Is 'charged black hole' an oxymoron?

[Dale]

How would charge escape from the hole if light cannot?

You can perform those experiments outside the EH.

 

[Dale]

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

It isn't eliminated, it is just a fictitious force in GR. I.e. the force of gravity is equivalent to the fictitious force in an accelerating reference frame.

 

[Q-reeus]

Update. Why not simply apply the well known notion of effective refractive index to this situation. At least for static gravity case, coordinate transverse light speed reduction and redshift in electrostatic/magnetostatic energy is obtained by having vacuum permittivity ε₀ and permeability μ₀ equally altered according to ε,μ = (1/√-g_tt)(ε₀,μ₀), with -g_tt = (1-2GM/(cr²)). This alteration is a coordinate determination and not locally observed. It naturally leads to the picture of charge being effectively screened by an increased vacuum dielectric constant, which gives a more consistent picture than that of simply depressing charge. Final result of applying this view is still that charge invariance globally fails and in particular no BH exterior electric field should exist.

 

[Dale]

I don't know Q-reeus. I have been going through the math for several days now, and I can show that the redshifting depends primarily on the rank and type of the tensor. Since charge and invariant mass are scalars (rank 0 tensor), the idea of either of them redshifting is, I think, incorrect. I am not quite done working on the math yet, but I don't think that any of the conclusions based on the idea that mass redshifts are going to be valid.

 

[Naty1]

but I don't think that any of the conclusions based on the idea that mass redshifts are going to be valid...

I'm unsure exactly what you are doing, but Marcus was able to offer some insights on this question which I raised.

Summary: the deBroglie wavelength [of a matter particle] redshifts just like light, except that with a matter particle it slows down while light maintains c locally while it redshifts.

Is the Universe leaking Energy?
http://www.physics.uq.edu.au/downloa...iAm_Energy.pdf

and she said the following [approximately]:

Photons traveling in an expanding universe appear to lose energy via cosmological redshift. What about matter: You find that the de Broglie wavelength of particles increases by exactly the same proportion as a photon’s wavelength does! Thus light and matter seem to behave in exactly the same way when it comes to 'energy loss'.

where Marcus noted:

...expansion causes things to lose momentum relative to the CMB...

'things' here meaning matter particles and photon redshift

the discussion begins here:
https://www.physicsforums.com/showthread.php?t=614297&page=2 with post #32...

As the discussion unfolds Marcus cites a Weinberg and Zang paper and notes "momentum goes as one over the scale factor", p proportional to 1/a...

Hope that helps.

edit: this is FLRW spacetime.

 

[Dale]

Interesting. I will check if the same holds true in Schwarzschild spacetime.

 

[Naty1]

Austin0:

How would charge escape from the hole if light cannot?

[side comment: I think that's a question lacking proper perspective. For another thing, beside the above answer, a charged black hole will have more gravitational curvature due to the energy of the electromagnetic field.]

One view from Roger Penrose:

There is no mass as we know it (inside); inside all particles have been destroyed and gravitational effects remain outside the event horizon along with a few characteristics (electric charge, spin, etc).

from a loooong discussion in these forums [sorry, I did not save a link]

...the "source" of the observed EM field around a charged BH is the charge-current density in the collapsing matter; i.e., the observed field at any event in the exterior region is entirely determined by field propagation from charge-current density in the past light cone of that event.

[sounds like a PeterDonis, maybe??]

 

[PeterDonis]

[sounds like a PeterDonis, maybe??]

Yes, that was me, and yes, I was taking the same position that Penrose is taking in the quote from him. AFAIK that is the standard answer to the question "how does the charge get out of the BH?" (i.e., it doesn't have to), similar to the standard answer to the question "how does gravity get out of the BH?" There would also be an analogous question, "how does the angular momentum get out of the BH?", but for some reason nobody ever asks that one.

 

[Q-reeus]

I don't know Q-reeus. I have been going through the math for several days now, and I can show that the redshifting depends primarily on the rank and type of the tensor. Since charge and invariant mass are scalars (rank 0 tensor), the idea of either of them redshifting is, I think, incorrect. I am not quite done working on the math yet, but I don't think that any of the conclusions based on the idea that mass redshifts are going to be valid.

Good to know you've been beavering away on it DaleSpam! You are surely aware I and one or two others have been using 'redshift' in a loose way wrt mass/energy and charge. We know lowering any form of mass/energy of coordinate rest energy m down the potential well of some isolated massive body of rest energy M results in a net increase in gravitating mass to M+m√-g_tt, less than M+m owing to 'redshift' of m. Which can be more or less directly linked to the mass 'redshift'by in principle annihilating m or rather m√-g_tt and recovering the requisite redshifted radiation to infinity. The factor is the same regardless of tensor rank. E=hf makes the linkage rather good. As for charge, the huge problem with standard view is to reconcile that field energy does depress by redshift factor (non-free-fall case), yet field strength supposedly experiences no diminution whatsoever. Which btw also logically demands charge field lines remain exactly flat spacetime Coulombic in form regardless of how gravitationally warped things are for everything else. Strange indeed. No way I can see a sensible reconciliation without allowing effective charge screening as per last post. :zzz:

 

[Q-reeus]

Summary: the deBroglie wavelength [of a matter particle] redshifts just like light, except that with a matter particle it slows down while light maintains c locally while it redshifts.

