In Reissner-Nordström black holes it's the electric field that contributes to the metric. shouldn't the elctrostatic repulsion energy contribute also? there is quite a bit involved when charges are brought together closely.
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$\begingroup$ it's the electric field that contributes to the metric What does that mean? $\endgroup$– G. SmithCommented May 31, 2021 at 0:23
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$\begingroup$ shouldn't the elctrostatic [sic] repulsion energy contribute also? The energy-momentum-stress tensor on the right side of the Einstein field equations contains the electrostatic energy density, so it determines the curvature and the metric. $\endgroup$– G. SmithCommented May 31, 2021 at 0:24
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This is briefly discussed in the following Physics SE Q&A : Does the Reissner-Nordstrom metric necessarily represent a charged black hole ?
The brief answer is that the R-N metric is the exterior solution but that a different metric applies to the interior and this interior metric does include the effect of repulsion as you ask. Bob Bee's nice answer to that question also gives a link to a relevant paper :
The Equilibrium of a Charged Sphere - W.B Bonner
The abstract states :
An equilibrium model is presented for a sphere of charged dust under it's own gravitational and electrical repulsion. The mass and radius are arbitrary. Equilibrium configurations of this sort suggest that electric repulsion may be able to halt the gravitational collapse of very large masses.
Now I would point out this is a static metric solution so it does not in fact represent a collapse, it just shows that in theory an equilibrium is possible, not that such an equilibrium can be reached in a dynamic collapse. However it does nicely illustrate that an interior metric is likely required to describe the interior region precisely because of the electric repulsion.
The exterior metric however is unaffected by any electric repulsion which is internal. This is essentially a no-hair theorem consequence - all we see on the outside is a massive charged object - it has no distribution of charge and in effect their is no difference (from the outside) between a point mass and charge at the center and a distribution.
answered May 29, 2021 at 11:24
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$\begingroup$ That's a nice answer! So if we squeeze two electrons together we will never get a black hole? $\endgroup$ Commented May 29, 2021 at 16:45
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$\begingroup$ We simply do not have a theory that would cover such a situations, and as I indicated there's a difference between a static idealized solution like the one in that paper and e.g. squeezing two (or even tillions) of electrons together. What is required is the elusive theory of quantum gravity which, even if such a thing exists, we can make no assumptions about what a purely classical (non-quantum) model like that says and what the mystery theory would. $\endgroup$ Commented May 29, 2021 at 17:47