## Is 'charged black hole' an oxymoron?

 Quote by stevendaryl For problems involving gravity, you can't compute the mass of a collection of objects by adding up the mass of the component objects.
Depending on exactly what that means, I probably agree - assuming it means assembly of matter originally dispersed, then naturally a more compact *and static* arrangement exhibits a reduced assembled mass. Which is entirely in accord with what I have been saying all along - energy is given off during assembly, and lowering a mass is part of an assembly process in this context. Assuming my above interpretation of what you meant is correct, actually it is you who previously claimed otherwise - e.g. from #9:
 You believe that lowering a mass into a black hole causes the mass to reduce? I don't think that's true.
stevendaryl - I take you as sincere in your position, but this kind of thing has me scratching head.
 Q-reeus: "That for reasons covered above and in #1 cannot be true unless conservation of energy fails (energy extracted to the outside with no loss of net system mass/energy)." No, that's not true. If you extract energy from a system of masses, the total mass will decrease by the same amount. The total mass of a collection of particles is not, as I said, equal to the sum of the masses of the individual particles.
I will put that down to a complete misunderstanding of what I was saying - which was refuting your own previous claim as per e.g. quote from #9 above!
 Charge, however, does add linearly in that way.
That's your's and others position here (apart from perhaps one), but I am to be convinced it amounts to more than dogmatic assertion.
 I think the paradox you're talking about has nothing specifically to do with black holes.
Agreed - I said as much back in #1, and used spherical capacitor as example. You drew a different conclusion though.
 You can get a similar paradox using charged plates. Suppose you have two positively charged flat plates that are touching each other. (Held together by clamps). This composite object will have a certain total charge, and a certain total mass. Now, allow the charged plates to do work by allowing them to separate because of mutual repulsion. Now, the total charge will be unchanged, but the total mass will be smaller. Any time a system does work, its total mass changes. I don't see why black holes are any different in this regard.
It's a case of yes and no. Yes clearly if the work extracted from charged plates is channelled exterior to the system there is net loss to that system. But in BH case involving purely free-fall, no there is no net change as everything stays inside. None of that is paradoxical. let's see if I can't turn a variant of your charged plates example around on you. Just consider slow lowering of a charged parallel-plate capacitor down to some floor level in a potential well. If we agree all forms of energy redshift the same, what happens to the field strength between those capacitor plates? Only one guess.

 Quote by jartsa Or does a falling object gain mass?
Depends on your frame of reference jartsa. Drop a mass from a high tower, then catch it in a spoked wheel. It spins around with KE - that means total mass/energy is greater than the rest mass observed up in the tower before the drop. But from pov of far away observer, the system earth+tower+mass has not changed in gravitating mass at all - provided no frictional loss results in heat escaping.

 Quote by PeterDonis ...I'll also comment on a couple of things in the OP that need clarification/correction...
Peter - splendid job of explaining various contributions to Reissner-Nordstrom metric - thanks. But first I comment on your comments on my OP.
 It should also be noted that the formula for the "redshift factor" has the extra term in Q in it; the OP does not include this.
And the reason is two-fold:
a: My scenario was that of a Schwarzschild BH (but more generally an uncharged central mass M) perturbed slightly by a vastly smaller mass/charge in the process of being lowered/dropped in. Which is not a pre-existing R-N system with central charge already in place. So no conflict there.
b: Being highly sceptical of the premise of finite exterior charge for a BH, my use of an R-N model would be rather inconsistent.
 So when the OP says that the "natural" case to consider is free fall, that is not correct for a charged object; the "natural" case there is the trajectory where a_e is the proper acceleration of the object's worldline. This also means that the object's energy at infinity will *not* be constant, unlike that of a neutral object.
Again, as I did not model things using a pre-existing 'charged' R-N BH, it boils down to whether free-fall is an appropriate description for infall of an assumed small charged mass toward a large neutral mass M. If one ignores the typically vanishingly small radiative back reaction (dependent on the q/m ratio for starters), I would say yes given the context of what we are about here. And while agreeing your energy argument may be strictly correct, any relatively infinitesimal radiative energy loss is surely not really germaine to the topic. Also isn't there continuing dispute about whether any radiative loss at all applies to charge in radial free-fall?

