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Dale
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You can perform those experiments outside the EH.Austin0 said:How would charge escape from the hole if light cannot?
You can perform those experiments outside the EH.Austin0 said:How would charge escape from the hole if light cannot?
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.Austin0 said:OK but I thought that the concept of downward force has been eliminated from GR ??
but I don't think that any of the conclusions based on the idea that mass redshifts are going to be valid...
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'.
'things' here meaning matter particles and photon redshift...expansion causes things to lose momentum relative to the CMB...
How would charge escape from the hole if light cannot?
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).
[sounds like a PeterDonis, maybe??]...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.
Naty1 said:[sounds like a PeterDonis, maybe??]
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√-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. 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:DaleSpam said: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.
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.Naty1 said: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.
DaleSpam said: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,
...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 said:Is this the issue where it was recently discussed 'why free fall has no redshift' or somesuch??
I don't know that, do you have a reference or derivation?Q-reeus said: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.
Q-reeus said: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.
DrGreg said: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.
PeterDonis said: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.
How would charge escape from the hole if light cannot?
Naty1 said:[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.]
Austin0 said:Could you elaborate on this concept of field propagation from past light cone?
PeterDonis said: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 [itex]F_{ab} n^{a} u^{b}[/itex] to an expression involving the charge-current 4-vector.
PeterDonis said: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,
Austin0 said: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.
Austin0 said: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.
Austin0 said: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.
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.PeterDonis said: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.
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.)
Q-reeus said: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.
Q-reeus said: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.
Q-reeus said: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.
DaleSpam said:The motion of charged vs uncharged particles.
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.PeterDonis said: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.)
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').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.
Q-reeus said:At question is the effective coordinate value of charge as a function of gravitational potential - a different matter entirely.
Q-reeus said: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'.
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.PeterDonis said: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.)
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: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.
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).Trenton said: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?
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.Trenton said:How indeed can a charged particle ever experience an electric field until it got to the singularity.
Q-reeus said: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.
Q-reeus said: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.
Q-reeus said: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:
Easy, virtual photons are off-shell, meaning they can travel faster than c and even have mass.GAsahi said:How would the electro(static) field propagate past the EH? Photons don't, how would virtual photons (the carriers of em field) ?
So, let me see if I correctly understand your chain of reasoning.Q-reeus said: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.
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.Q-reeus said: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.![]()
DaleSpam said:Easy, virtual photons are off-shell, meaning they can travel faster than c and even have mass.