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Suppose we have a material described by a dielectric tensor $\bar{\epsilon}$. In frequency domain, this tensor depends on the wave frequency $\omega$ and the wave vector $\vec{k}$.

Clearly not all $\bar{\epsilon}=f(\omega,\vec{k})$ are physically possible. Without considering the physics behind any particular $\bar{\epsilon}=f(\omega,\vec{k})$, is it possible to find general conditions that all $\bar{\epsilon}=f(\omega,\vec{k})$ must obey? I am mostly interested in Lorenz invariance : surely the most general description of any material must be parametrized by the material's velocity according to the observer ($\vec{v}_{mat}$), and then there must be some sort of symmetry between $\omega$ and $\vec{k}$ in $\bar{\epsilon}=f(\omega,\vec{k},\vec{v}_{mat})$ because the Lorentz transformation mixes up distances ($1/k$) and time intervals ($1/\omega$).

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Almost identical question, but with a much more specific title:… – Ron Maimon Aug 31 '11 at 5:21

You get constraints on the possible form of the dielectric since it is a "response function". The polarization $P$ in an applied field $E$ is given by $P(x,t) = \int \epsilon(x-x',t-t')E(x',t')dx'dt'$ up to various conventions. We must have $\epsilon(x,t) = 0$ when $t<0$, since an electric field can't change what the polarization was before the electric field was turned on. In fourier space this translates to $\epsilon(\omega)$ is analytic in the upper half complex $\omega$ plane. This in turn leads to something called Kramers-Kronig relations which you can look up on wikipedia under that name.

(I suppose we can get an even stricter condition since $\epsilon(x,t)$ must be zero outside of the forward light cone, $t > |\vec{x}|$.)

As for seeing how the dielectric changes with lorentz transformation - this should be a relatively straightforward task. Write down Maxwell's equation with everything expressed in terms of the fields and the dielectric and permitivity and perform a Lorentz transformation. This should take you back to same form of equations but with a different dielectric.

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It may be straightforward, but it's done nowhere. You get a four-index response tensor, and you need to give a symmetry decomposition, and it's annoying (but not hard). This would answer a bunch of upvoted problems on this site. – Ron Maimon Dec 30 '11 at 13:50

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