Anyway, back to your interesting R-N dissection. I concentrate on just the results for proper acceleration a of a neutral test mass, and ae for a charged test mass (agree with pretty much all your other pertinent observations as far as what R-N metric implies.):
 Outside r+, the spacetime works much the same as the exterior region of Schwarzschild spacetime, just with some extra terms in Q. For example, the proper acceleration of a "hovering" observer, who stays at the same radial coordinate r, is: $$a = \left( \frac{M}{r^{2}} - \frac{Q^{2}}{r^{3}} \right) \frac{1}{\sqrt{1 - \frac{2M}{r} + \frac{Q^{2}}{r^{2}}}}$$ You can see that this will be *less* than the corresponding quantity in Schwarzschild spacetime at the same r coordinate, because both of the terms in Q make it smaller. However, it will still diverge at the outer horizon, r+, because the "redshift factor" still goes to infinity there.
Which, along with the later expression you give for ae, is really baffling to me. While axial stress component for E field of Q is negative, the two transverse components are positive. So I would expect a net positive stress adds to a positive energy density, yet together appears as negative contribution in R-N expression?

Regardless though it is all based on there actually being a finite external electric field. We have this R-N expression, and yes one then derives specific interesting results as you have done in #14. My question is, what is the physical basis for that Q lying at the EH can project a finite E to the BH exterior? Still wondering if it comes down to, as I suspected, enforcing global charge invariance as axiom, and out comes R-N. Might be thought a jaundiced outlook, but I get back to scenarios given in #1. Here's another situation worth pondering. A balloon is tethered on a string to the centre of a large mat. Both are electrically charged with the same sign of charge - just sufficient to have the balloon float against earth's gravity. For a distant observer, the mass of earth and balloon are redshifted wrt to that locally observed (not equally of course - much of the earth's mass is already partially redshifted at the balloon location, but that's not important here). Now the balloon is still floating from that observer's pov, so what does that tell us about the value of balloon/mat charge and E fields in coordinate measure? Sleep on it maybe.

 Quote by pervect We can note that the product of the area of the enclosing sphere multiplied by the electric force is a constant, and that said constant is proportional to the enclosed charge, Q, of the black hole. This is one illustration that there's no problem with defining the electric charge of a black hole.
It's clear that statement of global charge invariance is the standard position, but does it amount to more than an axiom.
 Aside from working this out for yourself, I'm not sure if there's any quick proof unless you use the notion of differential forms, in which case the proof is trivial. (Finding it may not be trivial.)
Does use of said differential forms free one from reliance on the axiom of global charge invariance?
 The area of the enclosing sphere multiplied by the gravitational force is NOT a constant, but the area of the enclosing sphere multiplied by the "force at infinity" IS a constant. This is discussed somewhat in depth in the Wiki article http://en.wikipedia.org/w/index.php?...ldid=465638523.
Komar expression explicitly acknowledges the redshift of 'gravitational charge'. Said my piece in responding to PeterDonis on why a similar redshift of electric charge should imo be true. As done back in #1 really. What might really convince eitherway is a numerical GR simulation of infalling test charge E-field. Some chance.

 Quote by Q-reeus That's your's and others position here (apart from perhaps one), but I am to be convinced it amounts to more than dogmatic assertion.
You call something "dogmatic assertion" because someone did not present a textbook deriving it from first principles? What exactly are you expecting from a discussion forum? You can't just demand that people convince you--only you can convince yourself, one way or the other. The best you can do is ask questions.

I was mistaken in my first response--allowing a dropping mass to do work does change the total mass/energy. As far as the question of whether charge is unchanged by dropping it into a black hole, I'm not exactly sure what you would take as a convincing argument. Charge is a scalar; since charges come in discrete multiples of the charge on an electron, it's just a matter of counting. The charge of something can't change except by emitting or absorbing a charged particle.

