Canonical transformation

Discussion in 'Physics & Math' started by neelakash, Oct 14, 2008.

  1. neelakash Registered Senior Member

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    491
    Show that the time reversal transformation given by Q = q, P = − p and T = − t, is canonical, in the sense that the form of the Hamiltonian equations of motion is preserved. However, it does not satisfy the invariance of the fundamental Poisson Bracket relations. This is an example when the two criteria are not equal.

    This I have done...I just ask you to check if the procedure is correct.( ' denotes d/dt )

    Q'=(dQ/dT)=(dQ/dt)(dt/dT)= -(dQ/dt)= -[(∂Q/∂q)q' + (∂Q/∂p)p' + (∂Q/∂t)]= -[q']= -(∂H/∂p)

    Also, (∂K/∂P)=(∂H/∂p)(∂p/∂P)= -(∂H/∂p)...[we write (∂K/∂P)=(∂H/∂p) as Kamiltonian K is a

    function of Q,P,T and Hamiltonian H is a function of q,p,t]

    Thus, Q'= (∂K/∂P)...1st of Hamilton's canonical equations is proved.

    Similarly, P'= (dP/dT)= -(dP/dt)= -[(∂P/∂q)q' + (∂P/∂p)p' + (∂P/∂t)]= p'= -(∂H/∂q)

    Then, (∂K/∂Q)=(∂H/∂q)(∂q/∂Q)=(∂H/∂q)

    Thus, P'= -(∂K/∂Q)

    This shows that the given transformation leads (q,p) to canonically conjugate variables(Q,P)

    Evaluating the Poisson brackets it is easy to show that they do not satisfy fundamental Poisson bracket.

    Can anyone suggest why there is a mismatch between the two aspects?
     
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  3. neelakash Registered Senior Member

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    Are people not interested for I have not used TeX?

    My attempt:

    \(\dot {Q}= (dQ/dT) = (dQ/dt)(dt/dT) = -(dQ/dt) = -[(\partial Q / \partial q) \dot {q} + (\partial Q / \partial p) \dot {p} + (\partial Q / \partial t)] = -\dot {q} = -(\partial H / \partial p)\)


    \((\partial K / \partial P)=(\partial H / \partial p)(\partial\ p / \partial P)= - (\partial H / \partial p)\)

    Thus, \(\dot {Q}= (dQ/dT) = (\partial K / \partial P) \)

    Similarly,

    \(\dot {P}= (dP/dT) = (dP/dt)(dt/dT) = -(dP/dt) = -[(\partial P / \partial q) \dot {q} + (\partial P / \partial p) \dot {p} + (\partial P / \partial t)] = \dot {p} = -(\partial H / \partial q)\)


    \((\partial K / \partial Q)=(\partial H / \partial q)(\partial\ q / \partial Q)= (\partial H / \partial q)\)

    Thus, \(\dot {P}= (dP/dT) = -(\partial K / \partial Q) \)

    The most sensitive part of this observation is ofcourse the identification that \(\ K = \ H \) that I have

    assumed.In general,however,there will be a partial time derivative of a generating function F so that the equation lokks like

    \(\ K = \ H + (\partial F / \partial t)\)

    I have assumed that F has no explicit time dependence.So the later time derivative is zero and we have Hamilton's equations

    satisfied...

    My question is if my assumption is justified for this problem...Note that, the problem already ensures that the transformation does

    not satisfy the fundamental Poisson brackets.My calculation gave the value of Poissson Bracket -1 instead of 1 (the correct

    one). In that sense,it is not a canonical transformation...

    I am not sure, but it looks to me, my assumption that F does not have explicit time dependence may be somehow related to the fact of fundamnental Poisson Bracket not

    being satisfied...

    Can anyone please take interest?
     
    Last edited: Oct 15, 2008
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  5. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Sorry Neelakash, can you remind me what the variables are here?

    H is the hamiltonian, K is the transformed hamiltonian?

