Question about Black Holes (simple question)

[briduende]

Hi every1

My question is: we know nothing travel faster than light. Gravity travel at the speed of light, right? Now: light cannot escape from a black hole. So: gravity cannot escape from a Black Hole.

Then... why can we feel the gravitational field of a black hole if gravity cannot be transmitted outside the black hole??

Imagine that the black hole suffer some change: it eat some other black hole or ver big star... the gravity field will change. But if gravity travels at the speed of light and light cannot escape from a black hole... how could we feel the change of the system?

Thanks!

 

[Bill_K]

Gravity travel at the speed of light, right? So: gravity cannot escape from a Black Hole.

Gravitational waves travel at the speed of light. Gravity also includes static fields. The gravity field surrounding a black hole does not travel out of the hole, it was there to begin with. Just like the Coulomb field surrounding a charge does not travel out of the charge.

Imagine that the black hole suffer some change: it eat some other black hole or very big star... the gravity field will change. But if gravity travels at the speed of light and light cannot escape from a black hole... how could we feel the change of the system?

The star will have to fall in from infinity. When the star has reached some radius R say, the gravitational field for r > R will surround both the hole and the star, and will have a mass term equal to the sum of the masses of its components. This will be the case long before the star reaches the Schwarzschild radius.

 

[Q-reeus]

Gravitational waves travel at the speed of light. Gravity also includes static fields. The gravity field surrounding a black hole does not travel out of the hole, it was there to begin with. Just like the Coulomb field surrounding a charge does not travel out of the charge.

Back here https://www.physicsforums.com/showthread.php?t=507172&page=2, entry #17, I specifically challenged that last point about charge, and unsurprisingly from my experience here at PF, there were no takers. In QED interaction between charge/current via the EM field is an inherently dynamical continuous exchange of virtual photons. Well for an outside test charge to feel the field of a charge lying at the BH EH, it has by QED to *continuously exchange* virtual photons with the inner charge. The claim made in the link within the link above is that virtual photons can travel faster than c - thus 'there is no problem' with the electric field emanating 'normally' from a charged BH. This imo conveniently ignores that the clock rate has stopped at the EH, relative to the outside test charge. You cannot 'exchange' anything with a time-frozen corpse, unless one exchanges physics with physics fiction and postulates a truly infinite virtual photon propagation rate in order to somehow get around the problem of EH zero clock rate. But even then, merely giving the messenger particle infinite speed won't do - as a *continual stream* of virtual photons is required to account for a static electric field, how can a time-frozen corpse (the charge at the EH) accomplish this? It has no 'heartbeat' - there are no 'gears and wheels' in motion to direct such an exchange. Straw-man arguments I often see that claim the only valid perspective is proper time just won't do here - there is a *continuous* round-trip process to explain if any credence at all is given to virtual particle *exchange*.

The star will have to fall in from infinity. When the star has reached some radius R say, the gravitational field for r > R will surround both the hole and the star, and will have a mass term equal to the sum of the masses of its components. This will be the case long before the star reaches the Schwarzschild radius.

As far as mass is concerned, string/super-string/M-theory, as the 'leading candidate' for a TOE, treats gravity in part at least as a force field in that it is considered to be mediated *dynamically* by a *continual exchange* of graviton virtual particles, analogous to the electric field of a charged particle electric charge interaction being via virtual photon exchange. Hence by analogy the same issue arises as for charge above. Hence it seems we have either 'disproved' the leading candidate for a TOE in one stroke, or one might suspect the notion of BH has consistency issues! While not quite the simple issue the OP posed, he has raised a valid issue imo.

 

[Dale]

This imo conveniently ignores that the clock rate has stopped at the EH, relative to the outside test charge. You cannot 'exchange' anything with a time-frozen corpse

That is only an artifact of the Schwarzschild coordinate system. There is no timelike worldline for which it is true that the clock stops at the event horizon. So there is no reason that a charged object crossing the horizon would stop exchanging virtual photons.