 It's a case of yes and no. Yes clearly if the work extracted from charged plates is channelled exterior to the system there is net loss to that system. But in BH case involving purely free-fall, no there is no net change as everything stays inside. None of that is paradoxical.
I guess I don't understand what it is that you think is paradoxical. Draw a surface around the outside of the black hole a small distance from the event horizon. Inside this surface is a certain charge Q and a certain mass M. If you drop a charge through the surface, Q goes up and so does M. If you extract energy from the system, then M goes down (but Q doesn't). I'm not sure what you think is oxymoronic about any of this.

 let's see if I can't turn a variant of your charged plates example around on you. Just consider slow lowering of a charged parallel-plate capacitor down to some floor level in a potential well. If we agree all forms of energy redshift the same, what happens to the field strength between those capacitor plates? Only one guess.
The field strength doesn't change. The charge doesn't change.

 Quote by Q-reeus It's clear that statement of global charge invariance is the standard position, but does it amount to more than an axiom.
It's not at all clear what your point is. People are telling you what the theory says; conservation of charge is an inherent aspect of our current understanding of electromagnetism. Our current understanding could very well be wrong, of course. For example, if the photon has a very tiny mass, then charge is not conserved.

So it's not clear whether you are asking theoretical questions: what does the theory predict? Or are you asking what's really true? Nobody knows what's really true, they just have the current best guess.

 Quote by stevendaryl You call something "dogmatic assertion" because someone did not present a textbook deriving it from first principles?
Well so far that's what it amounts to - in so many words I am simply told it is so - take it or leave it. In #1 I gave reasons, right or wrong, for doubting it.
 What exactly are you expecting from a discussion forum? You can't just demand that people convince you--only you can convince yourself, one way or the other. The best you can do is ask questions.
Whoa on a little - have I made such demands? You perhaps just misread my style. Hoped for more than expected was that each scenario presented has a ready explanation according with global charge invariance. Numbered for easy reference, but needn't have bothered.
 I was mistaken in my first response--allowing a dropping mass to do work does change the total mass/energy. As far as the question of whether charge is unchanged by dropping it into a black hole, I'm not exactly sure what you would take as a convincing argument. Charge is a scalar; since charges come in discrete multiples of the charge on an electron, it's just a matter of counting. The charge of something can't change except by emitting or absorbing a charged particle.
Well that is an argument of sorts but you are probably aware that many here hold that otherwise sacred conservation of energy/momentum fails in GR. Why insist charge remains aloof necessarily?
 I guess I don't understand what it is that you think is paradoxical. Draw a surface around the outside of the black hole a small distance from the event horizon. Inside this surface is a certain charge Q and a certain mass M. If you drop a charge through the surface, Q goes up and so does M. If you extract energy from the system, then M goes down (but Q doesn't). I'm not sure what you think is oxymoronic about any of this.
Again, you simply assert there will be no change in externally observed charge. I'm not convinced and am not demanding you do convince me - just pointing out there is nothing in that statement to change my position of doubt.
 The field strength doesn't change. The charge doesn't change.
Probably not such a good example to give as coordinate perspective depends on orientation of the plates for one thing. However a physical fact is that if a remote observer discharges the capacitor, the extracted energy remotely received is for sure redshifted. And that energy resided in the field. Same redshift will apply to the force remotely applied if prising the charged plates apart - and F = qE. Something to think over.

 Quote by stevendaryl It's not at all clear what your point is. People are telling you what the theory says; conservation of charge is an inherent aspect of our current understanding of electromagnetism. Our current understanding could very well be wrong, of course. For example, if the photon has a very tiny mass, then charge is not conserved.
Conservation of charge is actually a different thing - that charge always comes in pairs with equal and opposite sign.
 So it's not clear whether you are asking theoretical questions: what does the theory predict? Or are you asking what's really true? Nobody knows what's really true, they just have the current best guess.
There is afaik absolutely no experimental/observational evidence that charge invariance globally holds. So it gets down as I have tried to test for internal consistency via gedanken experiments. And yes thought maybe someone has a neat first-principles theoretical explanation why GR forbids any violation. Might be hoping too much.