There is some similarity but clearly not a 1:1 correspondence. A light ray will always outrun and outlast an outgoing particle whose local deBroglie wavelength goes infinite at it's turn-around point for instance.

 

[PeterDonis]

Interesting. I will check if the same holds true in Schwarzschild spacetime.

It looks like marcus was talking about the change in scale factor with expansion in FRW spacetime, and how it affects photons vs. massive particles. I'm not sure how relevant that will be to Schwarzschild spacetime, except for the general point that the observed energy of an object is the contraction of the object's 4-momentum with the observer's 4-velocity. But the particular effect marcus was talking about as "losing momentum relative to the CMB" does not happen in Schwarzschild spacetime because there is no expansion.

 

[Naty1]

PeterDonis:

It looks like marcus was talking about the change in scale factor with expansion in FRW spacetime, and how it affects photons vs. massive particles. I'm not sure how relevant that will be to Schwarzschild spacetime,

Correct regarding Marcus. I edited my post above to highlight FLRW spacetime.

...the particular effect marcus was talking about as "losing momentum relative to the CMB" does not happen in Schwarzschild spacetime because there is no expansion.

exactly...I did not know Dalespam was working in Schwarzschild coordinates; On the other hand redshift is redshift, gravitational or 'expansion distance based', so it will be interesting to see if anything worthwhile appears.

Is this the issue where it was recently discussed 'why free fall has no redshift' or somesuch?? I just realized that might be where this discussion resides...I don't recall those coordinates. Sorry if I detoured the discussion.

 

[PeterDonis]

Is this the issue where it was recently discussed 'why free fall has no redshift' or somesuch??

No, that's this thread:

https://www.physicsforums.com/showthread.php?t=613902

We've covered some of the same ground here, but with a different focus.

 

[Dale]

We know lowering any form of mass/energy of coordinate rest energy m down the potential well of some isolated massive body of rest energy M results in a net increase in gravitating mass to M+m√-g_tt, less than M+m owing to 'redshift' of m.

I don't know that, do you have a reference or derivation?

I am also not sure of the relevance to most of the thought experiments proposed here, which have seemed to focus on test masses and charges interacting with each other rather than altering the field of the gravitating mass. I think that assuming the two things are equal is sketchy.

 

[PeterDonis]

We know lowering any form of mass/energy of coordinate rest energy m down the potential well of some isolated massive body of rest energy M results in a net increase in gravitating mass to M+m√-g_tt, less than M+m owing to 'redshift' of m. Which can be more or less directly linked to the mass 'redshift'by in principle annihilating m or rather m√-g_tt and recovering the requisite redshifted radiation to infinity.

I see DaleSpam has commented on this as well; this way of putting things conceals a lot of interpretation of ambiguous terms. We've gone into this in previous threads, and somewhat in this one, but perhaps it's worth some further comments to capture the thoughts I've come up with:

Suppose I am "hovering" at some large distance above a Schwarzschild black hole, and I measure its mass, M, by putting test objects into orbits about the hole, measuring their orbital parameters, and applying Kepler's Third Law. Then I drop an object of mass m, where I determine m locally by some similar procedure, into the hole. (All objects so far are electrically neutral; I'll talk about the charged BH/charged object case later in this post.) There are at least three possible ways I can do this:

(1) If I just let the object free-fall into the hole, and it doesn't give off any radiation, then the hole's mass, measured by me in the same way as before, will increase by m.

(2) If I slowly lower the object into the hole, extracting work from the process as I do so, then I can make the mass increase of the hole as small as I want by lowering the object closer and closer to the horizon before I finally have to release it and let it free-fall the rest of the way. In principle, how close to the horizon I can lower the object and still extract work depends on how I lower it and the strength of the materials I use to do so; ultimately it depends on what (finite) proper acceleration the lower end of the mechanism for lowering, the one attached to the object, can withstand. The final mass increase of the hole, for an idealized process where I extract the maximum amount of work possible while lowering the object to some radius r > 2M, will be $m \sqrt{1 - 2M / r}$.

(3) If I let the object free-fall towards the hole, but at some radius r > 2M, I capture all its kinetic energy and convert that to outgoing radiation (say, for example, it hits a large mirror which stops it, converts its kinetic energy into heat, and then reflects all the heat outward as it radiates away), I can in principle make the mass increase of the hole as small as I want, just as in #2 above, by moving the stopping point closer to the horizon. The only difference, from my point of view, is that I am now not capturing the difference between m and $m \sqrt{1 - 2M / r}$ as work; it's just radiating away as heat.

