## Is 'charged black hole' an oxymoron?

 Quote by stevendaryl 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".
Yes - I'm sceptical based on the implications as I see it for e.g. scenarios in #1 or say that charged-balloon-and-mat example in last para of #20. No higher maths, just following a comparative trend to it's logical conclusions. Others can come in and either translate or refute that using higher maths - and Internal cosistency is king.
 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.
As said last time, afaik there is no empirical evidence either way as to charge invariance holding when gravity is involved. So it's down to I suppose the correct calculus in curved spacetime yes. And I have no training in that arena - line of query is as presented earlier.
 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?
Agreed about charge counting. As to correctness of various integrals, it must imo be guided by or at least concur with logical conclusions from the sort of scenarios I have given.
 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. So, is your question really about Gauss's law: Are you asking whether we have proof that ∇.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?
Well your previous statement implies there is no effective difference particularly for static field case a la charged static BH or charged static mass in general. As per my closing comments in #21, on a straight basis (gravitational acceleration at constant r = 'gravitational flux density') the divergence law does fail for 'gravitational charge'. I again draw your attention back to the charged mat-and-ballon case in #20 - not noticing any obvious parallels?
 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.
Right - and we are stuck for the forseeable future with use of internal consistency arguments as said before. Which most here consider to be superfluous. You have made no comment on examples I gave you last time, nor attempted to detail where any previous scenarios go wrong. Which is hardly unique to yourself here but is a source of some personal chagrin. If my conclusions are wrong, it should not be too hard to show precisely how in each case. Still waiting.

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 Quote by Q-reeus 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 think that's right; I may not have been fully clear about it in earlier posts.

 Quote by Q-reeus So do we interpret R-N expression as that M = Mn+MQtot (Mn the strictly neutral mass, MQtot the net contribution from Q)?
I'm not sure you can separate out M_n, because the "contribution from Q" varies with radius. At any finite radius, there will be *some* amount by which the Q term in the metric reduces the effect of the M term. And the Q term continues to change all the way down to r = 0; so there's no point at which you could say that *all* of the Q contribution has been "subtracted out" and what's left is M_n.

 Quote by Q-reeus Well, ok, if 'fossil' EM field is exempt from definition of residue!
If a charged object were falling in, yes, the EM field "left behind" in the exterior vacuum region would not be "residue" in the sense I am using the term here. It would be analogous to the Weyl curvature left behind by the object's energy. Basically, the stuff that isn't "residue" is the stuff that is allowed to be "hair" on a BH, which means conserved quantities: mass-energy, charge, and angular momentum.

 Quote by Q-reeus Yes - I'm sceptical based on the implications as I see it for e.g. scenarios in #1 or say that charged-balloon-and-mat example in last para of #20. No higher maths, just following a comparative trend to it's logical conclusions. Others can come in and either translate or refute that using higher maths - and Internal cosistency is king.
I don't believe that you have pointed out any internal inconsistency in General Relativity or the theory of electromagnetism, or the two together. Charge and mass are actually different, because charge is a scalar, while mass is not. The total is just the sum of the charges of the pieces, but that's not true of mass, because interactions change the mass.

 As said last time, afaik there is no empirical evidence either way as to charge invariance holding when gravity is involved.
That's a weird way of putting things. I would say exactly the opposite---we have plenty of evidence of charge invariance in gravitational fields, and no evidence to the contrary.

 Agreed about charge counting. As to correctness of various integrals, it must imo be guided by or at least concur with logical conclusions from the sort of scenarios I have given.
I really don't think that the scenarios you have brought up have any bearing on
the invariance of charge. The current theory, which does have invariance of charge, works perfectly fine in all cases that we are capable of analyzing--there is no reason to believe that there is any problem. Of course, that doesn't mean that our current theories are correct, but there is no indication of any inconsistency in them (which is what I thought you were claiming).

 As per my closing comments in #21, on a straight basis (gravitational acceleration at constant r = 'gravitational flux density') the divergence law does fail for 'gravitational charge'.
There is no such thing as "gravitational charge", according to General Relativity. The source of gravity is not mass, it is energy-momentum. That makes a technical difference. In the nonrelativistic case, gravity seems a lot like the electric field, but they are very different in the relativistic case.

