesian components of the velocity of the particle, evaluate the Pois-</p><p>son brackets</p><p>[vi, vj], i ̸= j = 1, 2, 3,</p><p>The momentum of a charged particle in an electric field is given by</p><p>pi = mvi + qAi (15.1)</p><p>that we take us</p><p>vi =</p><p>pi − qAi</p><p>m</p><p>(15.2)</p><p>where A is our potential vector. Then, we can do</p><p>[vi, vj] =</p><p>1</p><p>m2</p><p>[pi − qAi, pj − qAj]</p><p>=</p><p>1</p><p>m2</p><p>([pi, pj]− [pi, qAj]− [qAi, pj] + q[Ai, Aj])</p><p>= − q</p><p>m2</p><p>([pi, Aj] + [Ai, pj])</p><p>=</p><p>q</p><p>m2</p><p>([pj, Ai]− [pi, Aj])</p><p>(15.3)</p><p>71</p><p>72 CHAPTER 15. EXERCISE 28 - JACKSON GALVÃO</p><p>For the potential vector in terms of Livi-Civita summation, can be</p><p>Ai = ϵiabBaxa (15.4)</p><p>Calculating the firt term, we have</p><p>[pj, Ai] =</p><p>1</p><p>2</p><p>[pj, ϵiabBaxb]</p><p>=</p><p>1</p><p>2</p><p>ϵiabBa[pj, xb]</p><p>=</p><p>1</p><p>2</p><p>ϵiabBaδjb</p><p>=</p><p>1</p><p>2</p><p>ϵiabBa</p><p>(15.5)</p><p>Then, for second terms, we have</p><p>[pi, Aj] =</p><p>1</p><p>2</p><p>ϵjaiBa (15.6)</p><p>Reacting to the terms, we must be have</p><p>[vi, vj] =</p><p>q</p><p>m2</p><p>([pj, Ai]− [pi, Aj])</p><p>=</p><p>q</p><p>m2</p><p>(</p><p>1</p><p>2</p><p>ϵiajBa −</p><p>1</p><p>2</p><p>ϵjaiBa</p><p>)</p><p>=</p><p>qBa</p><p>2m2</p><p>(ϵiaj − ϵjai)</p><p>(15.7)</p><p>Therefore,</p><p>[vi, vj] =</p><p>qBa</p><p>m2</p><p>ϵiaj (15.8)</p><p>(b) If pi is the canonical momentum conjugate to xi, also evaluate the Poisson brackets.</p><p>[xi, vj] , [pi, vj]</p><p>[xi, pj] , [pi, ṗj]</p><p>Being pi the canonical momentum conjugate to xi, the Poisson bracket get be written</p><p>like as</p><p>73</p><p>i) [xi, vj]</p><p>ii) [pi, vj]</p><p>iii) [xi, ṗj]</p><p>iv) [pi, ṗj]</p><p>Then, so for each, we have To first, i)</p><p>[xi, vj] =</p><p>1</p><p>m</p><p>[xi, pj − qAj]</p><p>=</p><p>1</p><p>m</p><p>([xi, pj]− q[xi, Aj])</p><p>=</p><p>1</p><p>m</p><p>(δij −</p><p>q</p><p>2</p><p>[xi, ϵjabBaxb])</p><p>=</p><p>1</p><p>m</p><p>(δij −</p><p>q</p><p>2</p><p>ϵjabBa[xi, xb])</p><p>=</p><p>δij</p><p>m</p><p>(15.9)</p><p>To second, ii)</p><p>[pi, vj] =</p><p>1</p><p>m</p><p>[pi, vj − qAj]</p><p>=</p><p>1</p><p>m</p><p>([pi, pj]− q[pi, Aj])</p><p>= − q</p><p>m</p><p>ϵjabBa[pi, xb]</p><p>= − q</p><p>m</p><p>ϵjaiBa</p><p>(15.10)</p><p>We observe that the Hamiltonian is given by, for a charge particle in a magnetic field</p><p>H =</p><p>1</p><p>m</p><p>(pi − qAi)2 (15.11)</p><p>So, we must be have to ṗi</p><p>ṗi = −</p><p>∂H</p><p>∂xi</p><p>= − 1</p><p>m</p><p>(pi − qAi)</p><p>∂Ai</p><p>∂xi</p><p>= − 1</p><p>m</p><p>(pi − qAi)</p><p>1</p><p>2</p><p>ϵijkBj</p><p>∂xk</p><p>∂xi</p><p>= − 1</p><p>2m</p><p>(pi − qAi)ϵijkBjδki</p><p>= − 1</p><p>2m</p><p>(pi − qAi)ϵijiBj</p><p>= 0</p><p>(15.12)</p><p>74 CHAPTER 15. EXERCISE 28 - JACKSON GALVÃO</p><p>Thereby, we must be have to the thirdth (iii) and too fourth (iv)</p><p>[xi, ṗj] = 0</p><p>[pi, ṗj] = 0</p><p>(15.13)</p><p>Exercise 30 - RODRIGO WEBER</p><p>PEREIRA</p><p>(a) Prove that the Poisson bracket of two constants of the motion is itself a con-</p><p>stant of the motion even when the constants depend upon time explicitly.</p><p>(b) Show that if the Hamiltonian and a quantity F are constants of the motion, then the</p><p>nth partial derivative of F with respect to t must also be a constant of the motion.</p><p>(c) As an illustration of this result, consider the uniform motion of a free particle of</p><p>mass m. The Hamiltonian is certainly conserved, and there exists a constant of the</p><p>motion</p><p>F = x− pt</p><p>m</p><p>Show by direct computation that the partial derivative of F with t, which is a constant</p><p>of the motion, agrees with [H,F ].</p><p>Solution: (a) To prove that the Poisson bracket of the constants of motion f and g</p><p>(ḟ = ġ = 0) is also a constant of motion, we take te total time derivative of the quantity:</p><p>d</p><p>dt</p><p>[f, g] =</p><p>∂</p><p>∂t</p><p>[f, g] + [H, [f, g]] (16.1)</p><p>Where we used the identity ẋ = ∂x</p><p>∂t</p><p>+ [H, x], where x(q⃗, p⃗, t) is any dynamical variable</p><p>and H is the Hamiltonian of the system. The above equation can be rewritten using</p><p>the Jacobi Identity:</p><p>[H, [f, g]] + [f, [g,H]] + [g, [H, f ]] = 0</p><p>And the following properties of the Poisson brackets:</p><p>∂</p><p>∂λ</p><p>[f, g] = [</p><p>∂f</p><p>∂λ</p><p>, g] + [f,</p><p>∂g</p><p>∂λ</p><p>],</p><p>75</p><p>76 CHAPTER 16. EXERCISE 30 - RODRIGO WEBER PEREIRA</p><p>Where λ is an arbitrary parameter. Using both properties in Eq. 16.1, we have:</p><p>d</p><p>dt</p><p>[f, g] = [</p><p>∂f</p><p>∂t</p><p>, g] + [f,</p><p>∂g</p><p>∂t</p><p>]−[f, [g,H]]︸ ︷︷ ︸</p><p>+[f; [H;g ]]</p><p>−[g, [H, f ]]︸ ︷︷ ︸</p><p>+[[H;f ];g]</p><p>,</p><p>Where the properties [f, g] = −[g, f ] and [f, λg] = λ[f, g] were used. So:</p><p>d</p><p>dt</p><p>[f, g] = [</p><p>∂f</p><p>∂t</p><p>, g] + [[H, f ], g] + [f,</p><p>∂g</p><p>∂t</p><p>] + [f, [H, g]]</p><p>Using the linearity of the Poisson brackets:</p><p>d</p><p>dt</p><p>[f, g] = [</p><p>∂f</p><p>∂t</p><p>+ [H, f ]︸ ︷︷ ︸</p><p>˙f =0</p><p>, g] + [f,</p><p>∂g</p><p>∂t</p><p>+ [H, g]︸ ︷︷ ︸</p><p>ġ=0</p><p>] = 0</p><p>In the last step we used the property [0, f ] = 0, concluding the proof.</p><p>(b) We start proving that the first partial derivative is a constant of motion. If F</p><p>is a constant of motion, Ḟ = 0, and we can write:</p><p>Ḟ︸︷︷︸</p><p>=0</p><p>=</p><p>∂F</p><p>∂t</p><p>+ [H,F ] ⇒ ∂F</p><p>∂t</p><p>= [F,H]</p><p>Taking the derivative, we immediately prove the result for the first partial derivative:</p><p>d</p><p>dt</p><p>(∂F</p><p>∂t</p><p>)</p><p>=</p><p>d</p><p>dt</p><p>[F,H] = 0</p><p>The last step</p><p>followed from (a), since both F and H are constants of motion. Now</p><p>we prove that the nth derivative of F are themselves constants of motion. Since the</p><p>Hamiltonian is a constant of motion, we may write:</p><p>Ḣ︸︷︷︸</p><p>=0</p><p>=</p><p>∂H</p><p>∂t</p><p>+ [H,H]︸ ︷︷ ︸</p><p>=0</p><p>⇒ ∂H</p><p>∂t</p><p>= 0</p><p>So, the secondth partial derivative with respect to time is:</p><p>∂2F</p><p>∂t2</p><p>=</p><p>∂</p><p>∂t</p><p>(∂F</p><p>∂t</p><p>)</p><p>︸ ︷︷ ︸</p><p>[F;H ]</p><p>= [</p><p>∂F</p><p>∂t</p><p>,H] + [F,</p><p>∂H</p><p>∂t︸︷︷︸</p><p>=0</p><p>] = [</p><p>∂F</p><p>∂t</p><p>,H]</p><p>The above manipulation may be done to partial derivative of higher orders, iterating</p><p>the nth partial derivative with the (n-1)th. So, it is clear that:</p><p>∂nF</p><p>∂tn</p><p>= [</p><p>∂n� 1F</p><p>∂tn� 1</p><p>, H] (16.2)</p><p>77</p><p>Now, by mathematical induction, if the statement holds for the (n−1)th partial deriva-</p><p>tive, that is,</p><p>d</p><p>dt</p><p>(∂n� 1F</p><p>∂tn� 1</p><p>)</p><p>= 0</p><p>we want to show that it still holds for the nth partial derivative. So, taking the deriva-</p><p>tive of Eq. 16.2:</p><p>d</p><p>dt</p><p>(∂nF</p><p>∂tn</p><p>)</p><p>=</p><p>d</p><p>dt</p><p>(</p><p>[</p><p>∂n� 1F</p><p>∂tn� 1</p><p>, H]</p><p>)</p><p>=</p><p>∂</p><p>∂t</p><p>[</p><p>∂n� 1F</p><p>∂tn� 1</p><p>, H] + [H, [</p><p>∂n� 1F</p><p>∂tn� 1</p><p>, H]]</p><p>d</p><p>dt</p><p>(∂nF</p><p>∂tn</p><p>)</p><p>= [</p><p>∂</p><p>∂t</p><p>∂n� 1F</p><p>∂tn� 1</p><p>, H] + [</p><p>∂n� 1F</p><p>∂tn� 1</p><p>,</p><p>∂H</p><p>∂t︸︷︷︸</p><p>=0</p><p>] + [[H,</p><p>∂n� 1F</p><p>∂tn� 1</p><p>], H]</p><p>d</p><p>dt</p><p>(∂nF</p><p>∂tn</p><p>)</p><p>=</p><p>[ ∂</p><p>∂t</p><p>∂n� 1F</p><p>∂tn� 1</p><p>+ [H,</p><p>∂n� 1F</p><p>∂tn� 1</p><p>]︸ ︷︷ ︸</p><p>d</p><p>dt</p><p>(</p><p>@n � 1F</p><p>@tn � 1</p><p>)</p><p>=0 , by hypothesis.</p><p>, H</p><p>]</p><p>= 0</p><p>And the proof is complete.</p><p>(c) In the given example, ṗ = 0 (free particle). The quantities are:</p><p>H =</p><p>p2</p><p>2m</p><p>F = x− pt</p><p>m</p><p>So, we can easily derive that Ḣ = Ḟ = 0 (using p = mẋ), which means that both</p><p>quantities are constants of motion. The Poisson brackets of of them is, therefore:</p><p>[H,F ] =</p><p>∂H</p><p>∂x</p><p>∂F</p><p>∂p</p><p>− ∂H</p><p>∂p</p><p>∂F</p><p>∂x</p><p>= − p</p><p>m</p><p>Since p and m are both constants of motion, [H,F ] also is. Still, if we were to take the</p><p>firs partial derivative, we would get the following result:</p><p>∂F</p><p>∂t</p><p>= − p</p><p>m</p><p>Which, again, is also a constant of motion. All the higher order partial derivatives</p><p>would obviously be constants of motion.</p><p>Exercise 32 - DOUGLAS OLIVEIRA NO-</p><p>VAES</p><p>Question: A system of two degrees of freedom is described by the Hamiltonian</p><p>H = q1p1 − q2p2 − aq21 + bq22</p><p>Show that</p><p>F1 =</p><p>p1 − aq1</p><p>q2</p><p>and F2 = q1q2</p><p>are constant of the motion. Are there any other independent algebraic constants of the</p><p>motion? Can any be constructed from Jacobi’s identity?</p><p>Response: For an F to be a constant of the motion (COM), it must have dF</p><p>dt</p><p>= 0.</p><p>But, for F = F (q, p) (time-independent)</p><p>dF</p><p>dt</p><p>=</p><p>�</p><p>�</p><p>��7</p><p>0</p><p>∂F</p><p>∂t</p><p>+</p><p>∂F</p><p>∂q</p><p>q̇ +</p><p>∂f</p><p>∂p</p><p>ṗ = 0 (17.1)</p><p>Since</p><p>∂H</p><p>∂qi</p><p>= −ṗi</p><p>∂H</p><p>∂pi</p><p>= q̇i</p><p>in terms of the Poisson brackets, the relation (17.1) results in</p><p>dF</p><p>dt</p><p>= [Fi, H] = 0</p><p>By Jacobi identity,</p><p>[F1, [F2, H]] + [F2, [H,F1]] + [H, [F1, F2]] = 0</p><p>79</p><p>80 CHAPTER 17. EXERCISE 32 - DOUGLAS OLIVEIRA NOVAES</p><p>If F2 is a COM and all are time-independent, so [F2, H] = 0</p><p>From F2 and H given</p><p>dF2</p><p>dt</p><p>=</p><p>∂F2</p><p>∂q1︸︷︷︸</p><p>q2</p><p>∂H</p><p>∂p1︸︷︷︸</p><p>q1</p><p>−</p><p>�</p><p>�</p><p>�</p><p>��></p><p>0</p><p>∂F2</p><p>∂p1</p><p>∂H</p><p>∂q1</p><p>+</p><p>∂F2</p><p>∂q2︸︷︷︸</p><p>q1</p><p>∂H</p><p>∂p2︸︷︷︸</p><p>−q2</p><p>−</p><p>�</p><p>�</p><p>�</p><p>��></p><p>0</p><p>∂F2</p><p>∂p2</p><p>∂H</p><p>∂q2</p><p>dF2</p><p>dt</p><p>= 0</p><p>This implies that F2 is a COM, and Jacobi’s identity come</p><p>[F2, [H,F1]] + [H, [F1, F2]] = 0</p><p>Now, for [H,F1]</p><p>[H,F1] =</p><p>∂H</p><p>∂q1</p><p>∂F1</p><p>∂p1</p><p>− ∂F1</p><p>∂p1</p><p>∂H</p><p>∂q1</p><p>+</p><p>�</p><p>�</p><p>�</p><p>��></p><p>0</p><p>∂F1</p><p>∂q2</p><p>∂H</p><p>∂p2</p><p>− ∂F1</p><p>∂p2</p><p>∂H</p><p>∂q2</p><p>= (p1 − 2aq1)</p><p>1</p><p>q2</p><p>+ q1</p><p>a</p><p>q2</p><p>+</p><p>��������:0</p><p>(−p2 + 2bq2) 0 − (−ZZq2 )</p><p>(</p><p>− 1</p><p>qA22</p><p>(p1 − aq1)</p><p>)</p><p>=</p><p>1</p><p>q2</p><p>(��p1 −HHH2aq1 +HHaq1 −��p1 +HHaq1 )</p><p>[H,F1] = 0</p><p>which implies that</p><p>[H, [F1, F2]] = 0</p><p>id est, the Poisson bracket is itself a COM.</p><p>So, it says if one has two COMs other than the Hamiltonian, then the mutual</p><p>Poisson bracket is another COM and, then, one can use that to create more Poisson</p><p>brackets and so on, and therefore an algebraic structure is evolved.</p><p>Exercise 33 - JHORDAN SILVEIRA DE</p><p>BORBA</p><p>Set up the magnetic monopole described in Exercise 28 (Chapter 3) in Hamil-</p><p>tonian formulation (you may want to use spherical coordinates). By means of Poison</p><p>Bracket formulation, show that the quantity D defined in that exercise is conserved.</p><p>Answer: In exercise 28 we have a magnetic field singularity defined by B = br</p><p>r3</p><p>,</p><p>and the quantity D is given by:</p><p>D = L− qb</p><p>c</p><p>r</p><p>r</p><p>By denoting the Poisson bracket simply as [u, v]q,p = {u, v} we want to demonstrate</p><p>that {H,Di} = 0. According to Liouville’s theorem, this result tells us that D is a</p><p>function of constants of motion, so it is itself conserved. From equation 8.35 in the</p><p>book, we have that the Hamiltonian of a single particle of massm, charge q and moving</p><p>at a nonrelativistic velocity v in an electromagnetic field is given (in the CGS system)</p><p>per:</p><p>H =</p><p>1</p><p>2m</p><p>(</p><p>p− q</p><p>c</p><p>A</p><p>)2</p><p>+ qϕ</p><p>=</p><p>1</p><p>2</p><p>mṙ2 + qϕ</p><p>Where we can see that ṙ = 1</p><p>m</p><p>(</p><p>p− q</p><p>c</p><p>A</p><p>)</p><p>. We also have ϕ (t, q1, q2, q3) and A (t, q1, q2, q3),</p><p>the scalar and vector potential specifically. And as the scalar potential ϕ describes</p><p>only the electric field generated by electric charges, which do not exist in the proposed</p><p>81</p><p>82 CHAPTER 18. EXERCISE 33 - JHORDAN SILVEIRA DE BORBA</p><p>problem, we are left with just a Hamiltonian describing a particle in a magnetic field:</p><p>H =</p><p>1</p><p>2</p><p>mṙ2</p><p>The Poisson bracket is defined as:</p><p>{u, v} =</p><p>N∑</p><p>i=1</p><p>(</p><p>∂u</p><p>∂qi</p><p>∂v</p><p>∂pi</p><p>− ∂u</p><p>∂pi</p><p>∂v</p><p>∂qi</p><p>)</p><p>So some useful and easily to verify properties are:</p><p>• {f, f} = 0</p><p>• {f, g} = −{g, f}</p><p>• {αf + βg, h} = α {f, h}+ β {g, h} , α, β ∈ R</p><p>• {fg, h} = f {g, h}+ g {f, h}</p><p>• {f, {g, h}}+ {h, {f, g}}+ {g, {h, f}} = 0</p><p>So we need to calculate some results. First some Poisson brackets involving only posi-</p><p>tion and/or velocity:</p><p>{mṙi,mṙj} =</p><p>{</p><p>pi −</p><p>q</p><p>c</p><p>Ai, pj −</p><p>q</p><p>c</p><p>Aj</p><p>}</p><p>=</p><p>{</p><p>pi −</p><p>q</p><p>c</p><p>Ai, pj</p><p>}</p><p>−</p><p>{</p><p>pi −</p><p>q</p><p>c</p><p>Ai, qAj</p><p>}</p><p>= {pi, pj} −</p><p>q</p><p>c</p><p>{Ai, pj} −</p><p>q</p><p>c</p><p>{pi, Aj}+</p><p>q2</p><p>c2</p><p>{Ai, Aj}</p><p>= {pi, pj} −</p><p>q</p><p>c</p><p>{Ai, pj}+</p><p>q</p><p>c</p><p>{Aj, pi}+</p><p>q2</p><p>c2</p><p>{Ai, Aj}</p><p>The first and last terms disappear, as they depend only on pi or qi respectively.