There's another subtle point about the above three scenarios: how do we define the "energy at infinity" present? The three applicable quantities are the Komar mass, the ADM mass, and the Bondi mass. Here's how I see those for the three scenarios above (all three scenarios assume that all masses except that of the hole and the object dropped/lowered in are negligible):

Starting state: all three masses are M + m. Mass M is the hole, mass m is the object we're about to drop/lower in, which is at some finite radius so all three masses will include it.

#1: All three are unchanged; the only change is that M + m is now all contained in the hole.

#2: All three are unchanged: the only change is that M + $m \sqrt{1 - 2M / r}$ is contained in the hole, and the remainder of m is still at our finite radius, where we recaptured it as work extracted from the lowering process.

#3: The ADM mass and Komar mass are still M + m; however, the Bondi mass is now decreased to M + $m \sqrt{1 - 2M / r}$, the new mass of the hole, because it will not include the mass (energy) of the radiation that escaped to infinity.

Now, suppose we do similar experiments to the above, but with a charged (R-N) black hole and a charged object. We have two possible cases, opposite charges and like charges. I am not presently working the math in detail as DaleSpam is, but it looks to me like the following are key points:

First, there should be a sort of ADM charge/Komar charge/Bondi charge for the spacetime as a whole, definable similar to the way the corresponding masses are defined. For the case we're considering, this total would include *both* the charge Q of the hole *and* the charge q of the object that's going to be dropped or lowered in. We, at a large finite radius, would measure the charge Q of the hole using charged test objects, similar to the way we measured its mass M by Keplerian orbits; we would also locally measure the charge q of the object to be dropped/lowered in similar to the way we measured its mass m locally.

Second, at the starting state, all three of the ADM/Komar/Bondi charges should be equal, just as the masses are; they should all be Q + q. However, at the end state for *any* of the three scenarios, *all three* charges should still be unchanged, *unlike* the case for masses above. This is because there is no way to radiate charge "away to infinity" as we can with mass by converting it into radiation. This is a key difference between mass and charge. (Note: it *is* possible for the charge q of the object dropped into cancel some or even all of the charge Q of the hole. However, if that is the case, all three of the ADM/Komar/Bondi charges *already* comprehend that; they already "see" the net charge, Q + q, which will be less than Q if q is of opposite sign.)

Third, if the object I am going to drop/lower into the hole is charged, then there is an extra energy e present in the spacetime as a whole, i.e., in the starting ADM/Komar/Bondi energy, due to the extra potential energy due to the charges Q and q being separated. So the total starting energy is M + m + e.

Note that as I've just defined it, e is positive if Q and q are of opposite sign--in this case, if I do a process like #2 or #3 above, I can in principle capture up to m + e of energy as work, or have that much energy radiate away to infinity, instead of it going into the hole. The final mass of the hole will be M + (m + e) * sqrt(1 - 2M/r), where r is the radius of the "stopping point", with the remainder of m + e captured as work or radiated away to infinity. The ADM/Komar mass will not change, and the Bondi mass will decrease to the new mass of the hole.

If Q and q are of the same sign, e will be *negative*. Depending on how large e is compared to m, it may be possible to do processes #1, #2, or #3 above, but capturing less energy than m (instead of more, as above); or it may not be possible to do them at all, because the electrical repulsion between the hole and the object is enough to require work to be *added* to the object to make it fall below the horizon.

Sorry for the long post, but I wanted to get all that down while it was fresh in my mind.

 

[Trenton]

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.

DrGreg,

I too have heard this but I am not sure quite how to view it. If, due to infall of space, nothing can be at rest inside the outer horizon because to do so would mean it was, in effect, traveling outwards at greater than C, then how can a charged particle do it? How indeed can a charged particle ever experience an electric field until it got to the singularity. Or is it that a charged black hole does not have a singularity but a dense entity bound by the inner horizon?

This last point arises from the idea that a spinning BH has a ring singularity. Spin distorts from a point to a ring. Charge if it distorts at all must be to a sphere - The only problem with that is it is no longer singular. One way at least to erradicate the naked singularity!

 

[PeterDonis]

there should be a sort of ADM charge/Komar charge/Bondi charge for the spacetime as a whole, definable similar to the way the corresponding masses are defined.

Having run some formulas, let me be more definite about what I was proposing in the above quote. I emphasize that I have not seen this in any standard literature in this form, although what I'm proposing for the "ADM charge" is basically Gauss's law in integral form, slightly rewritten and taken to an appropriate limit.