 

[Q-reeus]

That is only an artifact of the Schwarzschild coordinate system...

What physical meaning do you therefore give to the fact that at the EH specifically both radial distance measure and clock rate go to zero in Schwarzschild coordinates? If there is no valid physical significance here, why is it used at all? We are talking about *relative* measure of things, as the very name relativity implies.

There is no timelike worldline for which it is true that the clock stops at the event horizon. So there is no reason that a charged object crossing the horizon would stop exchanging virtual photons.

Not so sure. Quite often the claim by more or less GR experts here is that infinite redshift for an in-falling object at the EH is merely a Doppler shift optical effect - an 'illusion' with no physical significance. So consider please the following: A charge dipole is lying horizontally on the surface of and deep within the gravitational potential of a neutron star. In Schwarzschild coordinates the azimuthally oriented displacement distance between the charges is unaffected by the potential. Remove the toothpick keeping the charges apart - they come together and annihilate producing gamma rays. Suppose a perfect gamma ray mirror acts so both gamma rays exit to infinity. They are redshifted by the usual amount, hence the total energy of the original dipole also. Now since the dipole displacement distance was unaffected by the potential, one must conclude there is only one variable left to account for the reduced energy - the two charges were 'redshifted' in strength. Extrapolate that to the EH of a BH. What do we have? Sure the charge cannot be lying static on a surface at the EH, but explain please how free-fall makes it any better. Worse I would think. As is well known, the axial component of electric field for a relativistic charge tends to zero as v -> c. Charge invariance holds in the SR case, but my estimate would be the pancaked transverse field will in the BH case be bowed like a weeping willow - all field lines finish up diving straight down into that pit of dark doom. And that's the classical perspective of a purely static charge with none of the modern dynamical QED *continuous exchange* connotations. However one looks at it, I see no way a charge field can escape - given the properties a BH is supposed to have that is. And imo what goes for charge should consistently go for mass as source of gravity also.

 

[Dale]

What physical meaning do you therefore give to the fact that at the EH specifically both radial distance measure and clock rate go to zero in Schwarzschild coordinates?

Only that there can be no stationary observers in Schwarzschild coordinates at or below the EH.

 

[Q-reeus]

Only that there can be no stationary observers in Schwarzschild coordinates at or below the EH.

Fine, we agree it has to be a free-fall zone. That hardly answers it's relevance or otherwise when computing the *round trip* dynamics for any of a continuous stream of virtual photon exchange events. Unless there is some admission Schwarzschild coordinates are a useless fiction, one should surely acknowledge it is a predictive tool of whether two-way exchange, this side of infinity, is possible at all. And of course the other issues raised in #5 have not been addressed at all.

 

[Dale]

Fine, we agree it has to be a free-fall zone. That hardly answers it's relevance or otherwise when computing the *round trip* dynamics for any of a continuous stream of virtual photon exchange events.

It wasn't intended to answer any relevance nor aid in computing any dynamics. It was only intended to refute your fallacious "frozen corpse" objection to Bill_K's answer to the OP.

 

[Dale]

Quite often the claim by more or less GR experts here is that infinite redshift for an in-falling object at the EH is merely a Doppler shift optical effect - an 'illusion' with no physical significance.

Then I suggest you take it up with one of the people that make that claim quite often. I certainly have not made such a claim.

 

[cosmik debris]

Back here https://www.physicsforums.com/showthread.php?t=507172&page=2, entry #17, I specifically challenged that last point about charge, and unsurprisingly from my experience here at PF, there were no takers. In QED interaction between charge/current via the EM field is an inherently dynamical continuous exchange of virtual photons.

First point, there is no quantum theory of gravity so a comparison between electrodynamics and gravitational theory is not possible. Secondly the virtual particle exchange in quantum electrodynamics is model dependent, i.e they are a product of perturbation theory.

