In analogy to the electromagnetic tensor, with the components defined as the electric field $E$ and magnetic field $B$ as such:

$F^{ab} = \begin{bmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{bmatrix}$

Can one form a relativistic tensor from the electric displacement field $D$ and magnetizing field $H$ like so:

$\bar{F}^{ab} = \begin{bmatrix} 0 & -D_x/c & -D_y/c & -D_z/c \\ D_x/c & 0 & -H_z & H_y \\ D_y/c & H_z & 0 & -H_x \\ D_z/c & -H_y & H_x & 0 \end{bmatrix}$

and thus obtain an "electromagnetic tensor" that can be easily used to handle the effects of fields in materials?

Since $D = \epsilon_0 E + P$ and $H = \frac{1}{\mu_0}B - M$, this is equivalent to asking if $P$ and $M$ can be used to form a relativistic tensor.


3 Answers 3


Yes, you can form such a tensor with $D$ and $H$ but it is not as useful and natural as the tensor with $E$ and $B$ because $E$ and $B$ are the actual physical fields that naturally transform into each other by Lorentz transformations.

On the other hand, you may declare that $D$ and $H$ transform into each other by the Lorentz transformations but it is not too useful because these fields are only useful to describe fields inside a material, and the presence of this material spontaneously breaks the Lorentz symmetry, anyway.

This is related to some recent questions on this server about electrodynamics of moving media. $D$ and $H$ are normally used, together with the well-known equations they satisfy, only in the rest frame of a material. If the material is moving, the whole analysis has to be redone. To be sure that we do it right, we should return to the actual $B$ and $E$ field and if we define new ones to simplify our life, we shouldn't guess the equations for these new fields: we should derive them from the equations for the vacuum fields $E,B$ and the actual charge densities and currents, including the bound ones.

The resulting Maxwell's equations for a medium, which include $B,E,D,H$, are not Lorentz-covariant, anyway - because the material picks a preferred frame. To introduce new fields $D,H$ is just a bookeeping device to simplify some calculations - and it only works nicely in the medium's rest frame.


I'm not 100% sure, but I think the answer is yes. Here's my reasoning.

The physical effects of $\mathbf{P,M}$ result from the bound charges and currents they produce: $\rho_b=-\nabla\cdot{\bf P}$ and ${\bf J}_b=\nabla\times{\bf M}$. I'd expect $j\equiv (\rho_b,{\bf J}_b)$ to transform as a 4-vector (after all, they are "real" charges and currents). If you define an antisymmetric tensor $P^{\mu\nu}$ out of $\mathbf{P,M}$ in the natural way, then the definitions of $\rho_b,{\bf J}_b$ reduce to $j^\mu={P^{\mu\nu}}_{,\nu}$. If $P^{\mu\nu}$ is covariant, then $j^\mu$ is automatically covariant.

That's just a plausibility argument, not a proof, but I find it pretty convincing. Let me see if I can tighten up the argument a bit. In the local rest frame of the matter, define $P^{\mu\nu}$ to be the tensor built out of $\mathbf{P,M}$ in the natural way. In any other frame, define $P^{\mu\nu}$ via Lorentz transformations. Finally, define $j^\mu={P^{\mu\nu}}_{,\nu}$. Then the four-vector $j$ gives the correct bound charges and currents in the rest frame, and hence (since everything's covariant) it gives the correct bound charges and currents in all frames. Everything that matters about $\mathbf{P,M}$ is contained in the bound charges and currents, so we're done.

The correct answer is presumably contained in Landau & Lifschitz, Electrodynamics of Continuous Media, which I've never read. If what's in there contradicts what I've said, then I'm sure that they're right and I'm wrong.


The answer is "yes". One has \begin{equation} \bar{F}^{ab} = \bar{\epsilon}_{abcd} F^{cd} \end{equation} where \begin{equation} \bar{\epsilon}^{0i0j} = \epsilon^{ij} \end{equation} and \begin{equation} \bar{\epsilon}^{ijpq} = (1/\mu)^{kr} \end{equation} up to a factor $c$ here and there :-) . Here $i,j,k$ and $p,q,r$ are spatial indices such that $\hat{k}=\hat{i}\times\hat{j}$ and $\hat{r}=\hat{p}\times\hat{q}$. See A.O. Barut, Electrodynamics and Classical Theory of Fields and Particles.

  • $\begingroup$ Note that the most general $\bar{\epsilon}$ tensor will also have non-zero components of the form $\bar{\epsilon}^{0ijk}$; these will give rise to effects where $\vec{D}$ depends on both $\vec{E}$ $\vec{B}$ (and the same for $\vec{H}$.) Only in the rest frame of the medium do these effects generally vanish. $\endgroup$ Mar 21, 2018 at 20:30

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