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Canonical Transform proof

by Liquidxlax
Tags: canonical, proof, transform
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Liquidxlax
#1
Mar3-12, 04:09 PM
P: 322
Consider the following change of variables in phase space f: maps the reals and is smooth and invertable change of coordinates Q=f(q), q = f-1(Q). Given f, define a change of variables on phase space (q,p) -> (Q,P) by the pair of relations

Pj = (∇f-1)Tjk(f(q))pk

q runs from 1 to n

show that its canonical.



I know that for this to be canonical

(dQi/dqj)(q,p) = (dpj/dPi)(Q,P)

(dQi/dpj)(q,p) = -(dqj/dPi)(Q,P)


i'm having a couple problems, is f the generating function that i have to find explicitly?

Can i use the sympletic method such that MJM^T = J

what is the point of the transpose for the (∇f-1)Tjk part for? i thought f had to be symmetric.
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Liquidxlax
#2
Mar4-12, 11:11 AM
P: 322
for Pj it can equal

(∇f-1)TjkQpk

=(∇f)T *-1jkqpk

but for that to be true then (∇f-1)Tjk has to be symmetric therefore the transpose dissapears

looking at my notes, i think this is supposed to be the associated point transformation in phase space
Liquidxlax
#3
Mar5-12, 12:04 PM
P: 322
Guess i can answer my own question... i knew how to do it but i had an error in my notes

[dQ, dP)T = [{dQ/dq, dQ/dp}, {dP/dq, dP/dp}]*[dq, dp] = Mij*[dq, dp]

to prove its a canonical transformation

MJMT = J where J = [{0,I},{-I,0}] and T represents the transpose


If i do the matrix multiplication and say:


now i will let {a,b} where a and b are some arbitrary coordinates be the poisson brackets

MJMT = [(0,{Q,P}), (-{Q,P},0)]

It is known that {P,P} = 0 = {Q,Q} and {Q,P} = δij

Using the following vvvv


and there you have it


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