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Solutions to Problems in Goldstein, Classical Mechanics, Second Edition Homer Reid October 29, 2002

Chapter 9

Problem 9.1 One of the attempts at combining the two sets of Hamilton’s equations into one tries to take q and p as forming a complex quantity. Show directly from Hamilton’s equations of motion that for a system of one degree of freedom the transformation P = Q∗

Q = q + ip,

is not canonical if the Hamiltonian is left unaltered. Can you find another set of coordinates Q′ , P ′ that are related to Q, P by a change of scale only, and that are canonical? Generalizing a little, we put Q = µ(q + ip),

P = ν(q − ip).

The reverse transformation is   1 1 1 q= P , Q+ ν 2 µ

p=

1 2i



1 1 Q− P ν µ

(1)



.

The direct conditions for canonicality, valid in cases (like this one) in which the

1

Homer Reid’s Solutions to Goldstein Problems: Chapter 9

2

transformation equations do not depend on the time explicitly, are ∂Q ∂q ∂Q ∂p ∂P ∂q ∂P ∂p

∂p ∂P ∂q =− ∂P ∂p =− ∂Q ∂q = . ∂Q =

(2)

When applied to the case at hand, all four of these yield the same condition, namely 1 . µ=− 2iν For µ = ν = 1, which is the case Goldstein gives, these conditions are clearly not satisfied, so (1) is not canonical. But putting µ = 1, ν = − 12i we see that equations (1) are canonical.

Homer Reid’s Solutions to Goldstein Problems: Chapter 9

3

Problem 9.2

(a) For a one-dimensional system with the Hamiltonian p2 1 − 2, 2 2q

H=

show that there is a constant of the motion D=

pq − H t. 2

(b) As a generalization of part (a), for motion in a plane with the Hamiltonian H = |p|n − ar−n , where p is the vector of the momenta conjugate to the Cartesian coordinates, show that there is a constant of the motion D=

p·r − H t. n

(c) The transformation Q = λq, p = λP is obviously canonical. However, the same transformation with t time dilatation, Q = λq, p = λP, t ′ = λ2 t, is not. Show that, however, the equations of motion for q and p for the Hamiltonian in part (a) are invariant under the transformation. The constant of the motion D is said to be associated with this invariance.

(a) The equation of motion for the quantity D is dD ∂D = {D, H } + ∂t dT The Poisson bracket of the second term in D clearly vanishes, so we have 1 {pq, H } − H 2    1 1 1 = pq, p2 − pq, 2 − H. 4 q 4

=

(3)

The first Poisson bracket is 

 ∂(pq ) ∂(p2 ) ∂(pq) ∂(p2 ) pq, p2 = − ∂q ∂p ∂p ∂q = (p)(2p) − 0 = 2p2

(4)

4

Homer Reid’s Solutions to Goldstein Problems: Chapter 9

Next, 

pq,

1 q2



    1 1 ∂ ∂ 2 ∂(pq) ∂(pq ) q q2 − = ∂q ∂p ∂q ∂p   2 = 0− − 3 q q 2 = 2 q

(5)

Plugging (4) and (5) into (3), we obtain 1 dD p2 − 2 −H = 2 2q dt = 0. (b) We have H = (p12 + p22 + p23 )n/2 − a(x12 + x22 + x32)−n/2 so ∂H = anxi (x12 + x22 + x32)−n/2−1 ∂xi ∂H = 2npi(p21 + p22 + p32)n/2−1 . ∂pi Then 

∂(p1 x1 + p2 x2 + p3 x3 ) ∂H ∂(p1 x1 + p2 x2 + p3 x3 ) ∂H − ∂xi ∂xi ∂pi ∂pi i o Xn = np2i (p12 + p22 + p23 )n/2−1 − anx2i (x12 + x22 + x23 )−n/2−1

{p · r, H} =

X



i

= n(p21 + p22 + p23 )n/2 − an(x12 + x22 + x32)−n/2

so if we define D = p · r/n − Ht, then dD ∂D = {D, H } − ∂t dT ∂D 1 = {p · r, H} − ∂t n Substituting in from (6), = |p|n − ar−n − H = 0.

(6)

Homer Reid’s Solutions to Goldstein Problems: Chapter 9

5

(c) We put 

t′ Q(t ) = λq λ2 ′



,

 ′ t 1 P (t ) = p 2 . λ λ ′

(7)

Since q and p are the original canonical coordinates, they satisfy ∂H =p ∂p ∂H 1 p˙ = − = 3. q ∂q

q˙ =

(8)

On the other hand, differentiating (7), we have  ′ 1 t dQ = q ˙ dt′ λ λ2  ′ t 1 = p λ λ2

= P (t′ )  ′  t dP 1 p ˙ = λ3 λ2 dt′ 1 1 = 3  t′  λ q λ2 1 = 3 ′ Q (t )

which are the same equations of motion as (8).

Problem 9.4 Show directly that the transformation   1 sin p , Q = log p

P = q cot p

is canonical. The Jacobian of the transformation is ∂Q ∂q ∂P ∂q

M= =



− 1q cot p

∂Q ∂p ∂P ∂p

!

cot p −q csc2 p



.

