Is 'charged black hole' an oxymoron?

[PeterDonis]

Obviously, by the nature of parallel transport, the invariant mass (norm of the four-momentum) does not redshift, but that is about the only thing that I can say for sure.

If you assigned a charge-current 4-vector to the charged object falling into the hole, its norm would also be unchanged by parallel transport, correct? In the object's instantaneous rest frame, the components would always be (q, 0, 0, 0).

For everything else I would have to work out the parallel transport equation, but in my mind it is not clear what paths to use for the various cases that have been discussed.

There are three key timelike worldlines that I see as interesting:

(1) The worldline of a neutral object free-falling into the hole. Other than adding the Q term to the "redshift factor", these should work the same as they do in Schwarzschild spacetime.

(2) The worldline of an object with like charge to the hole, with no other forces acting except the EM force.

(3) The worldline of an object with opposite charge to the hole, with no other forces acting except the EM force.

For simplicity, I would use objects "at rest at infinity" as the initial condition for all three worldlines.

The first worldline would be an infalling geodesic. The second and third would have proper acceleration determined by the Lorentz force law.

In addition, the worldlines of infalling photons might be interesting, but they would just be ingoing null geodesics, and other than including the Q term in the "redshift factor" I don't see any significant differences from Schwarzschild spacetime for these.

One other approach that might be worth looking at: use an effective potential that includes a term for the potential energy in the EM field, as well as the usual one for "gravitational potential energy" in a static spacetime. Since the EM force is conservative, a conserved total energy, including both types of "potential energy", can be defined. I've seen something like this, for example, in this paper:

http://arxiv.org/pdf/1103.1807v3.pdf

 

[Dale]

You will need to elaborate - you mean I've somehow been getting it wrong on Maxwell's equations in flat spacetime?

Sorry, no that is not what I mean. I only meant that a lot of the scenarios you have proposed, like the one in 109, involve charges in the curved spacetime outside an uncharged planet. I.e. The metric is the Schwarzschild metric, not the RN metric. It makes things slightly easier.

Appreciate you're willing to tackle it, but how much of parallel transport in GR do I need to learn? And to what extent is parallel transport aloof from assumptions about how gravity interacts with EM - charge in particular?

I would recommend learning about parallel transport, it is a pretty fundamental part of the math of curved spaces. I don't know what you mean by "aloof", but parallel transport underlies a lot of the geometric concepts that GR is built on. It is central to things like curvature, the connection, covariant derivatives, etc., all of which are used by the physics.

For instance, there can be no disputing that say the arrangement in #109 demands the EM energy of a system of charges gravitationally redshifts

I dispute . I think that it is not even clear what is meant by the phrase "gravitationally redshifts" in this context. Since it is not clear what that means, IMO it is hard to say what can be concluded about "gravitational redshift" from 109.

When we say that the energy of light gravitationally redshifts in Schwarzschild coordinates what we mean, specifically, is that if you take a null four-momentum vector, and parallel transport it along an outgoing null geodesic, then the timelike component will be reduced by the usual factor. I don't know how to make a similar construction for the other quantities you have been discussing.

 

[PeterDonis]

Yes as such but that's owing to the moving-on-rails constraint imposing exact balance of opposing transverse forces as evaluated in any given frame. To see how each of those opposed forces transform we need to consider the situation of instantaneous free-fall of q along x (at the instant u_x = 0)

The whole system is in free fall for all time; it remains in the same inertial frame indefinitely (strictly speaking, one of two, depending on whether you pick the rest frame of the capacitor or q). The "instantaneous free-fall of q along x" is q's actual motion, so the force balance I wrote down *is* the one in the "instantaneous free fall frame".

You are confusing this situation with one in which there is actual proper acceleration. There is none in this scenario.

with rails constraint in place, transverse dp/dt (applying to each opposed electric and mechanical restraint force separately)

There is no electric force on any object in the problem except for q. All the other forces are mechanical. And there is no dp/dt anywhere; nothing is accelerating. There is relative motion in the y direction, but it's at a constant speed.

Let me put this another way: the "force transformation" equations for dp/dt that you keep throwing at me are *kinematic* equations. They tell you how to relate a dp/dt in one frame to a dp/dt in another frame, given that you already know what dp/dt is in the first frame. None of that applies here because dp/dt is *zero* for everything; zero transforms to zero. These kinematic equations do *not* tell you how to determine dp/dt in the original frame, from an original cause. For that, you need a dynamic equation, something that relates a dp/dt to a cause, such as an EM force.