 

[dm4b]

Also, as is discussed over in the quantum physics forum, it seems to be generally agreed upon that virtual particles have no reality. They are just mathematical artifacts that arise from solving an equation perturbatively.

 

[Q-reeus]

It wasn't intended to answer any relevance nor aid in computing any dynamics. It was only intended to refute your fallacious "frozen corpse" objection to Bill_K's answer to the OP.

Fallacious? Well I'm sure you're aware there are others here that also believe that according to GR things 'really' stop at the EH - 'relatively', as in relativity implies everything is relative, ok. You have studiously avoided answering my example in #5 of the electric dipole -> gamma rays scenario, which quite apart from extrapolating to the extremes at an EH, clearly 'suggests' a general failure of charge invariance in GR. To my admittedly layman's un-edjakated mind this would heuristically be attributed in part to a *physically real* slower *relative* clock rate (plus a *physically real* contracted *relative* radial length scale, which however would only show up in a before-after gravitational collapse sequence). Not to mention the matter of severe bending of field lines (something I cannot quantify properly, but surely is a factor also). Regardless of how the mix of those factors come together, my example in #5 is imo proof charge invariance *fails* - owing to the *physical reality* implicit in those SC transformations. And just to show you how 'whacko' I really am - don't even believe those SC's are valid! Point is you folks all do, so my thrust is merely to work from the perspective they are valid and imo point out their *physical consequences* which makes BH an inconsistent notion. Now of course if failure of charge invariance is in fact an acknowledged and well known property of the EFE's or whatever, please enlighten me. I am not personally acquainted with any material to that effect - http://en.wikipedia.org/wiki/Charge_invariance seems to be saying it always holds. And let me again give you a specific scenario that imo highlights the real physical consequences. We all know that by Gauss's law there is no field outside of a spherical capacitor, assuming of course equal and opposite uniformly distributed charges on the inner and outer conducting spherical shells. Well if the inner shell happens to be the surface of a neutron star, I say, as per scenario in #5, there will be a net electric field extending beyond the outer shell, owing to reduced effective charge on the inner shell. Is that clear enough statement that as far as I'm concerned charge invariance logically fails in GR? And further that will be true regardless of the validity of SC's - it follows imo as a general property of any decent gravitational theory; ie - gravitational redshift = charge non-invariance! And again as per my entries #3 and #5, it follows by analogy that one either 'disproves' string theory or finds the BH 'devours itself' - which just means an inconsistent mathematical monster imho. Ah, the joy of putting to shame a heretic - so go for it, don't be shy! But it would be nice to have *all* my points, and not just carefully excised snippets, answered in some detail. It is after all a package deal I'm presenting here - one thing follows from another as presented in #3 and #5.

 

[Q-reeus]

Then I suggest you take it up with one of the people that make that claim quite often. I certainly have not made such a claim.

Well good for you, but in relation to matters raised in my last and earlier postings, I'm sure you're aware others here have voiced the same general criticism (quite recently in other threads actually) - only they haven't taken the issue to the same 'radical' conclusion as myself. A minor comment in context, and I have no intention of dragging others as unwilling participants into this fray. What does faintly surprise is silence to date from the many GR experts re my 'radical' claims. A big yawn apparently. Or maybe my paranoia is justified - really am on some informal 'trouble maker' blacklist of sorts. That would be a buzz!

 

[Q-reeus]

First point, there is no quantum theory of gravity so a comparison between electrodynamics and gravitational theory is not possible. Secondly the virtual particle exchange in quantum electrodynamics is model dependent, i.e they are a product of perturbation theory.

True, as is said constantly, there is no final QG theory, but is it not also true, as stated in #3, "...string/super-string/M-theory, as the 'leading candidate' for a TOE, treats gravity in part at least as a force field in that it is considered to be mediated *dynamically* by a *continual exchange* of graviton virtual particles, analogous to the electric field of a charged particle electric charge interaction being via virtual photon exchange..."? Point being there is always a dynamical exchange process of some kind, and I'm simply saying this just breaks down when considering the properties at or 'inside' the EH that SC's are surely implying. Can you provide a physically plausible scenario that invalidates this viewpoint?