</p><p>And for those in the middle we simply have to:</p><p>{Ai, pj} =</p><p>N∑</p><p>i=1</p><p>(</p><p>∂Ai</p><p>∂qi</p><p>∂pj</p><p>∂pi</p><p>− ∂Ai</p><p>∂pi</p><p>∂pj</p><p>∂qi</p><p>)</p><p>=</p><p>N∑</p><p>i=1</p><p>∂Ai</p><p>∂qi</p><p>δij =</p><p>∂Ai</p><p>∂qj</p><p>(18.1)</p><p>Then:</p><p>{mṙi,mṙj} = −</p><p>q</p><p>c</p><p>∂Ai</p><p>∂qj</p><p>+</p><p>q</p><p>c</p><p>∂Aj</p><p>∂qi</p><p>=</p><p>q</p><p>c</p><p>ϵijk (∇×A)k</p><p>=</p><p>q</p><p>c</p><p>ϵijkBk</p><p>83</p><p>Where we use the Maxwell equation B = ∇ × A . And the fact that if i = j then</p><p>{Ai, pj} = 0 or {Ai, pj} = −{pj, Ai}, this is inserted through the Levi-Civita symbol.</p><p>Furthermore:</p><p>{ri,mṙj} =</p><p>{</p><p>ri, pj −</p><p>q</p><p>c</p><p>Aj</p><p>}</p><p>= {ri, pj} −</p><p>q</p><p>c</p><p>{ri, Aj}</p><p>=</p><p>∂ri</p><p>∂qj</p><p>=</p><p>∂qi</p><p>∂qj</p><p>= δij</p><p>Where we use the result obtained in the equation 18.1, and remembering the position</p><p>vector r = (q1, q2, q3) , we also have a null second Poisson bracket as both depend only</p><p>on position. So far we get:</p><p>{mṙi,mṙj} =</p><p>q</p><p>c</p><p>ϵijkBk {ri,mṙj} = δij (18.2)</p><p>Calculating now a brackets involving the angular momentum L = r ×mṙ and using</p><p>the result obtained in 18.2: And:</p><p>{Li,mṙa} =</p><p>{</p><p>3∑</p><p>j,k=1</p><p>ϵijkrjmṙk,mṙa</p><p>}</p><p>=</p><p>3∑</p><p>j,k=1</p><p>ϵijk {rjmṙk,mṙa}</p><p>=</p><p>3∑</p><p>j,k=1</p><p>ϵijk (rj {mṙk,mṙa}+mṙk {rj,mṙa})</p><p>=</p><p>3∑</p><p>j,k=1</p><p>ϵijk</p><p>(</p><p>rj</p><p>q</p><p>c</p><p>ϵkalBl +mṙkδja</p><p>)</p><p>=</p><p>3∑</p><p>j,k=1</p><p>ϵijkϵkal</p><p>q</p><p>c</p><p>rjBl +</p><p>3∑</p><p>j,k=1</p><p>ϵijkmṙkδja</p><p>=</p><p>3∑</p><p>j,k=1</p><p>ϵijkϵkal</p><p>q</p><p>c</p><p>rjBl + ϵiakmṙk</p><p>Now we have:</p><p>ϵijkϵlmn = det</p><p>∣∣∣∣∣∣</p><p>δil δim δin</p><p>δjl δjm δjn</p><p>δkl δkm δkn</p><p>∣∣∣∣∣∣</p><p>84 CHAPTER 18. EXERCISE 33 - JHORDAN SILVEIRA DE BORBA</p><p>And for our case, since we �xed an equal k on the two symbols, and that k is different</p><p>from all the others: i; j; a; l :</p><p>� ijk � kal = det</p><p>�</p><p>�</p><p>�</p><p>�</p><p>�</p><p>�</p><p>� ik � ia � il</p><p>� jk � ja � jl</p><p>� kk � ka � kl</p><p>�</p><p>�</p><p>�</p><p>�</p><p>�</p><p>�</p><p>= det</p><p>�</p><p>�</p><p>�</p><p>�</p><p>�</p><p>�</p><p>0 � ia � il</p><p>0 � ja � jl</p><p>1 0 0</p><p>�</p><p>�</p><p>�</p><p>�</p><p>�</p><p>�</p><p>= � ia � jl � � ja � il</p><p>So :</p><p>f L i ; m _rag =</p><p>3X</p><p>j;k =1</p><p>(� ia � jl � � ja � il )</p><p>q</p><p>c</p><p>r j B l + � iak m _r k</p><p>=</p><p>3X</p><p>j;k =1</p><p>� ia � jl</p><p>q</p><p>c</p><p>r j B l �</p><p>3X</p><p>j;k =1</p><p>� ja � il</p><p>q</p><p>c</p><p>r j B l + � iak m _r k</p><p>=</p><p>3X</p><p>j</p><p>� ia</p><p>q</p><p>c</p><p>r j B j �</p><p>q</p><p>c</p><p>B i ra + � iak m _r k</p><p>= � ia</p><p>q</p><p>c</p><p>3X</p><p>j</p><p>r j B j</p><p>!</p><p>�</p><p>q</p><p>c</p><p>B i ra + � iak m _r k</p><p>= � ia</p><p>q</p><p>c</p><p>(r � B ) �</p><p>q</p><p>c</p><p>B i ra + � iak m _r k</p><p>Or changing the notation, and using the �eld generated by the magnetic monopole</p><p>B = br</p><p>r 3 , we have:</p><p>f L i ; m _r j g = � ij</p><p>q</p><p>c</p><p>(r � B ) �</p><p>q</p><p>c</p><p>B i r j + � ijk m _r k (18.3)</p><p>= � ij</p><p>q</p><p>c</p><p>�</p><p>r �</p><p>br</p><p>r 3</p><p>�</p><p>�</p><p>q</p><p>c</p><p>bri</p><p>r 3</p><p>r j + � ijk m _r k (18.4)</p><p>= � ij</p><p>bq</p><p>r 3</p><p>(r � r ) �</p><p>q</p><p>c</p><p>bri</p><p>r 3</p><p>r j + � ijk m _r k (18.5)</p><p>= � ij</p><p>bq</p><p>cr3</p><p>r 2 �</p><p>q</p><p>c</p><p>bri</p><p>r 3</p><p>r j + � ijk m _r k (18.6)</p><p>=</p><p>bq</p><p>c</p><p>�</p><p>� ij</p><p>r</p><p>�</p><p>r i r j</p><p>r 3</p><p>�</p><p>+ � ijk m _r k (18.7)</p><p>85</p><p>Since r =</p><p>√</p><p>q2i + q2j + q2k, then:</p><p>{ri</p><p>r</p><p>,mṙj</p><p>}</p><p>=</p><p>{</p><p>r−1ri,mṙj</p><p>}</p><p>=</p><p>1</p><p>r</p><p>{ri,mṙj}+ ri</p><p>{(</p><p>q2i + q2j + q2k</p><p>)−1/2</p><p>,mṙj</p><p>}</p><p>=</p><p>δij</p><p>r</p><p>+ ri</p><p>{(</p><p>q2i + q2j + q2k</p><p>)−1/2</p><p>,m</p><p>1</p><p>m</p><p>(pj − qAj)</p><p>}</p><p>=</p><p>δij</p><p>r</p><p>+ ri</p><p>({(</p><p>q2i + q2j + q2k</p><p>)−1/2</p><p>, pj</p><p>}</p><p>− q</p><p>{(</p><p>q2i + q2j + q2k</p><p>)−1/2</p><p>, Aj</p><p>})</p><p>=</p><p>δij</p><p>r</p><p>+ ri</p><p>{(</p><p>q2i + q2j + q2k</p><p>)−1/2</p><p>, pj</p><p>}</p><p>=</p><p>δij</p><p>r</p><p>+ ri</p><p>n∑</p><p>l=1</p><p>(</p><p>∂</p><p>∂ql</p><p>(</p><p>q2i + q2j + q2k</p><p>)−1/2 ∂pj</p><p>∂pl</p><p>− ∂</p><p>∂pl</p><p>((</p><p>q2i + q2j + q2k</p><p>)−1/2</p><p>) ∂pj</p><p>∂ql</p><p>)</p><p>=</p><p>δij</p><p>r</p><p>+ ri</p><p>n∑</p><p>k=1</p><p>(</p><p>∂</p><p>∂qk</p><p>(</p><p>q2i + q2j + q2k</p><p>)−1/2</p><p>δjl</p><p>)</p><p>=</p><p>δij</p><p>r</p><p>+ ri</p><p>∂</p><p>∂qj</p><p>(</p><p>q2i + q2j + q2k</p><p>)−1/2</p><p>=</p><p>δij</p><p>r</p><p>− ri</p><p>1</p><p>2</p><p>(</p><p>q2i + q2j + q2k</p><p>)−3/2</p><p>2qj</p><p>=</p><p>δij</p><p>r</p><p>− riqj</p><p>r3</p><p>Where we use the fact that</p><p>{</p><p>1</p><p>r</p><p>, Aj</p><p>}</p><p>= 0 because it depend only on the position, so we</p><p>have simply</p><p>{</p><p>ri</p><p>r</p><p>,mṙj</p><p>}</p><p>=</p><p>δij</p><p>r</p><p>− riqj</p><p>r3</p><p>. Then replacing in 18.7:</p><p>{Li,mṙj} =</p><p>bq</p><p>c</p><p>{ri</p><p>r</p><p>,mṙj</p><p>}</p><p>+ ϵijkmṙk (18.8)</p><p>And calculating the Poisson bracket of mṙj with Di:</p><p>{Di,mṙj} =</p><p>{</p><p>Li −</p><p>qb</p><p>c</p><p>ri</p><p>r</p><p>,mṙj</p><p>}</p><p>= {Li,mṙj} −</p><p>qb</p><p>c</p><p>{ri</p><p>r</p><p>,mṙj</p><p>}</p><p>=</p><p>bq</p><p>c</p><p>{ri</p><p>r</p><p>,mṙj</p><p>}</p><p>+ ϵiakmṙk −</p><p>qb</p><p>c</p><p>{ri</p><p>r</p><p>,mṙj</p><p>}</p><p>= ϵijkmṙk</p><p>86 CHAPTER 18. EXERCISE 33 - JHORDAN SILVEIRA DE BORBA</p><p>Then calculating the Hamiltonian:</p><p>{H,Di} =</p><p>�</p><p>1</p><p>2</p><p>m _r 2, Di</p><p>�</p><p>=</p><p>�</p><p>1</p><p>2</p><p>mṙ21 +</p><p>1</p><p>2</p><p>mṙ22 +</p><p>1</p><p>2</p><p>mṙ23, Di</p><p>�</p><p>=</p><p>1</p><p>2</p><p>�</p><p>mṙ21, Di</p><p>+</p><p>1</p><p>2</p><p>�</p><p>mṙ22, Di</p><p>+</p><p>1</p><p>2</p><p>�</p><p>mṙ23, Di</p><p>=</p><p>1</p><p>2</p><p>�</p><p>mṙ21, Di</p><p>+</p><p>1</p><p>2</p><p>�</p><p>mṙ22, Di</p><p>+</p><p>1</p><p>2</p><p>�</p><p>mṙ23, Di</p><p>=</p><p>�</p><p>ṙ1</p><p>2</p><p>{mṙ1, Di}+</p><p>ṙ1</p><p>2</p><p>{mṙ1, Di}</p><p>�</p><p>+</p><p>�</p><p>ṙ2</p><p>2</p><p>{mṙ2, Di}+</p><p>ṙ2</p><p>2</p><p>{mṙ2, Di}</p><p>�</p><p>+</p><p>�</p><p>ṙ3</p><p>2</p><p>{mṙ3, Di}+</p><p>ṙ3</p><p>2</p><p>{mṙ3, Di}</p><p>�</p><p>= ṙ1 {mṙ1, Di}+ ṙ2 {mṙ2, Di}+ ṙ3 {mṙ3, Di}</p><p>= ṙ1ϵ1ikmṙk + ṙ2ϵ2ik + ṙ3ϵ3ik</p><p>So for any value of i , for example, choosing i = 1:</p><p>{H,D1} = ṙ1ϵ11kmṙk + ṙ2ϵ213ṙ3 + ṙ3ϵ312ṙ2</p><p>= −ṙ2ṙ3 + ṙ3ṙ2</p><p>= 0</p><p>We have analogous results for any choice of i, which is the result we were looking for.</p><p>Which brings us to:</p><p>{H,D } = 0</p><p>Since D is a constant of motion, we can then calculate r</p><p>r</p><p>· D to obtain that the trajec-</p><p>tory of the particle lies at a cone. This calculation goes beyond what was required,</p><p>but it simply follows from the fact that while we have r</p><p>r</p><p>· D = − qb</p><p>c</p><p>, we also have that</p><p>r</p><p>r</p><p>·D = D cos θ, then it remains to compare the results. An interesting demo of the trajec-</p><p>tory can be accessed in the Wolfram Demonstration Poject: <https://demonstrations.</p><p>wolfram.com/ClassicalElectronInTheFieldOfAMagneticMonopole/>.</p><p>Exercise 35 - JEFFERSON SANTANA</p><p>MARTINS</p><p>A particle moves in one dimension under a potential V = mk</p><p>x2 . Find x as a func-</p><p>tion of time by using the symbolic solution of the Poisson bracket form for the equation</p><p>of motion for the quantity y = x2 and initial conditions at t = 0 and v = 0.</p><p>Solution</p><p>The Hamiltonian for this one dimensional system is</p><p>H =</p><p>p2</p><p>2m</p><p>+</p><p>mk</p><p>x2</p><p>(19.1)</p><p>We want to find y(t), where y is the quantity x2, using the Poisson bracket oper-</p><p>ator formalism. The formal solution is</p><p>y(t) = y0 + (H y)0t+</p><p>1</p><p>2</p><p>(H 2y)0t</p><p>2 +</p><p>1</p><p>3!</p><p>(H 3y)0t</p><p>3 + · · · (19.2)</p><p>where H is the action of the operator H on y is defined as</p><p>H y = [y,H]· (19.3)</p><p>With y = x2, the explicit form of Eq.(19.3) is</p><p>H x2 =</p><p>[</p><p>x2, H</p><p>]</p><p>=</p><p>1</p><p>2m</p><p>[</p><p>x2, p2</p><p>]</p><p>=</p><p>2xp</p><p>m</p><p>· (19.4)</p><p>The next term in the series is</p><p>H x2 = H (H x2) =</p><p>[</p><p>H x2, H</p><p>]</p><p>=</p><p>2</p><p>m</p><p>[xp,H] · (19.5)</p><p>87</p><p>88 CHAPTER 19. EXERCISE 35 - JEFFERSON SANTANA MARTINS</p><p>Proceeding to evaluate this we �rst note that</p><p>[xp; H ] = x[p; H] + p[x; H ] = x[p;</p><p>1</p><p>x2</p><p>]mk +</p><p>p</p><p>2m</p><p>[x; p2]; (19.6)</p><p>and then that</p><p>[p;</p><p>1</p><p>x2</p><p>] = �</p><p>@</p><p>@x</p><p>1</p><p>x2</p><p>=</p><p>2</p><p>x3</p><p>; (19.7)</p><p>so that</p><p>[xp; H ] =</p><p>2mk</p><p>x2</p><p>+</p><p>p2</p><p>m</p><p>� (19.8)</p><p>If we compare this result with Eq.(19.1) we see that</p><p>[xp; H ] = 2H; (19.9)</p><p>and Eq.(19.5) results in</p><p>H 2x2 =</p><p>4</p><p>m</p><p>H � (19.10)</p><p>It should be clear from this result that H nx2 = 0 for n � 3. The series in Eq.(19.2) thus</p><p>terminates, and we can write the full solution as</p><p>x2(t) = x2</p><p>0 +</p><p>2x0p0</p><p>m</p><p>t +</p><p>2</p><p>m</p><p>H0t2 (19.11)</p><p>After applying the initial condition p0 = 0 to Eq.(19.11), we �nd the desired special</p><p>solution</p><p>x(t) = �</p><p>r</p><p>x2</p><p>0 +</p><p>2</p><p>m</p><p>H0t2 � (19.12)</p><p>Exercise 36 - EDGARD</p><p>KRETSCHMANN</p><p>For this problem, we can start by expanding the dot product.</p><p>[F � L ; G � L ] = [</p><p>X</p><p>i</p><p>Fi L i ;</p><p>X</p><p>j</p><p>Gj L j ] (20.1)</p><p>Using the linearity of the Poisson Brackets and the product between functions in the</p><p>Poisson Brackets properties we can expand the right-hand of the equation (1) as</p><p>X</p><p>i</p><p>X</p><p>j</p><p>Fi Gj [L i ; L j ] + Fi L j [L i ; Gj ] + L i Gj [Fi ; L j ] + L i L j [Fi ; Gj ] (20.2)</p><p>Using, now, the property of the angular momentum in the Poisson Brackets:</p><p>[Fi ; L j ] = � ijk Fk (20.3)</p><p>The expansion on equation (2) can become</p><p>X</p><p>i</p><p>X</p><p>j</p><p>Fi Gj � ijk L k � Fi L j � jik Gk + L i Gj � ijk Fk + L i L j [Fi ; Gj ] (20.4)</p><p>Rearranging the subscripts indices in order to group the equivalent terms in the sum,</p><p>this expansion becomes</p><p>X</p><p>i</p><p>X</p><p>j</p><p>� Fi Gj L k � ijk + L i L j [Fi ; Gj ] (20.5)</p><p>So, the �rst term of the sum is the triple product L � (G � F). So we can show that</p><p>[F � L ; G � L ] = L � (G � F) +</p><p>X</p><p>i</p><p>X</p><p>j</p><p>L i L j [Fi ; Gj ] (20.6)</p><p>89</p><p>90 CHAPTER 20. EXERCISE 36 - EDGARD KRETSCHMANN</p><p>If we use F and G as unit vectors in the direction of the perpendicular axes, the expan-</p><p>sion above will return that</p><p>[Lµ, Lν ] = L · (eν × eµ) +</p><p>∑</p><p>i</p><p>∑</p><p>j</p><p>LiLj[ei, ej] (20.7)</p><p>The last term will be null because of the anti-symmetric property of the Poisson Brack-</p><p>ets. So the expansion will become</p><p>[Lµ, Lν ] = ϵνµτLτ (20.8)</p><p>Where ϵνµτLτ = L · (eν × eµ)</p><p>Exercise 37 - GEOVANE NAYSINGER</p><p>FIGURE 21.1</p><p>The hamiltonian of the sistem is</p><p>H =</p><p>P 2</p><p>θ</p><p>2mb2</p><p>+</p><p>P 2</p><p>ψ</p><p>2mb2sen2θ</p><p>−mgbcosθ (21.1)</p><p>and there is two ways to solve this problem (that I know). One of them is using the</p><p>propertie {F,H} = ∂F</p><p>∂t</p><p>or using the simmetry of the problem as Julian Schwinger did.