    Specifically, how can you make the identification (∂K/∂P)=(∂H/∂p)(∂p/∂P)?
     
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  7. neelakash Registered Senior Member

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    491
    Hi Ben,

    yes...H is Hamiltonian: H(q,p,t) and K=K(Q,P,T) is the transformed Hamiltonian (often called Kamiltonian)

    Yes, apparently,I should hsve written

    (∂K/∂P)=(∂K/∂p)(∂p/∂P)

    But,as K is a function of (Q,P,T) we cannot take partial derivative w.r.t.

    p.However,I have assumed K=H (whether this assumption is justified will be

    cansidered later).And H=H(q,p,t) and we can take partal derivative w.r.t. p
     
  8. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    8,967
    Ok I see.

    The thing to realize is that P=P(p). That is, K is implicitly a function of p, if it is a function of P. This means that the expression

    (∂K/∂P)=(∂K/∂p)(∂p/∂P)

    Is a correct statement. Technically, you should write

    \(\frac{\partial K}{\partial P} = \frac{d K}{d p} + \frac{\partial K}{\partial p}\frac{\partial p}{\partial P}\)

    but the first term is zero automatically.
     
  9. neelakash Registered Senior Member

    Messages:
    491
    I could not understand the last part.

    However,I needed to show:

    (∂K/∂P)=(∂H/∂p)(∂p/∂P)

    and

    (∂K/∂Q)=(∂H/∂q)(∂q/∂Q)

    That is,I have replaced K by H...And here lies my problem:

    In general, K= H +(∂F/∂t)
     
  10. Guest254 Valued Senior Member

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    1,056
    Hi neelakash. I will try and read through your threads (this and contact transformation thread) tomorrow and answer your queries, assuming BenTheMan hasn't done so already.
     
  11. neelakash Registered Senior Member

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    491
    Hi Guest 254, I am waiting...
     
  12. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

    Messages:
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    I just realized I screwed up. I wrote:

    but I should have written:

    \(\frac{\partial K}{\partial p} = \frac{d K}{d p} + \frac{\partial K}{\partial P}\frac{\partial P}{\partial p}\)

    Sorry, I see why you were confused.

    A good reference for this is Goldstein "Classical Mechanics", 3 ed. You might check and see if your library has a copy, and look at chapter 9.

    You have to derive the generating function F. There is a table on pg. 373 that shows the four basic categories of the generating functions. I can't remember exactly how you pick your generating function, but I think that you choose it to make the statement true.

    I'd try to see if you could find a copy of that book and read the first few pages of chapter 9.

    But back to your original equation: why is there a discrepancy between the invariance under canonical transformations and the Poisson brackets? I'm not sure. From what I can see in Goldstein, it says "The invariance of the fundamental Poisson brackets is thus equivalent to the symplectic condition for a canonical transformation." So it seems that the transformation you gave above doesn't satisfy the "symplectic" condition for a canonical transformation.

    So my guess is that your transformation cannot be considered "canonical" in this sense. This is a guess, but Goldstein also says "Either the symplectic or the generator formaisms can be used to prove that canonical transformations satisfy the properties of a group." This seems to imply that you can't imagine your time reversal symmetry in the sense of following from some invariance in the Hamiltonian---i.e. you can't write the time reversal as an infinitessimal. Again, this is just a guess.
     
  13. neelakash Registered Senior Member

    Messages:
    491
    Hi Ben,I have tried to find the generating function but that is rather difficult. In fact,remember that F will be a function of time, in general.But from the set of equations you are referring to,it is not possible to find a good generating functions...Even if you neglect the condition time T= -t, F1 and F4 cannot simply be found...and F2 and F3 gives contradictory answers.

    Yes,the transformation truly does not satisfy symplectic condition.It is checked.

    It seems that the transformation IS a contact transformation,not a canonical one...That is reflected in symplectic condition or in fundamental Poisson bracket violation.However,by a brute force,we can make all the Fs to lie on the q-p plane---no time variation allowed in F.That makes Hamilton's equations satisfied.
     

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