The only dynamic equation that's relevant for this problem is the Lorentz force law, and it only acts on one object, the charge q. Everything else is electrically neutral. And because the Lorentz force is opposed by the force of the arm, the Lorentz force law does not actually translate into any nonzero dp/dt for this problem. So again, there's nothing to transform.

What the Lorentz force law does tell you is what static force from the arm is required to keep q motionless. But once you know that, everything is static and perpendicular to the relative motion. Nothing changes with time. That means that, once again, equations for dp/dt are irrelevant. All the gamma factors are irrelevant; they arise because things are changing with time, and "changing with time" is frame dependent. But a static 4-vector with its only nonzero component perpendicular to the relative motion is *not* frame-dependent; it looks the same in both frames, as I showed. And that static 4-vector, not any of the dp/dt's you showed, represents the observed forces O' and O.

If you still feel not, I invite you to solve that paradox of #188!

Which is not set in this scenario. As I've said before, if there really is something wrong with my answer, you should be able to show me a "paradox" set in *this* scenario, not some other one.

 

[Dale]

If you assigned a charge-current 4-vector to the charged object falling into the hole, its norm would also be unchanged by parallel transport, correct? In the object's instantaneous rest frame, the components would always be (q, 0, 0, 0).

Yes, the norm would always be unchanged by parallel transport, but I thought that the norm is ρ, not q.

There are three key timelike worldlines that I see as interesting:

(1) The worldline of a neutral object free-falling into the hole. Other than adding the Q term to the "redshift factor", these should work the same as they do in Schwarzschild spacetime.

I agree.

(2) The worldline of an object with like charge to the hole, with no other forces acting except the EM force.

(3) The worldline of an object with opposite charge to the hole, with no other forces acting except the EM force.

While those are interesting for studying the kinematics of charged particles in the RN solution, I don't think that they are useful for parallel transport. Since there is a force acting on these particles their four-momentum is not parallel transported but explicitly undergoes some change given by F=dp/dτ.

Also, I think that a spacelike path like $(t,r,\theta,\phi)=(0,r,90°,0°)$ may be useful for some of these "x gravitationally redshifts" statements. For example, when talking about lowering a mass on a rope, the worldline of the mass itself doesn't seem appropriate for discussing whether or not its timelike component redshifts since it is ingoing, nor does any outgoing timelike or null geodesic (since the mass doesn't follow those), so the spacelike path seems reasonable to me.

 

[PeterDonis]

Yes, the norm would always be unchanged by parallel transport, but I thought that the norm is ρ, not q.

By "q" I meant the charge (or charge density) as measured in the charged object's instantaneous rest frame. That should also be the norm of its charge-current 4-vector.

While those are interesting for studying the kinematics of charged particles in the RN solution, I don't think that they are useful for parallel transport. Since there is a force acting on these particles their four-momentum is not parallel transported but explicitly undergoes some change given by F=dp/dτ.

Yes, this is true for the 4-momentum. However, other 4-vectors associated with the object (such as a charge-current 4-vector) may still be parallel transported. In any case, I think the behavior of these worldlines is useful for comparison with what happens when, for example, a charged object is slowly lowered on a rope.

Also, I think that a spacelike path like (t,r,θ,ϕ)=(0,r,90°,0°) may be useful for some of these "x gravitationally redshifts" statements.

Do you mean a "purely radial" spacelike path--i.e., a spacelike curve in a slice of constant time in the purely radial direction?

 

[Dale]

By "q" I meant the charge (or charge density) as measured in the charged object's instantaneous rest frame. That should also be the norm of its charge-current 4-vector.

I didn't think that charge was the norm of any 4-vector. I thought only charge density was. If it is, then that makes that part very simple.

Yes, this is true for the 4-momentum. However, other 4-vectors associated with the object (such as a charge-current 4-vector) may still be parallel transported.

Good point, I hadn't considered that.

Do you mean a "purely radial" spacelike path--i.e., a spacelike curve in a slice of constant time in the purely radial direction?

Yes, due to the spherical symmetry I would like to keep everything purely radial wherever possible, just for simplicity.

 

[PeterDonis]

I didn't think that charge was the norm of any 4-vector. I thought only charge density was.

Technically, I think you're right. But for the case of an isolated "test object" with charge I think it will work. After all, for an isolated object we assign a 4-momentum, not a 4-momentum density; we just assume that we can use a 4-vector with units of mass, not mass density, as long as the object is isolated and we treat it like a "point particle".

