Exercises in Classical Mechanics CUNY GC, Prof. D. Garanin No.4 Solution ||||||||||||||||||||||||||||||||||||||||| 1 Hamiltonian formalism for the double pendulum (10 points) Consider a double pendulum that consists of two massless rods of length l1 and l2 with masses m1 and m2 attached to their ends. The ¯rst pendulum is attached to a ¯xed point and can freely swing about it. The second pendulum is attached to the end of the ¯rst one and can freely swing, too. The motion of both pendulums is con¯ned to a plane, so that it can be described in terms of their angles with respect to the vertical, '1 and '2: a) Write down the Lagrange function for this system. b) Introduce generalized momenta p1 and p2 and change to the Hamiltonian description. Find the transformation matrix that yields the velocities' _ 1 and' _ 2 in terms of the momenta p1 and p2: Write down the Hamilton function H('1; p1;'2; p2) using the transformation matrix. c) Obtain the Hamilton equations. Solution: a) Both kinetic and potential energy of the system are the sums of the contributions of the ¯rst and second masses: L = T ¡ U; T = T1 + T2;U = U1 + U2: (1) For the coordinates of the masses 1 and 2 one has x1 = l1 sin '1; y1 = l1 cos '1 (2) and x2 = l1 sin '1 + l2 sin '2; y2 = l1 cos '1 + l2 cos '2; (3) with the y-axis directed downwards. For the potential energies one has U1 = ¡m1gy1 = ¡m1gl1 cos '1;U2 = ¡m2gy2 = ¡m2g (l1 cos '1 + l2 cos '2) : (4) Kinetic energies are given by m T = 1 l2'_ 2 (5) 1 2 1 1 and m ³ ´ m h i T = 2 x_ 2 +y _2 = 2 l2'_ 2 + l2'_ 2 + 2l l cos (' ¡ ' )_' '_ : (6) 2 2 2 2 2 1 1 2 2 1 2 1 2 1 2 All together yields m + m m L = 1 2 l2'_ 2 + 2 l2'_ 2 + m l l cos (' ¡ ' )_' '_ 2 1 1 2 2 2 2 1 2 1 2 1 2 + (m1 + m2) gl1 cos '1 + m2gl2 cos '2: (7) b) The generalized momenta are given by @L 2 p1 = = (m1 + m2) l1'_ 1 + m2l1l2 cos ('1 ¡ '2)_'2 @'_ 1 @L 2 p2 = = m2l2'_ 2 + m2l1l2 cos ('1 ¡ '2)_'1: (8) @'_ 2 This can be written in the matrix form as à ! à ! à ! 2 p1 '_ 1 (m1 + m2) l1 m2l1l2 cos ('1 ¡ '2) = K ;K = 2 : (9) p2 '_ 2 m2l1l2 cos ('1 ¡ '2) m2l2 Note that the kinetic energy in Eq. (7) can be written as à ! à ! ³ ´ ³ ´ 1 '_ 1 1 p1 1 T = '_ 1 '_ 2 K = '_ 1 '_ 2 = (_'1p1 +' _ 2p2) (10) 2 '_ 2 2 p2 2 and that KT = K: The inverse transformation reads à ! à ! '_ p 1 = K¡1 1 ; (11) '_ 2 p2 where à ! 2 ¡1 1 m2l2 ¡m2l1l2 cos ('1 ¡ '2) K = 2 2 2 2 ; (12) m2l1l2 [m1 + m2 ¡ m2 cos ('1 ¡ '2)] ¡m2l1l2 cos ('1 ¡ '2)(m1 + m2) l1 ¡ ¢ again K¡1 T = K¡1: Using the transposed relation ³ ´ ³ ´ ³ ´T ³ ´ ¡1 ¡1 '_ 1 '_ 2 = p1 p2 K = p1 p2 K ; (13) one can write the kinetic energy, Eq. (10), in the form à ! ³ ´ 1 ¡1 p1 T = p1 p2 K : (14) 2 p2 Now the Hamilton function becomes, à ! ³ ´ 1 ¡1 p1 H = T + U = p1 p2 K ¡ (m1 + m2) gl1 