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 Quote by Q-reeus My scenario was that of a Schwarzschild BH (but more generally an uncharged central mass M) perturbed slightly by a vastly smaller mass/charge in the process of being lowered/dropped in.
Ah, I see. Yes, this is a different scenario, but I'm not sure how to model it in GR, since the geometry is changing from Schwarzschild to Reissner-Nordstrom as the charged object falls in. I suppose one could try to patch together a Schwarzschild region and an R-N region, with some kind of transition zone in between.

 Quote by Q-reeus Again, as I did not model things using a pre-existing 'charged' R-N BH, it boils down to whether free-fall is an appropriate description for infall of an assumed small charged mass toward a large neutral mass M. If one ignores the typically vanishingly small radiative back reaction (dependent on the q/m ratio for starters), I would say yes given the context of what we are about here.
In classical GR, there should be no radiative back reaction for a charged test object falling into a neutral BH. So I would agree that free fall is appropriate, since there is no electric force from the hole, at least not until the charged object has fallen in and the hole has taken on its charge. I would think conditions could be set up such that that wouldn't happen until late enough in the process that free fall would still apply until well inside the horizon.

If by "radiative back reaction" you are referring to radiation emitted due to the test object's own field, that would require the object to no longer be a test object; you would have to model its internal structure as well. I think that's way more complication than we need here.

 Quote by Q-reeus So I would expect a net positive stress adds to a positive energy density, yet together appears as negative contribution in R-N expression?
I haven't seen an actual derivation of the R-N metric from the solution of the EFE, which is what one would need to look at to see how the SET components contribute to the metric coefficients. MTW has it as a homework problem, but I've long since lost my notes from working it years and years ago.

I think the reason that the Q contribution is opposite in sign from the M contribution in the metric is that, if you are at a finite radius r, some of the stress-energy due to Q is above you, whereas all of M is below you, so to speak. Since the spacetime is spherically symmetric, stress-energy above you does not affect the metric at your location.

Another way of looking at this is to think of an object collapsing to form a BH, first a neutral object, then a charged object. As I've posted before, when a neutral object collapses to form a BH, the metric you see at a given radius r in the vacuum region, once the object has collapsed, is really due to the SET of the object in the past, before the horizon formed, which is still in your past light cone. It just so happens that that stress-energy leaves behind Weyl curvature, which is static at a given radius once the object has collapsed inside that radius. The M term in the metric reflects that Weyl curvature.

However, if a charged object collapses to form a BH, in addition to the Weyl curvature, it leaves a static electric field, which has nonzero stress-energy (Weyl curvature has zero stress-energy) and extends out to infinity. So at a finite radius r, some of that field stress-energy is above you. The Q contribution to the metric tells how much of that field energy is above you, and therefore how much needs to be "subtracted out" from the M contribution to get the actual metric that you see.

I emphasize that this is all heuristic and I have not seen or done the actual computation. But that's the sense I get from what I've read.

 Quote by Q-reeus My question is, what is the physical basis for that Q lying at the EH can project a finite E to the BH exterior?
This is the same sort of question as "how does gravity get out of the BH". The E field is not coming from inside the EH; it is coming from the collapsing object in the past light cone, which, if it is charged, leaves behind its static field as well as Weyl curvature as it collapses. The charge Q only "seems" to be located inside the hole; actually, as I noted in my previous post, Q is determined by experiments done far outside the hole, and the results of those experiments can be explained entirely by the regions of nonzero SET and nonzero charge density in the past light cone of the events where the experiments are done.
 The first equation we learned at school: E =F*S, that's very useful. Energy is redshifted in gravity field. When a vertical rope is tugged from it's lower end, at the upper end tugs of reduced energy are observed. The force F is reduced, not lenght S. If the lower end of the rope has a negative charge, and a posive charge is brought under the lower end of the rope, then at the upper end an increase of weight of rope is observed, and this increace is reduced by the redshift factor. So we conclude that the constant pulling force, transmitted by the rope, of that positive charge is reduced when observed at higher altitude. BUT the pulling force that the positive charge exerts on a negative charge at higher altitude is not reduced. Conservation of charge requires this. BUT at lower altitude it is observed that charges that are being lifted seem to gain more charge, and charges that are being lowered kind of lose charge. Seems the charge of outside universe may change as seen from a gravity well. Interesting. If this needs some justification, then I'll think up some justification.