 Right - and we are stuck for the forseeable future with use of internal consistency arguments as said before.
Well, there is no consistency problem with invariance of charge.

 Which most here consider to be superfluous. You have made no comment on examples I gave you last time,
I can't really make sense of them, but they seem to be motivated by a misunderstanding
of relativity theory. So it seems to me that it would be more productive for you to understand relativity better, rather than for me to understand your misunderstandings.

 nor attempted to detail where any previous scenarios go wrong. Which is hardly unique to yourself here but is a source of some personal chagrin. If my conclusions are wrong, it should not be too hard to show precisely how in each case. Still waiting.
I wish that you could put your argument in a form that is a logical deduction of whatever contradiction you think you've discovered. I didn't see a contradiction in anything you said. For a black hole with charge, the charge Q is equal to the sum of all the charges that have fallen into the black hole. The mass M is not equal to the sum of all the masses that have fallen in, because mass is not conserved, while charge is. But what do you think is contradictory about that?

 Quote by jartsa 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.
That's really not correct. Charge is not affected by being dropped into a black hole.
 I apologize if my response was a little grumpy. I guess I'm being a little impatient. I guess I should read what his been said more carefully before replying.

 Quote by Q-reeus 1: Matter of proper mass m gently lowered towards the EH reduces in coordinate measure as m' = fm, with f the usual redshift expression f = √(1-2GM/(rc2)). In keeping with that conservation of energy applies and work is being extracted in the lowering process. Now suppose that m also carries a charge q. it makes no difference to the net reduction in m as all forms of energy reduce the same. Locally there is no variation in the proper charge-to-mass ratio q/m. How can that local invariance of q/m (no free-fall case) not be also reflected as remotely observed - i.e. q' = fq? Certainly the locally invariant q/m will show remotely as a proportionately equally redshifted reduction in the Newtonian gravitational and Coulombic forces of attraction/repulsion between two adjacent such charged masses. The implication is obvious - as vanishes coordinate mass, so vanishes coordinate charge.
I guess the reason that I didn't respond specifically to this scenario is because it seems to start with a number of assumptions that I don't think are true at all. First, what does it mean to say that when you lower a mass, it "reduces in coordinate measure as m' = fm, with f the usual redshift expression". Where did you get that idea, that mass reduces by the redshift formula? I think that you are starting with an inconsistent notion of what GR says about mass.

 Quote by Q-reeus 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
Where are you getting this idea that mass is redshifted? That doesn't make any sense to me.

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 Quote by stevendaryl First, what does it mean to say that when you lower a mass, it "reduces in coordinate measure as m' = fm, with f the usual redshift expression". Where did you get that idea, that mass reduces by the redshift formula? I think that you are starting with an inconsistent notion of what GR says about mass.
If "mass" is interpreted to mean "energy at infinity", then it does "reduce" as described if you lower a mass slowly, by means of a rope, let's say, as opposed to letting it free-fall. I agree that "mass" is not a good term for this.

 Quote by stevendaryl That's really not correct. Charge is not affected by being dropped into a black hole.

Let me explain:

1: A hydrogen atom is lowered into a deep gravity well. Then a photon of visible light is dropped onto the atom, which becomes ionized, although visible light does not normally ionize hydrogen. That happened because the field that keeps the atom together weakened as the atom was lowered.

2: A hydrogen atom and a photon of visible light are both gently lowered into a gravity well. Now the photon does not ionize the atom. The reason for this is that the electric field of the photon weakened as the photon was being lowered, as did the electric field in the atom.

So a photon in a gravity well, observing an atom being lowered into the gravity well, will say that the force that keeps the atom together is weakenig, kind of like the charges in the atom were becoming smaller.