6

Homer Reid’s Solutions to Goldstein Problems: Chapter 9

Hence 

− q1 cot p

cot p − q1 cot p −q csc2 p    cot p −q csc2 p − 1q cot p = 1 − cot p cot p −q csc2 p q   0 csc2 p − cot2 p = 2 2 cot p − csc p 0   0 1 = −1 0

˜ MJM =

cot p −q csc2 p



0 1 −1 0





=J

so the symplectic condition is satisfied.

Problem 9.5 Show directly for a system of one degree of freedom that the transformation   αq αq 2 p2 Q = arctan , P = 1+ 2 2 α q p 2 is canonical, where α is an arbitrary constant of suitable dimensions. The Jacobian of the transformation is   M=

∂Q ∂q

∂Q ∂p

∂P ∂q

∂P ∂p

   α p

=

so 

˜ MJM =  

= =J

  α p

−

0



αq p2

1

1+( )

1

−1 0

1

2

1+( αq p )



1 2 1+( αq p )

−

αq

αq

αq 2 p



p α

  



so the symplectic condition is satisfied.

−



αq p2



1 2 1+( αq p )

p α

αq   α p



. p α

1

1+(

αq 2 p

)

+



αq p2



1 1+(

αq p

2

)





7

Homer Reid’s Solutions to Goldstein Problems: Chapter 9

Problem 9.6 The transformation equations between two sets of coordinates are Q = log(1 + q 1/2 cos p) P = 2(1 + q 1/2 cos p)q 1/2 sin p (a) Show directly from these transformation equations that Q, P are canonical variables if q and p are. (b) Show that the function that generates this transformation is F3 = −(eQ − 1)2 tan p. (a) The Jacobian of the transformation is   M=



=

∂Q ∂q

∂Q ∂p

∂P ∂q

∂P ∂p

1 2



q −1/2 cos p 1+q 1/2 cos p

1/2

q sin p − 1+q 1/2 cos p

2q 1/2 cos p + 2q cos2 p − 2q sin2 p     q 1/2 sin p q −1/2 cos p 1 − 1/2 cos p 1/2 1+q 1+q cos p 2 . = q −1/2 sin p + sin 2p 2q 1/2 cos p + 2q cos 2p q

−1/2

Hence we have    ˜ MJM =

sin p + 2 cos p sin p

−1/2 cos p 1 q 2 1+q 1/2 cos p 1/2 q sin p − 1+q 1/2 cos p

q −1/2 sin p + sin 2p 2q 1/2 cos p + 2q cos 2p









 q −1/2 sin p + sin 2p 2q 1/2 cos p + 2q cos 2p  ×   q−1/2 cos p q 1/2 sin p − 21 1+q 1/2 cos p 1+q 1/2 cos p   cos2 p+sin2 p+q 1/2 cos p cos 2p+q 1/2 sin p sin 2p 0 1/2 1+q cos p  = 2 2 1/2 cos p cos 2p+q 1/2 sin p sin 2p 0 − cos p+sin p+q 1+q 1/2 cos p   0 1  = −1 0 

=J

so the symplectic condition is satisfied.

Homer Reid’s Solutions to Goldstein Problems: Chapter 9

8

(b) For an F3 function the relevant relations are q = −∂F/∂p, P = −∂F/∂Q. We have F3 (p, Q) = −(eQ − 1)2 tan p so ∂F3 = 2eQ(eQ − 1) tan p P =− ∂Q ∂F3 q=− = (eQ − 1)2 sec2 p. ∂p The second of these may be solved to yield Q in terms of q and p: Q = log(1 + q 1/2 cos p) and then we may plug this back into the equation for P to obtain P = 2q 1/2 sin p + q sin 2p as advertised.

Problem 9.7

(a) If each of the four types of generating functions exist for a given canonical transformation, use the Legendre transformation to derive relations between them. (b) Find a generating function of the F4 type for the identity transformation and of the F3 type for the exchange transformation. (c) For an orthogonal point transformation of q in a system of n degrees of freedom, show that the new momenta are likewise given by the orthogonal transformation of an n−dimensional vector whose components are the old momenta plus a gradient in configuration space.

Problem 9.8 Prove directly that the transformation Q 1 = q1 ,

P1

Q 2 = p2 ,

P2

= p1 − 2p2

= −2q1 − q2

is canonical and find a generating function. After a little hacking I came up with the generating function F13(p1 , Q1 , q2 , Q2 ) = −(p1 − 2Q2 )Q1 + q2 Q2

9

Homer Reid’s Solutions to Goldstein Problems: Chapter 9

which is of mixed F3 , F1 type. This is Legendre-transformed into a function of the F1 type according to F1 (q1 , Q1 , q2 , Q2 ) = F13 + p1 q1 . The least action principle then says ˙ 2 − K (Qi , Pi ) + ∂F13p˙1 + ∂F13 Q˙ 1 p1q˙1 + p2q˙2 − H (qi , pi ) = P1Q˙ 1 + P2 Q ∂Q1 ∂p1 ∂F13 ˙ ∂F13 q˙2 + + Q2 + p1q˙1 + q1 p˙1 ∂Q2 ∂q2 whence clearly ∂F13 = Q1 ∂p1 ∂F13 = −p1 − 2Q2 P1 = − ∂Q1 q1 = −

X

= −p1 − 2p2 ∂F13 p2 = = Q2 ∂q2 ∂F13 P2 = − = −2Q1 − q2 ∂Q2

X X = −2q1 − q2

X.