 Quote by PeterDonis Ah, I see. Yes, this is a different scenario, but I'm not sure how to model it in GR, since the geometry is changing from Schwarzschild to Reissner-Nordstrom as the charged object falls in. I suppose one could try to patch together a Schwarzschild region and an R-N region, with some kind of transition zone in between.
Understand - no point busting a gut trying there.
 If by "radiative back reaction" you are referring to radiation emitted due to the test object's own field, that would require the object to no longer be a test object; you would have to model its internal structure as well. I think that's way more complication than we need here.
Just had in mind the case of a 'point' charge with mass so I guess no issue.
 I think the reason that the Q contribution is opposite in sign from the M contribution in the metric is that, if you are at a finite radius r, some of the stress-energy due to Q is above you, whereas all of M is below you, so to speak. Since the spacetime is spherically symmetric, stress-energy above you does not affect the metric at your location... ...However, if a charged object collapses to form a BH, in addition to the Weyl curvature, it leaves a static electric field, which has nonzero stress-energy (Weyl curvature has zero stress-energy) and extends out to infinity. So at a finite radius r, some of that field stress-energy is above you. The Q contribution to the metric tells how much of that field energy is above you, and therefore how much needs to be "subtracted out" from the M contribution to get the actual metric that you see. I emphasize that this is all heuristic and I have not seen or done the actual computation. But that's the sense I get from what I've read.
That explanation was basically my own initial conclusion but subsequently doubted it because the Q contributions both numerator and denominator just fade monatonically with increasing r as negative only influences. Weird but I must be missing something.
 This is the same sort of question as "how does gravity get out of the BH". The E field is not coming from inside the EH; it is coming from the collapsing object in the past light cone, which, if it is charged, leaves behind its static field as well as Weyl curvature as it collapses. The charge Q only "seems" to be located inside the hole; actually, as I noted in my previous post, Q is determined by experiments done far outside the hole, and the results of those experiments can be explained entirely by the regions of nonzero SET and nonzero charge density in the past light cone of the events where the experiments are done.
Ah this is indeed sounding familiar! Well I sort of anticipated that and I invite you Peter to read again last main paragraph in #1. This really does get into strange territory imo. Reminds me of some avante garde stuff maybe by Susskind where terms like holographic principle, 'duality', etc is bandied about a lot. IIRC one example was of an elephant that falls into a BH and is deemed to both sail on in to the singularity in finite proper time, and also remain at the EH to be smeared out as part of a hovering stringy mess, then thermalized and emitted as Hawking radiation. Yep.

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 Quote by Q-reeus That explanation was basically my own initial conclusion but subsequently doubted it because the Q contributions both numerator and denominator just fade monatonically with increasing r as negative only influences. Weird but I must be missing something.
The Q contributions decrease with increasing r because as r increases, less of the stress-energy of the static electric field is above you.

 Quote by Q-reeus Well I sort of anticipated that and I invite you Peter to read again last main paragraph in #1.
I did; it is not correct to view the field as "detaching itself" from the source. The "source" of the electric field at a given event is the region of nonzero charge density in the past light cone of that event, just as the source of the "gravitational field" at that event is the region of nonzero SET in the past light cone of that event. So there is no "detaching" required.

