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Bidimensional oscillator

  1. Jul 23, 2015 #1
    1. A bidimensional oscillator have energies:
    Sin título.png
    With C and K constants

    a) show by a transform of coordinates that this oscillator is equivalent to two isotropic harmonic oscillators.
    b) then find two independent constants of motion and verify this using:
    Sin título.png
    with "a" the constant.

    I tried to do this problem with a transform of kind
    Sin título.png
    So the Lagrangian is
    Sin título.png
    I wrote the Hamiltonian so in this case represent the energy and this will be a constant of motion, but I can't find the second constant in this problem... I don't Know if the transformation that I used is the correct, or if is another transformation to convert the equivalent problem to two isotropic harmonic oscillators.

    I hope you can help me
    Thanks
     
  2. jcsd
  3. Jul 24, 2015 #2

    jedishrfu

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    Don't you just have to show that L = L1 + L2 where L1 is for one oscillator and L2 is for the second oscillator?
     
  4. Jul 24, 2015 #3
    I have to apply some transformation that make the problem convert into a problem of two isotropic harmonic oscillators.
    I guess that the Lagrangian is the sum of two terms, just like you say, but I don't know if is right the way that I used.
     
  5. Jul 24, 2015 #4

    vela

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    How is the term isotropic supposed to apply in this problem? You've managed to transform the Lagrangian into two independent one-dimensional oscillators. But isotropic implies more than one dimension to me. Does the problem statement really say isotropic? Or is there some meaning of isotropic I'm not familiar with?
     
  6. Jul 25, 2015 #5

    jedishrfu

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    The OP shows a Lagrangian with an xy term that cross couples the x and y dimensions. The OP then transforms the x,y into q1,q2 coordinates in order to eliminate the xy term coupling.

    From there the Lagrangian looks like the sum of two independent harmonic oscillators so I'd say the OP has demonstrated the answer.

    I'm not versed enough to definitively say this is correct ( I took this course over four decades ago) so hopefully another mentor or science advisor will concur.
     
    Last edited: Jul 25, 2015
  7. Jul 25, 2015 #6
    Yes, just like you say that was the answer to the problem to separate in two oscillators, the part of isotropic was a mistake because when C=0 the system was isotropic and therefore Lz is a constant of motion.
    Summarizing, H=H1+H2 was the answer, so my two constants was E1 and E2, and obviously the condition exposed in the problem (da/dt) is correct for this constants. Now if C=0 appear another constant (Lz).
    Thanks you so much for your help.
    The final part of the exercise is write the symmetric matrix:
    A_ij=PiPj/2m +Kxixj/2
    In term of the constant that we found (E1, E2 and Lz), with these the matrix A is constant.
    I only can found the term in the diagonal which are in these case E1 and E2, but I don't know hoe to write A12 and of course A21.

    PS: I don't know how to edit the post to put these new information
     
  8. Jul 25, 2015 #7

    jedishrfu

    Staff: Mentor

    The Post editing feature is available to gold members. I think Greg gives out these memberships if you post an advertizement for PF at your school and you send him a photo of it on some bulletin board.

    Since you mentioned the matrix is symmetric, is it possible to show that the A values are zero? This is just a guess on my part and not based on any real knowledge of the problem.
     
  9. Jul 25, 2015 #8
    The terms in the diagonal of A are the energies of the two oscillators, and there are a constant of motion. So I have to find the other two terms of the matrix (A12 or A21) in terms of the constant of motion that I found early (E1, E2, Lz), I tired but I don't see how can I produce the term P1P2 and x1x2 from the constant of motion.
    In this case E1=P1^2/2m + Kx1/2, E2=P2^2/2m + Kx2/2, Lz=P2x1 - P1x2.
     
  10. Jul 26, 2015 #9

    vela

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    What you called the Lagrangian in the original post is actually the Hamiltonian. The Lagrangian is L = T-V. You have T+V.

    To evaluate the off-diagonal elements of A, start with explicit solutions for x and y:
    \begin{align*}
    x &= A_x \sin(\omega t + \theta_1) \\
    y &= A_y \sin(\omega t + \theta_2),
    \end{align*} find the corresponding momenta, and plug everything into the expression for ##A_{12}##. Express the amplitudes in terms of the oscillator energies and K.
     
  11. Jul 26, 2015 #10
    Thanks vela, I guess that work... But only if the terms of phase in the solution are equals or zero, because with that the element would be a constant.
     
  12. Jul 26, 2015 #11

    vela

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    ##\cos(\theta_1-\theta_2)## is a constant.
     
  13. Jul 26, 2015 #12
    Yes, I didn't see it, excuse me for that
     
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