</p><p>I’ll use the second option by simplicity and generality. So, the two coordinates trans-</p><p>formate is</p><p>x0= xcosψ + ysenψ</p><p>y0= −xsenψ + ycosψ,</p><p>As the vector L⃗ is a translation generator and a rotation around the z-axis is generate</p><p>by the Lz. So, assuming that the position vector is rotate by a dθ around the z-axis,</p><p>91</p><p>92 CHAPTER 21. EXERCISE 37 - GEOVANE NAYSINGER</p><p>substituting the ψ to dθ we have</p><p>X = x− ydθ</p><p>Y = xdθ + y</p><p>Z = z</p><p>then the variation is</p><p>δx = −ydθ</p><p>δy = xdθ</p><p>δz = 0.</p><p>We see that in sympletic notation that δz = ϵ {z, G}where G is a function of (q,p,t) and</p><p>is called as canonical transformation generator. Then, we can conclude that</p><p>{x, Lz} = −y</p><p>{y, Lz} = x</p><p>{z, Lz} = 0.</p><p>If we have some variables qi, qj, qk perpendicular which other and obeing the right-</p><p>hand rule, by simmetry we have {Li, Lj} = ϵijkLk. Looking to the previous three</p><p>equations, we can rewrite them as</p><p>{Lx, Lz} = −Ly</p><p>{Ly, Lz} = Lx</p><p>{Lz, Lz} = 0</p><p>and</p><p>{Lx, Ly} = Lz</p><p>{Lz, Lx} = Ly.</p><p>The arguement also show that, by geommetry, Pθ and Pψ are perpendicular compo-</p><p>nents of angular momenta in this problem.</p><p>Exercise 41 - GUILHERME SHOITI</p><p>YOSHIMATSU GIARDINI</p><p>We start with a time independent Hamiltonian H 0(q, p) and impose an external</p><p>oscillating field making the Hamiltonian</p><p>H = H 0(q, p)− ϵ sin(ω t) where (22.1)</p><p>ϵ and ω are given constants.</p><p>22.1 (a)</p><p>How are the canonical equations modified?</p><p>22.1.1 Solution (a)</p><p>We may write the Hamiltonian as</p><p>H (q⃗, p⃗, t) = ˙⃗q · p⃗− L (q⃗, ˙⃗q, t)</p><p>= q̇ipi − L (q⃗, ˙⃗q, t) where (22.2)</p><p>its derivative is</p><p>dH = q̇idpi − pidq̇i − dL (q⃗, ˙⃗q, t) then (22.3)</p><p>dL (q⃗, ˙⃗q, t) =</p><p>∂L</p><p>∂qi</p><p>dqi +</p><p>∂L</p><p>∂dq̇i</p><p>dq̇i +</p><p>∂L</p><p>∂t</p><p>dt where (22.4)</p><p>93</p><p>94 CHAPTER 22. EXERCISE 41 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>∂L</p><p>∂qi</p><p>= ṗi and (22.5)</p><p>∂L</p><p>∂q̇i</p><p>= pi , (22.6)</p><p>this way, the variation of the Hamiltonian may be written as</p><p>dH = q̇idpi − ṗidqi −</p><p>∂L</p><p>∂t</p><p>dt but (22.7)</p><p>the variation of the Hamiltonian can also be written as</p><p>dH =</p><p>∂H</p><p>∂qi</p><p>dqi +</p><p>∂H</p><p>∂pi</p><p>dpi +</p><p>∂H</p><p>∂t</p><p>dt , (22.8)</p><p>which results in the Hamilton dynamic equations</p><p>q̇i =</p><p>∂H</p><p>∂pi</p><p>(22.9)</p><p>−ṗi =</p><p>∂H</p><p>∂qi</p><p>(22.10)</p><p>−∂L</p><p>∂t</p><p>=</p><p>∂H</p><p>∂t</p><p>thus (22.11)</p><p>if we try to apply these equations above on the modified Hamiltonian we will obtain</p><p>∂H</p><p>∂pi</p><p>=</p><p>∂</p><p>∂pi</p><p>(H 0(q⃗, p⃗)− � � � � �</p><p>ϵ sin(ω t))</p><p>=</p><p>∂H 0(q⃗, p⃗)</p><p>∂pi</p><p>= q̇i , (22.12)</p><p>∂H</p><p>∂qi</p><p>=</p><p>∂</p><p>∂qi</p><p>(H 0(q⃗, p⃗)− � � � � �</p><p>ϵ sin(ω t))</p><p>=</p><p>∂H 0(q⃗, p⃗)</p><p>∂qi</p><p>(22.13)</p><p>= −ṗi and (22.14)</p><p>∂H</p><p>∂t</p><p>= −∂L</p><p>∂t</p><p>= −ϵ ω cos(ω t) . (22.15)</p><p>Seemingly, Hamilton’s equations are kept unmodified, but we must verify why</p><p>and if it is indeed true.</p><p>22.2. (B) 95</p><p>22.2 (b)</p><p>Find a canonical transformation that restores the canonical form of the equations</p><p>of motion and determine the "new" Hamiltonian.</p><p>22.2.1 Solution (b)</p><p>To find another set of canonical coordinates and also a "new" Hamiltonian, we</p><p>may use the principle of least action, where</p><p>δ</p><p>Z</p><p>(pidqi − H dt) = 0 where (22.16)</p><p>we read pidqi as a sum</p><p>P</p><p>i pidqi according to Einstein’s index notation. Thus, we know</p><p>that a different Hamiltonian K found through canonical transformations must also</p><p>follow this principle of least action</p><p>δ</p><p>Z</p><p>(PidQi − K dt) = 0 , (22.17)</p><p>in other words, it is the same as saying that both actions are related by a twice derivable</p><p>and continuous function F , in such a way that</p><p>pidqi − H dt = PidQi − K dt+ dF . (22.18)</p><p>If F = 0, then obviously K = H , now, we may represent the aforementioned</p><p>function as</p><p>dF = pidqi − PidQi − (H − K ) dt , (22.19)</p><p>according to this equation we found, the function depends on (qi, Qi, t). Thus, we may</p><p>write its variation as</p><p>dF (q,Q, t) =</p><p>∂F</p><p>∂qi</p><p>dqi +</p><p>∂F</p><p>∂Qi</p><p>dQi +</p><p>∂F</p><p>∂t</p><p>dt , (22.20)</p><p>which is equivalent to</p><p>pi =</p><p>∂F</p><p>∂qi</p><p>(22.21)</p><p>−Pi =</p><p>∂F</p><p>∂Qi</p><p>(22.22)</p><p>(K − H ) =</p><p>∂F</p><p>∂t</p><p>. (22.23)</p><p>96 CHAPTER 22. EXERCISE 41 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>It is also possible to find other "kinds" of F that depend on other variables, for</p><p>example F 0(q, P, t), it is possible through a Legendre transform of F (q,Q, t).</p><p>In the case of this question, we may suppose a function f(t) = dF , that depends</p><p>only on time, this way the equations that we derived above can be modified as</p><p>dF = pidqi − PidQi − (H −K) dt , (22.24)</p><p>but if dF (t) depends only on time, we get</p><p>dF (t) =</p><p>∂F</p><p>∂t</p><p>dt , (22.25)</p><p>which implies</p><p>K = H +</p><p>∂F</p><p>∂t</p><p>= H0 − ϵ sin(ω t) +</p><p>∂F</p><p>∂t</p><p>and (22.26)</p><p>pidqi = PidQi . (22.27)</p><p>This in turns implies that we may choose such a function F where</p><p>∂F</p><p>∂t</p><p>= ϵ sin(ω t) , (22.28)</p><p>in a way where</p><p>K = H0 and (22.29)</p><p>q → Q (22.30)</p><p>p → P or (22.31)</p><p>any other kind of transformation that respects</p><p>pidqi = PidQi . (22.32)</p><p>The correctness of this conclusion (that we may create a new Hamiltonian</p><p>"Kamiltonian" that simply ignores the time part −ϵ sin(ω t)) can be verified through</p><p>the use of the Legendre transform from H to L, such that</p><p>L = pq̇ −H</p><p>= pq̇ −H + f(t) . (22.33)</p><p>22.3. (C) 97</p><p>Where we know that when the Lagrangian is invariant to the addition of a function</p><p>that depends only on time, just like in item (a), that shows that the Hamilton equations</p><p>stay the same irrespective of the time part. It is as this, because the addition of an</p><p>arbitrary function f0(q, t) to the Lagrangian, will change the resulting action only at</p><p>the borders of the hyper-path of the path integral that gives the variation of action</p><p>δS = δ</p><p>Z 2</p><p>1</p><p>L(q, q̇, t)dt where (22.34)</p><p>q may be written as q(t, α) = q(t, 0) + αg(t) and α is one such parameter that when</p><p>varied, makes the path integral of L larger or smaller. Now for a different Lagrangian</p><p>L0, we will have</p><p>δS0 = δ</p><p>Z 2</p><p>1</p><p>�</p><p>L(q, q̇, t) +</p><p>∂F (q, t)</p><p>∂t</p><p>�</p><p>dt</p><p>= δ</p><p>�</p><p>S +</p><p>Z 2</p><p>1</p><p>∂F (q, t)</p><p>∂t</p><p>dt</p><p>�</p><p>= δS + δ (F (q2, t2)− F (q1, t1))</p><p>= δS . (22.35)</p><p>By definition, the endpoints q1, t1 and q2, t2, of the path integral, are kept fixed,</p><p>which implies that a variation of fixed terms is zero, proving that the Lagrangian is</p><p>invariant under the addition of a total derivative of a function that depends only on q</p><p>and t.</p><p>22.3 (c)</p><p>Give a possible interpretation of the imposed field.</p><p>22.3.1 solution (c)</p><p>A possible interpretation to the given field, could be any system with more than</p><p>one degree of freedom, where it is subject to an oscillatory motion in one direction</p><p>and has a dynamic given by a constant in time Hamiltonian H0 in the other and these</p><p>movement could be kept independent through some form of constraints, for example,</p><p>98 CHAPTER 22. EXERCISE 41 - GUILHERME SHOITI YOSHIMATSU GIARDINI</p><p>a ball constrained to a plate that goes up and down in a sinusoidal movement, the</p><p>ball’s motion, because it is forced to stay on the plate, all of its motions over the plate</p><p>will be independent of the vertical oscillating motion.</p><p>Bibliography</p><p>99</p><p>View publication stats</p><p>https://www.researchgate.net/publication/358249174</p><p>pbs@ARFix@1:</p><p>pbs@ARFix@2:</p><p>pbs@ARFix@3:</p><p>pbs@ARFix@4:</p><p>pbs@ARFix@5:</p><p>pbs@ARFix@6:</p><p>pbs@ARFix@7:</p><p>pbs@ARFix@8:</p><p>pbs@ARFix@9:</p><p>pbs@ARFix@10:</p><p>pbs@ARFix@11:</p><p>pbs@ARFix@12:</p><p>pbs@ARFix@13:</p><p>pbs@ARFix@14:</p><p>pbs@ARFix@15:</p><p>pbs@ARFix@16:</p><p>pbs@ARFix@17:</p><p>pbs@ARFix@18:</p><p>pbs@ARFix@19:</p><p>pbs@ARFix@20:</p><p>pbs@ARFix@21:</p><p>pbs@ARFix@22:</p><p>pbs@ARFix@23:</p><p>pbs@ARFix@24:</p><p>pbs@ARFix@25:</p><p>pbs@ARFix@26:</p><p>pbs@ARFix@27:</p><p>pbs@ARFix@28:</p><p>pbs@ARFix@29:</p><p>pbs@ARFix@30:</p><p>pbs@ARFix@31:</p><p>pbs@ARFix@32:</p><p>pbs@ARFix@33:</p><p>pbs@ARFix@34:</p><p>pbs@ARFix@35:</p><p>pbs@ARFix@36:</p><p>pbs@ARFix@37:</p><p>pbs@ARFix@38:</p><p>pbs@ARFix@39:</p><p>pbs@ARFix@40:</p><p>pbs@ARFix@41:</p><p>pbs@ARFix@42:</p><p>pbs@ARFix@43:</p><p>pbs@ARFix@44:</p><p>pbs@ARFix@45:</p><p>pbs@ARFix@46:</p><p>pbs@ARFix@47:</p><p>pbs@ARFix@48:</p><p>pbs@ARFix@49:</p><p>pbs@ARFix@50:</p><p>pbs@ARFix@51:</p><p>pbs@ARFix@52:</p><p>pbs@ARFix@53:</p><p>pbs@ARFix@54:</p><p>pbs@ARFix@55:</p><p>pbs@ARFix@56:</p><p>pbs@ARFix@57:</p><p>pbs@ARFix@58:</p><p>pbs@ARFix@59:</p><p>pbs@ARFix@60:</p><p>pbs@ARFix@61:</p><p>pbs@ARFix@62:</p><p>pbs@ARFix@63:</p><p>pbs@ARFix@64:</p><p>pbs@ARFix@65:</p><p>pbs@ARFix@66:</p><p>pbs@ARFix@67:</p><p>pbs@ARFix@68:</p><p>pbs@ARFix@69:</p><p>pbs@ARFix@70:</p><p>pbs@ARFix@71:</p><p>pbs@ARFix@72:</p><p>pbs@ARFix@73:</p><p>pbs@ARFix@74:</p><p>pbs@ARFix@75:</p><p>pbs@ARFix@76:</p><p>pbs@ARFix@77:</p><p>pbs@ARFix@78:</p><p>pbs@ARFix@79:</p><p>pbs@ARFix@80:</p><p>pbs@ARFix@81:</p><p>pbs@ARFix@82:</p><p>pbs@ARFix@83:</p><p>pbs@ARFix@84:</p><p>pbs@ARFix@85:</p><p>pbs@ARFix@86:</p><p>pbs@ARFix@87:</p><p>pbs@ARFix@88:</p><p>pbs@ARFix@89:</p><p>pbs@ARFix@90:</p><p>pbs@ARFix@91:</p><p>pbs@ARFix@92:</p><p>pbs@ARFix@93:</p><p>pbs@ARFix@94:</p><p>pbs@ARFix@95:</p><p>pbs@ARFix@96:</p><p>pbs@ARFix@97:</p><p>pbs@ARFix@98:</p><p>pbs@ARFix@99:</p><p>pbs@ARFix@100:</p><p>pbs@ARFix@101:</p><p>pbs@ARFix@102:</p><p>pbs@ARFix@103:</p><p>pbs@ARFix@104:</p><p>pbs@ARFix@105:</p><p>pbs@ARFix@106:</p><p>pbs@ARFix@107:</p><p>pbs@ARFix@108:</p><p>pbs@ARFix@109:</p><p>pbs@ARFix@110:</p><p>pbs@ARFix@111:</p><p>pbs@ARFix@112:</p><p>pbs@ARFix@113:</p><p>pbs@ARFix@114:</p><p>pbs@ARFix@115:</p><p>pbs@ARFix@116:</p><p>pbs@ARFix@117:</p><p>pbs@ARFix@118:</p><p>pbs@ARFix@119:</p><p>pbs@ARFix@120:</p><p>pbs@ARFix@121:</p><p>pbs@ARFix@122:</p><p>pbs@ARFix@123:</p><p>pbs@ARFix@124:</p><p>pbs@ARFix@125:</p><p>pbs@ARFix@126:</p><p>pbs@ARFix@127:</p><p>pbs@ARFix@128:</p><p>pbs@ARFix@129:</p><p>pbs@ARFix@130:</p><p>pbs@ARFix@131:</p><p>pbs@ARFix@132:</p><p>pbs@ARFix@133:</p><p>pbs@ARFix@134:</p><p>pbs@ARFix@135:</p><p>pbs@ARFix@136:</p><p>pbs@ARFix@137:</p><p>pbs@ARFix@138:</p><p>pbs@ARFix@139:</p><p>pbs@ARFix@140:</p><p>g</p><p>dv</p><p>dm</p><p>= − v</p><p>′</p><p>m</p><p>+</p><p>60g</p><p>m0</p><p>dv = − v</p><p>′</p><p>m</p><p>dm+</p><p>60g</p><p>m0</p><p>dm</p><p>and integrating ∫</p><p>dv = −</p><p>∫ me</p><p>m0</p><p>v′</p><p>m</p><p>dm+</p><p>∫ me</p><p>m0</p><p>60g</p><p>m0</p><p>dm</p><p>v = −v′ln</p><p>(</p><p>me</p><p>m0</p><p>)</p><p>+</p><p>60g</p><p>m0</p><p>(me −m0)</p><p>and m0 = me +mf , so</p><p>v = −v′ln</p><p>(</p><p>me</p><p>me +mf</p><p>)</p><p>+ 60g</p><p>(</p><p>mf</p><p>me +mf</p><p>)</p><p>and remember that me << mf . Then we can aproximate the equation to</p><p>v = −v′ln</p><p>(</p><p>me</p><p>mf</p><p>)</p><p>+ 60g</p><p>(</p><p>mf</p><p>mf</p><p>)</p><p>v + 60g</p><p>v′</p><p>= ln</p><p>(</p><p>mf</p><p>me</p><p>)</p><p>,</p><p>39</p><p>So the ratio of the mass is</p><p>mf</p><p>me</p><p>= e</p><p>v+60g</p><p>v′ (13.1)</p><p>and putting the values 11,2km/s for v, 9,8 for g and 2,1 km/s for v’, we have</p><p>mf</p><p>me</p><p>= 274,</p><p>The best value to v’ is 2,07 that given to us a ratio of 296.</p><p>Exercise 14 - Edgard Kretschmann</p><p>Considering that the center of the masses moves in the plane xy we can write</p><p>the cartesian coordinates of the two point mass in terms of α, θ and φ, where α is the</p><p>angle between the position vector of the center and x-axis, θ is the angle between the</p><p>projection of the rod in the plane xy and the x-direction and φ is the angle between the</p><p>rod and the z-direction.