 Quote by Q-reeus IIRC one example was of an elephant that falls into a BH and is deemed to both sail on in to the singularity in finite proper time, and also remain at the EH to be smeared out as part of a hovering stringy mess, then thermalized and emitted as Hawking radiation.
All that is more an issue of quantum field theory than GR, IMO. In classical GR, objects that fall through the horizon don't leave an "imprint" there. Even if the object is large enough to measurably increase the BH's mass and therefore expand the horizon, that is the only change it makes; there's no other "residue" of it left, classically.
 Blog Entries: 9 Recognitions: Gold Member Science Advisor After more cogitating, I have a few more items to throw out for consideration. (1) I'm wondering about the equation I posted for a_e, the proper acceleration on a charged test object due to the hole's electric field. As I've written it, that acceleration diverges at the outer horizon, r+; for like charges, the test object gets pushed outward with infinite acceleration, and for unlike charges, it gets pulled inward with infinite acceleration. That doesn't seem right, and in looking at some papers on arxiv that talk about charged particle motion in R-N spacetime, my equation for a_e may only apply if the test object is momentarily at rest relative to the hole, which would not be possible at the horizon and would allow my formula for a_e to diverge there without actually entailing the consequences I just described. For example, this paper: http://arxiv.org/pdf/1103.1807v3.pdf gives the following equation of motion for a charged test object (I've changed notation slightly from the paper to match my previous posts): $$\frac{dx^{a}}{d\tau} \nabla_{a} \frac{dx^{b}}{d\tau} = \frac{q}{m} F^{b}_{c} \frac{dx^{c}}{d\tau}$$ where $F^{b}_{c}$ is the electromagnetic field tensor due to the hole's charge, given by $$F^{t}_{r} = F^{r}_{t} = - \frac{Q}{r^{2}}$$ with all other components zero. My equation for a_e appears to be a valid special case of the above, if $\frac{dx^{0}}{d\tau}$, the "time" component, is the only nonzero component of the 4-velocity of the object; but once the object starts being either pulled in or pushed out, there will be a nonzero radial component of 4-velocity as well, and the formula becomes more complicated. I'm still working that out. (2) In thinking about how a charged object could collapse to a charged BH, I'm wondering if there is an analogue of Birkhoff's Theorem for the R-N case. If there is, it would follow that the "transition" I talked about in an earlier post, between Schwarzschild and R-N spacetime for the case where charged matter falls into a neutral hole, would have to occur much earlier than I thought; it would have to occur at a given radius as soon as the charged matter had fallen inside that radius. I'm also wondering whether there is an analogue of the Oppenheimer-Snyder solution for this case, which would also require an analogue of a collapsing FRW spacetime for charged matter. I haven't found anything useful yet by Googling on either of these items. (3) I'm also looking at trying to express quantities of interest in Painleve coordinates rather than Boyer-Lindquist coordinates, since the latter are singular at both horizons, r+ and r-. The Painleve chart for R-N spacetime looks similar to that for Schwarzschild, just with the extra terms in Q where you would expect them: $$ds^{2} = - \left( 1 - \frac{2M}{r} + \frac{Q^{2}}{r^{2}} \right) dT^{2} + 2 \sqrt{\frac{2M}{r} - \frac{Q^{2}}{r^{2}}} dT dr + dr^{2} + r^{2} d\Omega^{2}$$ So the "escape velocity" from a given radius r is $\sqrt{\frac{2M}{r} - \frac{Q^{2}}{r^{2}}}$. See, for example, this paper: http://www.saber.ula.ve/bitstream/12...1/a_family.pdf Another useful site for visualizing R-N spacetime is here: http://casa.colorado.edu/~ajsh/rn.html

 Quote by Q-reeus Conservation of charge is actually a different thing - that charge always comes in pairs with equal and opposite sign.
Okay, so you accept conservation of charge, but you are skeptical of what? That charge has the same value in every coordinate system? "Invariance" means "having the same value in every coordinate system".

I'm trying to tease apart what parts of your question are mathematical--having to do with how one does calculus in curved spacetime--and what parts of your question are empirical--what experiments tell us.