We could also look at it this way: the BH itself could be assigned a "charge-current" 4-vector in the asymptotically flat region, the norm of which is Q, similar to the way an overall 4-momentum can be assigned to it, the norm of which is M. Again, I haven't actually seen this done explicitly in the literature, but I don't see why it wouldn't make sense, since the definitions of Q and M are exactly analogous. If we can do it for the BH itself, we should be able to do it for a "test object".

 

[Q-reeus]

Sorry, no that is not what I mean. I only meant that a lot of the scenarios you have proposed, like the one in 109, involve charges in the curved spacetime outside an uncharged planet. I.e. The metric is the Schwarzschild metric, not the RN metric. It makes things slightly easier.

Fine - makes sense.

I dispute . I think that it is not even clear what is meant by the phrase "gravitationally redshifts" in this context. Since it is not clear what that means, IMO it is hard to say what can be concluded about "gravitational redshift" from 109.

To me not so unclear - EM static field energy is depressed in a gravitational potential well, equally as for any other form of energy. Note I have edited #208 somewhat.

When we say that the energy of light gravitationally redshifts in Schwarzschild coordinates what we mean, specifically, is that if you take a null four-momentum vector, and parallel transport it along an outgoing null geodesic, then the timelike component will be reduced by the usual factor. I don't know how to make a similar construction for the other quantities you have been discussing.

OK well redshift of light involves just source-free transverse fields, so that as such has nothing to say about possible global effective failure of [STRIKE]div.rho[/STRIKE] div.E = ρ/ε₀, which is what failure of global charge invariance implies. Maybe extra factor for BH (field lines that simply disappear into the interior.)

 

[Q-reeus]

The whole system is in free fall for all time; it remains in the same inertial frame indefinitely (strictly speaking, one of two, depending on whether you pick the rest frame of the capacitor or q). The "instantaneous free-fall of q along x" is q's actual motion, so the force balance I wrote down *is* the one in the "instantaneous free fall frame".
You are confusing this situation with one in which there is actual proper acceleration. There is none in this scenario.

I know that but I meant it's by allowing an instantaneous 'virtual' free-fall one obtains that relation I linked to which gives then the correct force acting on q and specifically shows it is not a function of velocity in lab frame. Which then demands that proper force in q's frame must read differently to lab frame value. No way out of that.

Q-reeus: "If you still feel not, I invite you to solve that paradox of #188!"
Which is not set in this scenario. As I've said before, if there really is something wrong with my answer, you should be able to show me a "paradox" set in *this* scenario, not some other one.

Well you just don't accept my explanations - so it really is important to deal with that #188 situation imo. Up to you - no gun to head. Just can't figure reluctance though; sure seems conceptually simple enough and may I suggest a game-changer if followed through to logical conclusions. :zzz:

 

[Austin0]

Obviously, by the nature of parallel transport, the invariant mass (norm of the four-momentum) does not redshift, but that is about the only thing that I can say for sure. For everything else I would have to work out the parallel transport equation, but in my mind it is not clear what paths to use for the various cases that have been discussed.

If a mass is transported to a lower altitude it then requires more energy to maintain a state of rest wrt the gravitating body . This would seem to be an increase in effective inertial/gravitational mass , or not?
What is this called as obviously the invariant mass is unchanged??

 

[Dale]

To me not so unclear - EM static field energy is depressed in a gravitational potential well, equally as for any other form of energy. Note I have edited #208 somewhat.

I agree that any energy which travels on null geodesics is redshifted as it comes up out of a potential well, but that doesn't in any way imply that the energy is depressed while it is down in a potential well.

EDIT: actually, having thought about it a bit I think I was hasty here.

Parallel transport is the method by which you compare tensors at different points on the manifold. The big problem with curved spaces, what distinguishes them from flat spaces, is that parallel transport is path dependent. So comparing two vectors at different points in the space will give different results depending on the path chosen. But that doesn't mean that we cannot arbitrarily choose parallel transport along null geodesics as our method of comparison and define the term redshifted to refer specifically to that.

 

[Dale]

If a mass is transported to a lower altitude it then requires more energy to maintain a state of rest wrt the gravitating body .

It doesn't require any energy to maintain a state of rest wrt the gravitating body. E.g. consider a book on a table.

What is this called as obviously the invariant mass is unchanged??