</p><p>x1 = acos(α) +</p><p>l</p><p>2</p><p>cos(θ)sin(φ)</p><p>y1 = asin(α) +</p><p>l</p><p>2</p><p>sin(θ)sin(φ)</p><p>z1 =</p><p>l</p><p>2</p><p>cos(φ)</p><p>x2 = acos(α)− l</p><p>2</p><p>cos(θ)sin(φ)</p><p>y2 = asin(α)− l</p><p>2</p><p>sin(θ)sin(φ)</p><p>z2 = −</p><p>l</p><p>2</p><p>cos(φ)</p><p>The time derivatives of each cartesian coordinate are</p><p>ẋ1 = −aα̇sin(α)−</p><p>l</p><p>2</p><p>(θ̇sin(θ)sin(φ)− φ̇cos(θ)cos(φ))</p><p>ẏ1 = aα̇cos(α) +</p><p>l</p><p>2</p><p>(θ̇cos(θ)sin(φ) + φ̇sin(θ)cos(φ))</p><p>ż1 = −</p><p>l</p><p>2</p><p>φ̇sin(φ)</p><p>ẋ2 = −aα̇sin(α) +</p><p>l</p><p>2</p><p>(θ̇sin(θ)sin(φ)− φ̇cos(θ)cos(φ))</p><p>41</p><p>42 CHAPTER 14. EXERCISE 14 - EDGARD KRETSCHMANN</p><p>ẏ2 = aα̇cos(α)− l</p><p>2</p><p>(θ̇cos(θ)sin(φ) + φ̇sin(θ)cos(φ))</p><p>ż2 =</p><p>l</p><p>2</p><p>φ̇sin(φ)</p><p>The kinetic energy in the cartesian coordinates of this system can be written as</p><p>T =</p><p>m</p><p>2</p><p>(ẋ1</p><p>2 + ẏ1</p><p>2 + ż1</p><p>2 + ẋ2</p><p>2 + ẏ2</p><p>2 + ż2</p><p>2)</p><p>So, replacing the expressions of the velocities in the equation above, we will have the</p><p>kinetic energy in therms of the generalized coordinates</p><p>T =</p><p>m</p><p>2</p><p>(a2α̇2 +</p><p>l2</p><p>4</p><p>(θ̇2 + φ̇2sin2(φ))</p><p>Exercise 15 - Guilherme S. Y. Giardini</p><p>15.1 a) Cartesian Coordinates</p><p>Knowing that we have an arbitrary potential</p><p>U(~r,~v) = V (r) + ~σ × ~L (15.1)</p><p>irrespective of the coordinate system, it is necessary to first determine the angu-</p><p>lar momentum ~L, whose formula is</p><p>43</p><p>44 CHAPTER 15. EXERCISE 15 - GUILHERME S. Y. GIARDINI</p><p>~L = (~r � ~p) (15.2)</p><p>= m(~r � ~v) (15.3)</p><p>or in the basis of equation (1.58) of Herbert B. Goldstein 3rd Edition (index nota-</p><p>tion)</p><p>~L = m</p><p>N∑</p><p>i=0</p><p>êi</p><p>N∑</p><p>j</p><p>N∑</p><p>k</p><p>εijkqj q̇k , (15.4)</p><p>where the symbol εijk is the Levi-Civita symbol that results in</p><p>εijk =</p><p></p><p>1 if the indexes are even</p><p>� 1 if the indexes are odd</p><p>0 if any two or more symbols are equal</p><p>. (15.5)</p><p>Now we proceed to find each component of the force over the particle under the</p><p>influence of potential (15.1). The equation that results in said force is</p><p>~Q = � ~r qU(~r,~v) +</p><p>d</p><p>dt</p><p>[</p><p>~r ~̇qU(~r,~v)</p><p>]</p><p>(15.6)</p><p>or in the index notation</p><p>Ql = �</p><p>∂U</p><p>∂ql</p><p>+</p><p>d</p><p>dt</p><p>[</p><p>∂U</p><p>∂q̇l</p><p>]</p><p>, (15.7)</p><p>where the letter l represents each component of a coordinate system.</p><p>Considering a Cartesian coordinate system, the position of a particle is given by</p><p>15.1. A) CARTESIAN COORDINATES 45</p><p>~r =</p><p>∑</p><p>i</p><p>riêi ;</p><p>∂êi</p><p>∂t</p><p>= 0 ∀ i ∈ [1, 2, 3] and (15.8)</p><p>its velocity by</p><p>~v =</p><p>∂~r</p><p>∂t</p><p>=</p><p>∑</p><p>i</p><p>∂ri</p><p>∂t</p><p>êi + ri</p><p>∂êi</p><p>∂t</p><p>=</p><p>∑</p><p>i</p><p>∂ri</p><p>∂t</p><p>êi . (15.9)</p><p>Now it only remains to substitute the velocity and position values into potential</p><p>(15.1) and subsequently into force (15.7)</p><p>Ql = −∂U</p><p>∂ql</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>∂U</p><p>∂q̇l</p><p>)</p><p>= − ∂</p><p>∂ql</p><p>[</p><p>V (r) +m</p><p>N∑</p><p>i=0</p><p>σi</p><p>N∑</p><p>j</p><p>N∑</p><p>k</p><p>εijkqj q̇k</p><p>]</p><p>+</p><p>d</p><p>dt</p><p>[</p><p>∂</p><p>∂q̇l</p><p>(</p><p>V (r) +m</p><p>N∑</p><p>i=0</p><p>σi</p><p>N∑</p><p>j</p><p>N∑</p><p>k</p><p>εijkqj q̇k</p><p>)]</p><p>. (15.10)</p><p>We know that in cartesian coordinates</p><p>r =</p><p>√</p><p>q2</p><p>1 + q2</p><p>2 + q2</p><p>3 , (15.11)</p><p>then</p><p>∂V (r)</p><p>∂q̇l</p><p>= 0 and (15.12)</p><p>46 CHAPTER 15. EXERCISE 15 - GUILHERME S. Y. GIARDINI</p><p>via the chain rule</p><p>∂V (r)</p><p>∂ql</p><p>=</p><p>∂V (r)</p><p>∂r</p><p>∂r</p><p>∂ql</p><p>=</p><p>∂V (r)</p><p>∂r</p><p>∂</p><p>√</p><p>q21 + q22 + q23</p><p>∂ql</p><p>=</p><p>∂V (r)</p><p>∂r</p><p>1</p><p>2</p><p>√</p><p>q21 + q22 + q23</p><p>(</p><p>2</p><p>∂q1</p><p>∂ql</p><p>+ 2</p><p>∂q2</p><p>∂ql</p><p>+ 2</p><p>∂q3</p><p>∂ql</p><p>)</p><p>. (15.13)</p><p>Due to the system being a cartesian system, we know that a coordinate direc-</p><p>tion or magnitude does not depend on one another, such that the derivative of one</p><p>coordinate with respect to another can be represented as</p><p>∂qi</p><p>∂qj</p><p>= δij , (15.14)</p><p>where δil is Kronecker’s delta and is valued as</p><p>δil =</p><p>{</p><p>1; i = l</p><p>0; i 6= l</p><p>. (15.15)</p><p>This way, the derivative of the radial part V (r) of potential U(~r,~v) with respect</p><p>to ql becomes</p><p>∂V (r)</p><p>∂ql</p><p>=</p><p>∂V (r)</p><p>∂r</p><p>(</p><p>q1δ1l + q2δ2l + q3δ3l√</p><p>q21 + q22 + q23</p><p>)</p><p>=</p><p>∂V (r)</p><p>∂r</p><p>(</p><p>q1δ1l + q2δ2l + q3δ3l</p><p>r</p><p>)</p><p>=</p><p>∂V (r)</p><p>∂r</p><p>(ql</p><p>r</p><p>)</p><p>. (15.16)</p><p>With this derivation in hands, we go back to the force ~Q(~r,~v) components</p><p>15.1. A) CARTESIAN COORDINATES 47</p><p>Ql = �</p><p>"</p><p>∂V (r)</p><p>∂r</p><p>ql</p><p>r</p><p>+m</p><p>∂</p><p>∂ql</p><p>NX</p><p>i =0</p><p>σi</p><p>NX</p><p>j</p><p>NX</p><p>k</p><p>εijk qj q̇k</p><p>#</p><p>+</p><p>d</p><p>dt</p><p>"</p><p>∂</p><p>∂q̇l</p><p>m</p><p>NX</p><p>i =0</p><p>σi</p><p>NX</p><p>j</p><p>NX</p><p>k</p><p>εijk qj q̇k</p><p>!#</p><p>. (15.17)</p><p>Now we proceed to derive the angular momentum dependant part of Ql . We</p><p>can assume that the components of vector ~σ are constant with respect to time and</p><p>space such that</p><p>∂σi</p><p>∂t</p><p>=</p><p>∂σi</p><p>∂ql</p><p>= 0 . (15.18)</p><p>If we wished to create a more generalized and maybe more interesting problem,</p><p>we could assume a higher order ~σ and angular momentum ~L in a way that the body in</p><p>subject to potential (15.1) would not be a particle anymore, but instead a non punctual</p><p>body like a cube for instance. For now we assume a punctual body as asked by the</p><p>exercise.</p><p>Also, the Levi-Civita symbol does not depend on x, y or z. Then</p><p>Ql = �</p><p>"</p><p>∂V (r)</p><p>∂r</p><p>ql</p><p>r</p><p>+m</p><p>NX</p><p>i =0</p><p>σi</p><p>NX</p><p>j</p><p>NX</p><p>k</p><p>εijk δjl q̇k</p><p>#</p><p>+</p><p>d</p><p>dt</p><p>"</p><p>∂</p><p>∂q̇l</p><p>m</p><p>NX</p><p>i =0</p><p>σi</p><p>NX</p><p>j</p><p>NX</p><p>k</p><p>εijk qj q̇k</p><p>!#</p><p>. (15.19)</p><p>The velocity components also do not have cross dependence, in other words</p><p>∂q̇i</p><p>∂q̇j</p><p>= δij , (15.20)</p><p>this way the force equation becomes</p><p>48 CHAPTER 15. EXERCISE 15 - GUILHERME S. Y. GIARDINI</p><p>Ql = −</p><p>[</p><p>∂V (r)</p><p>∂r</p><p>ql</p><p>r</p><p>+m</p><p>N∑</p><p>i=0</p><p>σi</p><p>N∑</p><p>j</p><p>N∑</p><p>k</p><p>εijkδjlq̇k</p><p>]</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>m</p><p>N∑</p><p>i=0</p><p>σi</p><p>N∑</p><p>j</p><p>N∑</p><p>k</p><p>εijkqjδlk</p><p>)</p><p>. (15.21)</p><p>Applying the time derivative over the second term of equation (15.21), it be-</p><p>comes</p><p>Ql = −</p><p>[</p><p>∂V (r)</p><p>∂r</p><p>ql</p><p>r</p><p>+m</p><p>N∑</p><p>i=0</p><p>σi</p><p>N∑</p><p>j</p><p>N∑</p><p>k</p><p>εijkδjlq̇k</p><p>]</p><p>+</p><p>[</p><p>m</p><p>N∑</p><p>i=0</p><p>σi</p><p>N∑</p><p>j</p><p>N∑</p><p>k</p><p>εijkq̇jδlk</p><p>]</p><p>, (15.22)</p><p>where with a swap of indexes, via the Levi-Civita index permutation rule (WIKI,</p><p>2021), it changes into</p><p>Ql = −</p><p>[</p><p>∂V (r)</p><p>∂r</p><p>ql</p><p>r</p><p>−m</p><p>N∑</p><p>i=0</p><p>σi</p><p>N∑</p><p>j</p><p>N∑</p><p>k</p><p>εijkδklq̇j</p><p>]</p><p>+</p><p>[</p><p>m</p><p>N∑</p><p>i=0</p><p>σi</p><p>N∑</p><p>j</p><p>N∑</p><p>k</p><p>εijkq̇jδlk</p><p>]</p><p>(15.23)</p><p>= −</p><p>[</p><p>∂V (r)</p><p>∂r</p><p>ql</p><p>r</p><p>− 2m</p><p>N∑</p><p>i=0</p><p>σi</p><p>N∑</p><p>j</p><p>N∑</p><p>k</p><p>εijkδklq̇j</p><p>]</p><p>.</p><p>We can observe that in the equation above, the Kronecker’s delta cancels every</p><p>k term except when k = leffectively cancelling the sum over k</p><p>Ql = −∂V (r)</p><p>∂r</p><p>ql</p><p>r</p><p>+ 2m</p><p>N∑</p><p>i=0</p><p>σi</p><p>N∑</p><p>j</p><p>εijlq̇j</p><p>= −∂V (r)</p><p>∂r</p><p>ql</p><p>r</p><p>+ 2m</p><p>N∑</p><p>i=0</p><p>N∑</p><p>j</p><p>εijlσiq̇j .</p><p>15.2. A) SPHERICAL COORDINATES 49</p><p>Because a sum is a linear operation, we can interchange σi, grouping it with the</p><p>other terms, and identify the resulting expression as a vector product between ~σ and</p><p>~v, where</p><p>(~σ × ~v)l =</p><p>∑</p><p>i</p><p>∑</p><p>j</p><p>εijlσiq̇j . (15.24)</p><p>This way, we can write the force as</p><p>Ql = −∂V (r)</p><p>∂r</p><p>ql</p><p>r</p><p>+ 2m(~σ × ~v)l</p><p>or in a vectorial notation</p><p>~Q = −~∇V (r) + 2m(~σ × ~v)</p><p>= −~∇V (r) + 2(~σ × ~p) (15.25)</p><p>15.2 a) Spherical Coordinates</p><p>In spherical coordinates, there is no way to make a general calculation in index</p><p>notation because each position versor (êr, êθ, êφ) depends differently on one another,</p><p>these dependencies are</p><p>êr = cos(θ)sin(φ)êx + sin(θ)sin(φ)êy + cos(φ)êz (15.26)</p><p>êθ = −sin(θ)êx + cos(θ)êy (15.27)</p><p>êφ = cos(θ)cos(φ)êx + sin(θ)cos(φ)êy − sin(φ)êz . (15.28)</p><p>In the spherical coordinate system shown above (15.28), the position of a particle</p><p>is</p><p>50 CHAPTER 15. EXERCISE 15 - GUILHERME S. Y. GIARDINI</p><p>~r = rêr and (15.29)</p><p>its velocity is</p><p>~v =</p><p>d~r</p><p>dt</p><p>=</p><p>dr</p><p>dt</p><p>êr + r</p><p>dêr</p><p>dt</p><p>= ṙêr + rθ̇sin(φ)êθ + rφ̇êφ . (15.30)</p><p>The value for dêr</p><p>dt</p><p>can be obtained by straightforward derivation of (15.26) and</p><p>comparing with the other two direction versors.</p><p>dêr</p><p>dt</p><p>=</p><p>d</p><p>dt</p><p>(cos(θ)sin(φ)êx + sin(θ)sin(φ)êy + cos(φ)êz)</p><p>=</p><p>(</p><p>−sin(θ)sin(φ)θ̇ + cos(θ)cos(φ)φ̇</p><p>)</p><p>êx</p><p>+</p><p>(</p><p>cos(θ)sin(φ)θ̇ + sin(θ)cos(φ)φ̇</p><p>)</p><p>êy</p><p>− sin(φ)φ̇êz , (15.31)</p><p>now</p><p>we regroup the terms in way that it is easier to identify the spherical coor-</p><p>dinates</p><p>dêr</p><p>dt</p><p>= −sin(θ)sin(φ)θ̇êx + cos(θ)sin(φ)θ̇êy</p><p>+ cos(θ)cos(φ)φ̇êx + sin(θ)cos(φ)φ̇êy − sin(φ)φ̇êz</p><p>= θ̇sin(φ) (−sin(θ)êx + cos(θ)êy) (15.32)</p><p>+ φ̇ (cos(θ)cos(φ)êx + sin(θ)cos(φ)êy − sin(φ)êz) , (15.33)</p><p>where now, it is easy to see that</p><p>15.2. A) SPHERICAL COORDINATES 51</p><p>dêr</p><p>dt</p><p>= θ̇ sin(φ)êθ + φ̇êφ . (15.34)</p><p>The next step is to find the angular momentum and use it to obtain the potential</p><p>(15.1) to afterwards find each component of the force actuating over the particle.</p><p>~L = m(~r × ~v)</p><p>= m</p><p>(</p><p>rêr × (ṙêr + rθ̇sin(φ)êθ + rφ̇êφ)</p><p>)</p><p>= m</p><p>(</p><p>r2êr × êr + r2θ̇sin(φ)êr × êθ + r2φ̇êr × êφ</p><p>)</p><p>= m</p><p>(</p><p>r2θ̇sin(φ)êr × êθ + r2φ̇êr × êφ</p><p>)</p><p>. (15.35)</p><p>It is important to note that the spherical coordinate system is not right-handed,</p><p>meaning that the cross product between two of its axis versors result in another posi-</p><p>tive versor. This can be illustrated more clearly as we calculate explicitly the relevant</p><p>cross products</p><p>êr × êθ = (cos(θ)sin(φ)êx + sin(θ)sin(φ)êy + cos(φ)êz)× (−sin(θ)êx + cos(θ)êy)</p><p>= cos2(θ)sin(φ)êx × êy − sin2(θ)sin(φ)êy × êx + cos(φ)êz × (���sin(θ)êx + cos(θ)êy)</p><p>= cos2(θ)sin(φ)êx × êy + sin2(θ)sin(φ)êx × êy + cos(φ) (−sin(θ)êy − cos(θ)êx)</p><p>= sin(φ)êx × êy − cos(θ)cos(φ)êx − sin(θ)cos(φ)êy</p><p>= sin(φ)êz − cos(θ)cos(φ)êx − sin(θ)cos(φ)êy</p><p>= −êφ and (15.36)</p><p>êr × êφ = (cos(θ)sin(φ)êx + sin(θ)sin(φ)êy + cos(φ)êz)</p><p>× (cos(θ)cos(φ)êx + sin(θ)cos(φ)êy − sin(φ)êz) ,</p><p>then multiplying all the terms</p><p>52 CHAPTER 15. EXERCISE 15 - GUILHERME S. Y. GIARDINI</p><p>êr × êφ =</p><p>( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( (</p><p>cos(θ)sin(θ)cos(φ)sin(φ)êx × êy +</p><p>( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( (</p><p>cos(θ)cos(φ)sin(θ)sin(φ)êy × êx</p><p>+ −cos(θ)sin2(φ)êx × êz − sin(θ)sin2(φ)êy × êz</p><p>+ cos(θ)cos2(φ)êz × êx + sin(θ)cos2(φ)êz × êy ,</p><p>here it is easy to see that the first two terms cancel each other, due to the cross product</p><p>êx × êy = −êy × êx. For the other four terms, we can use the trigonometric identity</p><p>cos2(α) + sin2(α) = 1, simplifying the equation to</p><p>êr × êφ = −cos(θ)sin2(φ)êx × êz − sin(θ)sin2(φ)êy × êz</p><p>+ cos(θ)cos2(φ)êz × êx + sin(θ)cos2(φ)êz × êy</p><p>= cos(θ)êy − sin(θ)êx = êθ . (15.37)</p><p>Going back to the angular momentum we have</p><p>~L = m</p><p>(</p><p>−r2θ̇sin(φ)êφ + r2φ̇êθ</p><p>)</p><p>. (15.38)</p><p>Then the potential (15.1) becomes</p><p>U = V (r) + ~σ · ~L</p><p>= V (r) +m~σ ·</p><p>(</p><p>−r2θ̇sin(φ)êφ + r2φ̇êθ</p><p>)</p><p>,</p><p>as for each force component, we get</p><p>Qr = −∂U</p><p>∂r</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>∂U</p><p>∂ṙ</p><p>)</p><p>= −</p><p>[</p><p>∂V (r)</p><p>∂r</p><p>+ 2mr~σ ·</p><p>(</p><p>−θ̇sin(φ)êφ + φ̇êθ</p><p>)]</p><p>+</p><p>d</p><p>dt</p><p>(0) . (15.39)</p><p>15.2. A) SPHERICAL COORDINATES 53</p><p>For now we will leave the radial component of the force as is, but later we will</p><p>modify it further to obtain the same result as in the Cartesian coordinates, as the coor-</p><p>dinate system chosen should not interfere with the particle’s trajectory.