(if photon does not have an electric field, then replace photon with a EM-wave pulse) (photon can be lowered using a mirror lined elevator)

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 Quote by jartsa 1: A hydrogen atom is lowered into a deep gravity well. Then a photon of visible light is dropped onto the atom, which becomes ionized, although visible light does not normally ionize hydrogen. That happened because the field that keeps the atom together weakened as the atom was lowered.
No, it happened because the photon was blueshifted as it dropped into the gravity well. A visible light photon emitted locally, at the same altitude as the atom, won't ionize it, so the field of the atom is not "weakened" at all according to local measurements. The difference that lowering the atom gently makes is that it is at rest deeper inside the well, so it "sees" the blueshift of the photon. To see why that's important, consider an alternate experiment where you let both a hydrogen atom and a visible light photon free-fall into the gravity well, in such a way that they meet up somewhere much deeper into the well than where you released them (you time the release of the atom and the photon from your much higher altitude to ensure this). Will the photon ionize the atom? No, because the atom is not at rest in the field; it is falling inward at a high speed, so there is a large Doppler redshift when it absorbs the photon that cancels the gravitational blueshift.

 Quote by jartsa 2: A hydrogen atom and a photon of visible light are both gently lowered into a gravity well. Now the photon does not ionize the atom. The reason for this is that the electric field of the photon weakened as the photon was being lowered, as did the electric field in the atom.
No, the reason is that, by hypothesis, the photon of visible light did *not* blueshift, because it was "gently lowered" instead of being allowed to free fall. (Actually, I'm not sure how you would "gently lower" a photon. What do you do, attach a rope to it? But for purposes of a thought experiment I'm fine with assuming you can somehow do it.)

 Quote by PeterDonis No, it happened because the photon was blueshifted as it dropped into the gravity well. A visible light photon emitted locally, at the same altitude as the atom, won't ionize it, so the field of the atom is not "weakened" at all according to local measurements. The difference that lowering the atom gently makes is that it is at rest deeper inside the well, so it "sees" the blueshift of the photon. To see why that's important, consider an alternate experiment where you let both a hydrogen atom and a visible light photon free-fall into the gravity well, in such a way that they meet up somewhere much deeper into the well than where you released them (you time the release of the atom and the photon from your much higher altitude to ensure this). Will the photon ionize the atom? No, because the atom is not at rest in the field; it is falling inward at a high speed, so there is a large Doppler redshift when it absorbs the photon that cancels the gravitational blueshift. No, the reason is that, by hypothesis, the photon of visible light did *not* blueshift, because it was "gently lowered" instead of being allowed to free fall. (Actually, I'm not sure how you would "gently lower" a photon. What do you do, attach a rope to it? But for purposes of a thought experiment I'm fine with assuming you can somehow do it.)

Well this is what I think:

A person that has spend some time in a deep gravity well will be observed having aged abnormally slowly, after he has been winched up from the well. He has produced some EM-waves by waving his slightly charged hands. Not very many EM-wave crests have been observed outside the gravity well, and the person does not report that he waved his hands very many times.

So the EM-waves that came up from the gravity well had just a few wave crests, because the person waved just a few times. And this is the reason the EM-waves that came up were slowly waving waves: Few crests during a long time, because of the slowness of the wave source.

The waves become longer at higher altitude, because light moves faster at higher altitude.

The case of free falling atom and photon I have to think about a bit.

 Quote by PeterDonis If "mass" is interpreted to mean "energy at infinity", then it does "reduce" as described if you lower a mass slowly, by means of a rope, let's say, as opposed to letting it free-fall. I agree that "mass" is not a good term for this.
I would not describe that situation with the terminology "Mass is reduced according to the redshift formula". I would say, rather, that if you extract energy from a system, then the total energy of that system becomes smaller. There's nothing surprising about that. In a situation in which energy is being extracted, but charge is not, I don't understand why anyone would believe that the ratio Total Energy/Total Charge would remain constant.

 Quote by jartsa Let me explain: 1: A hydrogen atom is lowered into a deep gravity well. Then a photon of visible light is dropped onto the atom, which becomes ionized, although visible light does not normally ionize hydrogen. That happened because the field that keeps the atom together weakened as the atom was lowered.
I don't think that's a very good explanation at all. I would not say that "the field that keeps the atom together weakened."

Look, the same thing happens in Newtonian gravity. If I throw a small stone at a car, it won't do much damage. But if I drop that stone from a great height, it can do a lot of damage. I would not describe this as "the forces holding the car together get weaker when the car is deep in a gravitational well".

