Electromagnetic stress–energy tensor
# Electromagnetic stress–energy tensor

In relativistic physics, the electromagnetic stress–energy tensor is the contribution to the stress–energy tensor due to the electromagnetic field. The stress–energy tensor describes the flow of energy and momentum in spacetime. The electromagnetic stress–energy tensor contains the negative of the classical Maxwell stress tensor that governs the electromagnetic interactions.

## Definition

### SI units

In free space and flat space–time, the electromagnetic stress–energy tensor in SI units is

    $T^{\mu \nu }={\frac {1}{\mu _{0}}}\left[F^{\mu \alpha }F^{\nu }_{\alpha }-{\frac {1}{4}}\eta ^{\mu \nu }F_{\alpha \beta }F^{\alpha \beta }\right]\,.$

where $F^{\mu \nu }$ is the electromagnetic tensor and where $\eta _{\mu \nu }$ is the Minkowski metric tensor of metric signature (− + + +). When using the metric with signature (+ − − −), the expression on the right of the equals sign will have opposite sign.

Explicitly in matrix form:

    $T^{\mu \nu }={\begin{bmatrix}{\frac {1}{2}}\left(\epsilon _{0}E^{2}+{\frac {1}{\mu _{0}}}B^{2}\right)&{\frac {1}{c}}S_{\text{x}}&{\frac {1}{c}}S_{\text{y}}&{\frac {1}{c}}S_{\text{z}}\\{\frac {1}{c}}S_{\text{x}}&-\sigma _{\text{xx}}&-\sigma _{\text{xy}}&-\sigma _{\text{xz}}\\{\frac {1}{c}}S_{\text{y}}&-\sigma _{\text{yx}}&-\sigma _{\text{yy}}&-\sigma _{\text{yz}}\\{\frac {1}{c}}S_{\text{z}}&-\sigma _{\text{zx}}&-\sigma _{\text{zy}}&-\sigma _{\text{zz}}\end{bmatrix}},$

where

    $\mathbf {S} ={\frac {1}{\mu _{0}}}\mathbf {E} \times \mathbf {B} ,$

is the Poynting vector,

    $\sigma _{ij}=\epsilon _{0}E_{i}E_{j}+{\frac {1}{\mu _{0}}}B_{i}B_{j}-{\frac {1}{2}}\left(\epsilon _{0}E^{2}+{\frac {1}{\mu _{0}}}B^{2}\right)\delta _{ij}$

is the Maxwell stress tensor, and c is the speed of light. Thus, $T^{\mu \nu }$ is expressed and measured in SI pressure units (pascals).

### CGS unit conventions

The permittivity of free space and permeability of free space in cgs-Gaussian units are

    $\epsilon _{0}={\frac {1}{4\pi }},\quad \mu _{0}=4\pi \,$

then:

    $T^{\mu \nu }={\frac {1}{4\pi }}\left[F^{\mu \alpha }F^{\nu }_{\alpha }-{\frac {1}{4}}\eta ^{\mu \nu }F_{\alpha \beta }F^{\alpha \beta }\right]\,.$

and in explicit matrix form:

    $T^{\mu \nu }={\begin{bmatrix}{\frac {1}{8\pi }}\left(E^{2}+B^{2}\right)&{\frac {1}{c}}S_{\text{x}}&{\frac {1}{c}}S_{\text{y}}&{\frac {1}{c}}S_{\text{z}}\\{\frac {1}{c}}S_{\text{x}}&-\sigma _{\text{xx}}&-\sigma _{\text{xy}}&-\sigma _{\text{xz}}\\{\frac {1}{c}}S_{\text{y}}&-\sigma _{\text{yx}}&-\sigma _{\text{yy}}&-\sigma _{\text{yz}}\\{\frac {1}{c}}S_{\text{z}}&-\sigma _{\text{zx}}&-\sigma _{\text{zy}}&-\sigma _{\text{zz}}\end{bmatrix}}$

where Poynting vector becomes:

    $\mathbf {S} ={\frac {c}{4\pi }}\mathbf {E} \times \mathbf {B} .$

The stress–energy tensor for an electromagnetic field in a dielectric medium is less well understood and is the subject of the unresolved Abraham–Minkowski controversy.