(1) ADM charge: the integral for the ADM mass can be expressed as follows (note that I am leaving out constant factors involving pi and so forth, I'm only trying to look at the structure of the formulas, not give exact results):

$$M = \lim_{S \rightarrow i^{0}} \int g^{uv} \left( g_{ua, v} - g_{uv, a} \right) n^{a} d S$$

as given, for example, in this paper (which I believe references some of the original ADM papers):

http://arxiv.org/pdf/gr-qc/0609079v1.pdf

But we can rewrite this in a more general form:

$$M = \lim_{S \rightarrow i^{0}} \int m_{a} n^{a} d S$$

$$m_{a} = g^{uv} \left( g_{ua, v} - g_{uv, a} \right)$$

where now we have clearly separated out two distinct things: (1) defining a 2-sphere surface dS with outward pointing normal $n^{a}$, which we will then "take to infinity" in the limit; and (2) defining a 1-form $m_{a}$ that represents "what we want to integrate" over the 2-surface. We will define "ADM charge" by keeping (1) the same but varying (2).

In the case of the ADM mass, what we are integrating over the 2-surface is basically the effect of the "mass" inside the surface on the metric at the surface. For charge, we want to integrate the effect of the charge inside the surface, over the surface. But we know how to do this: it's just Gauss's law. All we need to do is define a 1-form representing the "lines of force" going through the surface, which will be defined by the electric field normal to the surface. So we can write:

$$Q = \lim_{S \rightarrow i^{0}} \int F_{a0} n^{a} d S$$

This is basically Gauss's law, expressed in the "Schwarzschild" type coordinates we have been using for the R-N metric, taken to the limit at spatial infinity. But there's still one flaw: we've picked out a specific component of the EM field tensor, the "0" component. We can make the formula invariant by using the timelike Killing vector field of the spacetime, which will pick out the same "0" component in Schwarzschild coordinates, but which now allows the formula to be generalized to any chart, just like the ADM mass formula:

$$Q = \lim_{S \rightarrow i^{0}} \int F_{ab} n^{a} u^{b} d S$$

where $u^{b}$ is a timelike unit vector (in this case, the "time translation" unit vector of R-N spacetime). It looks to me like, modulo constant factors, the value of this integral should equal the "Q" parameter in the standard R-N metric. I'll work on checking that explicitly.

(2) Komar charge: Actually, what I just wrote down above should *be* the Komar charge, as well as the ADM charge, because the "use a timelike vector field to make the integral formula invariant" trick is the same trick that's used for the standard Komar mass formula. There is one wrinkle: I haven't written down any analogue to the stress-energy tensor for charge. I think there's a way to finesse that by using Maxwell's equations to equate the expression $F_{ab} n^{a} u^{b}$ to an expression involving the charge-current 4-vector. But I need to check that as well.

(3) Bondi charge: The formula for this would look the same as for the ADM charge, except that the limit would be taken as S goes to future null infinity instead of spatial infinity. Since there is no way for charge to radiate away to infinity, as I said before, this does not change the result at all, so Bondi charge = ADM charge.

Edit: Changed to timelike unit vector field in the above formulas (instead of timelike Killing vector field).

 

[Austin0]

Austin0:

How would charge escape from the hole if light cannot?

[side comment: I think that's a question lacking proper perspective. For another thing, beside the above answer, a charged black hole will have more gravitational curvature due to the energy of the electromagnetic field.]

You may be quite right about perspective but your comment was a little indefinite to be really helpful ;-)

One view from Roger Penrose:

There is no mass as we know it (inside); inside all particles have been destroyed and gravitational effects remain outside the event horizon along with a few characteristics (electric charge, spin, etc).



...the "source" of the observed EM field around a charged BH is the charge-current density in the collapsing matter; i.e., the observed field at any event in the exterior region is entirely determined by field propagation from charge-current density in the past light cone of that event.

Could you elaborate on this concept of field propagation from past light cone?
_
Thanks_________________

 

[PeterDonis]

Could you elaborate on this concept of field propagation from past light cone?

I believe Naty1 got this from me originally; it actually came up in another thread earlier today, and I posted this:

https://www.physicsforums.com/showpost.php?p=3965324&postcount=17

It talks about gravity, not charge, but the reasoning for charge is similar; the only difference is that the "source" that has to be somewhere in the past light cone is the charge-current density 4-vector (the source in Maxwell's equations) instead of the stress-energy tensor (the source in the EFE).

 

[PeterDonis]

I haven't written down any analogue to the stress-energy tensor for charge. I think there's a way to finesse that by using Maxwell's equations to equate the expression $F_{ab} n^{a} u^{b}$ to an expression involving the charge-current 4-vector.

I should have seen this one at once; it's just the generalized Stokes Theorem plus Maxwell's equation with source. We have:

$$\int F_{ab} n^{a} u^{b} dS = \int \nabla_{c} F_{ab} g^{ac} u^{b} dV = \int J_{b} u^{b} dV$$

where it's technically the charge-current 1-form instead of the vector appearing, but that's a minor point. The only other potential issue is how to properly define the integration measure; I think there should be a $\sqrt{-g}$ somewhere in there, but it may be that the appearance of the charge-current 1-form instead of the vector already implicitly takes that into account. But in any event I believe this shows how the "Komar charge" integral does in fact connect back to the source of the EM field, just as the Komar mass integral connects back to the source of gravity.

