Uniqueness of canonical transformations

  • Thread starter neelakash
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  • #1
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The following question seems to be simple enough...Anyway, I hope if someone could confirm what I am thinking.

Is canonical transformation in mechanics unique? We know that given [tex]\ (q, p)\rightarrow\ (Q, P)[/tex], [tex]\ [q,p] = [Q,P] = constant[/tex] and Hamilton's equations of motion stay the same in the new co-ordinates.

My question is: given [tex]\ q\rightarrow Q[/tex] in a canonical transoformation, is the map [tex]\ p\rightarrow P[/tex] uniquely determined? Seems yes to me, but I do not find an off-hand argument in favour.

Can anyone tell how to derive this map?

-Regards,
Neel
 

Answers and Replies

  • #2
vanhees71
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This is a simplified special case, i.e.,
[tex]Q=Q(q).[/tex]
Now we have
[tex]\{Q,P \}=Q'(q) \frac{\partial P}{\partial p} \stackrel{!}{=} 1.[/tex]
This means that
[tex]\frac{\partial P}{\partial p}=-\frac{1}{Q'(q)}.[/tex]
This means that
[tex]P(q,p)=-\frac{p}{Q'(q)}+\tilde{P}(q).[/tex]
with an arbitrary function [itex]\tilde{P}(q)[/itex] alone. Thus, the canonical transformation is determined up to this arbitrary function only.
 
  • #3
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Thanks for the reply...I think I see your point; in such a case, apparently [tex]p\rightarrow P[/tex] is a linear map...By the way, if it was intended, I could not understand the appearance of [tex]'!'[/tex] and [tex]'-'[/tex] sign.
 
  • #4
vanhees71
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Argh! The minus sign is simply wrong.

The exclamation mark over the equality sign indicates that this is a constraint to make the transformation a canonical one (i.e., a symplectomorphism on phase space). One can show that a transformation is canonical if and only if the canonical Poisson-bracket relations hold for the new variables. So the correct answer to your question is

[tex]Q=Q(q), \quad P(q,p)=+\frac{p}{Q'(q)}+\tilde{P}(q).[/tex]

By the way, this is the special case of a "point transformation", which is the same as changing from a generalized coordinate [itex]q[/itex] to an arbitrary new one, [itex]Q[/itex] within the Lagrangian formulation of analytical mechanics.

The important point to realize is that the Hamilton formulation admits a larger group of transformations, namely the canonical transformations!
 
  • #5
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I agree fully...The scope of canonical transformations is a larger than the so-called symmetry transformations...Thank you very much for the explanations...
 

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