The only meaningful notion of the "strength of forces holding an object together" is what you measure at the object, in the (local) inertial frame in which the object is at rest. That is not changed by dropping the object into a deep gravitational well.

What changes between high in the well and lower in the well is the translation of vectors. A velocity vector that is a small velocity high up in the well translates to a much larger velocity deeper down.

Of course, General Relativity understands gravity in a different way than Newtonian physics, but in neither case is it appropriate to say that "forces holding an object together get weaker when the object is lowered in a gravitational field".

 Quote by jartsa Well this is what I think: A person that has spend some time in a deep gravity well will be observed having aged abnormally slowly, after he has been winched up from the well. He has produced some EM-waves by waving his slightly charged hands. Not very many EM-wave crests have been observed outside the gravity well, and the person does not report that he waved his hands very many times.
General Relativity is about curved spacetime. There are two different ways to think about curved spaces, and each is useful for different purposes. The first way is to use curvilinear coordinates. For example, on the surface of the Earth, you can use longitude and latitude as coordinates. This is convenient, except that there are some weird effects: For example, in terms of longitude, objects at the North Pole seem "stretched out" compared with the same object near the equator, because the same object can cover more lines of longitude. Another effect is that if a plane attempts to fly as straight as possible, his path in terms of latitude will look "curved"; the shortest path from Montreal to Seattle is not to fly straight west, but to go northwest half the trip and southwest the second half.

An alternative way of thinking about curved spaces is in terms of gluing together lots of approximately flat spaces. Imagine taking the surface of the earth and breaking it up into lots of triangles of size 10 miles on a side. Now, within each little triangle, Euclidean geometry works fine. Straight lines look like straight lines on a map for that triangle. The "curvature" part is captured by how the various triangles are glued together. If a traveler leaves one triangle, you need to figure out which triangle he enters next, and what angle his trajectory makes in the second triangle.

So there is a "translation" process to translate vectors from one region of a curved space to another. In GR, a photon's momentum is a certain vector, that has to be translated when that photon moves from one region to another. That's what the redshift/blueshift formula is doing.

 Quote by stevendaryl I don't think that's a very good explanation at all. I would not say that "the field that keeps the atom together weakened." Look, the same thing happens in Newtonian gravity. If I throw a small stone at a car, it won't do much damage. But if I drop that stone from a great height, it can do a lot of damage. I would not describe this as "the forces holding the car together get weaker when the car is deep in a gravitational well". The only meaningful notion of the "strength of forces holding an object together" is what you measure at the object, in the (local) inertial frame in which the object is at rest. That is not changed by dropping the object into a deep gravitational well. What changes between high in the well and lower in the well is the translation of vectors. A velocity vector that is a small velocity high up in the well translates to a much larger velocity deeper down. Of course, General Relativity understands gravity in a different way than Newtonian physics, but in neither case is it appropriate to say that "forces holding an object together get weaker when the object is lowered in a gravitational field".

Nobody seems to know how the total energy of a falling rock changes. Obviously that energy that is responsible of car crushing does increase.

Light is more simple: When light enters a gravity field of a planet, the light slows down and the planet starts to move in the same direction as the light. So light loses energy when falling. Or as an extremely good aproximation the energy stays the same.

There was a long and technical discussion in this forum about the energy change of falling light, and those were the conclusion. The light losing energy is my own idea though.

So it must be the objects becoming weaker at lower altitude, which causes them to break more easily, when light from above hits them.

Hey I have one more scenario again: A charge in a gravity well is accelerated from 0 m/s to 100 m/s. Radiation energy is proportional to velocity change. As seen from higher altitude the velocity change was smaller than 100 m/s, and there is the reason why the radiation energy coming from the gravity well is smaller too.

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 Quote by Q-reeus So do we interpret R-N expression as that M = Mn+MQtot (Mn the strictly neutral mass, MQtot the net contribution from Q)?
Hi Q-reeus, I would not over-interpret the M and the Q as representing some particular mass or charge. I would think of them simply as parameters of the metric. The M term can include rest mass, energy (including energy in EM fields), pressure, stress, etc. And Q could be an E-field boundary condition at the edge of the manifold rather than some charged particles actually located in the manifold.

