# Thoughts on electrically charged black holes

1. Oct 4, 2011

### mrspeedybob

Suppose there exists a black hole with a large + electrical charge. The escape velocity for any given particle will depend on the net force on it, gravitational + electrical. It seems logical that the more positive a particles charge the lower its escape velocity will be from any given point. If the escape velocity for a neutral particle, such as a neutron or photon, would be > c then that particle is inside the black holes event horizon. But if a proton would have a lower escape velocity then it would seem possible for it to escape from a lower position then the photon could.

Is anything wrong with my logic so far? If not I have more questions about the implications of this but I don't want to get too speculative if there is some serious flaw with what I've said so far.

2. Oct 4, 2011

### Ben Niehoff

You are right that the electric force can counteract the gravitational pull to some degree. However, the horizon is still a fixed surface in space, the same for all observers, regardless of their charge. It represents the point at which gravity always "wins".

The reason for this is that physical charged black holes must always have their mass greater than their charge, M > Q in natural units. If one had a point mass with a charge Q > M, it would form a naked singularity, with no horizon at all. However, one can show that given a black hole with M > Q, there is no physical process that can raise Q > M...i.e., no matter what kind of stuff you can imagine throwing into the black hole (provided that stuff satisfies realistic physical conditions), you will never raise its charge-to-mass ratio to 1 or higher.

A charged black hole with M = Q is called "extremal", and is somewhat analogous to absolute zero (in that it is an ideal that cannot be reached by any physical process). In fact, the Hawking temperature of an extremal black hole is zero. A black hole with M = Q has its gravitational pull exactly balanced by its electric charge. In fact, one can easily find solutions to Einstein's equations for multiple extremal black holes, all of positive charge, distributed throughout space in any way you like: the reason this is possible is that the black holes, all having M = Q, exert no net force on each other. Such a solution is often called "BPS", after Bogomolnyi, Prasad, and Somerfeld, because it satisfies an extremal energy bound resembling those which BPS studied.

3. Oct 4, 2011

### mrspeedybob

Let me re-state what I think you said, and tell me if I got it right.

The definition of the event horizon is the surface from which nothing can escape. Since a charged black hole will repel like charges this brings the surface of the event horizon closer then it would otherwise be. The surface from which non-charged particles and light cannot escape would be unaffected but this surface would no longer be the event horizon.

4. Oct 4, 2011

### Ben Niehoff

No. The surface from which positively-charged, negatively-charged, and neutral particles cannot escape is the same surface for all kinds of particles.

5. Oct 4, 2011

### PAllen

The event horizon is defined in terms of null geodesics (light), period. It is simply the surface defined by null geodesics that can never escape to spatial infinity. Suppose the charged black hole is positive. A positive charge can still not escape from just below the horizon due to repulsion; despite repulsion, it still must follow a timelike path, which means once below a horizon by the tiniest amount it can never reach the horizion; it cannot overtake trapped light.

6. Oct 4, 2011

### DrGreg

Another way of looking at this is that it would require an infinite force (which is impossible) applied to a particle with non-zero mass to make it hover at the event horizon. Electrostatic repulsion can only supply a finite force, which is insufficient to prevent a charged particle fall through the horizon.

7. Oct 4, 2011

### mrspeedybob

If Q>M forms a black hole with en event horizon of radius zero then can I assume that event horizon radius is a continuous function of Q and M where 0<Q<M? and that as Q decreases radius increases?

If what you say is true the relationship between event horizon radius and charge cannot simply a matter of electrostatics because then the point-of-no-return would depend on the charge of the test particle. What then is the relationship between black hole charge and event horizon radius? How does charge effect radius?

My understanding of GR is somewhat limited and most of it is based on SR and Einsteins equivalence principal. Perhaps if you could explain how the equivalence principal would hold true in the gravitational field of a charged body it would give me a stepping stone toward understanding the more extreme black hole scenario.

8. Oct 4, 2011

### PAllen

Dr. Greg's explanation was based on SR, implicitly. GR includes as an axiom, that SR holds locally everywhere. So imagine just below the event horizon. Independent of charge, for a test particle to reach the event horizon is locally equivalent to catching up with light. The event horizon is a light-like surface. This is simply impossible, given SR. The principle of equivalence is irrelevant. This analysis explains that the electrostatics has nothing to do with point of no return. Imagine a negative charge just below the horizon of positive black hole, compared to a positive one. The finite difference in electrostatic force is irrelevant to the fact that it would take infinite force to reach the event horizon.

The event horizon does not shrink to zero as charge approaches the value for extremal black holes. Instead the event horizon simply disappears at this point (when charge exceeds the extremal value), leaving a naked singularity.

[For simplicity, I am deliberately leaving out discussion of Cauchy surfaces and ergospheres, sticking to outer event horizon.]

Last edited: Oct 4, 2011
9. Oct 4, 2011

### Passionflower

Not sure what you are trying to say here, even in the simple Schwarzschild solution a test observer will momentarily cross the event horizon with a local velocity of c.
Are you saying this is not the case?

