# Charged Black Holes

1. Feb 16, 2013

### gentsagree

The Schwarzschild metric describes spacetime around a spherically symmetric neutral object and, as such, it is considered as a vacuum solution, with zero contribution from the energy-momentum tensor that otherwise influences the space in the region 0<r<2GM.

The Schwarzschild metric can be analytically extended to cover the region beyond the horizon, obviously, but this remains a vacuum solution.

I was wondering why the Reissner-Nostrom metric, being the solution AROUND a spherically symmetric charged object, is not considered a vacuum solution as well.

What I mean is: aren't the actions of the gravitational field and of the electromagnetic field qualitatively the same? If the electromagnetic field can be "felt" outside the horizon (so that the metric outside $r_{+}$ is influenced by it), shouldn't it be the same for the gravitational field of the pointlike mass at the origin of the Schwarzschild solution, thus causing the latter to be influenced by some non-vanishing stress-energy tensor?

In the same way "there's no mass" outside the Schwarzschild radius, there is no charge outside $r_{+}$.

I'm sure I'm mistaken somewhere, but can't see why.
Thanks

2. Feb 16, 2013

### Mentz114

If I remember correctly, there is no charge outside r+ but there is an electric field. The stress-energy tensor ( from the Einstein tensor) is not zero but is the same as that of the field of a point charge.

3. Feb 16, 2013

### gentsagree

It makes sense, there is no charge, but only electromagnetic field. So, would you say that in the comparison with the Schwarzschild solution, outside the horizon there is no mass but only gravitational field? And this gravitational field is accounted for by the geometry of the space rather than by the stress energy tensor?

4. Feb 16, 2013

### Staff: Mentor

Because it isn't one. There is a non-zero stress-energy tensor present everywhere in the R-N metric, because there is energy stored in the electric field.

This is true, and there is in fact an important sense in which the electric field in the R-N metric *does* "work the same" as gravity in the Schwarzschild metric. The basic principle is that the "field" that is felt at a given event in spacetime is entirely due to the presence of a "source" for that field somewhere in the past light cone of the given event.

Take the gravity case first. A real Schwarzschild black hole will not have existed forever; it will have been formed by the collapse of some massive object at some time in the past. That means that there *is* a region of non-zero stress-energy tensor in the spacetime; it's just far in the past, from the viewpoint of someone who is outside the black hole long after it forms. This region of non-zero stress-energy tensor is the "source" for the hole's gravity; at any given event outside the hole's horizon, the gravity that is felt is due, ultimately, to the "source" (non-zero stress-energy tensor inside the collapsing mass) in the past light cone of that event. This means, of course, that the "source" of the hole's gravity, as it is felt outside the horizon, is not actually coming from inside the horizon; it is coming from a region of spacetime outside the horizon, but far in the past, from the massive object that collapsed to form the hole, during the period *before* its collapsed formed the horizon and created the hole.

Now suppose the collapsing object in the far past was charged. Then, in the past light cone of a given event outside the resulting R-N black hole, there will not only be a region of non-zero stress-energy tensor (which is the source of the hole's gravity); there will also be a region of non-zero charge density, which is the source of the hole's electric field. The field felt at the given event is entirely due to that non-zero charge density in the past light cone of the event; it is not coming from inside the horizon, any more than the hole's gravity is coming from inside the horizon.

The difference in the charged case is that there is energy stored in the electric field even after the original charged object has collapsed. In other words, the collapse of a charged object leaves behind stress-energy, stored in the electric field, whereas the collapse of an uncharged object leaves no stress-energy behind. Ultimately, the reason for this is the equivalence principle: different objects can "fall" with different accelerations in an electric field, but all objects fall with the same acceleration due to gravity. So gravity can be modeled entirely as the curvature of spacetime, but electromagnetism cannot. That's why the spacetime of an uncharged hole is vacuum, but the spacetime of a charged hole is not.

5. Feb 16, 2013

### gentsagree

Thank you PeterDonis for the very exhaustive answer.

6. Feb 16, 2013

### gentsagree

There is one thing I am unsure of. Why does the collapse of a charged object leave stress energy, whereas the other doesn't? If the answer is somehow encoded in your comparison with the equivalence principle, then I did not get it. :S

Last edited: Feb 16, 2013
7. Feb 16, 2013

### Staff: Mentor

Because electromagnetism can't be viewed as due to spacetime curvature alone, whereas gravity can. Spacetime curvature can be present in a vacuum, where there is no stress-energy; but electromagnetism can't. If there is an electromagnetic field present, it must store a non-zero amount of stress-energy everywhere it is present. The EM field of a charged black hole is present everywhere in the spacetime, so the non-zero stress-energy stored in it is also present everywhere.

8. Feb 16, 2013

### gentsagree

Awesome, thanks.