Parallel transport preserves the dot product. The norm squared of a vector is the dot product of a vector with itself. So parallel transport preserves the norm. Invariant mass is the norm of the four-momentum, so parallel transport preserves the invariant mass.

 

[PeterDonis]

I know that but I meant it's by allowing an instantaneous 'virtual' free-fall

There's no *need* to "allow" for any "instantaneous virtual free fall". Every object in the system is already *in* free fall, all the time.

Well you just don't accept my explanations - so it really is important to deal with that #188 situation imo.

Not until we've dealt with this one.

Just can't figure reluctance

And I can't figure your reluctance to stick to one scenario and to show me how my prediction for the relative scale readings leads to a "paradox" in that scenario. Instead you keep dragging in other scenarios and leave me to do the work of figuring out how they're supposed to be relevant.

 

[Austin0]

It doesn't require any energy to maintain a state of rest wrt the gravitating body. E.g. consider a book on a table.

Parallel transport preserves the dot product. The norm squared of a vector is the dot product of a vector with itself. So parallel transport preserves the norm. Invariant mass is the norm of the four-momentum, so parallel transport preserves the invariant mass.

`Sorry , I see I was really unclear on this one. By altitude I meant still up in the air. At rest was hovering through propulsion , hence more energy required than hovering at the initial higher altitude.
This is correct?
So this increase would seem to indicate an increase of some kind of mass or a decrease of energy output of the means of propulsion ?

 

[jartsa]

Yes. On the average, the lower clock runs slower, but not at every moment.

On the average, the lower clock, with charged clock hands, emits an EM-wave that waves slower, but not every moment.

Here you see, the slowness of the low clocks does cause some of the change of frequency of the light emitted from a gravity well.

 

[Austin0]

No, it isn't; it's the acceleration of the floor of the box. The boost only gives a constant relative velocity; but a constant relative velocity won't cause a *difference* in impacts on the upper and lower surfaces of the box. (If you doubt that, try analyzing the freely falling case, not in the MCIF as I did, but in a frame that is Lorentz boosted relative to the MCIF. In that frame, there will be a Doppler redshift at one end and a blueshift at the other, but they will cancel because the relative velocity is constant.) A difference in impacts, producing a net force, requires a *change in velocity* during the flight time of the photons, i.e., acceleration.



No, because there's only a single surface, and "locally" means I can get as close to the surface as I like to make the measurement. There has to be a change in velocity of the "box" *during the flight time of the photon*. If the measurement is purely local, there is no "flight time". The "box" has to have a finite distance between the upper and lower surfaces, to give the acceleration time to produce a measurable effect.

This does raise a point: I said previously that the box has to be small enough that tidal effects are negligible; but now I'm saying that the box can't be too small or acceleration effects are negligible! That's true; there has to be some finite range of "box size" where we can see the effects of acceleration, but can't yet see tidal effects. The reason there is in fact such a range of box size is that acceleration is a first-order effect but tidal effects are second order. More precisely, acceleration is due to the first derivative of the metric coefficients--the connection--while tidal effects are due to the second derivatives--curvature.

austin0
If we assume that the box is suspended by thrust and then the whole system moves to a lower potential under thrust do you think this changes conditions in the box .i.e. producing a different result from this scenario?

You missed this one.

Also do you think that a box suspended at a constant altitude (static) would lose energy in this manner?

 

[Austin0]

First of all, I didn't say anything about velocities. I was talking about coordinate transformations. Coordinate transformations change velocities, and clock rates.

Second, no, gravitational time dilation is not different from velocity time dilation--not in any absolute sense, anyway. What looks like gravitational time dilation in one coordinate system will look like velocity dependent time dilation in another coordinate system. "Gravitational time dilation" doesn't have anything specifically to do with gravity, it has to do with using curvilinear coordinates. If you use inertial coordinates, there is no gravitational time dilation.

Third, what you said is just not true. You can certainly find a coordinate system in which, at one particular time, the lower clock runs faster than the upper clock.

Let (x,t) be the Schwarzschild coordinates in which the two clocks are at rest. Then, according to those coordinates, Rate_{upper clock}/Rate_{lower clock} = 1+ gh/c². Now, switch coordinates to coordinates (X,T) defined by:

X = x
T = (1+Ax)t

Then in the coordinate system (X,T), the ratio of the clock rates will be given by:

Rate'_{upper clock}/Rate'_{lower clock} = (1+gh/c²)/ (1+Ah)

If A > g/c², then in this coordinate system, the lower clock runs faster than the upper clock.