</p><p>The other components are</p><p>Qφ = −∂U</p><p>∂φ</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>∂U</p><p>∂φ̇</p><p>)</p><p>= − ∂</p><p>∂φ</p><p>[</p><p>� � �V (r) +mr2~σ ·</p><p>(</p><p>−θ̇sin(φ)êφ + φ̇êθ</p><p>)]</p><p>+</p><p>d</p><p>dt</p><p>[</p><p>∂</p><p>∂φ̇</p><p>(</p><p>mr2~σ ·</p><p>(</p><p>−θ̇sin(φ)êφ + φ̇êθ</p><p>))]</p><p>, (15.40)</p><p>where the term V (r) is canceled because it is independent of φ, such that its derivative</p><p>with respect to φ is null.</p><p>It is harder to obtain the final form of the second force component because the</p><p>coordinates’ versors depend on φ and θ and so, their derivatives are not null</p><p>∂êφ</p><p>∂φ</p><p>=</p><p>∂</p><p>∂φ</p><p>(cos(θ)cos(φ)êx + sin(θ)cos(φ)êy − sin(φ)êz)</p><p>= −cos(θ)sin(φ)êx − sin(θ)sin(φ)êy − cos(φ)êz</p><p>= −êr , (15.41)</p><p>after finding the value for the derivative above (15.41), the force component Qφ</p><p>becomes</p><p>Qφ = −mr2~σ ·</p><p>[</p><p>−θ̇cos(φ)êφ − θ̇sin(φ)</p><p>∂êφ</p><p>∂φ</p><p>+ φ̇</p><p>�</p><p>�</p><p>�∂êθ</p><p>∂φ</p><p>]</p><p>+</p><p>d</p><p>dt</p><p>[</p><p>mr2~σ · êθ</p><p>]</p><p>= mr2θ̇~σ · [cos(φ)êφ − sin(φ)êr] + 2mrṙ~σ · êθ +mr2~σ · dêθ</p><p>dt</p><p>. (15.42)</p><p>To proceed further, we have to calculate the value of dêθ</p><p>dt</p><p>54 CHAPTER 15. EXERCISE 15 - GUILHERME S. Y. GIARDINI</p><p>dêθ</p><p>dt</p><p>=</p><p>d</p><p>dt</p><p>(−sin(θ)êx + cos(θ)êy)</p><p>= −cos(θ)θ̇êx − sin(θ)θ̇êφ</p><p>= −θ̇ (sin(φ)êr + cos(φ)êφ) , (15.43)</p><p>where the last step was achieved merely by comparing the result of the deriva-</p><p>tive to the coordinates axis versors (15.26), (15.28) and (15.27). Returning to the force</p><p>component equation we get</p><p>Qφ = mr2θ̇~σ · [� � � � �cos(φ)êφ − sin(φ)êr] + 2mrṙ~σ · êθ +mr2~σ ·</p><p>(</p><p>−θ̇ (sin(φ)êr + � � � � �cos(φ)êφ )</p><p>)</p><p>= 2mrṙ~σ · êθ − 2mr2θ̇sin(φ)~σ · êr</p><p>= 2mr~σ ·</p><p>(</p><p>ṙêθ − rθ̇sin(φ)êθ</p><p>)</p><p>. (15.44)</p><p>The third and last force component is</p><p>Qθ = −∂U</p><p>∂θ</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>∂U</p><p>∂θ</p><p>)</p><p>= − ∂</p><p>∂θ</p><p>[</p><p>� � �V (r) +mr2~σ ·</p><p>(</p><p>−θ̇sin(φ)êφ + φ̇êθ</p><p>)]</p><p>+</p><p>d</p><p>dt</p><p>[</p><p>∂</p><p>∂θ̇</p><p>(</p><p>� � �V (r) +mr2~σ ·</p><p>(</p><p>−θ̇sin(φ)êφ + φ̇êθ</p><p>))]</p><p>.</p><p>Here, we calculate the derivatives of the coordinate system versors with respect</p><p>to the coordinates and with respect to time. They are</p><p>∂êθ</p><p>∂θ</p><p>= −cos(θ)êx − sin(θ)êy</p><p>= − (sin(φ)êr + cos(φ)êφ) , (15.45)</p><p>∂êφ</p><p>∂θ</p><p>= −sin(θ)cos(φ)êx + cos(θ)cos(φ)êy</p><p>= cos(φ)êθ and (15.46)</p><p>15.2. A) SPHERICAL COORDINATES 55</p><p>dêφ</p><p>dt</p><p>= −sin(θ)cos(φ)θ̇êx − cos(θ)sin(φ)φ̇êx + cos(θ)cos(φ)θ̇êy</p><p>− sin(θ)sin(φ)φ̇êy − cos(φ)φ̇êz</p><p>= −φ̇êr + θ̇cos(φ)êθ . (15.47)</p><p>With all the derivatives known, we can obtain the third component of the force</p><p>actuating on the particle. The component becomes</p><p>Qθ = −mr2~σ ·</p><p>[</p><p>φ̇</p><p>∂êθ</p><p>∂θ</p><p>− θ̇sin(φ)∂êφ</p><p>∂θ</p><p>]</p><p>− d</p><p>dt</p><p>[</p><p>mr2sin(φ)~σ · êφ</p><p>]</p><p>= −mr2~σ ·</p><p>[</p><p>−φ̇ (sin(φ)êr + cos(φ)êφ)− θ̇sin(φ)cos(φ)êθ</p><p>]</p><p>− 2mrṙsin(φ)~σ · êφ +mr2~σ ·</p><p>(</p><p>−cos(φ)φ̇êφ − sin(φ) ˙̂e2</p><p>)</p><p>then (15.48)</p><p>we substitute the remaining derivative term (15.47) into Qθ equation</p><p>Qθ = mr2~σ ·</p><p>[</p><p>φ̇ (sin(φ)êr +���</p><p>��cos(φ)êφ ) +((((</p><p>((((</p><p>(</p><p>θ̇sin(φ)cos(φ)êθ</p><p>]</p><p>− 2mrṙsin(φ)~σ · êφ −mr2~σ ·</p><p>(</p><p>���</p><p>���cos(φ)φ̇êφ + sin(φ)</p><p>(</p><p>−φ̇êr +����</p><p>��</p><p>θ̇cos(φ)êθ</p><p>))</p><p>= 2mr2φ̇sin(φ)~σ · êr − 2mrṙsin(φ)~σ · êφ</p><p>= 2mrsin(φ)~σ ·</p><p>[</p><p>rφ̇êr − ṙêφ</p><p>]</p><p>. (15.49)</p><p>Consider all three components of the force ~Q, it may seem as we have obtained</p><p>a dead end, but if we analyse the results closer the result makes sense and reproduces</p><p>the same dynamics as the result when considering a Cartesian coordinate system. First,</p><p>we consider the cross product between two vectors in the spherical coordinate system</p><p>that is represented as the determinant of the matrix</p><p>~a×~b = Det</p><p>êr êθ êφ</p><p>ar aθ aφ</p><p>br bθ bφ</p><p></p><p>= (aθbφ − aφbθ) êr + (aφbr − arbφ) êθ + (arbθ − aθbr) êφ and (15.50)</p><p>56 CHAPTER 15. EXERCISE 15 - GUILHERME S. Y. GIARDINI</p><p>second, we investigate the velocity equation</p><p>~v = ṙêr + rθ̇sin(φ)êθ + rφ̇êφ such (15.51)</p><p>that we notice that the difference term in the force components correspond all to</p><p>a cross product between the velocity and the ~σ vector</p><p>(~σ × ~v) · êr = ~σ ·</p><p>(</p><p>φ̇êθ − θ̇sin(φ)êφ</p><p>)</p><p>, (15.52)</p><p>(~σ × ~v) · êφ = ~σ ·</p><p>(</p><p>ṙêθ − rθ̇sin(φ)êr</p><p>)</p><p>and (15.53)</p><p>(~σ × ~v) · êθ = ~σ ·</p><p>(</p><p>rφ̇êr − ṙêφ</p><p>)</p><p>. (15.54)</p><p>Taking all into account, the force vector can be represented as</p><p>~Q =</p><p>(</p><p>−∂V (r)</p><p>∂r</p><p>+ 2mêr · (~σ × ~v)</p><p>)</p><p>êr</p><p>+ (2mrêφ · (~σ × ~v)) êφ</p><p>+ (2mrsin(φ)êθ · (~σ × ~v)) êθ . (15.55)</p><p>15.3 b)</p><p>In item b) we are asked to show through equation (1.49) of Goldstein 3rd Ed. ,</p><p>shown below</p><p>15.3. B) 57</p><p>Qj =</p><p>∑</p><p>i</p><p>~Fi</p><p>∂~ri</p><p>∂qj</p><p>(15.56)</p><p>that the components of the Cartesian and spherical systems are related to one</p><p>another. To make the solution easier, we consider the force in Cartesian coordinates</p><p>~F (x, y, z) = −~r</p><p>r</p><p>∂V</p><p>∂r</p><p>+ 2m (~σ × ~v) , (15.57)</p><p>where</p><p>~r = xêx + yêy + zêz and (15.58)</p><p>r = |~r|</p><p>=</p><p>√</p><p>x2 + y2 + z2 also (15.59)</p><p>êr = cos(θ)sin(φ)êx + sin(θ)sin(φ)êy + cos(φ)êz . (15.60)</p><p>Furthermore the position vector can be expressed as</p><p>~r = rêr . (15.61)</p><p>With all considerations above, we proceed to find the relation between the forces</p><p>in both coordinates systems.</p><p>58 CHAPTER 15. EXERCISE 15 - GUILHERME S. Y. GIARDINI</p><p>~F · ∂~r</p><p>∂r</p><p>=</p><p>[</p><p>−~r</p><p>r</p><p>∂V</p><p>∂r</p><p>+ 2m (~σ × ~v)</p><p>]</p><p>· ∂</p><p>∂r</p><p>[r cos(θ)sin(φ)êx + r sin(θ)sin(φ)êy + r cos(φ)êz]</p><p>=</p><p>[</p><p>−~r</p><p>r</p><p>∂V</p><p>∂r</p><p>+ 2m (~σ × ~v)</p><p>]</p><p>· [cos(θ)sin(φ)êx + sin(θ)sin(φ)êy + cos(φ)êz]</p><p>=</p><p>[</p><p>−~r</p><p>r</p><p>∂V</p><p>∂r</p><p>+ 2m (~σ × ~v)</p><p>]</p><p>· ~r</p><p>r</p><p>=</p><p>[</p><p>−êr · êr</p><p>∂V</p><p>∂r</p><p>+ 2m (~σ × ~v) · êr</p><p>]</p><p>=</p><p>[</p><p>−∂V</p><p>∂r</p><p>+ 2m (~σ × ~v) · êr</p><p>]</p><p>= Qr next</p><p>we have the φ force component</p><p>~F · ∂~r</p><p>∂φ</p><p>=</p><p>[</p><p>−êr</p><p>∂V</p><p>∂r</p><p>+ 2m(~σ × ~v)</p><p>]</p><p>· ∂</p><p>∂φ</p><p>[r cos(θ)sin(φ)êx + r sin(θ)sin(φ)êy + r cos(φ)êz]</p><p>=</p><p>[</p><p>−êr</p><p>∂V</p><p>∂r</p><p>+ 2m(~σ × ~v)</p><p>]</p><p>· [r cos(θ)cos(φ)êx + r sin(θ)cos(φ)êy − r sin(φ)êz]</p><p>=</p><p>[</p><p>−êr</p><p>∂V</p><p>∂r</p><p>+ 2m(~σ × ~v)</p><p>]</p><p>· (rêφ)</p><p>=</p><p>[</p><p>−����</p><p>���êr · (φêφ)</p><p>∂V</p><p>∂r</p><p>+ 2m(~σ × ~v) · (rêφ)</p><p>]</p><p>= 2m(~σ × ~v) · (rêφ) = Qφ and (15.62)</p><p>finally</p><p>~F · ∂~r</p><p>∂θ</p><p>=</p><p>[</p><p>−êr</p><p>∂V</p><p>∂r</p><p>+ 2m(~σ × ~v)</p><p>]</p><p>· ∂</p><p>∂θ</p><p>[r cos(θ)sin(φ)êx + r sin(θ)sin(φ)êy + r cos(φ)êz]</p><p>=</p><p>[</p><p>−êr</p><p>∂V</p><p>∂r</p><p>+ 2m(~σ × ~v)</p><p>]</p><p>· [−r sin(θ)sin(φ)êx + r cos(θ)sin(φ)êy]</p><p>=</p><p>[</p><p>−êr</p><p>∂V</p><p>∂r</p><p>+ 2m(~σ × ~v)</p><p>]</p><p>· [r sin(φ) êθ]</p><p>=</p><p>[</p><p>−</p><p>���</p><p>���</p><p>���</p><p>r sin(φ)êr · êθ</p><p>∂V</p><p>∂r</p><p>+ 2m(~σ × ~v) · (r sin(φ) êθ)</p><p>]</p><p>= Qθ . (15.63)</p><p>15.4. C) 59</p><p>15.4 c)</p><p>In the third item we are asked to find the equations of movement of the particle</p><p>under the influence of potential (15.1). To do it, we first investigate the Euler Lagrange</p><p>equation, whose solution gives rise to the system’s movement equation. The Euler-</p><p>Lagrange equation has the form of</p><p>∂L</p><p>∂qj</p><p>− d</p><p>dt</p><p>(</p><p>∂L</p><p>∂q̇j</p><p>)</p><p>= 0 (15.64)</p><p>with this in mind, we use the force components of the previously solved items</p><p>Qj = −</p><p>∂U</p><p>∂qj</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>∂U</p><p>∂q̇j</p><p>)</p><p>, (15.65)</p><p>we see that the force depends on the potential U of the system and has a dif-</p><p>ferential equation format to the Euler-Lagrange equation. Due to it, we may write the</p><p>Euler-Lagrange equation as</p><p>∂L</p><p>∂qj</p><p>− d</p><p>dt</p><p>(</p><p>∂L</p><p>∂q̇j</p><p>)</p><p>=</p><p>∂</p><p>∂qj</p><p>(T − U)− d</p><p>dt</p><p>(</p><p>∂</p><p>∂q̇j</p><p>(T − U)</p><p>)</p><p>. (15.66)</p><p>Because the Lagrangian is written as L = T − U , the Euler-Lagrange equation</p><p>can be rewritten as a function of a Euler equation applied only to the kinetic energy</p><p>summed to the force components obtained in the previous exercise items.</p><p>∂L</p><p>∂qj</p><p>− d</p><p>dt</p><p>(</p><p>∂L</p><p>∂q̇j</p><p>)</p><p>=</p><p>∂T</p><p>∂qj</p><p>− ∂U</p><p>∂qj</p><p>− d</p><p>dt</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>∂U</p><p>∂q̇j</p><p>)</p><p>= Qj +</p><p>∂T</p><p>∂qj</p><p>− d</p><p>dt</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>= 0 (15.67)</p><p>60 CHAPTER 15. EXERCISE 15 - GUILHERME S. Y. GIARDINI</p><p>Also, because the Euler-Lagrange has the purpose to minimize the energy of a</p><p>system, its derivative form is equivalent to zero, such that its reorganized form also</p><p>has zero value.</p><p>Qj +</p><p>∂T</p><p>∂qj</p><p>− d</p><p>dt</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>= 0 . (15.68)</p><p>We can also write the obtained equation as</p><p>Qj = −</p><p>∂T</p><p>∂qj</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>. (15.69)</p><p>With this in mind, to find the system’s movement equations, all we have to do</p><p>is to obtain the kinetic energy and substitute it into the Euler equation (right side of</p><p>equation 15.69). The method to obtain the kinetic energy is straightforward, we just</p><p>have to use its definition T = mv2</p><p>2</p><p>T =</p><p>m(~v)2</p><p>2</p><p>=</p><p>m</p><p>2</p><p>(</p><p>ṙêr + r φ̇êφ + r θ̇sin(φ)êθ</p><p>)2</p><p>, (15.70)</p><p>after squaring the velocity we get</p><p>T =</p><p>m</p><p>2</p><p>(</p><p>ṙ2 + r2φ̇2 + r2θ̇2sin2(φ)</p><p>)</p><p>. (15.71)</p><p>Now we may apply the kinetic energy into (15.69), this way we have the radial</p><p>part of the Euler equation applied to the kinetic energy as</p><p>Qr = −</p><p>m</p><p>2</p><p>(</p><p>2 rφ̇2 + 2 rθ̇2sin2(φ)</p><p>)</p><p>+mr̈ , (15.72)</p><p>and the radial component of the force as</p><p>15.4. C) 61</p><p>Qr = −</p><p>∂V</p><p>∂r</p><p>− 2mr~σ ·</p><p>(</p><p>φ̇êθ − θ̇sin(φ)êφ</p><p>)</p><p>, (15.73)</p><p>this way we may equal both equations to obtain</p><p>−m</p><p>2</p><p>(</p><p>2 rφ̇2 + 2 rθ̇2sin2(φ)</p><p>)</p><p>+mr̈ = −∂V</p><p>∂r</p><p>− 2mr~σ ·</p><p>(</p><p>φ̇êθ − θ̇sin(φ)êφ</p><p>)</p><p>. (15.74)</p><p>To effectively obtain, we consider an arbitrary ~σ vector written as</p><p>~σ = σ1êr + σ2êφ + σ3êθ and (15.75)</p><p>use it on all three movement equations, this way, the movement equations be-</p><p>come</p><p>mr̈ =</p><p>m</p><p>2</p><p>(</p><p>2 rφ̇2 + 2 rθ̇2sin2(φ)</p><p>)</p><p>− ∂V</p><p>∂r</p><p>− 2mr</p><p>(</p><p>φ̇σ3 − θ̇sin(φ)σ2</p><p>)</p><p>, (15.76)</p><p>now for φ, we repeat what was done to r coordinate equations</p><p>Qφ = −</p><p>∂T</p><p>∂φ</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>∂T</p><p>∂φ̇</p><p>)</p><p>(15.77)</p><p>Qφ = −</p><p>m</p><p>2</p><p>(</p><p>2 r2θ̇sin(φ)cos(φ)</p><p>)</p><p>+</p><p>m</p><p>2</p><p>(</p><p>2φ̈r2 + 4rṙφ̇</p><p>)</p><p>(15.78)</p><p>Qφ = 2mr~σ</p><p>(</p><p>ṙêθ − rθ̇sin(φ)êr</p><p>)</p><p>(15.79)</p><p>62 CHAPTER 15. EXERCISE 15 - GUILHERME S. Y. GIARDINI</p><p>mφ̈ =</p><p>m</p><p>r2</p><p>[</p><p>−2 rṙφ̇+ r2θ̇sin(φ)cos(φ) + 2qq</p><p>(</p><p>ṙσ3 − rθ̇sin(φ)σ1</p><p>)]</p><p>. (15.80)</p><p>And finally for the θ coordinate equations</p><p>Qθ = −</p><p>∂T</p><p>∂θ</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>∂T</p><p>∂θ̇</p><p>)</p><p>(15.81)</p><p>Qθ = 0 +</p><p>d</p><p>dt</p><p>(</p><p>mθ̇r2sin2(φ)</p><p>)</p><p>(15.82)</p><p>= mθ̈r2sin2(φ) + 2m</p><p>(</p><p>rṙθ̇sin2(φ) + r2φ̇θ̇sin(φ)cos(φ)</p><p>)</p><p>(15.83)</p><p>= 2mrsin(φ)</p><p>(</p><p>rφ̇σ1 − ṙσ2</p><p>)</p><p>(15.84)</p><p>(15.85)</p><p>m θ̈ =</p><p>−2m</p><p>r2sin2(φ)</p><p>[</p><p>rṙθ̇sin2(φ) + r2φ̇θ̇sin(φ)cos(φ)</p><p>]</p><p>+</p><p>2m</p><p>r2sin2(φ)</p><p>r sin(φ)</p><p>(</p><p>r φ̇σ1 − ṙσ2</p><p>)</p><p>(15.86)</p><p>Exercise 16 - Lucas Bourscheidt</p><p>Solution:</p><p>Starting from the definition of generalized force:</p><p>Qr = −</p><p>∂U</p><p>∂r</p><p>+</p><p>d</p><p>dt</p><p>∂U</p><p>∂ṙ</p><p>=</p><p>1</p><p>r2</p><p>− 1</p><p>c2</p><p>(</p><p>ṙ2</p><p>r2</p><p>− 2r̈</p><p>r</p><p>)</p><p>=</p><p>1</p><p>r2</p><p>− 1</p><p>c2</p><p>ṙ2</p><p>r2</p><p>+</p><p>2</p><p>c2</p><p>dṙ</p><p>dt</p><p>1</p><p>r</p><p>=</p><p>1</p><p>r2</p><p>− 1</p><p>c2</p><p>ṙ2</p><p>r2</p><p>+</p><p>2</p><p>c2</p><p>[</p><p>d</p><p>dt</p><p>(</p><p>ṙ</p><p>r</p><p>)</p><p>− ṙ d</p><p>dt</p><p>(</p><p>1</p><p>r</p><p>)]</p><p>=</p><p>1</p><p>r2</p><p>− 1</p><p>c2</p><p>ṙ2</p><p>r2</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>2ṙ</p><p>c2r</p><p>)</p><p>− 2ṙ</p><p>c2</p><p>dr</p><p>dt</p><p>d</p><p>dr</p><p>(</p><p>1</p><p>r</p><p>)</p><p>=</p><p>1</p><p>r2</p><p>− 1</p><p>c2</p><p>ṙ2</p><p>r2</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>2ṙ</p><p>c2r</p><p>)</p><p>+</p><p>2</p><p>c2</p><p>ṙ2</p><p>r2</p><p>63</p><p>64 CHAPTER 16. EXERCISE 16 - LUCAS BOURSCHEIDT</p><p>=</p><p>1</p><p>r2</p><p>+</p><p>1</p><p>c2</p><p>ṙ2</p><p>r2</p><p>+</p><p>d</p><p>dt</p><p>(</p><p>1</p><p>c2r</p><p>∂ṙ2</p><p>∂ṙ</p><p>)</p><p>=</p><p>1</p><p>r2</p><p>+</p><p>1</p><p>r2</p><p>ṙ2</p><p>c2</p><p>+</p><p>d</p><p>dt</p><p>∂</p><p>∂ṙ</p><p>(</p><p>1</p><p>r</p><p>ṙ2</p><p>c2</p><p>)</p><p>Then</p><p>−∂U</p><p>∂r</p><p>=</p><p>1</p><p>r2</p><p>+</p><p>1</p><p>r2</p><p>ṙ2</p><p>c2</p><p>= − ∂</p><p>∂r</p><p>(</p><p>1</p><p>r</p><p>+</p><p>1</p><p>r</p><p>ṙ2</p><p>c2</p><p>)</p><p>(A)</p><p>and</p><p>∂U</p><p>∂ṙ</p><p>=</p><p>∂</p><p>∂ṙ</p><p>(</p><p>1</p><p>r</p><p>ṙ2</p><p>c2</p><p>)</p><p>=</p><p>∂</p><p>∂ṙ</p><p>(</p><p>1</p><p>r</p><p>+</p><p>1</p><p>r</p><p>ṙ2</p><p>c2</p><p>)</p><p>(B)</p><p>From equations (A) and (B) we conclude that</p><p>U (r, ṙ) =</p><p>1</p><p>r</p><p>+</p><p>1</p><p>r</p><p>ṙ2</p><p>c2</p><p>=</p><p>1</p><p>r</p><p>(</p><p>1 +</p><p>ṙ2</p><p>c2</p><p>)</p><p>Thus, the Lagrangian function is given by</p><p>L = T − U =</p><p>1</p><p>2</p><p>m</p><p>(</p><p>ṙ2 + r2θ̇2 + r2sin2θ φ̇2</p><p>)</p><p>− 1</p><p>r</p><p>(</p><p>1 +</p><p>ṙ2</p><p>c2</p><p>)</p><p>Exercise 17 - Jenifer A. de Matos</p><p>A nucleus originally at rest, decays radioactively by emitting an eletron of mo-</p><p>mentum 1.73MeV/c, and at right angles, to the direction of the electron a neutrino</p><p>with momentum 1.00MeV/c. (The MeV, million electron volt, is a unit of energy used</p><p>in modern physics equal to 1.60 x 10-13 J. Correspondingly, MeV/c is a unit of linear</p><p>momentum equal to 5.34 x 10 -22 kg.m/s.) In what direction does the nucleus recoil?</p><p>What is its momentum in MeV/c? If the mass of the residual nucleus is 3.90 x 10-25 kg</p><p>what is its kinectic energy in electron volts?</p><p>To solve this problem, we are initially going to draw the situation using a frame</p><p>of reference:</p><p>65</p><p>66 CHAPTER 17. EXERCISE 17 - JENIFER A. DE MATOS</p><p>Using momentum conservation, we can say that the nucleus must acquire the</p><p>following momentum:</p><p>px = 1.00 MeV/c</p><p>py = 1.73 MeV/c</p><p>Therefore we can determine the total momentum of the nucleus and its direc-</p><p>tion:</p><p>tan(θ) =</p><p>py</p><p>px</p><p>(17.1)</p><p>θ = tan−1</p><p>(</p><p>py</p><p>px</p><p>)</p><p>(17.2)</p><p>θ = tan−1</p><p>(</p><p>1,73</p><p>1,00</p><p>)</p><p>(17.3)</p><p>θ = 60,00o (17.4)</p><p>And for the momentum,</p><p>p =</p><p>√</p><p>p2x + p2y (17.5)</p><p>p =</p><p>√</p><p>1,002 + 1,732 (17.6)</p><p>p = 2,00</p><p>MeV</p><p>c</p><p>(17.7)</p><p>67</p><p>To determine the kinectic energy of the nucleus, we can use:</p><p>p = 10.68 x 10 -22 kg.m/s</p><p>m = 3.90 x 10-25 kg</p><p>Therefore:</p><p>T =</p><p>p2</p><p>2m</p><p>(17.8)</p><p>T =</p><p>(10,68× 10−22)2</p><p>2× (3,90× 10−25)</p><p>(17.9)</p><p>T = 1,46× 10</p><p>−18</p><p>J = 9, 125 eV (17.10)</p><p>Exercise 18 - Gustavo C. G. Kessler</p><p>69</p><p>Exercise 19 - Douglas O. Novaes</p><p>Obtain the Lagrange equations of motion for a spherical pendulum, i.e., a mass</p><p>point suspended by a rigid weightless rod.</p><p>The Cartesian coordinates of system are:</p><p>x = l sin θ cosφ</p><p>y = l sin θ sinφ</p><p>z = l cos θ</p><p>where, l is the constant length of rod, θ is the inclination angle, and φ is the azimuthal</p><p>angle.</p><p>Since the kinetic and potential energy are given by:</p><p>T =</p><p>1</p><p>2</p><p>mv2 =</p><p>1</p><p>2</p><p>m(ẋ2 + ẏ2 + ż2)</p><p>U = mgz</p><p>(19.1)</p><p>The velocities components are:</p><p>ẋ = l (θ̇ cos θ cosφ− φ̇ sin θ cosφ)</p><p>ẏ = l (θ̇ cos θ sinφ+ φ̇ sin θ cosφ)</p><p>ż = −l θ̇ sin θ</p><p>71</p><p>72 CHAPTER 19. EXERCISE 19 - DOUGLAS O. NOVAES</p><p>Then,</p><p>ẋ2 + ẏ2 + ż2 = l2(θ̇2 cos2 θ cos2 φ+ φ̇2 sin2 θ sin2 φ−</p><p>((((</p><p>(((</p><p>((((</p><p>((</p><p>2θ̇φ̇ cos θ cosφ sin θ sinφ</p><p>+ θ̇2 cos2 θ sin2 φ+ φ̇2 sin2 θ cos2 φ+</p><p>(((</p><p>((((</p><p>(((</p><p>(((</p><p>2θ̇φ̇ cos θ cosφ sin θ sinφ</p><p>− θ̇2 sin2 θ)</p><p>= l2(θ̇2 cos2 θ + φ̇2 sin2 θ − θ̇2 sin2 θ)</p><p>v2 = l2 (θ̇2 + φ̇2 sin2 θ)</p><p>Then, replacing at (19.1),</p><p>T =</p><p>1</p><p>2</p><p>m l2 (θ̇2 + φ̇2 sin2 θ)</p><p>U = mgl cos θ</p><p>Since the Lagrangian is L = T − U , come:</p><p>L (θ, θ̇, φ̇) =</p><p>1</p><p>2</p><p>m l2 (θ̇2 + φ̇2 sin2 θ)−mgl cos θ</p><p>The Lagrange equations are:</p><p>∂L</p><p>∂θ</p><p>− d</p><p>dt</p><p>∂L</p><p>∂θ̇</p><p>= 0 (19.2)</p><p>∂L</p><p>∂φ</p><p>− d</p><p>dt</p><p>∂L</p><p>∂φ̇</p><p>= 0 (19.3)</p><p>Solving the derivatives of the Lagrangian for the polar angle θ, from (19.2), result</p><p>in:</p><p>∂L</p><p>∂θ</p><p>=</p><p>1</p><p>�2</p><p>m l2 φ̇2</p><p>�2 sin θ cos θ +mgl sin θ = m (l2 φ̇2 cos θ + gl) sin</p><p>θ</p><p>∂L</p><p>∂θ̇</p><p>= m l2 θ̇</p><p>d</p><p>dt</p><p>∂L</p><p>∂θ̇</p><p>= m l2 θ̈</p><p>Then, replacing at (19.2):</p><p>��m (l�2 φ̇2 cos θ + gl) sin θ −��m l�2 θ̈ = 0</p><p>l φ̇2 cos θ sin θ + g sin θ − l θ̈ = 0</p><p>73</p><p>And the first equation of motion is:</p><p>θ̈ = φ̇2 sin 2 θ +</p><p>g</p><p>l</p><p>sin θ</p><p>Similarly, solving the Lagrange equation for the azimuth φ, result in:</p><p>∂L</p><p>∂φ</p><p>= 0</p><p>∂L</p><p>∂φ̇</p><p>=</p><p>1</p><p>�2</p><p>m l2 �2 φ̇ sin</p><p>2 θ = m l2 φ̇ sin2 θ</p><p>d</p><p>dt</p><p>∂L</p><p>∂φ̇</p><p>= m l2 φ̈ sin2 θ +m l2 φ̇ θ̇ 2 sin θ cos θ</p><p>Then, returning to (19.3):</p><p>��m ��l2 φ̈ sin�2 θ + ��m ��l2 φ̇ θ̇ 2 � � �sin θ cos θ = 0</p><p>And, finally, the second equation of motion is:</p><p>φ̈ sin θ = −2 φ̇ θ̇ cos θ</p><p>Exercise 20 - Jefferson S. Martins</p><p>A particle of mass m moves in one dimension such that it has the Lagrangian</p><p>L =</p><p>m2ẋ4</p><p>12</p><p>+mẋ2V (x)− V 2(x)</p><p>where V is some differentiable function of x. Find the equation of motion for x(t) and</p><p>describe the physical nature of the system on the basis of this equation.</p><p>Solution: Applying the Euler-Lagrange equation to the Lagrangian:</p><p>d</p><p>dt</p><p>∂L</p><p>∂ẋ</p><p>− ∂L</p><p>∂x</p><p>= 0</p><p>∂L</p><p>∂ẋ</p><p>=</p><p>∂[m</p><p>2ẋ4</p><p>12</p><p>+mẋ2V (x)− V 2(x)]</p><p>∂ẋ</p><p>=</p><p>m2ẋ3</p><p>3</p><p>+ 2mẋV (x)</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ẋ</p><p>) =</p><p>d</p><p>dt</p><p>[</p><p>m2ẋ3</p><p>3</p><p>+ 2mẋV (x)] = m2ẋ2ẍ+ 2mẍV (x)</p><p>∂L</p><p>∂x</p><p>= mẋ2V ′(x)− 2V (x)V ′(x)</p><p>d</p><p>dt</p><p>∂L</p><p>∂ẋ</p><p>− ∂L</p><p>∂x</p><p>= m2ẋ2ẍ+ 2mẍV (x) +mẋ2V ′(x) + 2V (x)V ′(x) (20.1)</p><p>Factoring (20.1) and dividing everything by two, we get:</p><p>mẍ[</p><p>mẋ2</p><p>2</p><p>+ V (x)] + V ′(x)[</p><p>mẋ2</p><p>2</p><p>+ V (x)] = 0</p><p>[mẍ+ V ′(x)][</p><p>mẋ2</p><p>2</p><p>+ V (x)] = 0 (20.2)</p><p>75</p><p>76 CHAPTER 20. EXERCISE 20 - JEFFERSON S. MARTINS</p><p>Considering the relationships F = −dV</p><p>dx</p><p>and K = mẋ2</p><p>2</p><p>, we can rewrite the rela-</p><p>tionship (20.2) as follows:</p><p>−[−mẍ+ F ][K + V (x)] = 0 (20.3)</p><p>Equation (20.3) is satisfied if F −mẍ = 0 or K + V (x) = 0. We note that the first</p><p>term is derivative of the second term with respect to x:</p><p>d</p><p>dx</p><p>[K + V (x)] =</p><p>dV</p><p>dx</p><p>+</p><p>d[m ẋ2</p><p>2</p><p>]</p><p>dx</p><p>= −F +</p><p>2</p><p>2</p><p>mẋ</p><p>dẋ</p><p>dx</p><p>d</p><p>dx</p><p>[K + V (x)] = −F +mẋ</p><p>dẋ</p><p>dt</p><p>dt</p><p>dx</p><p>= −F +mẍ</p><p>Making E = K + V (x), we can write the relation (20.3) as follows:</p><p>−EE ′ = 0⇒ 2EE ′ = 0⇒ d[E2]</p><p>dx</p><p>= 0⇒ E2 = const</p><p>Therefore, we have that E2 = const. So, if E = 0 into t = 0, E = 0 for all</p><p>t > 0. If E 6= 0 for t = 0, E 6= 0 for all other times greater than zero (t > 0). Whereas</p><p>mẍ + V ′(x) = 0. Thus, K + V (x) = const. With Lagrangian, d</p><p>dt</p><p>∂L</p><p>∂ẋ</p><p>− ∂L</p><p>∂x</p><p>= 0, we obtain</p><p>the same equations of motion for a particle of mass m under the action of a potential</p><p>V (x).</p><p>Exercise 21 - Lucas Bourscheidt</p><p>Solution:</p><p>The following figure illustrates the situation described in the problem:</p><p>77</p><p>78 CHAPTER 21. EXERCISE 21 - LUCAS BOURSCHEIDT</p><p>We choose point A, where y=0, so that r=y whatever the length of the string.</p><p>For m1:</p><p>T1 =</p><p>1</p><p>2</p><p>m1(ṙ</p><p>2 + r2θ̇2)</p><p>=</p><p>1</p><p>2</p><p>m1(ẏ</p><p>2 + y2θ̇2)</p><p>V1 = m1gh</p><p>For m2:</p><p>T2 =</p><p>1</p><p>2</p><p>m2ẏ</p><p>2</p><p>V2 = m2gy</p><p>The generalized coordinates are y and θ.</p><p>Lagrangian:</p><p>L = T1 + T2 − V1 − V2</p><p>=</p><p>1</p><p>2</p><p>m1(ẏ</p><p>2 + y2θ̇2) +</p><p>1</p><p>2</p><p>m2ẏ</p><p>2 −m1gh−m2gy</p><p>=</p><p>1</p><p>2</p><p>Mẏ2 +</p><p>1</p><p>2</p><p>m1y</p><p>2θ̇2 −m1gh−m2gy</p><p>where</p><p>M = m1 +m2</p><p>Equations of motion:</p><p>For θ:</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂θ̇</p><p>)</p><p>− ∂L</p><p>∂θ</p><p>= 0</p><p>d</p><p>dt</p><p>(m1y</p><p>2θ̇) = 0</p><p>m1y</p><p>2θ̇ = const = l (A)</p><p>For y:</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ẏ</p><p>)</p><p>− ∂L</p><p>∂y</p><p>= 0</p><p>Mÿ −m1yθ̇</p><p>2 +m2g = 0 (B)</p><p>Replacing (A) in the second term in (B):</p><p>Mÿ − l2</p><p>m1y3</p><p>+m2g = 0</p><p>79</p><p>As the Lagrangian function does not explicitly depend on time:∑</p><p>i</p><p>∂L</p><p>∂q̇i</p><p>q̇i − L = const = E (Energy)</p><p>Then</p><p>E =Mẏẏ +m1y</p><p>2θ̇θ̇ −</p><p>(</p><p>1</p><p>2</p><p>Mẏ2 +</p><p>1</p><p>2</p><p>m1y</p><p>2θ̇2 −m1gh−m2gy</p><p>)</p><p>=</p><p>1</p><p>2</p><p>Mẏ2 +</p><p>1</p><p>2</p><p>m1y</p><p>2θ̇2 +m1gh+m2gy</p><p>Employing (A):</p><p>E =</p><p>1</p><p>2</p><p>Mẏ2 +</p><p>1</p><p>2</p><p>l2</p><p>m1y2</p><p>+m1gh+m2gy</p><p>Exercise 22 - Douglas O. Novaes</p><p>Obtain the Lagrangian and equations of motion for the double pendulum illus-</p><p>trated in Fig. 1,4, where the lengths of the pendulum are l1 and l2 with corresponding</p><p>masses m1 and m2.