The element $T^{\mu \nu }\!$ of the stress–energy tensor represents the flux of the μth-component of the four-momentum of the electromagnetic field, $P^{\mu }\!$, going through a hyperplane ($x^{\nu }$ is constant). It represents the contribution of electromagnetism to the source of the gravitational field (curvature of space–time) in general relativity.

## Algebraic properties

The electromagnetic stress–energy tensor has several algebraic properties:

* It is a symmetric tensor: $T^{\mu \nu }=T^{\nu \mu }$
* The tensor $T^{\nu }_{\alpha }$ is traceless: $T^{\alpha }_{\alpha }=0.$ Proof Starting with $T_{\mu }^{\mu }=\eta _{\mu \nu }T^{\mu \nu }$ Using the explicit form of the tensor, $T_{\mu }^{\mu }={\frac {1}{4\pi }}\left[\eta _{\mu \nu }F^{\mu \alpha }F^{\nu }_{\alpha }-\eta _{\mu \nu }\eta ^{\mu \nu }{\frac {1}{4}}F^{\alpha \beta }F_{\alpha \beta }\right]$ Lowering the indices and using the fact that $\eta ^{\mu \nu }\eta _{\mu \nu }=\delta _{\mu }^{\mu }$ $T_{\mu }^{\mu }={\frac {1}{4\pi }}\left[F^{\mu \alpha }F_{\mu \alpha }-\delta _{\mu }^{\mu }{\frac {1}{4}}F^{\alpha \beta }F_{\alpha \beta }\right]$ Then, using $\delta _{\mu }^{\mu }=4$, $T_{\mu }^{\mu }={\frac {1}{4\pi }}\left[F^{\mu \alpha }F_{\mu \alpha }-F^{\alpha \beta }F_{\alpha \beta }\right]$ Note that in the first term, μ and α and just dummy indices, so we relabel them as α and β respectively. $T_{\alpha }^{\alpha }={\frac {1}{4\pi }}\left[F^{\alpha \beta }F_{\alpha \beta }-F^{\alpha \beta }F_{\alpha \beta }\right]=0$
* The energy density is positive-definite: $T^{00}\geq 0$

The symmetry of the tensor is as for a general stress–energy tensor in general relativity. The trace of the energy–momentum tensor is a Lorentz scalar; the electromagnetic field (and in particular electromagnetic waves) has no Lorentz-invariant energy scale, so its energy–momentum tensor must have a vanishing trace. This tracelessness eventually relates to the masslessness of the photon.

## Conservation laws

The electromagnetic stress–energy tensor allows a compact way of writing the conservation laws of linear momentum and energy in electromagnetism. The divergence of the stress–energy tensor is:

    $\partial _{\nu }T^{\mu \nu }+\eta ^{\mu \rho }\,f_{\rho }=0\,$

where $f_{\rho }$ is the (4D) Lorentz force per unit volume on matter.

This equation is equivalent to the following 3D conservation laws

    ${\begin{aligned}{\frac {\partial u_{\mathrm {em} }}{\partial t}}+\mathbf {\nabla } \cdot \mathbf {S} +\mathbf {J} \cdot \mathbf {E} &=0\\{\frac {\partial \mathbf {p} _{\mathrm {em} }}{\partial t}}-\mathbf {\nabla } \cdot \sigma +\rho \mathbf {E} +\mathbf {J} \times \mathbf {B} &=0\ \Leftrightarrow \ \epsilon _{0}\mu _{0}{\frac {\partial \mathbf {S} }{\partial t}}-\nabla \cdot \mathbf {\sigma } +\mathbf {f} =0\end{aligned}}$

respectively describing the flux of electromagnetic energy density

    $u_{\mathrm {em} }={\frac {\epsilon _{0}}{2}}E^{2}+{\frac {1}{2\mu _{0}}}B^{2}\,$

and electromagnetic momentum density

    $\mathbf {p} _{\mathrm {em} }={\mathbf {S} \over {c^{2}}}$

where J is the electric current density, ρ the electric charge density, and $\mathbf {f}$ is the Lorentz force density.