 

[Austin0]

I believe Naty1 got this from me originally; it actually came up in another thread earlier today, and I posted this:

https://www.physicsforums.com/showpost.php?p=3965324&postcount=17

It talks about gravity, not charge, but the reasoning for charge is similar; the only difference is that the "source" that has to be somewhere in the past light cone is the charge-current density 4-vector (the source in Maxwell's equations) instead of the stress-energy tensor (the source in the EFE).

so we would have to go a billion years into the Earth's past light cone to find the "source" of the gravity the Earth feels at this instant--but the gravity itself, the effect, would be the same if the Sun's mass were the same, because the way the field from the collapsing object "propagates" through the empty vacuum region outside it is static--it stays the same for all time (again,

The basic past light cone analogy is self evident wrt the Earth-Sol system.
But if an electromagnetic or gravitational field is dependent on propagation, renewal from charged particles or mass, then the idea of a static propagation is a little obscure.
this would seem to imply that if the mass was moved somehow, the field could remain behind independent of a necessary source.
In the case of gravity it makes sense that the field would still be there. Possible that gravity itself is somehow exempt and still could emanate from the mass. But with charge, as Penrose said the source of charge , particles no longer exist at that point so the idea of a field remaining, extending to infinity with no source whatever is hard to understand.
As you might guess I am just starting to pay attention to black holes ;-)

 

[PeterDonis]

But if an electromagnetic or gravitational field is dependent on propagation, renewal from charged particles or mass, then the idea of a static propagation is a little obscure.
this would seem to imply that if the mass was moved somehow, the field could remain behind independent of a necessary source.

If the mass is moved, its "arrangement" in everybody's past light cone will change as the information about the move propagates to them. That changes the observed field.

In the particular case we've been discussing, that effect is obscured because the spacetime is static; nothing changes with time. In a more realistic, dynamic spacetime, the field observed at a given point would change with time, as dynamic information about the movement of masses elsewhere propagated around.

In the case of gravity it makes sense that the field would still be there. Possible that gravity itself is somehow exempt and still could emanate from the mass.

No, gravity isn't exempt. But the mass it's "emanating" from isn't inside the black hole.

If you want to view the field at a given point in space, outside the hole, as "emanating" from something, then it is emanating from portions of the collapsing matter far in the past, closer and closer to the horizon; as the collapsing matter far in the past gets closer and closer to the horizon, the time delay for light signals from it to get out to some fixed radius increases without bound. So the gravity you are sensing now "emanates", if you want to look at it that way, from some small piece of the collapsing matter's worldline, say, 10 meters above the horizon; the gravity you will sense some time in the future will have emanated from a small piece of the collapsing matter's worldline, say, 9 meters above the horizon; and so on.

In the case of a static spacetime (after the hole forms), the "emanating" view just above is equivalent to the "static" view, where once the collapsing matter falls inward past a certain radius, the field at that radius becomes fixed for all future time. But in a dynamic spacetime, there is no "static" view, so yes, you would need to look for dynamic information about the movement of mass traveling around.

But with charge, as Penrose said the source of charge , particles no longer exist at that point so the idea of a field remaining, extending to infinity with no source whatever is hard to understand.

The same thing I said above goes for charge; it emanates from charge-current density in the past just as gravity emanates from stress-energy in the past. Gravity "extends to infinity" just as charge does; the mass of any gravitating body leaves an "imprint" on the spacetime far away. Even if the gravitating body collapses to a black hole, the "imprint" of its mass on the spacetime remains the same, because the imprint doesn't come from inside the hole; it comes from the past, before the object disappeared behind the horizon, as described above. The "imprint" of charge on the spacetime works the same way. Even if there is no charge-current density inside the black hole (or any nonzero stress-energy), since it's all been destroyed in the singularity, there is still charge-current density (and stress-energy) in the past.

The ADM, Komar, and Bondi masses I talked about in recent posts are ways of describing the "imprint" of mass on the spacetime far away; the similar stuff I proposed for charges captures the same thing.

 

[Q-reeus]

Q-reeus: "We know lowering any form of mass/energy of coordinate rest energy m down the potential well of some isolated massive body of rest energy M results in a net increase in gravitating mass to M+m√-gtt, less than M+m owing to 'redshift' of m. Which can be more or less directly linked to the mass 'redshift'by in principle annihilating m or rather m√-gtt and recovering the requisite redshifted radiation to infinity."

#3: The ADM mass and Komar mass are still M + m; however, the Bondi mass is now decreased to M + m√(1−2M/r), the new mass of the hole, because it will not include the mass (energy) of the radiation that escaped to infinity.