Problem 9.14 By any method you choose show that the following transformation is canonical: 1 p ( 2P1 sin Q1 + P2 ), α 1 p y = ( 2P1 cos Q1 + Q2 ), α

x=

px py

α p ( 2P1 cos Q1 − Q2 ) 2 α p = − ( 2P1 sin Q1 − P2 ) 2 =

where α is some fixed parameter. Apply this transformation to the problem of a particle of charge q moving in a plane that is perpendicular to a constant magnetic field B. Express the Hamiltonian for this problem in the (Qi , Pi ) coordinates, letting the parameter α take the form α2 =

qB . c

From this Hamiltonian obtain the motion of the particle as a function of time. We will prove that the transformation is canonical by finding a generating function. Our first step to this end will be to express everything as a function

Homer Reid’s Solutions to Goldstein Problems: Chapter 9

10

of some set of four variables of which two are old variables and two are new. After some hacking, I arrived at the set {x, Q 1 , py , Q2 }. In terms of this set, the remaining quantities are   1 1 1 y= x − 2 py cot Q1 + Q2 (9) α 2 α  2  α 1 α px = x − py cot Q1 − Q2 (10) 4 2 2   2 2 1 1 α x (11) − xpy + 2 p2y csc2 Q1 P1 = 2 8 2α α 1 (12) P 2 = x + py α 2 We now seek a generating function of the form F (x, Q1 , py , Q2 ). This is of mixed type, but can be related to a generating function of pure F1 character according to F1 (x, Q1 , y, Q2 ) = F (x, Q1 , py , Q2 ) − ypy .

Then the principle of least action leads to the condition

∂F ˙ ∂F ∂F ˙ ∂F Q1 + p˙y + Q2 + yp˙y + pyy˙ pxx˙ + py y˙ = P1 Q˙ 1 + P2Q˙ 2 + x˙ + ∂Q2 ∂Q1 ∂py ∂x from which we obtain ∂F ∂x ∂F y=− ∂py ∂F P1 = − ∂Q1 ∂F . P2 = − ∂Q2 px =

(13) (14) (15) (16)

Doing the easiest first, comparing (12) and (16) we see that F must have the form α 1 F (x, Q1 , py , Q2 ) = − xQ2 − py Q2 + g(x, Q1 , py ). (17) α 2 Plugging this in to (14) and comparing with (14) we find   1 1 (18) g(x, Q1 , py ) = − xpy + 2 p2y cot Q1 + ψ(x, Q1 ). 2α 2 Plugging (17) and (18) into (13) and comparing with (10), we see that ∂ψ α2 x cot Q1 = 4 ∂x

Homer Reid’s Solutions to Goldstein Problems: Chapter 9

11

or

α 2 x2 cot Q1 . (19) 8 Finally, combining (19), (18), (17), and (15) and comparing with (11) we see that we may simply take φ(Q1 ) ≡ 0. The final form of the generating function is then     2 2 α 1 1 1 α x F (x, Q1 , py , Q2 ) = − − xpy + 2 py2 cot Q1 x + py Q 2 + 2 8 2 α 2α ψ(x, Q1 ) =

and its existence proves the canonicality of the transformation. Turning now to the solution of the problem, we take the B field in the z k, and put direction, i.e. B = B0 ˆ  B0  A= − y iˆ+ x ˆj . 2 Then the Hamiltonian is

1  q 2 p− A c 2m "  2  #  1 qB0 2 qB0 = + py − y x px + 2c 2c 2m " 2 # 2  α2 1 α2 = px + x y + py − 2m 2 2

H (x, y, px , py ) =

where we put α2 = qB/c. In terms of the new variables, this is  2  p 2  1  p α 2P1 cos Q1 + α 2P1 sin Q1 H (Q1 , Q2 , P1 , P2 ) = 2m α2 = P1 m = ωc P1 where ωc = qB/mc is the cyclotron frequency. From the Hamiltonian equations of motion applied to this Hamiltonian we see that Q 2 , P1 , and P2 are all constant, while the equation of motion for Q1 is ∂H Q˙ 1 = = ωc −→ Q 1 = ωc t + φ ∂P1 √ for some phase φ. Putting r = 2P1 /α, x0 = P2 /α, y0 = Q2 /α we then have x = r(sin ωc t + φ) + x0 ,

px

y = r(cos ωc t + φ) + y0 ,

py

mωc [r cos(ωc t + φ) − y0 ] 2 mωc = [r sin(ωc t + φ) + x0 ] 2 =

in agreement with the standard solution to the problem....