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EP222: Classical Mechanics
Tutorial Sheet 8: Solution
Thistutorialsheetcontainsproblemsrelatedtocanonicaltransformations, Poissonbrack-
ets etc.
1. One of the attempts at combining 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
∗
Q=q+ip, P =Q
is not canonical if the Hamiltonian is left unaltered. Can you find another set of
coordinates Q′ and P′ that are related to Q, P by a change of scale only, and that are
canonical?
Soln: A given transformation is canonical if the Hamilton’s equations are satisfied in
the transformed coordinate system. Therefore, let us evaluate ∂H and ∂H
∂Q ∂P
∂H = ∂H ∂q + ∂H ∂p
∂Q ∂q ∂Q ∂p ∂Q
∂H = ∂H ∂q + ∂H ∂p
∂P ∂q ∂P ∂p ∂P
Using the fact that canonical variables (q,p) satisfy Hamilton’s equations, we obtain
∂H =−p˙∂q +q˙∂p
∂Q ∂Q ∂Q
∂H =−p˙∂q +q˙∂p
∂P ∂P ∂P
Given the fact that
q = 1(P +Q)
2
p = i(P −Q),
2
we have
∂q = ∂q = 1
∂Q ∂P 2
∂p =−∂p =−i
∂Q ∂P 2
Substituting these above, we obtain
∂H 1 i i i ˙
∂Q =−2p˙− 2q˙ = −2(q˙−ip˙) = −2P
∂H 1 i i i ˙
∂P =−2p˙+ 2q˙ = 2(q˙+ip˙) = 2Q
1
Thus, Hamiltonian H expressed in terms of Q and P does not satisfy the Hamilton’s
equations, making the transformation non-canonical. Let us scale these variables to
define Q′ = λQ, and P′ = µP, so that
˙
∂H ∂H ∂Q iP i ˙′
∂Q′ = ∂Q∂Q′ =−2λ =−2λµP
˙
∂H ∂H ∂P iQ i ˙′
∂P′ = ∂P ∂P′ = 2µ = 2λµQ.
If we choose λ and µ such that λµ = i, the Hamilton’s equations will be satisfied in
′ ′ 2
variables Q and P , and the transformation will become canonical. One choice which
will achieve that is
1/2
i (1 +i)
λ=µ=√ = 2
2
2. Show that the transformation for a system of one degree of freedom,
Q=qcosα−psinα
P =qsinα+pcosα,
satisfies the symplectic condition for any value of the parameter α. Find a generating
function for the transformation. What is the physical significance of the transformation
for α = 0? For α = π/2? Does your generating function work for both the cases?
Soln: We will check the symplectic conditions using the order of variables
η = q
p
ζ = Q ,
P
with this
∂Q ∂Q !
M= ∂q ∂p
∂P ∂P
∂q ∂p
= cosα −sinα
sinα cosα
Now we check the two symplectic conditions
MTJM= cosα sinα 0 1 cosα −sinα
−sinα cosα −1 0 sinα cosα
2 2
= sinαcosα−sinαcosα sin α+cos α
−sin2α−cos2α sinαcosα−sinαcosα
= 0 1 =J
−1 0
2
Thus, symplectic condition 1 is satisfied. Similarly, it is easy to verify that the second
symplectic condition MJMT = J is also satisfied for all values of α, making the
transformation canonical. Let us try to find a generating function of the first type,
i.e., F1(q,Q) for the transformation. The governing equations for F1 are
p = ∂F1
∂q
P =−∂F1
∂Q
Using the transformation equations, we can express both p and P in terms of q and
Q, as follows
p = qcotα−Qcscα
P =qsinα+pcosα=qsinα+(qcotα−Qcscα)cosα
2
=⇒ P =q(cos α +sinα)−Qcotα=qcscα−Qcotα.
sinα
Now we integrate the generating equations
∂F1 = p = qcotα−Qcscα
∂q
q2
=⇒ F1 = 2 cotα−Qqcscα+f(Q).
Using this in the second generating equation for F , ∂F1 = −P, we obtain
1 ∂Q
−qcscα+ df =−qcscα+Qcotα
dQ
=⇒ df =Qcotα
dQ
2
=⇒ f(Q)= Q cotα,
2
leading to the final expression for generating function
F (q,Q) = 1 q2 +Q2cotα−Qqcscα.
1 2
Let us consider α = 0, which is nothing but the identity transformation, and our
F1 is indeterminate for that case. This is understandable because we know that this
transformation is generated by F2 = qP. We would have got the correct limiting
behavior for this case if we had instead used F2 generating function. For α = π/2, we
have the interchange transformation, and our generating function becomes F1 = −qQ,
which is the correct result.
3. Show directly that the transformation
Q=log1sinp, P =qcotp
q
3
is canonical.
Soln: We need to just check one of the symplectic conditions, with
∂Q ∂Q !
M= ∂q ∂p
∂P ∂P
∂q ∂p
−1 cotp
= q .
cotp −qcsc2p
Now we check the symplectic condition
T −1 cotp 0 1 −1 cotp
M JM= q q
cotp −qcsc2p −1 0 cotp −qcsc2p
−1 cotp cotp −qcsc2p
= q 1
cotp −qcsc2p −cotp
q
cotp − cotp csc2 p − cot2 p 0 1
= q q = =J
−(csc2p−cot2p) −qcsc2pcotp+qcsc2pcotp −1 0
Because the symplectic condition is satisfied, the transformation is canonical.
4. Show directly that for a system of one degree of freedom the transformation
Q=arctanαq, P = αq2 1+ p2
p 2 α2q2
is canonical, where α is an arbitrary constant of suitable dimensions.
Soln: We will just check one of the symplectic conditions, with
∂Q ∂Q !
M= ∂q ∂p
∂P ∂P
∂q ∂p
αp − αq
2 2 2 2 2 2
= p +α q p +α q .
αq p
α
Let us check the symplectic condition
αp αq 0 1 αp − αq
T 2 2 2 2 2 2 2 2 2
M JM= p +α q p +α q p +α q
− αq p −1 0 αq p
2 2 2
p+αq α α
αp αq αq p/α
2 2 2
= p +α q αp αq
− αq p −
2 2 2 2 2 2 2 2 2
p+αq α !p +α q p +α q
2 2 2 2 2
α pq−α pq p +α q 0 1
2 2 2 2 2 2
= p +α q p +α q = =J.
2 2 2
−p +α q pq−pq −1 0
2 2 2 2 2 2
p +α q p +α q
Thus the transformation is canonical.
5. The transformation between two sets of coordinates are
Q=log(1+q1/2cosp),
P =2(1+q1/2cosp)q1/2sinp.
4
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