Well I had hoped it would have been evident this definition is what was being implied in bit quoted - we are obviously discounting as gone from considered system energy extracted in lowering (and 'lowering' was deliberately the term used) mass m. All btw covered in #1.

Now, suppose we do similar experiments to the above, but with a charged (R-N) black hole and a charged object. We have two possible cases, opposite charges and like charges. I am not presently working the math in detail as DaleSpam is, but it looks to me like the following are key points:

First, there should be a sort of ADM charge/Komar charge/Bondi charge for the spacetime as a whole, definable similar to the way the corresponding masses are defined. For the case we're considering, this total would include *both* the charge Q of the hole *and* the charge q of the object that's going to be dropped or lowered in. We, at a large finite radius, would measure the charge Q of the hole using charged test objects, similar to the way we measured its mass M by Keplerian orbits; we would also locally measure the charge q of the object to be dropped/lowered in similar to the way we measured its mass m locally.

Second, at the starting state, all three of the ADM/Komar/Bondi charges should be equal, just as the masses are; they should all be Q + q. However, at the end state for *any* of the three scenarios, *all three* charges should still be unchanged, *unlike* the case for masses above. This is because there is no way to radiate charge "away to infinity" as we can with mass by converting it into radiation. This is a key difference between mass and charge. (Note: it *is* possible for the charge q of the object dropped into cancel some or even all of the charge Q of the hole. However, if that is the case, all three of the ADM/Komar/Bondi charges *already* comprehend that; they already "see" the net charge, Q + q, which will be less than Q if q is of opposite sign.)

All that and your later points pretty well makes perfect sense *if* one starts off with the assumption of an RN BH that acts externally as a charged object - which however is what this thread set out to seriously question. I gave an updated argument involving effective charge screening in #248 which imo provides a level of conceptual consistency lacking in standard RN picture, as one or two of the bizarre consequences pointed out in #254 were meant to highlight. In fact apart from that added thought in #248 nothing substantially new has been added since #1 really. So I don't know, in the end it may simply come down to another fizzle/fadeout which at bottom amounts to "concensus/majority opinion ruulz - OK!" But no - sour grapes premature at this point.

 

[PeterDonis]

All that and your later points pretty well makes perfect sense *if* one starts off with the assumption of an RN BH that acts externally as a charged object - which however is what this thread set out to seriously question.

Well, everything I said would apply equally well to a charged massive object like a planet or star; DaleSpam recommended earlier that we start with that case, which avoids potential issues with what happens at or inside the R-N BH horizon, and if you're having problems with the BH case I would agree with his recommendation. The only difference with a planet or star vs. a BH is that the minimum radius r that you can lower something to is quite a bit larger than the horizon radius. But the ADM/Komar/Bondi charges I defined, just like their mass counterparts, are perfectly well-defined for a charged massive object, so they don't depend on having an R-N BH; they only depend on the exterior vacuum spacetime geometry of a charged massive object being R-N, i.e., only on R-N geometry outside some radius r greater than the horizon radius for the given mass and charge. I think the latter assumption is pretty safe.

 

[PeterDonis]

a level of conceptual consistency lacking in standard RN picture, as one or two of the bizarre consequences pointed out in #254 were meant to highlight.

In #254, you say:

the huge problem with standard view is to reconcile that field energy does depress by redshift factor (non-free-fall case), yet field strength supposedly experiences no diminution whatsoever.

I'm not sure what you mean by "field strength...experiences no diminution". Go back to scenario #2 in my previous post with a charged object, charged oppositely to the hole. The ADM/Komar/Bondi charge "at infinity" is Q + q always; but by hypothesis, you, at your finite radius, before you lower the charged object, see a field strength based on Q, *not* Q + q. After the object is lowered, you now see a reduced field strength, based on Q + q, which is less than Q, because that is now the charge you see on the hole. (Again, for "hole" read "central charged massive object" if that works better, see my previous post.)

Also remember that when you say "field energy does depress by redshift factor", what you really mean is that you are *extracting* energy from the lowering process. Where is that energy coming from? From the energy "at infinity" m + e of the charged object you are lowering. It does *not* come from the "energy of the hole", so it does not come from the "field energy" associated with the hole's charge Q. It comes only from "field energy" that is present because there is a second, opposite charge, q, which started out separated from Q, and then you brought them together. I don't see any problem or inconsistency anywhere in this.

 

[GAsahi]

The motion of charged vs uncharged particles.

How would the electro(static) field propagate past the EH? Photons don't, how would virtual photons (the carriers of em field) ?

 

[Q-reeus]

Q-reeus: "the huge problem with standard view is to reconcile that field energy does depress by redshift factor (non-free-fall case), yet field strength supposedly experiences no diminution whatsoever."