 Quote by stevendaryl I guess the reason that I didn't respond specifically to this scenario is because it seems to start with a number of assumptions that I don't think are true at all. First, what does it mean to say that when you lower a mass, it "reduces in coordinate measure as m' = fm, with f the usual redshift expression". Where did you get that idea, that mass reduces by the redshift formula? I think that you are starting with an inconsistent notion of what GR says about mass.
I'm ignoring your earlier posting #37 and later one in #41 (but accept the nice sentiments in #39), as it all hinges on getting right what you say here. Maybe you have already changed again - I say 'again' because in #22 there is "I was mistaken in my first response--allowing a dropping mass to do work does change the total mass/energy." - which was a seeming backflip from your earlier position, and amazingly you seem to have done another 180 and it's back to the original stance. [Stop press: just now read your #46. Evidently then you believe the energy gain/loss is to be considered a wholly delocalized affair - shared amongst the entire system more or less equally?] Then consider the following:

Suppose that mass/energy in a slowly lowered mass m is in the form of an unstable matter/antimatter doublet that self-annihilates and escapes entirely 'to infinity' as gamma rays. We surely agree that in escaping the gravitational potential well of central mass M, those rays - which carry all the energy tied up originally in m(r), are redshifted in coordinate measure. Annihilate the same matter/antimatter doublet out there in distant space, and obviously the gamma rays are not redshifted at all. This little experiment of the mind imo nicely indicates it is proper to consider the energy loss/gain in lowering/raising matter/energy of mass m(r) in a grav well as essentially confined to just that mass m(r) - provided m(r) << M. Unless that is one wishes to argue transporting matter/energy from a resting position at one potential to a resting position at another potential can be anything but a path independent process - assuming of course central mass M is taken as static. Again if one argues those gamma rays are not 'resting' just remember we are free to have them absorbed by nuclei in a massive block of matter etc. which then constitutes resting position.

You might also care to look again at the parallel-plate capacitor situation of last para in #24. We agree that locally, it only makes sense to ascribe energy loss (transfer of energy to higher up via electrical line) upon capacitor discharge to be confined essentially to the capacitor field-charge system, not say the planet of mass M as a whole. Again, haul the charged capacitor up to higher potential, then discharge. More power available - gained via the hauling process - and to be located in the capacitor field. It's only when masses are roughly comparable that delocalization/sharing becomes significant.

There has been some nitpickery over fine distinctions between mass and energy but let's recall the scenario 1: in #1 to which you have at least specifically addressed in #40 as quoted above - it's one of slow lowering. For which it's then perfectly appropriate to treat the mass/charge as momentarily resting at a given height. What then is the distinction between rest mass and total stress-energy? Sure there must be stress in the matter involved (it feels 'g' forces) and yes technically that comes under a different part of the stress-energy tensor contribution to total gravitating mass m(r). It will though be typically extremely small compared to the rest energy part, and anyway acts here simply as a scalar additive term. So can we please just take it that m(r) is inclusive of rest energy and stress without further fuss and ado?

Re your's and to some extent PeterDonis's criticism of jartsa over his description of atoms being 'weaker' further down. The local perspective is not the only legitimate one and from a coordinate viewpoint I would agree with his thrust. There is something similar in SR. A slab of dielectric lies immersed in an E field of a parallel plate capacitor. Both dielectric and capacitor are stationary in lab frame S', and E' there is below dielectric breakdown value. Now propel that slab to a relativistic velocity in S', normal to direction of E' such that in proper frame S of slab, E exceeds breakdown value and there is catastrophic failure and discharge. It is not legitimate to say that in S' the slab can be viewed as having 'weakened', since E has not changed in S'? Notice too that in jartsa's scenario if an atom spontaneously decays radioactively releasing a gamma ray, it is redshifted as seen from outside of potential well. That can legitimately qualify it as a 'weaker' atom with weaker internal EM and nuclear fields in my book - it's all a matter of pov.

There is a potential can of worms I'm uncovering or rather rediscovering in working through all the ways of looking at how mass, energy, momentum, field etc. should be evaluated in coordinate measure - more later. Maybe much later.