Relativistic harmonic oscillator

[gerald V]

I have some difficulties in viewing the literature on the topic. In textbooks on analytical mechnics the procedure given for Special relativistic motion is to write the kinetic term relativistically and attach the unchanged potential term. So, for a harmonic oscillator the Lagrangian is $L = -m\sqrt{1-\dot{x}^2} - \frac{m\omega^2}{2}x^2$. But in this expression, the symmetry between momentum and position known from the nonrelativistic theory got lost (or didn't it?).
The article by Aldaya et.al. (Phys. Lett. A 156, 381) gives the dispersion relation $E^2 = m^2 + p^2 + m\omega^2x^2$, and as one can easily compute the according Lagrangian is $L = -m \sqrt{1-\dot{x}^2 + \omega^2x^2}$. As far as I understood, there is a vast further literature using this relation implicetely, but I have nowhere found the Lagrangian written out explicetly. Rather, the literature deals with quantum aspects while the non-quantum theory only is touched on.

As the quoted paper points out, the underlying symmetry is so(2,1) (it speaks about an "affine version" of this algebra(?)), what appears as plausible to me. Nevertheless, there seems to be some explosive in it, since in particular the singular velocity is not constantly unity.

My questions: Does the said Lagrangian explicitely appear somewhere in the literature? Has anyone extensively discussed the non-quantum so(2,1) oscillator? Or is it so trivial that no one bothered? Why do standard textbooks give the other expression? And last not least, which one is the appropriate one to describe nature?

Thank you very much in advance for any answer.

 

[sweet springs]

Hi.
I am afraid potential energy that is the energy of function of coordinate U(x,y,z) does not stand for relativity because it implies instantaneous propagation of force. I am sorry if I miss your point.

 

[gerald V]

Thank you so far. Meanwhile I fear that I miscalculated something. The Lagrangian I computed seems to be wrong. Sorry. I shall come back when I have checked and corrected.

 

[gerald V]

If my discalculia hasn't tricked me again, the correct Lagriangian to reproduce the dispersion relation $E^2=m^2+p^2+mω^2x^2$ (which looks highly sensible to me) is $L = - m\sqrt{(1-\dot{x}^2)(1+\omega^2x^2) }$. This is fine what regards the speed of light, but it is yet not the same as the textbook Lagrangian $L = - m\sqrt{1-\dot{x}^2} - \frac{m^2\omega^2}{2}x^2$. Since there can only be one, I would again be grateful for reply to my original questions.

 

[sweet springs]

Hi. They are same to the first order of x^2 in Taylor expansion.

 

[gerald V]

Thank you very much. I made a further (rather clerical) error: In the last term of the dispersion relation, there should be $m^2$ in place of $m$.

Correctness up to first order is not very much for such a fundamental object as a harmonic oscillator. In fact, upon quantization, the so(2,1) oscillator has a very different ground state energy than what is known from the nonrelativistic case. Well, one can argue the ground state energy has to be renormalized away anyway, but is this the last word?

I had expected that a thorough discussion of the non-quantum relativistic oscillator was done more than 100 years ago, and of the quantum h.o. at least 80 years ago. But it seems, this was not the case. Surprising.

 

[vanhees71]

The question is what you want to describe with your Lagrangian. You cannot say the one Lagrangian is more correct than the other, as long as you don't specify the problem. One thing is clear immediately: It's some effective Lagrangian of an open system since obviously the action is no Poincare invariant for both Lagrangians.

 

[gerald V]

vanhees71, thank you. In the meantime I also consulted some literature, in particular Goldstein’s „Classical Mechanics“.

Now it appears to me that the situation is really intriguing. The fundamental fields can be treated in a covariant way without problems, and lead to the $\frac{1}{r}$ potential. The $r^2$ potential, in contrast, does not have such a basis. It is produced by something like a spring, which is far from being a fundamental object and breaks Lorentz symmetry (and translational symmetry in space) by its mere existence.

My problem now is that first the oscillator potential is a preferred one due to Bertrand’s theorem, like the $\frac{1}{r}$ potential. Second, and much more grave, mass is just an oscillator frequency. How can mass and the lovely quadratic coupling term be such a non-fundamental entity?

Is this among the reasons why mass of gauge field is produced by the Higgs mechanism? But what about scalar or spinor fields like the electron?

I am aware that mass breaks gauge symmetry. But is this actually the correspondance to breaking of Lorentz symmetry by a spring in case of particle motion? Or what else is the correspondance? Is there literature on this subject?

Many thanks in advance!