I'm not sure what you mean by "field strength...experiences no diminution". Go back to scenario #2 in my previous post with a charged object, charged oppositely to the hole. The ADM/Komar/Bondi charge "at infinity" is Q + q always; but by hypothesis, you, at your finite radius, before you lower the charged object, see a field strength based on Q, *not* Q + q. After the object is lowered, you now see a reduced field strength, based on Q + q, which is
less than Q, because that is now the charge you see on the hole. (Again, for "hole" read "central charged massive object" if that works better, see my previous post.)

Chalk and cheese comparison imo. Your point here amounts to simply that opposite charges cancel which of course I agree with. At question is the effective coordinate value of charge as a function of gravitational potential - a different matter entirely.

Also remember that when you say "field energy does depress by redshift factor", what you really mean is that you are *extracting* energy from the lowering process. Where is that energy coming from? From the energy "at infinity" m + e of the charged object you are lowering. It does *not* come from the "energy of the hole", so it does not come from the "field energy" associated with the hole's charge Q. It comes only from "field energy" that is present because there is a second, opposite charge, q, which started out separated from Q, and then you brought them together. I don't see any problem or inconsistency anywhere in this.

This is now a very long thread so won't bother to try and quote previous posts but one can get the point without resorting to a lowering process per se. Start off with closely spaced charges already at a lower potential and separate then transversely to form a dipole p. We do I trust agree that the energy required to form said p is less by the redshift factor than if same operation is performed 'out there'. That's reduced stored field energy, and yet if any electric field whatsoever can be perceived externally to an EH, the infinite redshift existing there logically demands no gravitational diminution of field strength is possible - period. As I have said repeatedly in earlier posts. So Peter - kindly offer please your synthesis of those two facts (or at least one firm fact and one supposed 'fact').

 

[PeterDonis]

At question is the effective coordinate value of charge as a function of gravitational potential - a different matter entirely.

It's not unrelated, because whatever geometric object describes the charge of the object being lowered, it must behave differently in some respects as a function of radius than the object's 4-momentum, which describes its energy, because the energy can be extracted as work or radiated to infinity, but the charge can't. I suggested in an earlier post that this would be reflected by the charge-current 4-vector always being parallel transported along the object's worldline, while the 4-momentum would only be parallel transported if the object traveled on a geodesic, i.e., if it was freely falling; for my case #2, an object being slowly lowered with work extracted, the 4-momentum would *not* be parallel transported. (My case #3 would ultimately work the same--the 4-momentum would be parallel transported while the object was freely falling, but would then undergo a very sudden non-parallel-transport change when the object hit the mirror at the bottom and its kinetic energy was converted into outgoing radiation.)

Start off with closely spaced charges already at a lower potential and separate then transversely to form a dipole p. We do I trust agree that the energy required to form said p is less by the redshift factor than if same operation is performed 'out there'.

No, I'm not sure we agree on that. For energy measured locally, it's the same regardless of where you do the experiment, by the equivalence principle. For energy measured "at infinity", I'm not sure; I would have to work through the math.

 

[Q-reeus]

It's not unrelated, because whatever geometric object describes the charge of the object being lowered, it must behave differently in some respects as a function of radius than the object's 4-momentum, which describes its energy, because the energy can be extracted as work or radiated to infinity, but the charge can't. I suggested in an earlier post that this would be reflected by the charge-current 4-vector always being parallel transported along the object's worldline, while the 4-momentum would only be parallel transported if the object traveled on a geodesic, i.e., if it was freely falling; for my case #2, an object being slowly lowered with work extracted, the 4-momentum would *not* be parallel transported. (My case #3 would ultimately work the same--the 4-momentum would be parallel transported while the object was freely falling, but would then undergo a very sudden non-parallel-transport change when the object hit the mirror at the bottom and its kinetic energy was converted into outgoing radiation.)

Agreed that unlike energy ('gravitating charge' in effect) one cannot radiate away charge, but consider my commentary in #17 to be an appropriate reflection on that.

No, I'm not sure we agree on that. For energy measured locally, it's the same regardless of where you do the experiment, by the equivalence principle. For energy measured "at infinity", I'm not sure; I would have to work through the math.

Sorry I neglected to specify the energy redshift was a coordinate measurement, but thought that would be taken for granted. We have worked through numbers of equivalent such situations here and can honestly see no room for doubt over what is a pretty simple and straightforward scenario. It is nothing more than an application of what we both agreed to surely in e.g. #254 or situation #3: in your #260 - adapted to finite potential drop. As the 'test charges' are clearly to be treated as a minor perturbation to a far greater central mass there surely is no thought of any significant alteration to Schwarzschild metric. Why would one then expect other than that the rather localized dipople electrostatic field energy would redshift any differently to a lump of neutral matter? :zzz:

 

[DrGreg]

If, due to infall of space, nothing can be at rest inside the outer horizon because to do so would mean it was, in effect, traveling outwards at greater than C, then how can a charged particle do it?

