# Lorentz invariance of Maxwell's equations in matter

I know that Maxwell's equations of electromagnetism are Lorentz invariant in a vacuum. But what about in a generalized medium, e.g. a metal, a rubber, a dielectric, a magnet? I have read it comes down to whether the electric and magnetic polarizations, $$M$$ and $$P$$, are themselves Lorentz invariant. (Note: I am ignoring gravity.) My feeling is that they must be, albeit some researchers might use approximations that are not. So can anyone answer my question:

Are Maxwell's equations in a medium Lorentz invariant?

Maxwell is not Lorentz invariant in matter because the material selects a preferred reference frame — the one in which the lump of matter is at rest.

Of course you can make it all look Lorentz covariant by including the local four velocity $$u^\mu$$ of the piece of matter in the equations for the dielectric constant and the magnetic permeability, and by defining $$E_\mu = F_{\mu\nu}u^{\nu}, \quad B_\mu = \frac 12 \epsilon_{\mu\nu\sigma\tau} u^\nu F^{\sigma\tau}.$$ to be the $${\bf E}$$ and $${\bf B}$$ fields in the frame moving with the matter — but that extra $$u^\mu$$ makes everything rather complicated. It's best to avoid all this unless you really want to to do relativistic fluid/continuum mechanics such as investigating the magnetic field on a neutron star or the accretion disc of a black hole.

• Can you elaborate a little on why the material can select a preferred reference frame? For instance, Earth does not get to choose rest when it witness relativistic particles whizzing by. What is going on at a molecular level that means that there somehow a preferred reference frame when witnessing inter-molecular events? May 8, 2019 at 16:14
• The material is the preferred (in the sense of being different from any other) frame. Light travels at set speed $c/n$ with respect to the material. If you move past the material at speed $v$ the speed of light with respect to you is different depending on how fast you go. In vaccum the speeed of light you measure is always $c=3\times 10^8$ m/s independent of your speed. May 8, 2019 at 17:09
• @RoryCornish An analysis of relativistic particles whizzing through Earth's atmosphere is going to have Earth as a preferred reference frame. For instance, cosmologists speak of "microwave radiation" and "gamma rays", even though for any photon, whether it's is in the "microwave" or "gamma ray" part of the spectrum depends on the reference frame. "microwave radiation" is understood to mean "radiation that is, in the Earth's reference frame, in the microwave part of the spectrum", and similarly for gamma rays, and microwave and gamma rays will exhibit different phenomena. May 8, 2019 at 18:39
• aside from boosts, doesn't the material also break rotational symmetry? May 9, 2019 at 1:08

Are Maxwell's equations in a medium Lorentz invariant?

No, they're not. An electromagnetic wave can have a speed $$v in a medium. That means you can choose a frame in which the wave's velocity is zero. A zero-velocity wave is a solution in that frame, but not in the rest frame of the medium. Therefore the equations are not form-invariant under a Lorentz boost.

• Apologies if I am being thick. But couldn't you use the same argument with a water wave of speed v. There is a frame at which the water waves velocity is zero. A zero velocity wave is a solution in that frame but not in the rest frame. But this does not mean that the fluid equations are not invariant under Galilean transformation. So I am not understand how your thought experiment implies Maxwell's equations are not form-invariant under Lorentz boosts. What am I missing in your reasoning? May 8, 2019 at 16:23
• Ben: this is wrong. The Maxwell equations are perfectly covariant in matter: $dF=j$, where $j$ denotes the current density. What may not be covariant is the usual approximation $\vec P=\epsilon\vec E$, but that is an issue of the approximation, not of the Maxwell equations themselves. I'm sure you already know all this, but your answer is very misleading. The answer to the question in the OP is a categorical Yes: light moves at the speed of light, $c$, also in a medium. The emergent/macroscopic description may not, but the fundamental/microscopic description does. May 8, 2019 at 16:40
• @AccidentalFourierTransform: It seems like we have different interpretations of the question. Your interpretation seems to make the question vacuous.
– user4552
May 8, 2019 at 18:02
• @RoryCornish: As far as I'm concerned, the fluid equations are variant under Galilean transformations. As an example, sound waves in a fluid obey the wave equation $$\frac{\partial^2 p}{\partial t^2} = c_s^2 \nabla^2 p$$ when you're in the rest frame of the fluid. If you then transform to a set of coordinates $t' = t$, $\vec{r}' = \vec{r} + \vec{v} t$, you'll get terms of the form $\vec{v} \cdot \nabla \dot{p}$ (I think) that are not present in the rest frame. May 8, 2019 at 18:27
• This answer has a point, but AccidentalFourierTransform is right as well. It is because the term "Maxwell's equations" is ambiguous in this case. Ben's Maxwell's equations in medium here mean "equations that macroscopic fields $\mathbf E,\mathbf B$ obey in medium". These equations resemble the fundamental Maxwell's equations in vacuo, with permittivity and permeability having different values, but only in case the medium is at rest. When we try to express equations for macroscopic fields in medium when the medium moves, the result is more complicated than the standard-form Maxwell's equations. May 9, 2019 at 21:58

