Suppose then you're asked to find the metric of a charged black hole. Given a generic, static and spherically symmetric metric tensor,

$\text{d}s^{2} = A(r)\text{d}t^{2} + B(r)\text{d}r^{2} +r^{2}(\text{d}\theta ^{2} + \sin^{2}\theta \text{d}\phi^{2}) \tag{1},$

the one should use the stress-energy-momentum tensor:

$T^{\mathrm{(EM)}}_{\mu\nu} =\frac{1}{4\pi}\big(F_{\mu \lambda} F_{\nu}\hspace{0.1mm}^{\lambda} - g_{\mu\nu}F_{\alpha \beta}F^{\alpha \beta}\big)\tag{2},$

to solve the Einstein Field Equations (EFE), $G_{\mu\nu} = 8\pi T^{\mathrm{(EM)}}_{\mu\nu}$. The final result yields the Reissner-Nordström geometry:

$\text{d}s^{2} = -\Big(1-\frac{2m}{r}+\frac{q^2}{r^2}\Big)\text{d}t^{2} + \frac{\text{d}r^{2}}{\Big(1-\frac{2m}{r}+\frac{q^2}{r^2}\Big)} +r^{2}(\text{d}\theta ^{2} + \sin^{2}\theta \text{d}\phi^{2}) \tag{3}.$

My Question

Suppose now you are starting with metric tensor $(3)$ in your hands. Then you calculate the Einstein tensors $G^{(c)}_{\mu\nu}$(*); with the calculated Einstein tensors, you give, by hand, a matter tensor (in a suitable tetrad basis):

$T^{(F)}_{\mu\nu} = \mathrm{Diag}[\rho(r),p_{r},p_{\theta},p_{\phi}].\tag{4}$

Now, my question is: the EFE will be $G^{(c)}_{\mu\nu} = 8\pi\big[T^{(F)}_{\mu\nu} + T^{\mathrm{(EM)}}_{\mu\nu} \big]$ or just $G^{(c)}_{\mu\nu} = 8\pi T^{(F)}_{\mu\nu}$? Since the information about the charge $q$ is already in the spacetime (in the metric).


(*) the $^{(c)}$ is for "the components were calculated with the given metric $(3)$"


1 Answer 1


Once you have the metric, you have already set the stress-energy tensor automatically. You'll have $G_{\mu\nu}^{(\text{c})} = 8 \pi T_{\mu\nu}^{(\text{EM})}$. What you might then attempt to do is to set $T_{\mu\nu}^{(\text{EM})} = T_{\mu\nu}^{(\text{F})}$ and try to solve for the densities and pressures, with the possibility of failure. Otherwise, you'd need to solve the full EFE's with $G_{\mu\nu}^{(\text{c})} = 8\pi\left(T_{\mu\nu}^{(\text{EM})} + T_{\mu\nu}^{(\text{F})}\right)$, but then they wouldn't need to give rise to the Reissner–Nordström solution. For example, if the fluid you are adding behaves as a cosmological constant, you might end with the charged version of the Schwarzschild–De Sitter solution.


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