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Uniqueness of canonical transformations

  1. Sep 11, 2012 #1
    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
     
  2. jcsd
  3. Sep 11, 2012 #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.
     
  4. Sep 11, 2012 #3
    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.
     
  5. Sep 12, 2012 #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!
     
  6. Sep 12, 2012 #5
    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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