A charged particle approaching a charged black hole would experience repulsion before it reached the outer horizon. If it was going fast enough to cross the horizon it would never escape, as by then the gravity would be stronger than the repulsion (to borrow Newtonian language).

How indeed can a charged particle ever experience an electric field until it got to the singularity.

Other posts in this thread (e.g. #264, #267) have been discussing that the charged particles that originally fell into the hole in the past (to charge it) still leave an electromagnetic field behind them, outside the horizon, which "can't escape fast enough", to put it rather crudely.

 

[PeterDonis]

Agreed that unlike energy ('gravitating charge' in effect) one cannot radiate away charge, but consider my commentary in #17 to be an appropriate reflection on that.

But your commentary in #17 does not even address the key difference: the *reason* that energy "redshifts" in the scenario where an object is slowly lowered is that energy is being *extracted* during the process. There is no way to correspondingly "extract charge" from a charged object that you are slowly lowering.

As the 'test charges' are clearly to be treated as a minor perturbation to a far greater central mass there surely is no thought of any significant alteration to Schwarzschild metric.

So your intent in this scenario was to create a dipole from a pair of oppositely charged test objects hovering above a *neutral* BH? I had thought you meant to have them above a charged BH (or a charged planet/star).

Why would one then expect other than that the rather localized dipople electrostatic field energy would redshift any differently to a lump of neutral matter? :zzz:

It would seem to me that the following would be true, anyway: if the work needed to create the dipole is done locally, the amount of work will be the same regardless of where in spacetime the dipole is created. Call that work W. If, however, the work is done elsewhere and "transmitted" to where the dipole is formed (say, for example, that the charged objects are pushed apart by a gear mechanism, and the mechanism is driven by a linkage that reaches down from a higher altitude), then at the higher altitude where the work is generated, the amount of work needed will indeed be "redshifted"; it will be less than W. But if the work done locally by the gear mechanism is measured, it will be W; as the work is transmitted down the linkage, it "blueshifts" back to its local value. That would be true of any scenario where work is transmitted from a higher to a lower altitude.

As far as "charge redshifting" is concerned, I see what you are saying: viewed from the higher altitude, the energy stored in the dipole is less than W, but the separation of charges is the same (since it's tangential), so it looks like the charges must be reduced. But this can equally well be interpreted as the effect of the curved spacetime in between, rather than anything intrinsic to the charge. After all, locally the dipole has stored energy W, so whatever change is seen from a distance would seem to be due to the intervening spacetime.

We've had this discussion before, of course; I don't see that this scenario is any different from the other ones where we've had it. You are trying to assign a meaning to "how things look from a distance" that is not necessary. It's not that it's "wrong", exactly; just that it's not necessary, you can do all the physics and interpret all the physics without it, so it's just your personal way of viewing things, IMHO.

 

[Dale]

How would the electro(static) field propagate past the EH? Photons don't, how would virtual photons (the carriers of em field) ?

Easy, virtual photons are off-shell, meaning they can travel faster than c and even have mass.

 

[Dale]

Well I had hoped it would have been evident this definition is what was being implied in bit quoted - we are obviously discounting as gone from considered system energy extracted in lowering (and 'lowering' was deliberately the term used) mass m. All btw covered in #1.

So, let me see if I correctly understand your chain of reasoning.

1) There exist three global measures of mass that could be considered in an asymptotically flat and static metric. Out of those three you equate the Bondi mass with the M parameter of the Schwarzschild metric.
2) You note that the Bondi mass "redshifts" in the specific case where energy is radiated to null infinity. You then generalize that claim to mean that mass itself "redshifts" in changes to the Schwarzschild M parameter.
3) You then further generalize that claim to mean that the redshift in the global measure of mass (Bondi) and the global mass parameter of the spacetime can be localized to the mass that is added to the system, and therefore an additional mass m is "redshifted" wrt local interactions.
4) You then further generalize that since local interactions are unchanged charge must also be redshifted.
5) You then note that no charge is radiated to null infinity. Therefore, since charge is redshifted and not radiated away it must not be conserved.

Is that an accurate representation of your line of reasoning?

 

[Dale]

In fact apart from that added thought in #248 nothing substantially new has been added since #1 really. So I don't know, in the end it may simply come down to another fizzle/fadeout which at bottom amounts to "concensus/majority opinion ruulz - OK!" But no - sour grapes premature at this point.

Q-reeus, from #1 you have never even rigorously defined let alone justified your key "mass redshifts" concept. This has nothing to do with intellectual "mob rule" but instead, a lack of rigor in your idea. You are absolutely convinced by your handwaving arguments that there is yet another flaw in GR, but when pushed to clarify your key concepts you cannot do so.

 

[GAsahi]

Easy, virtual photons are off-shell, meaning they can travel faster than c and even have mass.

True. How does this "help" in propagating past the EH? For example, having "mass" seems more like hindrance, not help.