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Canonical transformation

by Logarythmic
Tags: canonical, transformation
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Logarythmic
#1
Dec17-06, 04:49 AM
P: 282
1. The problem statement, all variables and given/known data

Consider a canonical transformation with generating function

[tex]F_2 (q,P) = qP + \epsilon G_2 (q,P)[/tex],

where [tex]\epsilon[/tex] is a small parameter.
Write down the explicit form of the transformation. Neglecting terms of order [tex]\epsilon^2[/tex] and higher,find a relation between this transformation and Hamilton's equations of motion, by setting [tex]G_2=H[/tex] (why is this allowed?) and [tex]\epsilon = dt[/tex].


2. The attempt at a solution

I think the transformation equations are

[tex]\delta p = P - p = -\epsilon \frac{\partial G_2}{\partial q}[/tex]

and

[tex]\delta q =Q-q=\epsilon \frac{\partial G_2}{\partial q}[/tex]


Quote Quote by vanesch
I guess there's a typo here:
[tex]\delta q =Q-q=\epsilon \frac{\partial G_2}{\partial P}[/tex]
but how can I solve the last part? Can I just say that with the use of H and dt the equations can be written as

[tex]\dot{p}=-\frac{\partial H}{\partial q}[/tex]

and

[tex]\dot{q}=\frac{\partial H}{\partial P}[/tex]

which are the Hamiltonian equations of motion? And why is this allowed?

Quote Quote by vanesch
The idea is that we work in first order in [tex]\epsilon[/tex], and that you can hence replace everywhere [tex]P[/tex] by [tex]p[/tex] as the difference will introduce only second-order errrors.
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vanesch
#2
Dec17-06, 08:21 AM
Emeritus
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P: 6,236
Ooops !

I'm terribly sorry, instead of replying, I erroneously edited your post !
Logarythmic
#3
Dec17-06, 09:13 AM
P: 282
Yes it should be a P, not a q. I get the relations

[tex]Q = q + \delta q[/tex]

and

[tex]P = p + \delta p[/tex]

but why is it allowed to use G = H?

vanesch
#4
Dec17-06, 10:04 AM
Emeritus
Sci Advisor
PF Gold
P: 6,236
Canonical transformation

Quote Quote by Logarythmic View Post
Yes it should be a P, not a q. I get the relations

[tex]Q = q + \delta q[/tex]

and

[tex]P = p + \delta p[/tex]

but why is it allowed to use G = H?
You have the choice ! Every function G(q,P) (which is smooth enough) will generate a canonical transformation. So you may just as well use H(q,P), and then - that's the whole point - the transformation equations from (q,p) into (Q,P) give you simply the genuine time evolution where epsilon is the small time step. For an arbitrary G that isn't the case of course, you've just transformed your coordinates (q,p) in some other (Q,P). But for G = H, you've transformed the coordinates (q,p) in what they will be, a small moment later !
The reason for that is that your transformation equations you've found for an arbitrary G are what they are, and become the Hamilton equations of motion when you pick G to be equal to H.

Now, strictly speaking we should write H(q,P) instead of H(q,p), but we can replace the P by p here, because they are only a small amount different.
Marco_84
#5
Dec18-06, 03:24 PM
P: 173
and it also preserve the fundamental poisson brackets....

its just a quick calcolus {Q,P}


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