JimWhoKnew said:
Can you demonstrate it yourself
I'll take a stab at it here. I'll be making use of some equations from two of my Insights articles, on the EFE and Maxwell's Equations in a spherically symmetric spacetime:
https://www.physicsforums.com/insig...-in-a-static-spherically-symmetric-spacetime/
https://www.physicsforums.com/insig...-in-a-static-spherically-symmetric-spacetime/
Note that those articles focus on the static case, but the equations I'll be using from them don't assume that the spacetime is static, only that it's spherically symmetric. As with the standard Birkhoff's Theorem, the presence of an additional Killing vector field will be part of what we derive.
We adopt Schwarzschild coordinates, which are adapted to the spherical symmetry of the spacetime. The line element in these coordinates for a general spherically symmetric spacetime is:
$$
ds^2 = - J(r) dt^2 + \frac{1}{1 - \frac{2m(r)}{r}} dr^2 + r^2 d\Omega^2
$$
where ##J(r)## and ##m(r)## are functions of ##r## that we will try to determine. The most general stress-energy tensor for a general spherically symmetric spacetime has only three nonzero components:
$$
T^t{}_t = - \rho(r)
$$
$$
T^r{}_r = p(r)
$$
$$
T^\theta{}_\theta = T^\varphi{}_\varphi = s(r)
$$
which gives us three more functions of ##r## to determine.
We also can derive the following equations by computing the Einstein tensor of the metric above and applying the EFE, as shown in the first article referenced above:
$$
\frac{dm}{dr} = 4 \pi r^2 \rho
$$
$$
\frac{1}{2J} \frac{dJ}{dr} = \frac{m + 4 \pi r^3 p}{r \left( r - 2m \right)}
$$
$$
\frac{dp}{dr} = - \left( \rho + p \right) \frac{1}{2J} \frac{dJ}{dr} - \frac{2}{r} \left( p - s \right)
$$
Finally, we have from the second article that the electromagnetic 4-potential, given the constraint of spherical symmetry, must be
$$
A_t = \Phi(r)
$$
with all other components zero.
Now we make use of two further constraints that are imposed by spherical symmetry plus the fact that the only stress-energy present is a source-free EM field (i.e., the charge-current 4-vector ##j## vanishes). These are that the SET must be traceless, and that we must have ##J(r) = 1 - 2m(r) / r## (i.e., the ##g_{tt}## metric coefficient must be minus the reciprocal of the ##g_{rr}## component). (I have not been able to find a proof of the latter fact, but we know it's true for the R-N metric, and it seems reasonable given that there are no sources anywhere, so there's no way for the "redshift factor" to vary from its electrovacuum value.)
Now we start applying our constraints:
The trace of the SET being zero means ##- \rho + p + 2 s = 0##, or ##\rho = p + 2 s##.
Knowing that ##J = 1 - 2m(r) / r## lets us compute
$$
\frac{1}{2J} \frac{dJ}{dr} = \frac{m - r \frac{dm}{dr}}{r \left( r - 2m \right)}
$$
Plugging in the equation we have above for ##dm / dr## gives
$$
\frac{1}{2J} \frac{dJ}{dr} = \frac{m - 4 \pi r^3 \rho}{r \left( r - 2m \right)}
$$
We now have two equations for the same quantity (the LHS of the equation above), and setting the RHS of the two gives ##p = - \rho##. From the tracelessness equation above that gives us ##s = \rho##.
Now we plug all of this into the equation for ##dp / dr##, to get:
$$
\frac{dp}{dr} = - 4 \frac{p}{r}
$$
This has the obvious solution
$$
p = - \frac{Q^2}{r^4}
$$
where the constant in the numerator has been chosen to match what we already know of the form of the R-N solution, and the minus sign is because the sign of ##p## must be negative (because the sign of ##\rho##, the energy density, must be positive, and ##p = - \rho##).
Note that it is not possible to add another term, as in the hypothesis in post #17. Note also that I made no assumption whatever regarding the form of the SET in terms of the EM 4-potential or EM field tensor. But we can show, of course, that given the 4-potential above, the electric field ##E = - d \Phi / dr##, and from the source-free Maxwell Equations, as shown in the second article referenced above, we have
$$
\frac{dE}{dr} = - 2 \frac{E}{r}
$$
which of course has the solution ##E = Q / r^2##. So the energy density we found above is just ##E^2##, as expected.