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Canonical Transform proof 
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#1
Mar312, 04:09 PM

P: 321

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
P_{j} = (∇f^{1})^{T}_{jk}(f(q))p_{k} q runs from 1 to n show that its canonical. I know that for this to be canonical (dQ_{i}/dq_{j})_{(q,p)} = (dp_{j}/dP_{i})_{(Q,P)} (dQ_{i}/dp_{j})_{(q,p)} = (dq_{j}/dP_{i})_{(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})^{T}_{jk} part for? i thought f had to be symmetric. 


#2
Mar412, 11:11 AM

P: 321

for P_{j} it can equal
(∇f^{1})^{T}_{jk}Qp_{k} =(∇f)^{T *1}_{jk}qp_{k} but for that to be true then (∇f1)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 


#3
Mar512, 12:04 PM

P: 321

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] = M_{ij}*[dq, dp] to prove its a canonical transformation MJM^{T} = 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 MJM^{T} = [(0,{Q,P}), ({Q,P},0)] It is known that {P,P} = 0 = {Q,Q} and {Q,P} = δ_{ij} Using the following vvvv and there you have it 


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