# Lorent-invariance of the Maxwell's equations in the medium

1. May 28, 2012

### sergiokapone

The Maxwell's equations in vacuum leads to the wave equations for the fields of the form
$\nabla^2 \vec E = \frac{1}{c^2} \frac{\partial ^2 \vec E}{\partial t^2}$
(the same for the magnetic field)
Such equations are Lorentz-invariant.
Let's consider now the electromagnetic field in a homogeneous medium.
Field in a medium subject to a rate lower than in vacuum $v=c/n$, where $n=\sqrt{\epsilon\mu}$ and the equations have the form:
$\nabla^2 \vec E = \frac{\epsilon\mu}{c^2} \frac{\partial ^2 \vec E}{\partial t^2}$
But such equations are obviously not Lorentz-invariant. Why is this a paradox?

2. May 28, 2012

### Staff: Mentor

It's not.

But I expect that you asking why it's not a paradox, and the answer is that the particles of the medium are interacting with and influencing the electrical and magnetic fields that make up the light.

Lorentz invariance only applies to the speed of light in vacuum, when there's nothing to influence the propagation of light - and even then we needed experiments such as the Michelson-Morley to suggest that there's really nothing there.

BTW: yes, it is possible to travel in such a medium at a speed greater than the speed with which light propagates in the medium. It's still not possible to exceed c.

Edit: sorry, the bit in bold isn't what I meant to say. It's the invariance of the speed light that applies in vacuum, and I surely do I hope that the laws of E&M in a medium transform properly like everything else.

Last edited: May 28, 2012
3. May 28, 2012

### sergiokapone

Well, can we say that Maxwell's equations in the medium are invariant under Lorentz transformations, in which the speed of light in vacuum is replaced by the speed of light in the medium.

4. May 28, 2012

### Staff: Mentor

No; what you can say is that Maxwell's equations with a non-zero source (because the medium will have a non-zero charge and current density) are Lorentz invariant, provided we properly transform the source terms as well as the fields. Those equations will still use the speed of light in vacuum.

The equation you posted requires an extra step of derivation from Maxwell's equations in a medium; you have to restrict to a single frame (and I believe you also have to adopt a particular gauge, but I would have to look some things up to be sure). It's not the same as deriving the vacuum wave equation from the vacuum Maxwell equations.

5. May 28, 2012

### Staff: Mentor

D'oh.... Just edited my post.

6. May 28, 2012

### Bill_K

Plus the index of refraction is always to some extent frequency dependent, so it's not just Maxwell's Equations with a different velocity. If you know of a substance in which n is not frequency dependent you can make a fortune making dispersion-free optical lenses.

7. May 28, 2012

### Dickfore

The frame of reference in which the medium is at rest is a preferential reference frame in which permeability and permittivity are defined. You may use Lorentz transformations to find the effective dielectric and magnetic constants for a moving medium.

8. May 31, 2012

### Meir Achuz

You can Lorentz transform the E and B fields, but not matter dependent quantities, which apply only in the rest frame.

9. May 31, 2012

### Dickfore

Electric displacement $\mathbf{D}$, and magnetic induction $\mathbf{H}$ are given by:
$$\mathbf{D} = \epsilon_0 \, \mathbf{E} + \mathbf{P}$$
where $\mathbf{P}$ is the polarization (dipole moment per unit volume) of the dielectric medium,

and
$$\mathbf{H} = \frac{1}{\mu_0} \, \mathbf{B} - \mathbf{H}$$
where $\mathbf{M}$ is the magnetization (magnetic moment per unit volume) of the magnetic medium.

In pure vacuum, P = H = 0. Therefore, $\mathbf{D}$ has the same transformation properties as $\varepsilon_0 \, \mathbf{E}$, and $\mathbf{H}$ has the same transformation properties as $\frac{1}{\mu_0} \, \mathbf{B}$. But, we know that the E and B field are elements of the antisymmetric field tensor $F_{\mu \nu} = -F_{\nu \mu}$, with $F_{0 i} = \frac{E^{i}}{c}$, and $F_{i k} = -\epsilon_{i k l} \, B^{l}$.

Therefore, the antisymmetric quantity with:
$$\begin{array}{rcl} \mu_0 \, G_{0 i} & = & \frac{1}{\epsilon_0 \, c} \, D^{i} \\ \mu_0 \, G_{i k} & = & -\epsilon_{i k l} \mu_0 \, H^{l} \end{array}$$
should behave as a two-fold tensor.

The material Maxwell equations can be written in a covariant form as:
$$\partial_{\nu} G^{\mu \nu} = J^{\mu}_{\mathrm{free}}$$
where $J^{\mu}_{\mathrm{free}}$ is the 4-current density due to the free charges only.

Last edited: May 31, 2012
10. Jun 1, 2012

### Meir Achuz

Yes. After this, everything you do is for pure vacuum with no polarizable matter,
where E and D, and B and H, are the same fields except for a confusion of units in SI.

11. Jun 1, 2012

### Dickfore

The "material" Maxwell Equations
$$\nabla \cdot \mathbf{D} = \rho_{\mathrm{free}$$
$$\nabla \times \mathbf{H} = \frac{\partial \mathbf{D}}{\partial t} + \mathbf{J}_\mathrm{free}$$
hold in every reference frame, i.e. they are laws of Nature. They hold for arbitrary materials as well. Specifically, they hold for vacuum. This allows us to find the transformation properties of D and H by knowing them for E and B. If they are known for vacuum, they are known for any material.

What is not true in every reference frame is:
$$D_i = \epsilon_{0} \, \epsilon_{i k} \, E_{k}$$
and
$$H_{i} = \frac{1}{\mu_0} \mu^{-1}_{i k} \, B_{k}$$

These constitutive relations hold only in a reference frame where the medium is stationary. However, by knowing the transformation properties of the fields, we can generalize them for moving media.