</p><p>This pendulum oscillates in the x-y plane and the origin of the system is at the</p><p>pivot of pendulum 1. The Cartesian coordinates for the two masses are:</p><p>x1 = l1 sin θ1, x2 = l1 sin θ1 + l2 sin θ2</p><p>y1 = −l1 cos θ1, y2 = −l1 cos θ1 − l2 cos θ2</p><p>Since the kinetic and potential energy is given by:</p><p>T =</p><p>1</p><p>2</p><p>(m1v</p><p>2</p><p>1 +m2v</p><p>2</p><p>2)</p><p>U = m1 g y1 +m2 g y2</p><p>(22.1)</p><p>and v2i = ẋ2i + ẏ2i , (i = 1, 2), the velocity components are:</p><p>ẋ1 = l1θ̇1 cos θ1, ẋ2 = l1θ̇1 cos θ1 + l2θ̇2 cos θ2</p><p>ẏ1 = −l1θ̇1 sin θ1, ẏ2 = −l1θ̇1 sin θ1 − l2θ̇2 sin θ2</p><p>Then,</p><p>ẋ1</p><p>2 = l21 θ̇</p><p>2</p><p>1 cos</p><p>2 θ1, ẋ2</p><p>2 = l21 θ̇</p><p>2</p><p>1 cos</p><p>2 θ1 + l22 θ̇</p><p>2</p><p>2 cos</p><p>2 θ2 + 2l1l2θ̇1θ̇2cos θ1 cos θ2</p><p>ẏ1</p><p>2 = l21 θ̇</p><p>2</p><p>1 sin</p><p>2 θ1, ẏ2</p><p>2 = l21 θ̇</p><p>2</p><p>1 sin2 θ1 + l22 θ̇</p><p>2</p><p>2 sin</p><p>2 θ2 + 2l1l2θ̇1θ̇2sin θ1 sin θ2</p><p>Adding them, cos2 θi+sin2 θi = 1 and sin θ1 sin θ2+cos θ1 cos θ2 = cos(θ1 − θ2) and</p><p>81</p><p>82 CHAPTER 22. EXERCISE 22 - DOUGLAS O. NOVAES</p><p>returning to (22.1)</p><p>T =</p><p>1</p><p>2</p><p>m1l</p><p>2</p><p>1θ̇</p><p>2</p><p>1 +</p><p>1</p><p>2</p><p>m2</p><p>[</p><p>l21θ̇</p><p>2</p><p>1 + l22θ̇</p><p>2</p><p>2 + 2l1l2θ̇1θ̇2 cos(θ1 − θ2)</p><p>]</p><p>U = −m1 g l1 cos θ1 −m2 g (l1 cos θ1 + l2 cos θ2)</p><p>Rearranging the terms, one has to</p><p>T =</p><p>1</p><p>2</p><p>(m1 +m2)l</p><p>2</p><p>1θ̇</p><p>2</p><p>1 +</p><p>1</p><p>2</p><p>m2</p><p>[</p><p>l22θ̇</p><p>2</p><p>2 + 2l1l2θ̇1θ̇2 cos(θ1 − θ2)</p><p>]</p><p>U = −(m1 +m2) g l1 cos θ1 −m2 g l2 cos θ2)</p><p>Since L = T − U , come:</p><p>L (θ1, θ̇1, θ2, θ̇2) =</p><p>1</p><p>2</p><p>(m1 +m2)l</p><p>2</p><p>1θ̇</p><p>2</p><p>1 +</p><p>1</p><p>2</p><p>m2</p><p>[</p><p>l22θ̇</p><p>2</p><p>2 + 2l1l2θ̇1θ̇2 cos(θ1 − θ2)</p><p>]</p><p>+ (m1 +m2)g l1 cos θ1 +m2 g l2 cos θ2)</p><p>The Lagrangian is:</p><p>L =</p><p>1</p><p>2</p><p>(m1 +m2)l</p><p>2</p><p>1θ̇</p><p>2</p><p>1 +</p><p>1</p><p>2</p><p>m2 l</p><p>2</p><p>2θ̇</p><p>2</p><p>2</p><p>+m2 l1l2θ̇1θ̇2 cos(θ1 − θ2)</p><p>+ (m1 +m2)g l1 cos θ1 +m2 g l2 cos θ2</p><p>Since the Lagrange equations are</p><p>∂L</p><p>∂θ1</p><p>− d</p><p>dt</p><p>∂L</p><p>∂θ̇1</p><p>= 0 (22.2)</p><p>∂L</p><p>∂θ2</p><p>− d</p><p>dt</p><p>∂L</p><p>∂θ̇2</p><p>= 0 (22.3)</p><p>Solving the derivatives of the Lagrangian for θ1 and θ̇1, result in:</p><p>∂L</p><p>∂θ1</p><p>= −m2l1l2θ̇1θ̇2 sin (θ1 − θ2)− (m1 +m2)g l1 sin θ1</p><p>∂L</p><p>∂θ̇1</p><p>= m2l1l2θ̇2 cos (θ1 − θ2) + (m1 +m2)l</p><p>2</p><p>1θ̇1</p><p>d</p><p>dt</p><p>∂L</p><p>∂θ̇1</p><p>= m2l1l2θ̈2 cos (θ1 − θ2)−m2l1l2θ̇2 sin (θ1 − θ2)(θ̇1 − θ̇2) + (m1 +m2)l</p><p>2</p><p>1θ̈1</p><p>(22.4)</p><p>83</p><p>For θ2 and θ̇2</p><p>∂L</p><p>∂θ2</p><p>= m2l1l2θ̇1θ̇2 sin (θ1 − θ2)−m2 g l2 sin θ2</p><p>∂L</p><p>∂θ̇2</p><p>= m2l1l2θ̇1 cos(θ1 − θ2) +m2 l</p><p>2</p><p>2θ̇2</p><p>d</p><p>dt</p><p>∂L</p><p>∂θ̇2</p><p>= m2l1l2θ̈1 cos(θ1 − θ2)−m2l1l2θ̇1 sin(θ1 − θ2)(θ̇1 − θ̇2) +m2 l</p><p>2</p><p>2θ̈2</p><p>(22.5)</p><p>Replacing (22.4) in (22.2) for the angle θ1:</p><p>m2��l1l2 sin (θ1 − θ2)θ̇1θ̇2 + (m1 +m2)g ��l1 sin θ1 +m2��l1l2 cos (θ1 − θ2)θ̈2</p><p>−m2��l1l2 sin (θ1 − θ2)(θ̇1θ̇2 − θ̇22) + (m1 +m2)l�</p><p>2</p><p>1 θ̈1 = 0</p><p>and</p><p>l2</p><p>m2</p><p>(m1 +m2)</p><p>�</p><p>sin (θ1 − θ2)(��</p><p>�θ̇1θ̇2 −��</p><p>�θ̇1θ̇2 + θ̇22) + cos (θ1 − θ2)θ̈2</p><p>�</p><p>+ g sin θ1 + l1θ̈1 = 0</p><p>Therefore:</p><p>θ̈1 +</p><p>l2</p><p>l1</p><p>m2</p><p>(m1 +m2)</p><p>�</p><p>cos (θ1 − θ2) θ̈2 + sin (θ1 − θ2) θ̇22</p><p>�</p><p>+</p><p>g</p><p>l1</p><p>sin θ1 = 0 (22.6)</p><p>Replacing (22.5) in (22.3) for the angle θ2:</p><p>��m2 l1��l2 θ̇1θ̇2 sin (θ1 − θ2)−��m2g ��l2 sin θ2 −��m2 l1��l2 θ̈1 cos (θ1 − θ2)</p><p>+��m2 l1��l2 θ̇1 sin (θ1 − θ2)(θ̇1 − θ̇2)−��m2 l�</p><p>2</p><p>2 θ̈2 = 0</p><p>and</p><p>l1</p><p>�</p><p>− sin (θ1 − θ2)(��</p><p>�θ̇1θ̇2 + θ̇21 −��</p><p>�θ̇1θ̇2 ) + cos (θ1 − θ2)θ̈1</p><p>�</p><p>+ g sin θ2 + l2θ̈2 = 0</p><p>Therefore, dividing for l1:</p><p>l2</p><p>l1</p><p>θ̈2 + cos (θ1 − θ2) θ̈1 − sin (θ1 − θ2) θ̇21 +</p><p>g</p><p>l1</p><p>sin θ2 = 0 (22.7)</p><p>Let</p><p>l2</p><p>l1</p><p>= λ</p><p>m2</p><p>(m1 +m2)</p><p>= µ</p><p>g</p><p>l1</p><p>= ω2</p><p>0</p><p>84 CHAPTER 22. EXERCISE 22 - DOUGLAS O. NOVAES</p><p>From 22,6 and 22,7 the equations of motion can be written as:</p><p>θ̈1 + λ µ</p><p>�</p><p>cos (θ1 − θ2) θ̈2 + sin (θ1 − θ2) θ̇22</p><p>�</p><p>+ ω2</p><p>0 sin θ1 = 0</p><p>λ θ̈2 + cos (θ1 − θ2) θ̈1 − sin (θ1 − θ2) θ̇21 + ω2</p><p>0 sin θ2 = 0</p><p>To small angles, sin θ ≈ θ, sin (θ1 − θ2) ≈ 0 and cos (θ1 − θ2) ≈ 1. Then,</p><p>θ̈1 + λ µ θ̈2 + ω2</p><p>0 θ1 = 0</p><p>θ̈1 + λ θ̈2 + ω2</p><p>0 θ2 = 0</p><p>Exercise 23 - Guilherme S. Y. Giardini</p><p>23. Obtain the equation of motion of a particle falling vertically under the influ-</p><p>ence of gravity when frictional forces obtainable from a dissipation function 1</p><p>2</p><p>kv2 are</p><p>present. Integrate the equation to obtain the velocity as a function of time and show</p><p>that the maximum possible velocity for a fall from the rest is m = mg/k.</p><p>23.1 Non Conservative Euler-Lagrange Equation</p><p>To solve the exercise, first we need to learn how to treat a system under the influ-</p><p>ence of dissipative forces. Usually, a potential only exists when the force field ~F (~r) is</p><p>conservative, in other words ~∇× ~F = 0. If this assumption wasn’t true, the work done</p><p>to move a block in a plane with friction would result in a change of potential energy,</p><p>even though the net position in relation to the potential didn’t change, which would</p><p>imply in a zero change in potential energy, giving</p><p>raise to a contradiction, thus a poten-</p><p>tial cannot be defined when a force is non conservative. For us to obtain the movement</p><p>equations of a dissipative system, we will use the Rayleigh dissipative function.</p><p>First, we consider a dissipative force in cartesian coordinates</p><p>~Fi = −hi(vi)</p><p>~vi</p><p>vi</p><p>; i = 1, . . . , N . (23.1)</p><p>Now, we consider equation (1.52) from Herbert Goldstein 3rd Edition</p><p>85</p><p>86 CHAPTER 23. EXERCISE 23 - GUILHERME S. Y. GIARDINI</p><p>∑</p><p>j=1</p><p>{[</p><p>d</p><p>dt</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>− ∂T</p><p>∂qj</p><p>]</p><p>−Qj</p><p>}</p><p>δqj = 0 . (23.2)</p><p>If we assume that if we had a conservative force, the equation above would</p><p>result in the Euler-Lagrange equation like usual</p><p>d</p><p>dt</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>− ∂T</p><p>∂qj</p><p>−Qj =</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂q̇j</p><p>)</p><p>− ∂L</p><p>∂qj</p><p>, (23.3)</p><p>then we can separate Qj into a conservative and a non conservative part.</p><p>The conservative would give rise to a usual Euler-Lagrange equation and the non-</p><p>conservative part would remain separated</p><p>d</p><p>dt</p><p>(</p><p>∂T</p><p>∂q̇j</p><p>)</p><p>− ∂T</p><p>∂qj</p><p>−</p><p>(</p><p>Qconservative</p><p>j +Qnon-conservative</p><p>j</p><p>)</p><p>=</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂q̇j</p><p>)</p><p>− ∂L</p><p>∂qj</p><p>−Qnon-conservative</p><p>j ,</p><p>then</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂q̇j</p><p>)</p><p>− ∂L</p><p>∂qj</p><p>= Qnon-conservative</p><p>j . (23.4)</p><p>Knowing that Qj is an arbitrary force in the generalized coordinates, we can</p><p>assume a dissipative force ~F and find its generalized counterpart</p><p>Q</p><p>dissipative</p><p>j =</p><p>∑</p><p>i</p><p>~Fi ·</p><p>∂~r</p><p>∂qj</p><p>= −</p><p>∑</p><p>j</p><p>∇vF ·</p><p>∂~ri</p><p>∂qj</p><p>= −</p><p>∑</p><p>j</p><p>∇vF ·</p><p>∂~̇ri</p><p>∂q̇j</p><p>= −∂F</p><p>∂q̇j</p><p>, (23.5)</p><p>23.2. SOLUTION 87</p><p>where F is the Rayleigh dissipation function. Now we proceed to solve the</p><p>exercise.</p><p>23.2 Solution</p><p>Now we proceed to solve the exercise, where F = 1</p><p>2</p><p>kv2. We know that the</p><p>exercise considers a particle free-falling and that it happens only in one dimension. We</p><p>will call this dimension z.</p><p>To find the dissipative part of the system we use equation (23.5)</p><p>Qdissipative = −∂F</p><p>∂ż</p><p>= − ∂</p><p>∂ż</p><p>(</p><p>1</p><p>2</p><p>kv2</p><p>)</p><p>= −kv ,</p><p>here we know that ż = v. Now we have to find the conservative part Euler-</p><p>Lagrangian equation. The kinetic energy is written as</p><p>T =</p><p>mv2</p><p>2</p><p>and (23.6)</p><p>the potential energy written as</p><p>U = −mgz , (23.7)</p><p>then the Lagrangian becomes</p><p>L =</p><p>mv2</p><p>2</p><p>+mgz and (23.8)</p><p>88 CHAPTER 23. EXERCISE 23 - GUILHERME S. Y. GIARDINI</p><p>the Euler-Lagrange equation</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ż</p><p>)</p><p>− ∂L</p><p>∂z</p><p>= Qnon-conservative</p><p>d</p><p>dt</p><p>(mv)−mg = −kv</p><p>a =</p><p>mg − kv</p><p>m</p><p>. (23.9)</p><p>Now we say that a = dv</p><p>dt and solve the resulting differential equation for v</p><p>dv</p><p>dt</p><p>= g − kv</p><p>m</p><p>. (23.10)</p><p>The process to solving this differential equation consists in solving separately</p><p>the constant part and the velocity dependent one v(t) = v1(t) + v2(t).</p><p>dv1</p><p>dt</p><p>= −kv1</p><p>m</p><p>v1(t) = v1(0) e</p><p>�t ; α = − k</p><p>m</p><p>v1(t) = v1(0) e</p><p>� k</p><p>m t and</p><p>v2(t) = v2(0) ; ∀t (23.11)</p><p>because v2 is a constant, by substitution into (23.10), we get v2 = mg</p><p>k . Then the</p><p>complete solution becomes</p><p>v(t) = v1(0) e</p><p>� k</p><p>m t +</p><p>mg</p><p>k</p><p>. (23.12)</p><p>Now we proceed to find the initial condition (t = 0)</p><p>23.2. SOLUTION 89</p><p>v(0) = v1(0) +</p><p>mg</p><p>k</p><p>v1(0) = v(0)− mg</p><p>k</p><p>then (23.13)</p><p>the complete solution becomes</p><p>v(t) =</p><p>(</p><p>v(0)− mg</p><p>k</p><p>)</p><p>e−</p><p>k</p><p>m</p><p>t +</p><p>mg</p><p>k</p><p>(23.14)</p><p>Now we see that when t→∞, we get the terminal velocity</p><p>v(t→∞) =</p><p>mg</p><p>k</p><p>. (23.15)</p><p>We could have just used the derived velocity equation to find an inflection point</p><p>of the velocity and that would also be the terminal velocity.</p><p>Exercise 24 - Geovane Naysinger</p><p>24.1 Questão 1.24</p><p>FIGURE 24.1</p><p>Consider a movement in the x-y plane such that the acceleration of gravity is in</p><p>the positive direction of the y axis.</p><p>24.1.1 a)</p><p>In this section, we will find the Lagrangian of the problem. To start, we have the</p><p>bond to the spring at rest</p><p>91</p><p>92 CHAPTER 24. EXERCISE 24 - GEOVANE NAYSINGER</p><p>FIGURE 24.2</p><p>mg � k(L0 � L � ) = 0</p><p>where L � is the spring length at rest and L0 the spring length at static equilib-</p><p>rium (elastic force modulus is equal to gravitational force modulus).</p><p>We write the position vector</p><p>~r = r (sin � i + cos� j ) = r êr (24.1)</p><p>and its time derivative</p><p>_~r = _r êr + r</p><p>dêr</p><p>dt</p><p>dêr</p><p>dt</p><p>= _� (cos� i + sin � j ) = _� ê�</p><p>_~r = _r êr + r _� ê� (24.2)</p><p>So we are able to write the kinetic term of Lagrangian. The kinetic energy is</p><p>T =</p><p>mv2</p><p>2</p><p>=</p><p>m</p><p>2</p><p>�</p><p>_r 2 + r 2 _� 2</p><p>�</p><p>; (24.3)</p><p>24.1. QUESTÃO 1.24 93</p><p>The relating term to potential energy can be written as a function of r and � as</p><p>V(r; � ) = � mgr cos� +</p><p>k</p><p>2</p><p>(r � L � )2 ; (24.4)</p><p>Therefore, using equations (1.3) and (1.4) we can write the Lagrangian of the system</p><p>L = T � V =</p><p>m</p><p>2</p><p>�</p><p>_r 2 + r 2 _� 2</p><p>�</p><p>+ mgr cos� �</p><p>k</p><p>2</p><p>(r � L � )2 ; (24.5)</p><p>To analyze the movement in r, we must analyze the variations around the equilibrium</p><p>point, that is, we analyze the r � L0. Therefore, it is convenient to define a generalized</p><p>coordinate � such that</p><p>� � r � L0</p><p>and we rewrite the Lagrangian (1.5) in terms of the new coordinate</p><p>L =</p><p>m</p><p>2</p><p>�</p><p>_� 2 + ( � + L0)2 _� 2</p><p>�</p><p>+ mg(� + L0) cos� �</p><p>k</p><p>2</p><p>(� + mg=k)2 (24.6)</p><p>and remembering that mg</p><p>k = L0 � L � .</p><p>Substituting expression L in the Euler-Lagrange equation</p><p>@L</p><p>@q</p><p>�</p><p>d</p><p>dt</p><p>�</p><p>@L</p><p>@_q</p><p>�</p><p>= 0</p><p>we find the equations of motion for the coordinates � and �</p><p>m•� � m(� + L0) _� 2 � mgcos� + k</p><p>�</p><p>� +</p><p>mg</p><p>k</p><p>�</p><p>= 0</p><p>m(� + L0)2•� + 2m(� + L0) _� _� + mg(� + L0) sin � = 0</p><p>(24.7)</p><p>24.1.2 b)</p><p>If � << L 0 and � << 1 ( _� � 0), the equations described above remain</p><p>m•� + k� = 0</p><p>mL 2</p><p>0</p><p>•� + mgL0� = 0</p><p>(24.8)</p><p>94 CHAPTER 24. EXERCISE 24 - GEOVANE NAYSINGER</p><p>representing simple harmonic oscillators frequency</p><p>ωη =</p><p>√</p><p>k</p><p>m</p><p>ωθ =</p><p>√</p><p>g</p><p>L0</p><p>,</p><p>The solutions is well know</p><p>η = A cos (ωηt+ φη)</p><p>θ = B cos (ωθt+ φθ),</p><p>24.1.3 c)</p><p>For the next order, we change the variable η to analyze the deviation in the equi-</p><p>librium length L0. We do</p><p>x = η − mg</p><p>k</p><p>replacing in (1.8), as we keep the first order terms, so</p><p>ẍ = − k</p><p>m</p><p>x</p><p>θ̈ = − g</p><p>Lα +</p><p>m</p><p>k</p><p>g</p><p>θ,</p><p>(24.9)</p><p>And the solutions are similar to the previous case. Therefore,</p><p>η =</p><p>mg</p><p>k</p><p>+ A cos</p><p>(√</p><p>k</p><p>m</p><p>t+ φη</p><p>)</p><p>θ = B cos</p><p>(√</p><p>kg</p><p>kLα +mg</p><p>t+ φθ</p><p>)</p><p>Resonance is rarely (or never) seen in this system. The spring-mass pendulum system</p><p>is a good example to be studied in chaos theories.</p><p>24.1. QUESTÃO 1.24 95</p><p>24.1.4 d)</p><p>In this section we will make a model for a spring-mass pendulum whose spring</p><p>has mass (m’). Like any good model, what will be done is an approximation and we</p><p>will discuss what seems coherent with reality and its obvious problems.