# Electromagnetic stress-energy tensor

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 7; Problem 7.8; Chapter 20, Problem 20.5.

Since energy is the time component of the four-momentum and the density of a scalar quantity such as mass or charge is the time component of a four-vector (four-current in the case of charge), the energy density of a collection of particles would appear to transform as the ${tt}$ component of a second-rank tensor (since we have one Lorentz transformation for the energy and another for the density). If we then take a slight leap in logic and assume that the energy density of electromagnetic fields transforms in the same way, then we can assume that this energy density is the ${tt}$ component of a tensor known as the stress-energy tensor.

To deduce the form of this tensor, we use the usual technique of trying to convert an expression from classical physics into tensor form. In this case the classical expression for energy density is

$\displaystyle \rho_{E}=\frac{1}{8\pi k}\left(E^{2}+B^{2}\right) \ \ \ \ \ (1)$

The idea, then, is to find a second-rank tensor from the electromagnetic field tensor that has this as its ${tt}$ component. We can get a clue as to how to do this by looking at the quantity ${F_{ij}F^{ij}=2\left(B^{2}-E^{2}\right)}$. During the derivation of this, we found that the ${tt}$ (that is, the 00 component) component of the tensor product ${F^{0j}F_{j0}=E^{2}}$. Combining this with ${B^{2}=\frac{1}{2}F_{ij}F^{ij}+E^{2}=\frac{1}{2}F_{ij}F^{ij}+F^{0j}F_{j0}}$ we get

$\displaystyle E^{2}+B^{2}=2F^{0j}F_{j0}+\frac{1}{2}F_{ij}F^{ij} \ \ \ \ \ (2)$

If we raise the 0 index in the first term on the RHS, we get

$\displaystyle E^{2}+B^{2}=-2F^{0j}F_{jk}\eta^{k0}+\frac{1}{2}F_{ij}F^{ij} \ \ \ \ \ (3)$

where we’ve introduced the minus sign in the first term since raising the index changes the sign in the 0 column of ${F_{jk}}$.

This looks promising, but the expression on the RHS is the sum of the ${tt}$ component of a tensor (the first term) and a scalar. We can convert this to be the ${tt}$ component of a tensor by multiplying the last term by ${-\eta^{00}=+1}$ (the Kronecker delta):

$\displaystyle E^{2}+B^{2}=-2F^{0j}F_{jk}\eta^{k0}-\frac{1}{2}F_{ij}F^{ij}\eta^{00} \ \ \ \ \ (4)$

If we now propose this is the ${tt}$ term of the stress-energy tensor ${8\pi kT^{ml}}$ we can postulate

 $\displaystyle T^{ml}$ $\displaystyle =$ $\displaystyle -\frac{1}{8\pi k}\left(2F^{mj}F_{jk}\eta^{kl}+\frac{1}{2}F_{ij}F^{ij}\eta^{ml}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\epsilon_{0}F^{mj}F_{jk}\eta^{kl}-\frac{\epsilon_{0}}{4}F_{ij}F^{ij}\eta^{ml} \ \ \ \ \ (6)$

The other elements of ${T^{ml}}$ have meanings that we’ll get to eventually, but we can note that for ${i\ne0}$

$\displaystyle T^{0i}=-\epsilon_{0}F^{0j}F_{jk}\eta^{ki} \ \ \ \ \ (7)$

For ${i=1}$ for example, we get

 $\displaystyle T^{01}$ $\displaystyle =$ $\displaystyle -\epsilon_{0}F^{0j}F_{j1}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\left(E_{y}B_{z}-E_{z}B_{y}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\left(\mathbf{E}\times\mathbf{B}\right)_{x} \ \ \ \ \ (10)$

This is the ${x}$ component of the Poynting vector, which describes the rate of energy transfer per unit area of an electromagnetic field. Again, something we’ll get to in EM theory eventually.