cos '1 ¡ m2gl2 cos '2: (15) 2 p2 In the standard form this reads 2 2 2 2 m2l2p1 + (m1 + m2) l1p2 ¡ 2m2l1l2 cos ('1 ¡ '2) p1p2 H = 2 2 2 ¡ (m1 + m2) gl1 cos '1 ¡ m2gl2 cos '2: (16) 2m2l1l2 [m1 + m2 ¡ m2 cos ('1 ¡ '2)] c) The Hamilton equations read @H @H '_ 1 = ; p_1 = ¡ @p1 @'1 @H @H '_ 2 = ; p_2 = ¡ : (17) @p2 @'2 The task to work out these equations is left to the reader. The equations forp _1 andp _2 are pretty cumbersome since one has to di®erentiate the denominator. It is best to do with a mathematical software. The whole system of Hamiltonian equations for the double pendulum is much more cumbersome than the system of Lagrange equations. The only purpose to consider the Hamilton equations here is to show that the Hamiltonian formalism is not well suited for engineering-type problems with constraints. 2 Canonical transformations (10 points) a) The canonical transformations between two sets of variables are p p p Q = ln (1 + q cos p) ;P = 2 (1 + q cos p) q sin p: (18) Show directly that this transformation is canonical. Show that ³ ´2 Q FpQ(p; Q) = ¡ e ¡ 1 tan p is the generating function of this transformation. b) For what values of ® and ¯ do the equations Q = q® cos (¯p) ;P = q® sin (¯p) represent a canonical transformation? What is the form of the generating function FpQ(p; Q) in this case? Solution: a) To check whether a transformation is canonical, one can show that the fundamental Poisson brackets are invariant: fQ; P gq;p = fQ; P gQ;P = 1: (19) Explicit calculation is below: @Q @P @Q @P fQ; P g = ¡ q;p @q @p @p @q 1 1 p p p p = p p cos p £ 2 [¡ q sin p q sin p + (1 + q cos p) q cos p] 1 + q cos p 2 q " # 1 p 1 + p q sin p £ p sin p + 2 cos p sin p 1 + q cos p q 1 h p p p i = p ¡ q cos p sin2 p + cos2 p + q cos3 p + sin2 p + 2 q cos p sin2 p 1 + q cos p 1 h p p i = p 1 + q cos p sin2 p + q cos3 p 1 + q cos p 1 h p ³ ´i = p 1 + q cos p sin2 p + cos2 p = 1: (20) 1 + q cos p Now check that ³ ´2 Q FpQ(p; Q) = ¡ e ¡ 1 tan p is the generating function of this transformation. that is, one has to check the relations @F @F q = ¡ pQ ;P = ¡ pQ : (21) @p @Q One obtains @F ³ ´2 1 ¡ pQ = eQ ¡ 1 : (22) @p cos2 p On the other hand, from Eq. (18) follows p eQ ¡ 1 = q cos p; (23) so that indeed @F ¡ pQ = q: (24) @p Further with the help of Eq. (23) one obtains @F ³ ´ p p ¡ pQ = 2 eQ ¡ 1 eQ tan p = (2 q cos p) (1 + 2 q cos p) tan p @Q p p = 2 (1 + 2 q cos p) q sin p = P; (25) as it should be. b) Let us calculate the Poisson brackets @Q @P @Q @P fQ; P g = ¡ q;p @q @p @p @q = ®q®¡1 cos (¯p) £ q®¯ cos (¯p) + q®¯ sin (¯p) ®q®¡1 sin (¯p) h i = ®¯q2®¡1 cos2 (¯p) + sin2 (¯p) = ®¯q2®¡1: (26) For the transformation to be canonical, this Poisson bracket should be identically equal to 1 that requires ® = 