If charge is conserved, then that means that the charge inside of a closed surface is a matter of counting. Counting can't give different answers in different coordinate systems. So of course it's an invariant. So I think the real question is about the various integrals that are used to compute the charge?

The following equivalence is provable, for any vector field E:

∫E . dS = ∫(∇.E) dV

(The flux of E through a surface is equal to the integral of the diverge of E over the volume enclosed by that surface)

In relativity, the electric field is not actually a vector, but 3 components of a 6 component tensor Fαβ, but there is a corresponding fact about tensor fields. The equivalence of the two integrals is a coordinate-independent fact that's provable using calculus, and it works the same in curved spacetime.

∇.E = 4∏ ρ

where ρ = the charge density? Or are you asking whether the electric field E is really part of a tensor field Fαβ?

There are a lot of interrelated concepts that are involved in the conservation of charge in curved spacetime, some of them are provable using calculus, and some are empirical.

The empirical part I think is captured by Maxwell's equations, together with the Lorentz force law. If those hold (or rather, their generalization to curved spacetime), then conservation of charge and invariance under coordinate changes is a matter of pure mathematics--if they are true, then it's provable using calculus in curved spacetime. I think that they are true, but the detailed proof is not something I have personally have worked through.

Turning that claim around, if charge is not conserved, or if it is not an invariant, then I believe that is only possible if Maxwell's equations or the Lorentz force law are false. Those have only been tested near Earth, which only has fairly mild gravity, so there is no guarantee that they hold in regions of very strong gravity.

 Quote by PeterDonis After more cogitating, I have a few more items to throw out for consideration.
Those are interesting facts about charged black holes and the trajectories of test particles, but it seems to me that Q-reeus is asking a more fundamental question, which is whether total charge inside a surface is an invariant under changes of coordinate systems. The details of what happens to a test particle near a charged black hole doesn't seem relevant to that question. Or does it?

 Quote by jartsa The first equation we learned at school: E =F*S, that's very useful. Energy is redshifted in gravity field. When a vertical rope is tugged from it's lower end, at the upper end tugs of reduced energy are observed. The force F is reduced, not lenght S. If the lower end of the rope has a negative charge, and a posive charge is brought under the lower end of the rope, then at the upper end an increase of weight of rope is observed, and this increace is reduced by the redshift factor. So we conclude that the constant pulling force, transmitted by the rope, of that positive charge is reduced when observed at higher altitude. BUT the pulling force that the positive charge exerts on a negative charge at higher altitude is not reduced. Conservation of charge requires this. BUT at lower altitude it is observed that charges that are being lifted seem to gain more charge, and charges that are being lowered kind of lose charge. Seems the charge of outside universe may change as seen from a gravity well. Interesting. If this needs some justification, then I'll think up some justification.
Keep at it - we seem to be on about the same wavelength!

 Quote by PeterDonis The Q contributions decrease with increasing r because as r increases, less of the stress-energy of the static electric field is above you.
That makes sense if M does not just represent the neutral mass but is in some way inclusive of the total mass/energy contribution from Q. I may have misunderstood what you were intending in that regard in #14:
 This means that the "mass" M of the hole does *not* represent the entire "source" of gravity associated with the hole, as it does for Schwarzschild spacetime; there is an additional effect due to "stress" associated with the charge Q, which is not accounted for in M.
So do we interpret R-N expression as that M = Mn+MQtot (Mn the strictly neutral mass, MQtot the net contribution from Q)? Kind of a funny way of then including effect of Q by hiding net contribution in M then deducting the exterior portion as func{Q(r)} but there seems little other way to see it.
 All that is more an issue of quantum field theory than GR, IMO. In classical GR, objects that fall through the horizon don't leave an "imprint" there. Even if the object is large enough to measurably increase the BH's mass and therefore expand the horizon, that is the only change it makes; there's no other "residue" of it left, classically.
Well, ok, if 'fossil' EM field is exempt from definition of residue! Best to avoid further discussion on that one - likely to side-track.