Maxwell's Equations for a static, spherically symmetric spacetime
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
[tex]F_{ab} = \partial_b A_a - \partial_a A_b[/tex]
where [itex]A_a[/itex] is the electromagnetic 4-potential.
In covariant form, Maxwell's Equations in general are:
[tex]\partial_c F_{ab} + \partial_b F_{ca} + \partial_a F_{bc} = 0[/tex]
[tex]\nabla_a F^{ab} = \partial_a F^{ab} + \Gamma^a{}_{ac} F^{cb} = 4 \pi J^b[/tex]
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 [itex]F_{ab}[/itex] is antisymmetric. Similarly, in the second equation, the second connection coefficient term that would normally appear, [itex]\Gamma^b{}_{ac} F^{ac}[/itex], is identically zero for an antisymmetric tensor.
The general form of the stress-energy tensor for an electromagnetic field is
[tex]T^a{}_b = \frac{1}{4 \pi} \left( F^{ac} F_{bc} - \frac{1}{4} \delta^a{}_b F^{cd} F_{cd} \right)[/tex]
which requires us to raise two indexes on the EM field tensor, as follows:
[tex]F^{ab} = g^{ac} g^{bd} F_{cd}[/tex]
To specialize the above to the static, spherically symmetric case, we note first that, since the only coordinate on which anything can depend is [itex]r[/itex], the only possible nonzero components of the EM field tensor are
[tex]F_{rt} = - F_{tr} = - \partial_r A_t = E[/tex]
[tex]F_{r \theta} = - F_{\theta r} = - \partial_r A_\theta = B[/tex]
[tex]F_{r \phi} = - F_{\phi r} = - \partial_r A_\phi = B[/tex]
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:
[tex]T^\theta{}_\phi = \frac{1}{4 \pi} F^{\theta r} F_{\phi r}[/tex]
(where the second term in the SET is identically zero because of the [itex]\delta^a{}_b[/itex] in front). So the angular terms in the EM field tensor must be zero, and we have [itex]F_{rt} = - F_{tr} = E[/itex] as the only nonzero terms. This EM field tensor describes a static, spherically symmetric, radial electric field, where we have
[tex]A_t = \varphi[/tex]
for the 4-potential, with all other components zero, and
[tex]F_{rt} = - F_{tr} = E = - \partial_r \varphi[/tex]
for the EM field tensor, as above, and
[tex]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)[/tex]
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 [itex]T^t_t[/itex] 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 [itex]t[/itex] component, which is:
[tex]\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}[/tex]
Expanding this out gives:
[tex]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)[/tex]
Substituting and tedious but straightforward algebra gives:
[tex]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[/tex]
We can re-arrange the terms slightly in a way that will often be useful:
[tex]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)[/tex]
where we have also made use of the fact that, for the portion of the SET that is due to the EM field, [itex]\rho[/itex] and [itex]p[/itex] cancel each other (they have equal magnitude but opposite signs, as above). So only the [itex]\rho[/itex] and [itex]p[/itex] due to any other matter present will appear in the last term, and we designate these as [itex]\rho_m[/itex] and [itex]p_m[/itex].
The electromagnetic field tensor is given in general by
[tex]F_{ab} = \partial_b A_a - \partial_a A_b[/tex]
where [itex]A_a[/itex] is the electromagnetic 4-potential.
In covariant form, Maxwell's Equations in general are:
[tex]\partial_c F_{ab} + \partial_b F_{ca} + \partial_a F_{bc} = 0[/tex]
[tex]\nabla_a F^{ab} = \partial_a F^{ab} + \Gamma^a{}_{ac} F^{cb} = 4 \pi J^b[/tex]
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 [itex]F_{ab}[/itex] is antisymmetric. Similarly, in the second equation, the second connection coefficient term that would normally appear, [itex]\Gamma^b{}_{ac} F^{ac}[/itex], is identically zero for an antisymmetric tensor.
The general form of the stress-energy tensor for an electromagnetic field is
[tex]T^a{}_b = \frac{1}{4 \pi} \left( F^{ac} F_{bc} - \frac{1}{4} \delta^a{}_b F^{cd} F_{cd} \right)[/tex]
which requires us to raise two indexes on the EM field tensor, as follows:
[tex]F^{ab} = g^{ac} g^{bd} F_{cd}[/tex]
To specialize the above to the static, spherically symmetric case, we note first that, since the only coordinate on which anything can depend is [itex]r[/itex], the only possible nonzero components of the EM field tensor are
[tex]F_{rt} = - F_{tr} = - \partial_r A_t = E[/tex]
[tex]F_{r \theta} = - F_{\theta r} = - \partial_r A_\theta = B[/tex]
[tex]F_{r \phi} = - F_{\phi r} = - \partial_r A_\phi = B[/tex]
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:
[tex]T^\theta{}_\phi = \frac{1}{4 \pi} F^{\theta r} F_{\phi r}[/tex]
(where the second term in the SET is identically zero because of the [itex]\delta^a{}_b[/itex] in front). So the angular terms in the EM field tensor must be zero, and we have [itex]F_{rt} = - F_{tr} = E[/itex] as the only nonzero terms. This EM field tensor describes a static, spherically symmetric, radial electric field, where we have
[tex]A_t = \varphi[/tex]
for the 4-potential, with all other components zero, and
[tex]F_{rt} = - F_{tr} = E = - \partial_r \varphi[/tex]
for the EM field tensor, as above, and
[tex]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)[/tex]
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 [itex]T^t_t[/itex] 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 [itex]t[/itex] component, which is:
[tex]\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}[/tex]
Expanding this out gives:
[tex]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)[/tex]
Substituting and tedious but straightforward algebra gives:
[tex]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[/tex]
We can re-arrange the terms slightly in a way that will often be useful:
[tex]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)[/tex]
where we have also made use of the fact that, for the portion of the SET that is due to the EM field, [itex]\rho[/itex] and [itex]p[/itex] cancel each other (they have equal magnitude but opposite signs, as above). So only the [itex]\rho[/itex] and [itex]p[/itex] due to any other matter present will appear in the last term, and we designate these as [itex]\rho_m[/itex] and [itex]p_m[/itex].
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