 

[Q-reeus]

Also, as is discussed over in the quantum physics forum, it seems to be generally agreed upon that virtual particles have no reality. They are just mathematical artifacts that arise from solving an equation perturbatively.

Yes, quite aware of that - but also aware as I assume you are that not every QFT expert agrees. Regardless though, virtual particle exchange a la Feynman diagrams is supposed to be a valid and long established methodology for describing and computing things in QFT - agreed? Then if it utterly fails a la BH EH 'time freeze' argument, isn't that news? It gets down here to whether one believes that relative to an outside observer/test-particle/whatever, time *effectively* halts at the EH, re any possible continuous dynamic exchange process. Not that I'm only relying on that argument, as earlier postings will show - gravitational redshift has imo interesting consequences quite apart from the extremes of a notional BH.
EDIT: A clarification - 'utterly fails' above is not meant to imply any disproof of QFT, virtual particle exchange picture etc - rather the logic of two way dynamical communication fails a la BH EH re 'outside' contact.

 

[pervect]

Well, you'll have to find someone who understand virtual particles better than I do if you seriously want to use them and be sure you're actually getting senisble results. And your best bet on finding such a person would be on some other forum, because classical GR doesn't use them. You might get lucky and find someone on this forum who can do it, and if you're even luckier it'll be understandable.

I suspect about 99% of the people who use the virtual particle model expect them to behave exactly like real particles, and it's obvious they don't - for instance all the usual arguments about abberation.

However, from my limited understanding from just reading the FAQ's, virtual particles are unphysical enough to be faster than light, i.e. the lines on the Feynman diagram may apparently be spacelike lines. What voodoo makes the diagrams work with space-like lines isn't clear to me (assuming I"ve understood correctly in the first place), but since it is clear that the virtual particles don't actually carry any information, it wouldn't violate any physical laws.

Meanwhile, there are lots of subtle aspects to the classical picture, including the whole idea of "force".

 

[Q-reeus]

Well, you'll have to find someone who understand virtual particles better than I do if you seriously want to use them and be sure you're actually getting senisble results. And your best bet on finding such a person would be on some other forum, because classical GR doesn't use them. You might get lucky and find someone on this forum who can do it, and if you're even luckier it'll be understandable.

I suspect about 99% of the people who use the virtual particle model expect them to behave exactly like real particles, and it's obvious they don't - for instance all the usual arguments about abberation.

However, from my limited understanding from just reading the FAQ's, virtual particles are unphysical enough to be faster than light, i.e. the lines on the Feynman diagram may apparently be spacelike lines. What voodoo makes the diagrams work with space-like lines isn't clear to me (assuming I"ve understood correctly in the first place), but since it is clear that the virtual particles don't actually carry any information, it wouldn't violate any physical laws.

Meanwhile, there are lots of subtle aspects to the classical picture, including the whole idea of "force".

Sure pervect, I'm not at all claiming anything like a deep understanding of the ins and outs of QFT, Feynman diagrams etc. However there is afaik a common theme to all TOE candidate theories - dynamical exchange processes of one kind or another underly all interactions. And GR (or similar gravitational theories) claims any and all dynamical processes are subject to the effects of spacetime curvature - and if a patch over there is subject to extreme curvature (ie BH), then ANY kind of two-way interaction with over here seems a logical inconsistency with those extreme curvature claims. This seems so general I don't feel expertise in quantum theory is needed here, but will stand corrected if that can be shown untrue.
To buttress that, here's a further example of where redshift-is-just-optical-effect just doesn't cut it imo. Suppose we drop towards a BH a beacon having a given rotational speed about an axis coincident with the radial ordinate, and let's say it emits a linearly polarized beam of light radially outwards - back to some fixed external observer. On the beacon's way down to doom, the returned beam is not only undergoing redshift, but the rotation speed of the polarization vector is also slowing. I say both redshift goes to infinity AND polarization rotation cease at the EH. The latter in particular is speaking loudly to me re the consistency of *any* kind of exchange process. It ceases re external observer - period. Now if I have this all wrong then fine I will eat humble pie, but that would require some penetrating and detailed logical counter-argument.

