In an infinite plane elastic plate of thickness $d$, it is shown that the modes of oscillation corresponding to a fixed time-frequency $\omega$ have wave-numbers given by solutions of the Rayleigh-Lamb equations

$$\frac{\tan (pd)}{\tan (qd)}=-\left[\frac{4k^2pq}{\left(k^2-q^2\right)^2}\right]^{\pm 1}$$

where the +1 exponent corresponds to symmetric modes and the -1 exponent to antisymmetric modes, and

$$p^2=\frac{\omega ^2}{c_L^2}-k^2$$

$$q^2=\frac{\omega ^2}{c_T^2}-k^2$$

where $c_L^2=\frac{2\mu +\lambda }{\rho }$ is the longitudinal wave-speed and $c_T^2=\frac{\mu }{\rho }$ is the transverse wave-speed ($\mu$ and $\lambda$ being the Lame constants). From this equation one obtains certain dispersion curves, relating the wave-number of each mode to the frequency. My question is: For each value of the frequency, the Rayliegh-Lamb equations give a discrete number of wave-numbers. How do you decide which wave-number belongs to which mode? That is, if I have a set of wave-numbers corresponding to one value of the frequency, say $\omega_1$, and then I get a new set of wave-numbers corresponding to a different frequency, say $\omega_2$, then how do I identify wave-numbers corresponding to different frequencies as belonging to the same mode?

I hope I've explained myself adequately. You can see that my question can also be framed for many other trascendental equations with multiple roots that show up in mathematical physics. I chose this particular problem to make my point. One might think that it is just a matter of ordering the wave-numbers in increasing order. But I've seen dispersion curves in which the curves for different modes intersect, so an ordering criterion is not applicable (not sure if this happens for the Rayleigh-Lamb modes, though)

Thanks. And sorry for the brick of text.

  • $\begingroup$ Good question. This isn't my area of expertise so I don't know anything about it, but I hope you get a good answer. $\endgroup$ – David Z Jun 8 '11 at 0:25

Let me see if I understand the question correctly: in general our solutions for a system will involve for every wavevector k a set of frequencies $\omega(k)$. When these curves giving $\omega(k)$ do not cross, there is in obvious sense in which we can separate our solutions into different modes. But when they do cross how do we assign modes?

I believe the answer is in general we don't. (Or we do but we shouldn't.) Unless the solutions have different symmetry properties, which is common enough, our label don't have meaning except convenience. I don't think this is not entirely about definitions. Consider adiabatic motion: we think that if we perturb our solution slowly enough it will just change its frequency and wavenumber, but stay on the same mode. But in the vicinity of a crossing of two modes our solution will evolve to have components in both modes, barring symmetry. So is there is no sense to think of them as separate modes.

(The only case I can think where the labeling of modes actually matters is in topological insulators, a case which requires no [bulk] band touching. Other than that it is just a convenience or it labels symmetry properties)

Apologies if I misunderstood your question

  • $\begingroup$ If I understood your answer correctly, then I think you understood my question correctly (sorry if it was not very clear). Your answer is that there's no definite way to assign modes if the dispersion curves intersect. I think I agree with you, but I've seen dispersion curves where the authors insist in splitting the modes (NOT by symmetry properties, of this much I'm sure). Can't cite any papers as I've only seen this on conferences. Perhaps it's only for convenience, as you say. $\endgroup$ – becko Jun 10 '11 at 20:37
  • $\begingroup$ @becko: Yes, it seems like we agree. As for examples, the labeling of bands in solid-state physics is ubiquitous, because it is extremely convenient. But nothing physical should depend on being able to unambiguously assign a mode to a frequency, except when there is symmetry or something... $\endgroup$ – BebopButUnsteady Jun 10 '11 at 21:48

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