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# Maxwell's Equations for a static, spherically symmetric spacetime

Posted Aug15-12 at 01:35 PM by PeterDonis
Updated Mar17-13 at 02:50 PM by PeterDonis

Following on from my previous post, here is a quick reference for Maxwell's Equations in a static, spherically symmetric spacetime.

The electromagnetic field tensor is given in general by

$$F_{ab} = \partial_b A_a - \partial_a A_b$$

where $A_a$ is the electromagnetic 4-potential.

In covariant form, Maxwell's Equations in general are:

$$\partial_c F_{ab} + \partial_b F_{ca} + \partial_a F_{bc} = 0$$

$$\nabla_a F^{ab} = \partial_a F^{ab} + \Gamma^a{}_{ac} F^{cb} = 4 \pi J^b$$

Note that we don't need covariant derivatives in the first equation, even in a general curved spacetime, because the connection coefficient terms all cancel since $F_{ab}$ is antisymmetric. Similarly, in the second equation, the second connection coefficient term that would normally appear, $\Gamma^b{}_{ac} F^{ac}$, is identically zero for an antisymmetric tensor.

The general form of the stress-energy tensor for an electromagnetic field is

$$T^a{}_b = \frac{1}{4 \pi} \left( F^{ac} F_{bc} - \frac{1}{4} \delta^a{}_b F^{cd} F_{cd} \right)$$

which requires us to raise two indexes on the EM field tensor, as follows:

$$F^{ab} = g^{ac} g^{bd} F_{cd}$$

To specialize the above to the static, spherically symmetric case, we note first that, since the only coordinate on which anything can depend is $r$, the only possible nonzero components of the EM field tensor are

$$F_{rt} = - F_{tr} = - \partial_r A_t = E$$

$$F_{r \theta} = - F_{\theta r} = - \partial_r A_\theta = B$$

$$F_{r \phi} = - F_{\phi r} = - \partial_r A_\phi = B$$

where we have defined useful symbols for the components based on their obvious physical interpretations. The last two components must be equal by spherical symmetry; however, it would be really nice if we could prove, what seems intuitively correct, that they must be zero.

We can, as it turns out, do this by looking at the stress-energy tensor components. We know, from my previous post, that the SET for a static, spherically symmetric spacetime must be diagonal (since the Einstein tensor is diagonal). But if the angular components of the EM field tensor are nonzero, then we would have nonzero off-diagonal terms in the SET such as this one:

$$T^\theta{}_\phi = \frac{1}{4 \pi} F^{\theta r} F_{\phi r}$$

(where the second term in the SET is identically zero because of the $\delta^a{}_b$ in front). So the angular terms in the EM field tensor must be zero, and we have $F_{rt} = - F_{tr} = E$ as the only nonzero terms. This EM field tensor describes a static, spherically symmetric, radial electric field, where we have

$$A_t = \varphi$$

for the 4-potential, with all other components zero, and

$$F_{rt} = - F_{tr} = E = - \partial_r \varphi$$

for the EM field tensor, as above, and

$$T^t{}_t = T^r{}_r = - T^\theta{}_\theta = - T^\phi{}_\phi = - \rho_{EM} = - \frac{E^2}{8 \pi} \left( \frac{r - 2m}{r J} \right)$$

for the stress-energy tensor, where we have used the general metric given in my previous blog post on the EFE in a static, spherically symmetric spacetime to raise indexes on the EM field tensor components, per the formula given above. Note that this describes a "substance" with positive energy density (remember the sign switch in $T^t_t$ due to the metric signature), a negative radial stress (i.e., a radial tension), and a positive tangential stress (i.e., a tangential pressure), all with the same magnitude.

We can now write down the second Maxwell Equation taking into account all of the above; the only nonzero component is the $t$ component, which is:

$$\nabla_a F^{at} = 4 \pi J^t = \partial_a F^{at} + \Gamma^a{}_{ac} F^{ct} = \partial_r F^{rt} + \Gamma^a{}_{ar} F^{rt}$$

Expanding this out gives:

$$4 \pi J^t = \left[ \partial_r + \Gamma^t{}_{tr} + \Gamma^r{}_{rr} + \Gamma^\theta{}_{\theta r} + \Gamma^\phi{}_{\phi r} \right] \left( g^{tt} g^{rr} F_{rt} \right)$$

Substituting and tedious but straightforward algebra gives:

$$4 \pi J^t = - \frac{r - 2m}{rJ} \frac{dE}{dr} - \frac{1}{J}\left[ \frac{2 \left( r - 2m \right)}{r^2} + 4 \pi r \left( \rho + p \right) \left( 1 - \frac{2 \left( r - 2m \right)}{r J} \right) \right] E$$

We can re-arrange the terms slightly in a way that will often be useful:

$$J^t = - \frac{1}{4 \pi} \frac{r - 2m}{rJ} \left( \frac{dE}{dr} + \frac{2}{r} E \right) - \left( \rho_m + p_m \right) E \left( \frac{r}{J} - \frac{2 \left( r - 2m \right)}{J^2} \right)$$

where we have also made use of the fact that, for the portion of the SET that is due to the EM field, $\rho$ and $p$ cancel each other (they have equal magnitude but opposite signs, as above). So only the $\rho$ and $p$ due to any other matter present will appear in the last term, and we designate these as $\rho_m$ and $p_m$.
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