# Eigen-Energy of Vibration on a Loop

I recently started this recreational project to find the energy of the modes on a loop. To obtain the eigen-energies $E_n$, I decided to solve the 1D wave equation on a loop of circumference $2\pi l$ with periodic boundary conditions as follows using separation of variables $u(x,t)=X(x)T(t)$:

$$u_{tt}=c^2u_{xx}$$$$u(0,t)=u(2\pi l,t)$$$$u_x(0,t)=u_x(2\pi l,t)=0$$

Now by solving for $X$ for non trivial solutions we get a single cosine term with wavenumber $k_n=n/l$. Using this to solve for $T$ we get that the time dependence has a frequency $\omega_n=ck_n=\frac{nc}{l}$.

Now next, assuming that the loop is sufficiently small, I hand-wavingly set $E_n=\frac{\hbar c}{l}n$ and also since $c=\tau/\rho$ (tension over mass per unit length) it becomes $E_n=\frac{\hbar \tau}{m}n$

Is this a good way to approach such a problem? In particular would it be right to assume that for a sufficiently small loop $E=\hbar\omega$? If not what would be the best approach for such a problem?

Thank you for your time in helping me make sense of what I'm doing.

As a side note, given the almost quantum-mechanical nature of the problem, I tried also solving the Klein-Gordon equation $u_{tt}=c^2 u_{xx} - \left(\frac{mc^2}{\hbar}\right)^2u$ with the same spatial boundary conditions and the same procedure. From this I got that $E_n=\sqrt{\left(\frac{\hbar c n}{l}\right)^2+\left(mc^3\right)^2}$. I am not too sure this approach makes any sense but if one considers the propagating waves as massless particles (or negligible mass, keep in mind that for the wave equation, $m$ was the mass of the loop, here I think is the mass of the "particle", I should say that I am not very familiar with the KGE) it simplifies exactly to the solution obtained by using the wave equation

• This sounds like the rigid rotor? hyperphysics.phy-astr.gsu.edu/hbase/molecule/rotqm.html – Pieter Jun 21 '18 at 22:28
• @Pieter So vibration on a loop can be described by a rotating loop? – Jepsilon Jun 21 '18 at 22:50
• A loop has circular symmetry, so angular momentum is a good quantum number. "Vibrating" sounds like a standing wave, which can be described as the sum (or difference) of two wave functions with opposite angular momentum (counter-propagating traveling waves). – Pieter Jun 22 '18 at 7:52
• @Pieter Oh ok I'll look into it and see if I can get a more meaningful result – Jepsilon Jun 22 '18 at 12:37