Clock rates are a coordinate-dependent quantity. There is no absolute sense in which the upper clock always runs faster than the lower clock.

Could you explain a little the coordinate system you are describing here.
Are you talking about arbitrary scaling of clock rates? mechanically adjusting the rates?

 

[stevendaryl]

Could you explain a little the coordinate system you are describing here.
Are you talking about arbitrary scaling of clock rates? mechanically adjusting the rates?

Yes, it's an arbitrary coordinate system, involving a distance-dependent scaling of the time. General Relativity allows absolutely any coordinate system to be used, and I just picked one. There's no particular physical significance to it.

 

[Dale]

`Sorry , I see I was really unclear on this one. By altitude I meant still up in the air. At rest was hovering through propulsion , hence more energy required than hovering at the initial higher altitude.
This is correct?

Yes.

So this increase would seem to indicate an increase of some kind of mass or a decrease of energy output of the means of propulsion ?

Or an increase in the gravitational force on the same mass.

 

[stevendaryl]

On the average, the lower clock, with charged clock hands, emits an EM-wave that waves slower, but not every moment.

Here you see, the slowness of the low clocks does cause some of the change of frequency of the light emitted from a gravity well.

I'm not sure I understand what the significance of "charged clock hands" is. But in GR, we can define the notion of a "standard clock". A standard clock has an elapsed time $\tau$ that satisfies: d$\tau$² = g_{$\alpha\beta$} dx^$\alpha$ dx^$\beta$
where g is the metric tensor. If this is the case, then whether that clock "runs slow" or "runs fast" is a coordinate-dependent fact.

Whether a nonstandard clock is running slow or fast compared with a standard clock at the same location is a coordinate-independent fact.

 

[Q-reeus]

And I can't figure your reluctance to stick to one scenario and to show me how my prediction for the relative scale readings leads to a "paradox" in that scenario. Instead you keep dragging in other scenarios and leave me to do the work of figuring out how they're supposed to be relevant.

Peter - on this lengthy side issue we evidently have reached a stalemate with neither prepared to concede anything to the other side. Best then imo to end it with an internet-mediated firm, sincere, warm and friendly handshake :!), and concentrate on the main theme of this thread. Do we agree on that?

 

[jartsa]

I'm not sure I understand what the significance of "charged clock hands" is. But in GR, we can define the notion of a "standard clock". A standard clock has an elapsed time $\tau$ that satisfies: d$\tau$² = g_{$\alpha\beta$} dx^$\alpha$ dx^$\beta$
where g is the metric tensor. If this is the case, then whether that clock "runs slow" or "runs fast" is a coordinate-dependent fact.

Whether a nonstandard clock is running slow or fast compared with a standard clock at the same location is a coordinate-independent fact.

Pop science writers chooce a coordinate system where a clock at the top of The Empite State Building runs faster than a clock at the basement of The Empire State building, when they write that "a clock at the top of The Empite State Building runs faster than a clock at the basement of The Empire State building".

In aforementioned coordinate system a photon, dropped from the top of the aforementioned building, keeps a constant frequency.

 

[PeterDonis]

You missed this one.

The box being lowered by a rope loses energy because work is being extracted from it. How is work being extracted from the box being held up by rocket thrust? If there is no work being extracted, then yes, the results will be different.

Also do you think that a box suspended at a constant altitude (static) would lose energy in this manner?

Again, how would you extract work from a box that's not changing altitude? If there is no work being extracted, then its energy is not changing. For a box at rest a constant altitude, this should be obvious--kinetic energy zero, potential energy constant, therefore total energy constant.

 

[PeterDonis]

Peter - on this lengthy side issue we evidently have reached a stalemate with neither prepared to concede anything to the other side. Best then imo to end it with an internet-mediated firm, sincere, warm and friendly handshake :!), and concentrate on the main theme of this thread. Do we agree on that?

No problem. I agree it's a side issue.

 

[Dale]

Pop science writers chooce a coordinate system where a clock at the top of The Empite State Building runs faster than a clock at the basement of The Empire State building, when they write that "a clock at the top of The Empite State Building runs faster than a clock at the basement of The Empire State building".

Yes.

In aforementioned coordinate system a photon, dropped from the top of the aforementioned building, keeps a constant frequency.

No, in such a coordinate system such a photon's frequency increases, it does not stay constant.