</p><p>We will approximate the spring-mass pendulum to a massive spring for the case</p><p>of the Physical Pendulum. So we will start to change our Lagrangian. Let’s make some</p><p>important considerations for the model:</p><p>• The mass is said to be m >> m0. Therefore, we can approximate that the</p><p>position of the center of mass of this system is L0. And we’ll use M as</p><p>M = m+m0.</p><p>• This system does NOT feature a rigid body due to oscillation of position in</p><p>the massive spring. Remembering that the distance between two points is</p><p>constant on the rigid body. However, we will model the system as if it were</p><p>rigid.</p><p>Once that’s done, let’s write the kinetic and potential energy of this new system.</p><p>Kinetic energy is written as</p><p>T =</p><p>Mṙ2</p><p>2</p><p>+</p><p>Iθ̇2</p><p>2</p><p>where I is the moment of inertia of the system. Using Steiner’s Theorem (parallel axes</p><p>theorem)</p><p>I = Icm +Mr2 =ML2</p><p>0 +Mr2</p><p>we can write the kinetic energy as</p><p>T =</p><p>Mṙ2</p><p>2</p><p>+</p><p>θ̇2</p><p>2</p><p>(</p><p>ML2</p><p>0 +Mr2</p><p>)</p><p>,</p><p>Remember that we have the link mg − k(L0 − Lα) = 0 and we can write the L2</p><p>0 as</p><p>L2</p><p>0 = L2</p><p>α+</p><p>(</p><p>Mg</p><p>k</p><p>)2</p><p>+2Lα</p><p>Mg</p><p>k</p><p>and since everything is constant in this case, let’s define this</p><p>expression as L2</p><p>α +</p><p>(</p><p>Mg</p><p>k</p><p>)2</p><p>+ 2Lα</p><p>Mg</p><p>k</p><p>≡ A to facilitate future bills.</p><p>96 CHAPTER 24. EXERCISE 24 - GEOVANE NAYSINGER</p><p>So we write kinetic energy as</p><p>T =</p><p>Mṙ2</p><p>2</p><p>+</p><p>Mθ̇2</p><p>2</p><p>(</p><p>A+ r2</p><p>)</p><p>,</p><p>The potential energy is</p><p>V (r, θ) = −Mgr cos θ +</p><p>k</p><p>2</p><p>(r − Lα)2</p><p>and is written similarly to the case dealt with in (a).</p><p>Once the kinetic and potential energies are found, the system’s Lagrangian en-</p><p>ergy is then:</p><p>L =</p><p>Mṙ2</p><p>2</p><p>+</p><p>Mθ̇2</p><p>2</p><p>(</p><p>A+ r2</p><p>)</p><p>+Mgr cos θ − k</p><p>2</p><p>(r − Lα)2 , (24.10)</p><p>Solving for r, we have</p><p>∂L</p><p>∂r</p><p>=Mθ̇2r +Mg cos θ − k (r − Lα)</p><p>∂L</p><p>∂ṙ</p><p>=Mṙ</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂ṙ</p><p>)</p><p>=Mr̈</p><p>the equation of motion is then</p><p>Mr̈ −Mrθ̇2 −Mg cos θ + k (r − Lα) = 0, (24.11)</p><p>For θ,</p><p>we have</p><p>∂L</p><p>∂θ</p><p>= −Mgr sin θ</p><p>∂L</p><p>∂θ̇</p><p>=Mθ̇</p><p>(</p><p>A+ r2</p><p>)</p><p>d</p><p>dt</p><p>(</p><p>∂L</p><p>∂θ̇</p><p>)</p><p>=Mθ̈</p><p>(</p><p>A+ r2</p><p>)</p><p>+ 2Mθ̇ṙr</p><p>and the equation of motion for θ is</p><p>Mθ̈</p><p>(</p><p>A+ r2</p><p>)</p><p>+ 2Mθ̇ṙr +Mgr sin θ = 0, (24.12)</p><p>24.1. QUESTÃO 1.24 97</p><p>Notemos que para m >> m′, podemos escrever M = m e a contribuição da</p><p>massa da mola deixa apenas um "rastro" geométrico em A. Assim sendo,</p><p>mr̈ −mrθ̇2 −mg cos θ + k (r − Lα) = 0</p><p>mθ̈</p><p>(</p><p>A+ r2</p><p>)</p><p>+ 2mθ̇ṙr +mgr sin θ = 0,</p><p>(24.13)</p><p>Note that for m >> m′, we can write M = m and the contribution of the spring mass</p><p>leaves just a geometric "trail" in A. Consequently,</p><p>mr̈ + kr = 0</p><p>Aθ̈ + gL0θ = 0</p><p>and the angular frequencies are</p><p>ωr =</p><p>√</p><p>k</p><p>m</p><p>ωθ =</p><p>√</p><p>gL0</p><p>A</p><p>=</p><p>√</p><p>gL0</p><p>L2</p><p>α +</p><p>(</p><p>Mg</p><p>k</p><p>)2</p><p>+ 2Lα</p><p>Mg</p><p>k</p><p>=</p><p>√</p><p>g</p><p>L0</p><p>(24.14)</p><p>which means that, for small amplitudes of oscillation, the mass of the spring is not</p><p>relevant as long as the mass is much smaller than the mass of the hanging object.</p><p>Bibliography</p><p>PARIS, P.; ZHANG, L. A disk rolling on a horizontal surface without slip. Mathematical</p><p>and Computer Modelling, v. 36, n. 7, p. 855–860, 2002. ISSN 0895-7177. Disponível em:</p><p><https://www.sciencedirect.com/science/article/pii/S0895717702002327>.</p><p>WIKI, L.-C. 2021. Disponível em: <https://en.wikipedia.org/wiki/Levi-Civita_</p><p>symbol>.</p><p>99</p><p>UNIVERSIDADE FEDERAL DO RIO GRANDE DO SUL</p><p>INSTITUTO DE FÍSICA</p><p>SOLUTIONS MANUAL HERBERT B. GOLDSTEIN 3RD ED.</p><p>L IST 2 OF CLASSICAL MECHANICS</p><p>CLASS : FIP00001</p><p>TEACHER : MARIA BEATRIZ DE LEONE GAY DUCATI</p><p>PORTO ALEGRE , RS</p><p>2021/01</p><p>Contents</p><p>1 Exercise 1 - Jefferson Santana Martins 3</p><p>2 Exercise 2 - Jhordan Silveira De Borba 7</p><p>3 Exercise 3 - Geovane Naysinger 11</p><p>4 Exercise 4 - Douglas Oliveira Novaes 13</p><p>5 Exercise 5 - Jackson Galvão 17</p><p>6 Exercise 6 - Francisco Augusto Ferreira Almeida 21</p><p>7 Exercise 7 - Lucas Bourscheidt 31</p><p>8 Exercise 8 - Guilherme Shoiti Yoshimatsu Giardini 37</p><p>9 Exercise 9 - Rodrigo Weber Pereira 49</p><p>10 Exercise 10 - Edgard Kretschmann 51</p><p>11 Exercise 11 - Gustavo Cortazzi Garcia Kessler 53</p><p>12 Exercise 12 - Jenifer Andrade De Matos 55</p><p>13 Exercise 13 - Jhordan Silveira De Borba 59</p><p>14 Exercise 14 - Douglas Oliveira Novaes 63</p><p>15 Exercise 15 - Jenifer Andrade De Matos 67</p><p>16 Exercise 16 - Geovane Naysinger 71</p><p>16.1 Rascunho Questao 11 cap 8 . . . . . . . . . . . . . . . . . . . . . . . . . . 74</p><p>16.1.1 Rascunho Questao 24 cap 8 . . . . . . . . . . . . . . . . . . . . . . 75</p><p>16.1.2 Rascunho Questao 34 cap 8 . . . . . . . . . . . . . . . . . . . . . . 75</p><p>17 Exercise 17 - Gustavo Cortazzi Garcia Kessler 77</p><p>18 Exercise 18 - Rodrigo Weber Pereira 79</p><p>iii</p><p>CONTENTS 1</p><p>19 Exercise 19 - Jackson Galvão 81</p><p>20 Exercise 20 - Lucas Bourscheidt 85</p><p>21 Exercise 21 - Edgard Kretschmann 95</p><p>22 Exercise 22 - Jefferson Santana Martins 97</p><p>23 Exercise 23 - Francisco Augusto Ferreira Almeida 101</p><p>24 Exercise 24 - Guilherme Shoiti Yoshimatsu Giardini 103</p><p>25 Exercise 25 - Jackson Galvão 107</p><p>26 Exercise 26 - Jhordan Silveira de Borba 113</p><p>Bibliography 117</p><p>Exercise 1 - Jefferson Santana Mar tins</p><p>Complete the solution of the brachistochrone problem begun in the text and</p><p>show the desired curve is a cycloid with a cusp at the initial point at which the particle</p><p>is released. Show also that if the particle is projected with an initial kinetic energy,</p><p>1/2mv20 that the solution is still a cycloid passing through the two points with a cusp at</p><p>a height z above the initial point given by v20 = 2gz.</p><p>R. a)The classical problem in calculus of variation is the so-called brachis-</p><p>tochrone problem, posed (and solved) by Bernoulli in 1696. Given two points 1 and</p><p>2, �nd the path along which an object would slide (disregarding any friction) in the</p><p>shortest possible time from 1 to 2, if it starts at 1 in rest and is only accelerated by</p><p>gravity (see Figure 1).</p><p>Figure 1. Sketch of the brachistochrone problem.</p><p>The �rst (large) step will be the derivation of a formula for the travel time of an</p><p>object from 1 to 2 given the function y. We note �rst that the velocity v(x) of the object at</p><p>the point (x, y(x)) is already determined by y(x): Conservation of energy implies that</p><p>3</p><p>4 CHAPTER 1. EXERCISE 1 - JEFFERSON SANTANA MARTINS</p><p>the sum of the kinetic and the potential energy of the object always remains constant.</p><p>Since it is at rest at the point (0, 0), and the difference in potential energy between the</p><p>point (0, 0) and (x, y(x)) is equal to mgy(x) (m being the mass of the object and g the</p><p>gravitational acceleration), it follows that</p><p>t12 =</p><p>∫ 2</p><p>1</p><p>ds</p><p>v</p><p>=</p><p>∫ 2</p><p>1</p><p>√</p><p>1 +</p><p>(</p><p>dy</p><p>dx</p><p>)2</p><p>v</p><p>dx</p><p>1</p><p>2</p><p>mv2 = mg∆h = mgy → v =</p><p>√</p><p>2gy</p><p>t12 =</p><p>∫ 2</p><p>1</p><p>√</p><p>1 + ẏ2</p><p>2gy</p><p>dx→ ẏ =</p><p>dy</p><p>dt</p><p>(1.1)</p><p>The brachistochrone problem was de�ned by the function F given by</p><p>F =</p><p>√</p><p>1 + ẏ2</p><p>2gy</p><p>(1.2)</p><p>Now, we applied the Euler�Lagrange equations</p><p>d</p><p>dx</p><p>(</p><p>∂F</p><p>∂ẏ</p><p>)</p><p>− ∂F</p><p>∂y</p><p>= 0 (1.3)</p><p>∂F</p><p>∂ẏ</p><p>=</p><p>1</p><p>2</p><p>2ẏ</p><p>√</p><p>2gy</p><p>√</p><p>1 + ẏ2</p><p>→ d</p><p>dx</p><p>(</p><p>∂F</p><p>∂ẏ</p><p>)</p><p>= 0→ ∂F</p><p>∂ẏ</p><p>= c</p><p>ẏ2 = c22gy</p><p>(</p><p>1 + ẏ2</p><p>)</p><p>→ ẏ2</p><p>(</p><p>1− c22gy</p><p>)</p><p>= c22gy</p><p>ẏ =</p><p>dy</p><p>dx</p><p>=</p><p> y(</p><p>1</p><p>2gc2</p><p>)</p><p>− y</p><p>1/2</p><p>=</p><p>(</p><p>y</p><p>k − y</p><p>)1/2</p><p>where k =</p><p>1</p><p>2gc2</p><p>dx =</p><p>(</p><p>k − y</p><p>y</p><p>)1/2</p><p>dy (1.4)</p><p>5</p><p>doing y = k</p><p>2</p><p>[1− cos (φ)]</p><p>dy =</p><p>k</p><p>2</p><p>sin (φ) dφ</p><p>dx =</p><p>[</p><p>k/2 (1 + cos(φ))</p><p>k/2 (1− cos(φ))</p><p>]1/2</p><p>[ksin(φ)dφ] =</p><p>[</p><p>(1 + cos(φ))</p><p>(1− cos(φ))</p><p>]1/2</p><p>[ksin(φ)dφ]</p><p>dx =</p><p>[</p><p>(1 + cos(φ))2</p><p>1− (cos(φ))2</p><p>]1/2</p><p>k</p><p>2</p><p>sin(φ)dφ = ±k</p><p>2</p><p>[1 + cos(φ)] dφ</p><p>x = x0 ±</p><p>k</p><p>2</p><p>[φ+ sin(φ)] and y =</p><p>k</p><p>2</p><p>[1− cos(φ)] (1.5)</p><p>these are the equations of a cycloid, which has a cusp in k = 1</p><p>2gc2</p><p>.</p><p>b) when the initial velocity is different from 0, it is necessary to add a term to the</p><p>energy conservation equation:</p><p>1</p><p>2</p><p>mv2 − 1</p><p>2</p><p>mv20 = mg∆h = mgy → v =</p><p>√</p><p>2gy + v20</p><p>∂F</p><p>∂ẏ</p><p>=</p><p>ẏ√</p><p>2gy + v20</p><p>√</p><p>1 + ẏ2</p><p>and</p><p>d</p><p>dy</p><p>(</p><p>∂F</p><p>∂ẏ</p><p>)</p><p>= 0→ ẏ2 = c2</p><p>(</p><p>2gy + v20</p><p>) (</p><p>1 + ẏ2</p><p>)</p><p>ẏ =</p><p>dy</p><p>dx</p><p>=</p><p>(</p><p>y + v20/2g</p><p>1/2gc2 − y − v20/2g</p><p>)</p><p>=</p><p>(</p><p>k2 + y</p><p>k1 − y</p><p>)1/2</p><p>(1.6)</p><p>with k1 = 1</p><p>2gc2</p><p>− v2</p><p>2g</p><p>and k2 =</p><p>v20</p><p>2g</p><p>.</p><p>dx</p><p>dy</p><p>=</p><p>(</p><p>k1 − y</p><p>k2 + y</p><p>)1/2</p><p>(1.7)</p><p>Doing</p><p>k1 − y = R [1− cos(φ)] and k2 + y = R [1 + cos(φ)] (1.8)</p><p>dx =</p><p>(</p><p>1− cos(φ)</p><p>1− cos(φ)</p><p>)1/2</p><p>[−Rsin(φ)] dφ</p><p>6 CHAPTER 1. EXERCISE 1 - JEFFERSON SANTANA MARTINS</p><p>dx =</p><p>(</p><p>1− cos(φ)</p><p>1− cos(φ)</p><p>)1/2</p><p>[−Rsin(φ)] dφ (1.9)</p><p>x = x1 ±R [φ− sin (φ)] and y = R [1− cos (φ)] (1.10)</p><p>Exercise 2 - Jhordan Silveira De Borba</p><p>Show that if the potential in the Lagrangian contains velocity-dependent terms,</p><p>the canonical momentum corresponding to a coordinate of rotation θ of the entire sys-</p><p>tem is no longer the mechanical angular momentum Lθ but is given by</p><p>pθ = Lθ −</p><p>∑</p><p>i</p><p>n · ri ×∇viU, (2.1)</p><p>where ∇v is the gradient operator in which the derivative are with respect to the ve-</p><p>locity components and n is a unit vector in the direction of rotation. If the forces are</p><p>electromagnetic in character, the canonical momentum is therefore</p><p>pθ = Lθ −</p><p>∑</p><p>i</p><p>n · ri ×</p><p>qi</p><p>c</p><p>Ai, (2.2)</p><p>Answer :</p><p>We have that the generalized moment associated with a coordinate qj is de�ned</p><p>as:</p><p>pj =</p><p>∂L</p><p>∂q̇j</p><p>If the potential is dependent of speed, then for θ:</p><p>pθ =</p><p>∂T</p><p>∂θ̇</p><p>− ∂U</p><p>∂θ̇</p><p>=</p><p>∑</p><p>i</p><p>∂T</p><p>∂ṙi</p><p>∂ṙi</p><p>∂θ̇</p><p>−</p><p>∑</p><p>i,j</p><p>∂U</p><p>∂ṙij</p><p>∂ṙij</p><p>∂θ̇</p><p>7</p><p>8 CHAPTER 2. EXERCISE 2 - JHORDAN SILVEIRA DE BORBA</p><p>Where we have that rij is the component j of the position of the particle i. Remember-</p><p>ing that T = 1</p><p>2</p><p>∑</p><p>imiṙ</p><p>2</p><p>i . Then we can rewrite as:</p><p>pθ =</p><p>∑</p><p>i</p><p>miṙi ·</p><p>∂ṙi</p><p>∂θ̇</p><p>−</p><p>∑</p><p>i</p><p>(∇viU) · ∂ṙi</p><p>∂θ̇</p><p>=</p><p>∑</p><p>i</p><p>[miṙi − (∇viU)] · ∂ṙi</p><p>∂θ̇</p><p>Using spherical coordinatesr = (x, y, z) = (r cosφ sin θ, r sinφ sin θ, r cos θ) , since θ =</p><p>θ (t):</p><p>ṙi =</p><p>(</p><p>ri cosφ cos θθ̇, ri sinφ cos θθ̇,−ri sin θθ̇</p><p>)</p><p>Then:</p><p>∂ṙi</p><p>∂θ̇</p><p>= (ri cosφ cos θ, ri sinφ cos θ,−ri sin θ)</p><p>So, as:</p><p>φ̂× ri = riθ̂</p><p>= (ri cosφ cos θ, ri sinφ cos θ,−ri sin θ)</p><p>=</p><p>∂ṙi</p><p>∂θ̇</p><p>In the coordinate system that we wrote, rotation happens around the axis de�ned by</p><p>the direction φ̂, but without loss of generality we can denote that φ̂ = n, since θ is an</p><p>arbitrary rotation of the entire system. Then:</p><p>pθ =</p><p>∑</p><p>i</p><p>[miṙi − (∇viU)] · n× ri</p><p>=</p><p>∑</p><p>i</p><p>miṙi · (n× ri)−</p><p>∑</p><p>i</p><p>(∇viU) · (n× ri)</p><p>Using a · (b× c) = c · (a× b) = b · (c× a), then:</p><p>pθ =</p><p>∑</p><p>i</p><p>[miṙi − (∇viU)] · n× ri</p><p>=</p><p>∑</p><p>i</p><p>n · (ri ×miṙi)−</p><p>∑</p><p>i</p><p>n · ri ×∇viU</p><p>Then we can rewrite:</p>