PeterDonis said:
If you're talking about my #1, yes, I didn't mean to imply that all photons in cavities or other such setups still have zero rest mass. But I can imagine an idealized container where they would.
And #2 as well. To see why (waveguide) modes acquire rest mass, consider the wave equation, and assume that the transversal cordinate dependence separates from the longitudinal coordinate dependence:
<br />
\psi(x, y, z, t) = \phi(x, t) \, \Phi(y, z)<br />
<br />
\frac{1}{c^2} \, \frac{\partial^2 \phi}{\partial t^2} - \frac{\partial^2 \phi}{\partial x^2} + \lambda^{2}_{n} \phi = 0<br />
<br />
\frac{\partial^2 \Phi_{n}}{\partial y^2} + \frac{\partial^2 \Phi_{n}}{\partial z^2} + \lambda^{2}_{n} \, \Phi_{n} = 0<br />
The discrete eigenvalues for the transversal Laplace operator are determined by the geometry of the waveguide and the boundary conditions.
Then, performing a Fourier transform of the longitudinal wave equation we get the dispersion relation:
<br />
\omega^2 = c^2 k^2 + c^2 \lambda^2_{n}<br />
Then, quantizing these propagating modes, we may assign a mass:
<br />
m_{n} = \frac{\hbar \lambda_{n}}{c}<br />
As for the propagation of em waves in a dielectric, the dispersion relation becomes:
<br />
\mathrm{det} \left[ \left(\frac{\omega}{c} \right)^{2} \varepsilon_{i k}(\omega, \mathbf{k}) - k^{2} \delta_{i k} - k_{i} k_{k} \right] = 0<br />
where \varepsilon_{i k}(\omega, \mathbf{k}) is a frequency and wave vector dependent dielectric tensor that describes the coupling of the system to the external em field.
I'm not sure I would say it "loses its meaning", but I agree things get a lot more complicated.
It means that the presence of a medium introduces a preferred reference frame (that where the medium is at rest). We no longer require Lorentz invariant expressions as dispersion relations.