 

[Dale]

On the average, the lower clock, with charged clock hands, emits an EM-wave that waves slower, but not every moment.

Even this isn't true in all coordinate systems. In fact, a particularly simple system is obtained by the transform:
$$T= t \sqrt{1-\frac{R}{r}}$$

In this coordinate system no stationary clock anywhere ever runs fast or slow.

 

[jartsa]

Yes.

No, in such a coordinate system such a photon's frequency increases, it does not stay constant.

Well how does it work then? Clock has an important role in frequency measurement.

If clock is slow by time dilation factor, and photon is fast by time dilation factor, then frequency measurement of the photon, using the clock, says that photon is fast by twice the time dilation factor. Or should I say fast by time dilation factor squared. It's wrong result anyway.

 

[GAsahi]

In aforementioned coordinate system a photon, dropped from the top of the aforementioned building, keeps a constant frequency.

GR and the Pound-Rebka experiment say that you are wrong.

 

[Dale]

If clock is slow by time dilation factor

Perhaps this is your problem. In GR, clocks don't slow down in any coordinate-independent sense, they just measure proper time along their worldline, which is an invariant quantity similar to path length along a curve in geometry.

What slows down or speeds up is coordinate time, and that is obviously a coordinate dependent quantity. You can make coordinates, like Schwarzschild coordinates, where many seconds of coordinate time pass for every second of proper time for clocks near the horizon. You can make other coordinates, like the one I proposed, where that doesn't happen.

Does that help?

 

[Trenton]

An oxymoron quite possible. According to most articles I can find, eg Wikipeadia, a charged black hole has an inner horizon. The in-fall of spacetime is arrested by charge.

But other accounts of the formation of a (non-charged) black hole have it that once a concetration of mass falls within Rs it must nessesarily collapse to a point singularity because otherwise matter would exceed C. This is because the infall of spacetime occurs at the Newtonian escape velocity which reaches C at Rs and exceeds C at r < Rs.

These two accounts appear to contradict. What is so special about charge? Could it be that the infall of spacetime can be slowed by anything with sufficient force to resist it? If a nutron star (for which the EoS is not known) were able to withstand enough force, it might form an event horizon but not actually collapse to a singularity until a much greater mass had been reached?

Of course it could be the other way round and that a charged black hole can not exist or at least, does not have an inner horizon. It depends on what a singularity might consist of. One theory is that all the mass of the black hole becomes pure energy in order to accommodate the collapse. In this case it is possible charge can be destroyed maybe.

It would be very nice to park an ion-engine in orbit round a black hole and pump a few tones of protons at it to see what would happen. If there is an inner horizon (spacetime infall is arrested by sufficient force) then when this horizon meets the outer horizon a mini big bang will occur.

 

[DrGreg]

It would be very nice to park an ion-engine in orbit round a black hole and pump a few tones of protons at it to see what would happen. If there is an inner horizon (spacetime infall is arrested by sufficient force) then when this horizon meets the outer horizon a mini big bang will occur.

As I understand it, it's actually very difficult for a black hole to acquire a significant charge. A positively charged black hole would repel any protons near it, so those protons would have to be fired inwards with a very high kinetic energy to overcome the electrical repulsion. If absorption is achieved, the kinetic energy contributes to the mass of the black hole (via E=mc²), and eventually you reach a point where each proton adds more mass than charge. Therefore it's postulated (I think) that the two horizons might never meet.

 

[GAsahi]

As I understand it, it's actually very difficult for a black hole to acquire a significant charge. A positively charged black hole would repel any protons near it, so those protons would have to be fired inwards with a very high kinetic energy to overcome the electrical repulsion. If absorption is achieved, the kinetic energy contributes to the mass of the black hole (via E=mc²), and eventually you reach a point where each proton adds more mass than charge. Therefore it's postulated (I think) that the two horizons might never meet.

What type of measurement would distinguish between a charged and an uncharged black hole?

 

[Dale]

What type of measurement would distinguish between a charged and an uncharged black hole?

The motion of charged vs uncharged particles.

 

[Austin0]

Originally Posted by Austin0

So this increase would seem to indicate an increase of some kind of mass or a decrease of energy output of the means of propulsion ?

Or an increase in the gravitational force on the same mass.

OK but I thought that the concept of downward force has been eliminated from GR ??

Unless it was an upward acceleration without actual motion. Stationary on the ground eg.

 

[Austin0]

The motion of charged vs uncharged particles.

How would charge escape from the hole if light cannot?