10. Oct 4, 2011

### PAllen

Only as seen by an observer at infinity. In their own frame, they are obviously not moving at all. However, the moment they cross the event horizon, an infinite force would be required to get back to the horizon. Actually, though, I'm not sure analogy is quite precise. Really, their light cones have rotated as they cross the horizon such that the singularity is in their inevitable future and the horizon is in their past. They can no more get back to the horizon than they can go back in time. (This light cone description is for the Schwarzschild solution. I have not studied the light cone structure of Kerr-Newman black holes; it may be similar, but I don't know for sure).

11. Oct 5, 2011

### Ben Niehoff

The same light-cone structure carries over to Kerr-Newmann and Reissner-Nordstrom black holes. In rotating black holes, the light cone first leans over (at the ergosphere), forcing co-rotation, before finally leaning inward (at the horizon).

In every kind of black hole, the horizon is a null surface where the spacelike vs. timelike roles of two basis vectors switch (except in extremal cases where the horizon is a double root, causing the two to immediately switch back again).

12. Oct 5, 2011

### Passionflower

Here is a graph showing various local velocities for test observers in a Schwarzschild solution. Clearly at r=rs they are all 1.

[PLAIN]http://img40.imageshack.us/img40/1793/velocities.png [Broken]

Last edited by a moderator: May 5, 2017
13. Oct 5, 2011

### Ben Niehoff

The phrase "local velocity of a test observer" makes no sense, as it is by definition zero.

14. Oct 5, 2011

### Passionflower

No, that is not correct, as dr/dt in Schwarzschild coordinates goes to zero when r->2M, and besides this is not a local velocity but a velocity at a distance.

It is also very easy to see that the local velocity of a test observer free falling from infinity at r=2M is c in Gullstrand–Painlevé coordinates. In fact for both the Schwarzschild and Gullstrand–Painlevé coordinates the local velocity is -sqrt(2M/r).

15. Oct 5, 2011

### PAllen

What on earth do you mean by local velocity? Velocity of what relative to what? I thought I was quite clear in my scenario about what is measured relative to what:

- a charge has just free fallen through the event horizon.
- sufficiently locally, its physics can be described as if it were at rest in a Minkowski inertial frame (this is axiomatically true in GR, at all points of the manifold; noting that the true singularity is not in the manifold).
- in this frame, getting back the event horizon would be equivalent to overtaking light (well, actually, worse; a spacelike path would be required to get back to the event horizon).

What is not clear about this? It is certainly true. I think you are giving unusual meaning to coordinate quantities.

16. Oct 5, 2011

### Passionflower

Let me ask you this, do you think that velocity is clearly defined in GR if it is not measured locally?

Relative to what? For instance relative to stationary or other observers. However when the velocity becomes c it is obviously c relative to all observers.

17. Oct 5, 2011

### PAllen

Velocity is clearly defined as 4-vector. A 4-vector is quantity with an infinite number of coordinate representations. A local statement about a velocity measurement must entail two 4-velocities - that of the measuring device and that of the object being measured, both at the same event. Then you can say, unambiguously, that the object's velocity vector expressed in the frame of the instrument is what the instrument would measure.

I agree that velocity 'at a distance' is not well defined in GR. In that sense, I agree one of my statements was inaccurate: velocity observed by a distant observer is never well defined.

Velocity of a free fall body relative to a coincident hovering observer would be well defined, but it is not remotely the scenario I was describing.

The locally measured speed of a material body can never reach c. It is mathematically impossible, as this would mean there is a frame (of an instrument) in which a unit vector becomes a null vector.

I'm not sure what it is you think you are calculating, but it certainly can't be speed of free falling body relative to a hovering observer at the horizon because there is simply no such thing as a hovering observer at the horizon.

18. Oct 5, 2011

### Passionflower

We can have a hovering body momentarily at the EH but aside from that, we can easily calculate the velocity at the EH between a free falling observer falling at escape velocity and compare those to the velocities of a free falling observer from a given r value and from a free falling observer with an initial velocity at infinity. In both cases we calculate that the velocity is c.

But if you disagree and do not think it is c, I suppose you would not mind telling me the velocity between a free falling observer at escape velocity and an observer dropped at say R=10 at the EH?

We can even continue past the EH, while the spacetime is no longer stationary we could still calculate the local velocity between two observers.

Last edited: Oct 5, 2011
19. Oct 5, 2011

### PAllen

A path even momentarily stationary at the event horizon has a lightlike tangent rather than a timelike tangent. Thus it would represent the frame of light, which is normally considered undefined in relativity.

No material instrument will ever locally measure the velocity of a body to be equal or greater than c. As I have explained, this is mathematically impossible in GR and SR (conventionally interpreted; let's not sidetrack into tachyons or the OPERA measurement).

20. Oct 5, 2011

### PAllen

Sure you can continue timelike paths past the event horizon. Sure you can compute relative velocity of two objects passing the horizon at the same event with different infall histories. You will never get c for the relative velocity. To claim you do amounts to the claim that timelike unit vector expressed in some local frame becomes a null vector. This is just mathematically impossible. Whatever you are calculating it cannot be a velocity of material body measured locally by a material instrument.