Non-Rotating Charged Black Holes: Mass, Electric Field, & Quantum Charges

[JustinLevy]

Here is the metric for an uncharged non-rotating black hole:
http://en.wikipedia.org/wiki/Schwarzschild_black_hole

Here is the metric for a charged non-rotating black hole:
http://en.wikipedia.org/wiki/Reissner-Nordström_black_hole

Note that this is the same as plugging M^* = M - Q²/2r into the Schwarzschild metric.


My question is: If there is energy in the electric field of a charged particle, why doesn\'t this increase the effective mass?

Is this true for any quantum \"charge\"? For instance, if the black hole has excess \"weak charge\" or \"color\", does this decrease the radius of its event horizon/effective mass?

 

[Giulio B.]

don't take it as a bible but you can't consider an electric field like such and mass both, otherwise you would have got its energy doubled...

 

[pervect]

Here is the metric for an uncharged non-rotating black hole:
http://en.wikipedia.org/wiki/Schwarzschild_black_hole

Here is the metric for a charged non-rotating black hole:
http://en.wikipedia.org/wiki/Reissner-Nordström_black_hole

Note that this is the same as plugging M^* = M - Q²/2r into the Schwarzschild metric.My question is: If there is energy in the electric field of a charged particle, why doesn\'t this increase the effective mass?

Is this true for any quantum \"charge\"? For instance, if the black hole has excess \"weak charge\" or \"color\", does this decrease the radius of its event horizon/effective mass?

The charged black hole is not a vacuum solution. What the solution is saying is that the electric field of the black hole does contribute to its mass.

Look at the M* at r=infinity. This is equal to be 'M', the mass of the black hole.

At r < infinity, M* < M. You've excluded part of the electric field of the black hole. By this exclusion, you've reduced its "mass".

To really make this more formal, we need a couple of things,.

1) Birkhoff's theorem tells us that any spherically symmetric solution is going to be a Schwarzschild solution. The agreement of the R-N solution with the Schwarzschild solution with a variable substitution is therefore expected.

2) We can't (quite!) apply Gauss's law to a black hole unmodified. With a static metric, we can define a "gravitational field". We can multiply this falue of the "field" by the area of an enclosing sphere. This gives us our first shot at Gauss's law for gravity - but we find that this result (field * area) is not independent of r. Remember that in electromagnetism, the intergal of the field (E-field) times the area IS a constant, and equal to the charge - this is Gauss's law.

But we can "fix up" Gauss's law so that it does work for gravity, by adding a correction factor that's related to the time-dilation factor at the value of 'r' that our enclosing sphere is at (relative to an obsever at infinity).

This "fixed up" Gauss-law mass is known as the Komar mass. I wrote a short piece about it https://www.physicsforums.com/showthread.php?t=116769" which is still just a very brief overview compared to the typical textbook treatment.

This "fixed up" gauss-law mass, i.e the Komarr mass, is just the parameter M that is constant in the Schwarzschild solution, and is not constant in the R-N solution.

Therefore the solution is telling us that the R-N solution is a spherically symmetric non-vacuum solution, where the electric field contributes to the mass, thus the solution has its full value of mass only at r=infinity. The Gass-law intergal excludes some of the electric field at r<infinity, and hence excludes some of the mass of the black hole.

The remaining issue is that GR is a classical theory. Therefore it is running into the same problem that classical E&M has with the self-energy of a point charge - it's infinite :-(.

Probably the R-N solution breaks down at some point because of this issue, which isn't solved in GR (yet another issue for quantum gravity).

Note that the field from a static black hole is likely to exceed the "Schwinger limit", the maximum field that can exist without pair creation if you take the R-N equations too literally. So remember that the R-N solution is purely classical, because GR is a classical theory, therefore such quantum mechanical effects are not modeled correctly. The problem with the self-energy of an infinite point charge is only solved in QM, not classical mechanics.