1=2; ¯ = 2; (27) i.e., Q = q1=2 cos (2p) ;P = q1=2 sin (2p) : (28) The generating function FpQ(p; Q) should satisfy Eq. (21). To use Eq. (21) to ¯nd FpQ(p; Q); one should ¯rst express q and P via the arguments p; Q: From Eq. (28) one obtains @F @F q = ¡ pQ ;P = ¡ pQ : (29) @p @Q Q2 q = ;P = Q tan (2p) : cos2 (2p) Now integrating the equation @F P = Q tan (2p) = ¡ pQ (30) @Q on Q one obtains Q2 F = ¡ tan (2p) + f(p): (31) pQ 2 Here the integration Q-constant f(p) can be obtained from another relation Q2 @F Q2 df(p) q = = ¡ pQ = + : (32) cos2 (2p) @p cos2(2p) dp This yields df(p) = 0 ) f(p) = const; (33) dp an irrelevant constant that can be dropped. Thus FpQ is given by Eq. (31) with f(p) = 0: 3 Vortex dynamics (10 points) Consider the equations of motions describing vortices of strength γi with positions ri = (xi; yi) in the plane X y ¡ y X x ¡ x x_ = ¡ γ i j ; y_ = + γ i j : (34) i j jr ¡ r j2 i j jr ¡ r j2 j6=i i j j6=i i j Consider the Hamiltonian H and the following Poisson brackets: µ ¶ 1 X Xn 1 @f @g @f @g H = ¡ γ γ ln jr ¡ r j; ff; gg = ¡ : (35) 2 i j i j γ @x @y @y @x j6=i i i i i i i (a) Check that Hamilton equations x_ i = fxi; Hg; y_i = fyi; Hg (36) reproduce the equations of motions. (Check that the standard Hamilton equations can be written in this form, too) (b) Show that the following quantities are conserved: X X Px = γiyi;Py = ¡ γixi: (37) i (c) Find the Poisson brackets fPx; Hg, fPy; Hg, and fPx;Pyg, as de¯ned above. (d) Find the solution of the equations of motion for a system of two vortices. Solution: a) The Poisson brackets yield the Hamiltonian equations à ! X 1 @x @H @x @H 1 @H x_ = fx ; Hg = k ¡ k = k k γ @x @y @y @x γ @y j j j j j j k k à ! X 1 @y @H @y @H 1 @H y_ = fy ; Hg = k ¡ k = ¡ : (38) k k γ @x @y @y @x γ @x j j j j j j k k Using Eq. (35) one obtains 1 @H 1 X γ γ @ ln jr ¡ r j 1 X γ γ 1 @jr ¡ r j x_ = = ¡ i j i j = ¡ i j i j (39) k γ @y 2 γ @y 2 γ jr ¡ r j @y k k j6=i k k j6=i k i j k and 1 @H 1 X γ γ @ ln jr ¡ r j 1 X γ γ 1 @jr ¡ r j y_ = ¡ = i j i j = i j i j (40) k γ @x 2 γ @x 2 γ jr ¡ r j @x k k j6=i k k j6=i k i j k In these expressions q 2 2 @jri ¡ rjj @ (xi ¡ xj) + (yi ¡ yj) yi ¡ yj @ (yi ¡ yj) yi ¡ yj = = q = (±ik ¡ ±jk) (41) @yk @yk 2 2 @yk jri ¡ rjj (xi ¡ xj) + (yi ¡ yj) and @jri ¡ rjj xi ¡ xj = (±ik ¡ ±jk) : (42) @xk jri ¡ rjj Substituting these results in the equations above one obtains 1 X γ γ y ¡ y 1 X y ¡ y 1 X y ¡ y x_ = ¡ i j i j (± ¡ ± ) = ¡ γ k j + γ i k (43) k 2 γ jr ¡ r j2 ik jk 2 j jr ¡ r j2 2 i jr ¡ r j2 j6=i k i j j6=k k j k6=i i k that is, X y ¡ y x_ = ¡ γ k i (44) k i jr ¡ r j2 i6=k k i and, similarly, X x ¡ x y_ = γ k i : (45) k i jr ¡ r j2 i6=k k i These equations indeed coincide with those given in the formulation of the problem.