The electromagnetic energy tensor

Consider a continuous volume distribution of charged matter in the presence of an electromagnetic field. Let there be $n_0$ particles per unit proper volume (unit volume determined in the local rest frame), each carrying a charge $e$. Consider an inertial frame in which the 3-velocity field of the particles is ${\bf u}$. The number density of the particles in this frame is $n=\gamma(u) n_0$. The charge density and the 3-current due to the particles are $\rho = e n$ and ${\bf j} = e n {\bf u}$, respectively. Multiplying Eq. (1564) by the proper number density of particles, $n_0$, we obtain an expression

$$f_\mu = c^{-1} F_{\mu\nu} J^\nu$$ (1568)

for the 4-force $f_\mu$ acting on unit proper volume of the distribution due to the ambient electromagnetic fields. Here, we have made use of the definition $J^\mu = e n_0 U^\mu$. It is easily demonstrated, using some of the results obtained in the previous section, that

$$f^\mu = \left( \rho {\bf E} + {\bf j}\times{\bf B}, \frac{{\bf E}\!\cdot \!{\bf j}}{c}\right).$$ (1569)

The above expression remains valid when there are many charge species (e.g., electrons and ions) possessing different number density and 3-velocity fields. The 4-vector $f^\mu$ is usually called the Lorentz force density.

We know that Maxwell's equations reduce to

$$\partial_\mu F^{\mu\nu} = \frac{J^\nu}{c \epsilon_0},$$ (1570)

$$\partial_\mu G^{\mu\nu} = 0,$$ (1571)

where $F^{\mu\nu}$ is the electromagnetic field tensor, and $G^{\mu\nu}$ is its dual. As is easily verified, Eq. (1571) can also be written in the form

$$\partial_\mu F_{\nu\sigma} + \partial_\nu F_{\sigma\mu} + \partial_\sigma F_{\mu\nu} = 0.$$ (1572)

Equations (1568) and (1570) can be combined to give

$$f_\nu = \epsilon_0 F_{\nu\sigma} \partial_\mu F^{\mu\sigma}.$$ (1573)

This expression can also be written

$$f_\nu = \epsilon_0 \left[\partial_\mu(F^{\mu\sigma}F_{\nu\sigma}) -F^{\mu\sigma} \partial_\mu F_{\nu\sigma}\right].$$ (1574)

Now,

$$F^{\mu\sigma} \partial_\mu F_{\nu\sigma} = \frac{1}{2} F... ...ma} (\partial_\mu F_{\nu\sigma} + \partial_\sigma F_{\mu\nu}),$$ (1575)

where use has been made of the antisymmetry of the electromagnetic field tensor. It follows from Eq. (1572) that

$$F^{\mu\sigma} \partial_\mu F_{\nu\sigma} = - \frac{1}{2} ... ...u} = \frac{1}{4} \partial_\nu (F^{\mu\sigma} F_{\mu\sigma}).$$ (1576)

Thus,

$$f_\nu =\epsilon_0 \left[\partial_\mu(F^{\mu\sigma}F_{\nu\sig... ...frac{1}{4} \partial_\nu (F^{\mu\sigma}F_{\mu\sigma})\right].$$ (1577)

The above expression can also be written

$$f_\nu = - \partial_\mu {T^\mu}_\nu,$$ (1578)

where

$${T^\mu}_\nu = \epsilon_0\left[ F^{\mu\sigma}F_{\sigma\nu} + ... ...1}{4} \delta_\nu^\mu (F^{\rho\sigma}F_{\rho\sigma})\right]$$ (1579)

is called the electromagnetic energy tensor. Note that ${T^\mu}_\nu$ is a proper-4-tensor. It follows from Eqs. (1481), (1484), and (1516) that

$${T^i}_j = \epsilon_0 E^i E_j + \frac{B^i B_j}{\mu_0} - \delta_j^i \frac{1}{2} \! \left(\epsilon_0 E^k E_k +\frac{B^k B_k}{\mu_0}\right),$$ (1580)

$${T^i}_4 = - {T^4}_i = \frac{\epsilon^{ijk} E_j B_k}{\mu_0 c},$$ (1581)

$$T^4_4 = \frac{1}{2} \left(\epsilon_0 E^k E_k +\frac{B^k B_k}{\mu_0}\right).$$ (1582)

Equation (1578) can also be written

$$f^\nu =-\partial_\mu T^{\mu\nu},$$ (1583)

where $T^{\mu\nu}$ is a symmetric tensor whose elements are

$$T^{ij} = -\epsilon_0 E^i E^j - \frac{B^i B^j}{\mu_0} +\delta^{ij} \frac{1}{2} \!\left(\epsilon_0 E^2 +\frac{B^2}{\mu_0}\right),$$ (1584)

$$T^{i4} = T^{4i} = \frac{({\bf E}\times{\bf B})^i}{\mu_0 c},$$ (1585)

$$T^{44} = \frac{1}{2} \left(\epsilon_0 E^2 +\frac{B^2}{\mu_0}\right).$$ (1586)

Consider the time-like component of Eq. (1583). It follows from Eq. (1569) that

$$\frac{{\bf E}\!\cdot \!{\bf j}}{c} = - \partial_i T^{i4} - \partial_4 T^{44}.$$ (1587)

This equation can be rearranged to give

$$\frac{\partial U}{\partial t} + \nabla\!\cdot \!{\bf u} = - {\bf E}\!\cdot\!{\bf j},$$ (1588)

where $U=T^{44}$ and ${\bf u}^i = c T^{i4}$, so that

$$U = \frac{\epsilon_0 E^2}{2} + \frac{B^2}{2 \mu_0},$$ (1589)

and

$${\bf u} = \frac{{\bf E}\times{\bf B}}{\mu_0}.$$ (1590)

The right-hand side of Eq. (1588) represents the rate per unit volume at which energy is transferred from the electromagnetic field to charged particles. It is clear, therefore, that Eq. (1588) is an energy conservation equation for the electromagnetic field (see Sect. 8.2). The proper-3-scalar $U$ can be identified as the energy density of the electromagnetic field, whereas the proper-3-vector ${\bf u}$ is the energy flux due to the electromagnetic field: i.e., the Poynting flux.

Consider the space-like components of Eq. (1583). It is easily demonstrated that these reduce to

$$\frac{\partial{\bf g}}{\partial t}+\nabla\cdot\! {\bf G}= -\rho {\bf E} - {\bf j}\times{\bf B},$$ (1591)

where $G^{ij} = T^{ij}$ and $g^i = T^{4i}/c$, or

$$G^{ij} = -\epsilon_0 E^i E^j - \frac{B^i B^j}{\mu_0} +\del... ...frac{1}{2} \!\left(\epsilon_0 E^2 +\frac{B^2}{\mu_0}\right),$$ (1592)

and

$${\bf g} = \frac{\bf u}{c^2} = \epsilon_0 {\bf E}\times{\bf B}.$$ (1593)

Equation (1591) is basically a momentum conservation equation for the electromagnetic field (see Sect. 8.4). The right-hand side represents the rate per unit volume at which momentum is transferred from the electromagnetic field to charged particles. The symmetric proper-3-tensor $G^{ij}$ specifies the flux of electromagnetic momentum parallel to the $i$th axis crossing a surface normal to the $j$th axis. The proper-3-vector ${\bf g}$ represents the momentum density of the electromagnetic field. It is clear that the energy conservation law (1588) and the momentum conservation law (1591) can be combined together to give the relativistically invariant energy-momentum conservation law (1583).