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Peculiar solution

  1. Dec 13, 2006 #1
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    [tex]q_{c}(t)=\frac{1}{\sinh \omega \tau} }[ q' \sinh{\omega(t''-t)} + q'' \sinh{\omega(t-t')} ] [/tex]

    is the solution to the classical equation of motion for the harmonic oscillator

    [tex]\ddot {q}_{c}(t)-\omega^{2}q_{c}(t)=0[/tex]

    where [tex]q_{c}(t)[/tex] is the position vector, [tex]\tau = t'' - t'[/tex], [tex]q_{c}(t')=q'[/tex], [tex]q_{c}(t'')=q''[/tex] and [tex]\ddot {q}_{c}(t)[/tex] is the double time derivative of [tex]q_{c}(t)[/tex]

    I can differentiate [tex]q_{c}(t)[/tex] twice with respect to time easily enough to show that that the first equation is a solution to the equation of motion but there are many functions that satisfy this.

    How do I show that it's the solution?

    I've tried solving the equation of motion directly but I can't find any way to solve it so that I end up with [tex]q_{c}(t)[/tex] as a funciton of t, t', t'', q' and q''. (For example the way everyone is usually taught to solve it, you simply end up with a sin function of t multiplied by an amplitude..)

    Any help/suggestions would be very much appreciated. Thank you for taking the time to read this :)
    Last edited: Dec 13, 2006
  2. jcsd
  3. Dec 13, 2006 #2


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    Do you really think that is what it is asking you to do?

    I think that if you have shown that q_c(t) is a valid solution, then you are done.
  4. Dec 13, 2006 #3
    hmm.. the book I'm following (Jean Zinn Justins QFT and critical phenomena) is "calculating the classical action explicitly" and one of the steps along the way is finding that [tex]q_{c}(t)[/tex] is as given.

    perhaps you are right.. strictly speaking I have found that this form of [tex]q_{c}(t)[/tex] is valid but shouldn't I really understand the motivation behind using that particular solution? To be honest I think the book should be telling me that, but there's no reason given other than we find the solution to be thus.. I can't help feeling I'm missing something..

    EDIT: perhaps the motivation is that this form has all the bits that define the path integral? (In the overall scheme of things we're applying a path integral with the action for a harmonic oscillator)
    Last edited: Dec 13, 2006
  5. Dec 13, 2006 #4

    Physics Monkey

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    Hi alfred, it is appropriate to say "the" instead of "a" because there are boundary conditions specified. In other words, you require that q_c satisfy the differential equation and have the value q' at t' and q'' at t''. As you no doubt recall from the theory of linear second order differential equations, solutions to such equations contain two constants of integration. These are fixed by the two boundary conditions. If you had not specified initial data or boundary conditions, then it would only make sense to say you had found a solution to the differential equation. The application of boundary conditions turns the "a" into a "the".

    Hope this helps.
  6. Dec 14, 2006 #5
    ohh i see now! how stupid of me. I should have at least thought of putting the values t' and t'' into the solution and thus seen that q_c satisfies those boundary conditions. doh!

    grr i really need to develop these basic instincts when looking at equations and learn to play with them, and understand what they are telling me :D

    oh well, thank you physics monkey!
    Last edited: Dec 14, 2006
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