0
$\begingroup$

I'm currently playing with a toy model given by the Lagrangian $$L=\frac{m\dot{x}^2}{2}+\frac{m\dot{y}^2}{2}+\frac{1}{2}m\omega^2x^2+x y,$$ which is basically a free particle (described by $y(t)$) and a harmonic oscillator (described by $x(t)$) coupled in a simple way.

My question is if you guys know any system that could correspond to such a Lagrangian? Basically this is a free particle that is forced to oscillate by a harmonic oscillator that pulls it...

$\endgroup$
  • 3
    $\begingroup$ Is there a factor in front of the $xy$ term? Otherwise a problem with dimensions. $\endgroup$ – jim Apr 7 '16 at 14:04
  • $\begingroup$ This is very similar to the Lagrange an of an electron in the beam of a free electron laser $\endgroup$ – Lewis Miller Apr 7 '16 at 14:07
  • $\begingroup$ Dimensionally its incorrect. $\endgroup$ – Anubhav Goel Apr 7 '16 at 14:17
  • $\begingroup$ This question (v1) seems like a list question. $\endgroup$ – Qmechanic Apr 7 '16 at 15:21
  • $\begingroup$ @jim I just encountered this Lagrangian in a course on path integration, where it was a toy model to solve analytically. It might be that in a physical system there can be a factor in front of it (depending on what units you use I guess ;-) )! $\endgroup$ – Nick Apr 8 '16 at 9:45
5
$\begingroup$

First of all I guess that what you wrote is the Hamiltonian and not the Lagrangian of the system and $\dot{x}$ stays for $p_x$ and $\dot{y}$ stays for $p_y$.

You can decouple the problem redefining $$(X,Y)^t = R(x,y)^t$$ for a suitable $R\in O(2)$ diagonalizing the symmetric matrix in the potential part of your Hamiltonian. This way you see the final Hamiltonian is $$\left(\frac{p^2_X}{2m} + \lambda_+ X^2\right) + \left(\frac{p^2_Y}{2m} + \lambda_- Y^2\right)$$ where $\lambda_\pm$ are the eigenvalues of the above symmetric matrix.

In the considered case (up to the dimensional problem already stressed) you find that $ \lambda_+ \lambda_-<0$ (because the determinant of the symmetric matrix is negative nomatter the sign in front of $xy$).

So you have a pair of 1D non-mutually interacting particles, one subjected to a standard harmonic potential and the other subjected to a repulsive harmonic potential (like the one of centrifugal force for a constant angular speed).

$\endgroup$
1
$\begingroup$

enter image description here

For example, if the first particle is moving on a spring and its position sets the electric potential that controls the second, electrically charged, particle. This way you'd have the potential energy of the coupling in the form of the product of coordinates.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.