The aspect of Maxwell's equations that fails to be "fully covariant" in the presence of matter are the constitutive relations. Specifically, if you write out Maxwell's equations in the presence of matter, \begin{align*} \nabla \cdot \vec{D} &= \rho_f & \nabla \times \vec{E} &= - \frac{\partial \vec{B}}{\partial t} \\ \nabla \cdot \vec{B} &= 0 & \nabla \times \vec{H} &= \vec{J}_f + \frac{\partial \vec{D}}{\partial t} \end{align*} you need to supplement these with a set of relationships between the auxiliary fields $$\vec{D}$$ & $$\vec{H}$$ and the "real" fields $$\vec{E}$$ and $$\vec{B}$$. For example, we commonly assume that $$\vec{D} = \epsilon \vec{E}$$ and $$\vec{H} = \frac{1}{\mu} \vec{B}$$; from this, we can show that electromagnetic waves will travel through the material at the same speed $$v = c/\sqrt{\epsilon \mu}$$ in all directions; the fields all obey the wave equation with a characteristic speed $$v$$.

But these simple-looking relationships ($$\vec{D} \propto \vec{E}$$, $$\vec{H}\propto \vec{B}$$) are a frame-dependent statement, and they will not necessarily hold in another reference frame in which the medium is moving. As an example of this, let's assume that the new "primed" reference frame is moving in the positive $$x$$-direction with respect to our original frame. The fields in the two reference frames are related to each other by (for example) \begin{align*} E_y &= \gamma (E'_y + v B'_z) & D'_y &= \gamma(D_y - v H_z) & B_z &= \gamma (B'_z + v E'_y) \end{align*} (note that the transformation among $$\vec{D}$$ and $$\vec{H}$$ is the same as the transformation between $$\vec{E}$$ and $$\vec{B}$$.) We then have \begin{align*} D'_y &= \gamma(D_y - vH_z) \\ &= \gamma \left(\epsilon E_y - v \frac{1}{\mu} B_z \right) \\ &= \gamma \left[\epsilon \gamma (E'_y + v B'_z) - v \frac{1}{\mu} \gamma (B'_z + v E'_y) \right] \\ &= \gamma^2 \left(\epsilon - \frac{v^2}{\mu}\right) E'_y + \gamma v \left( \epsilon - \frac{1}{\mu} \right) B'_z \neq \epsilon E'_y. \end{align*}

Thus, even if $$\vec{D}$$ is proportional to $$\vec{E}$$ in the rest frame of the material, this does not imply that $$\vec{D}'$$ will be proportional to $$\vec{E}'$$ in a non-rest frame.1 Instead, the constitutive relations are more complicated in this frame; $$\vec{D}$$ will depend on both $$\vec{E}$$ and $$\vec{B}$$, as will $$\vec{H}$$.

We could in principle still use Maxwell's equations and these new constitutive relations to write down a second-order differential equation for $$\vec{E}$$ or $$\vec{B}$$ alone. But what we'll find is a wave-like equation in which the speed of wave propagation differs in different directions. And if we take a valid propagation velocity vector in the rest frame and transform it into our new frame, it'll line up exactly with a valid propagation velocity of light according to our new wave-like equation.

1 Unless $$\epsilon = 1/\mu$$, but then we're talking about a medium in which all waves travel at the speed of light anyhow.

Maxwell's equations in vacuum are Lorentz covariant, not invariant. In a moving medium they are also covariant, but as you stated, mostly they are not written in a covariant manner. Any physical system is Lorentz covariant else special relativity would fail.

• Yes. That is what I was thinking. Sorry, I should have said Covariant, but was being sloppy. My feeling was that any apparent violation of Lorentz, would most likely be down to the material models being approximate. I was worried about the apparent contradictions otherwise. E.g. the Standard Model is Lorentz Covariant and in principle should describe the whole material behaviour, but that would contradict the "Not Lorentz Covariant" position. May 8, 2019 at 18:39
• Also, light entering one end of a material, passing through and exiting the other side can be timed. That external timing WILL be Lorentz covariant. But if what happens inside the material isn't Lorentz covariant, the timing outside couldn't be. Another contradiction. May 8, 2019 at 18:40
• @Rory You made a good observation. Perhaps you could edit the question accordingly. May 8, 2019 at 22:40

I'm going to pick on your choice of word.

The optical axis or axes of a crystal are only Lorentz invarant if the direction of the boost is parallel, antiparallel, or perpendicular to the optical axis/axes. (In fact, birefringence was the first phenomenon to come to mind when I read your question.)

Electromagnetics in a material is an example of spontaneous symmetry breaking.

Consider a crystal in a rest frame. The lattice structure breaks the rotation symmetry, and indeed, crystals are often birefringent. Similarly, a ferromagnet has a preferred direction determined by the bulk magnetization.

The underlying governing equations for the crystal and the ferromagnet are Maxwell's equations and a many-particle Schrödinger equation for nuclei and electrons, and they certainly have rotation symmetry, but the crystalline or ferromagnetic ground state does not, and the nearby (in energy) excitations do not either. The ground state has spontaneously broken the symmetry.