 

[Dale]

What does faintly surprise is silence to date from the many GR experts re my 'radical' claims.

I didn't even notice the one that you had linked to earlier, and I only noticed this one because it was early in a thread that I clicked on. Nobody reads every post in every thread, so it shouldn't surprise you that things slip through.

If you think one of your claims has merit and that it is being ignored then you should start your own thread. Otherwise don't take a lack of response as indicative of anything.

 

[Q-reeus]

I didn't even notice the one that you had linked to earlier, and I only noticed this one because it was early in a thread that I clicked on. Nobody reads every post in every thread, so it shouldn't surprise you that things slip through.

Fine, understand that.

If you think one of your claims has merit and that it is being ignored then you should start your own thread. Otherwise don't take a lack of response as indicative of anything.

If this thread fizzles out without anything resolved properly, will probably take that suggestion up, but let's first see if momentum already underway here leads anywhere useful.

 

[Dale]

Fallacious? Well I'm sure you're aware there are others here that also believe that according to GR things 'really' stop at the EH - 'relatively', as in relativity implies everything is relative, ok.

Yes, the frozen-corpse argument was fallacious regardless of how many other people here hold similar opinions.

You have studiously avoided answering my example in #5

Yes, and I have already explained why. Why should I refute an argument against a premise which I don't believe? That makes no sense whatsoever.

The whole example is based on the premise that the gravitational Doppler shift is an illusion. If the Doppler shift is an illusion then by your chain of logic the charge must be "redshifted" by an unspecified amount. You have asserted that this premise is believed to be true by many people on this forum, but I am not one of them.

 

[Q-reeus]

Yes, and I have already explained why. Why should I refute an argument against a premise which I don't believe? That makes no sense whatsoever.
The whole example is based on the premise that the gravitational Doppler shift is an illusion...

Huh? Let's get this straight. Time and again I have seen arguments by GR buffs that claim SC's are misleading re physical consequences - that infinite redshift is basically Doppler, that one cannot consider it indicative of 'time freeze' at the EH, that one should focus on the 'reality' of proper time as seen by the infalling entity in KS coordinates or similar. I have *never* implied Doppler shift here is an illusion - quite the opposite! Rather that the inference it is illusory re physical consequences for an outside observer is wrong. Are we now clear on that one?

If the Doppler shift is an illusion then by your chain of logic the charge must be "redshifted" by an unspecified amount. You have asserted that this premise is believed to be true by many people on this forum, but I am not one of them.

Huh? Please go back and read carefully what I actually wrote in #5 - I took redshift there as very physically real, and simply drew a logical conclusion from that - namely that SC's actually predict charge non-invariance from the very physically real phenomenon of gravitational redshift. Try and get my actual arguments straight please!

 

[Dale]

Huh? Please go back and read carefully what I actually wrote in #5

Quite often the claim by more or less GR experts here is that infinite redshift for an in-falling object at the EH is merely a Doppler shift optical effect - an 'illusion' with no physical significance.

[sarcasm]I can't imagine how I could have so eggregiously misconstrued your argument.[/sarcasm]

I took redshift there as very physically real, and simply drew a logical conclusion from that - namely that SC's actually predict charge non-invariance from the very physically real phenomenon of gravitational redshift. Try and get my actual arguments straight please!

Then your conclusion doesn't follow from your premise. If the Doppler shift is "very physically real" then the decreased energy from the two gamma rays is entirely accounted for by the Doppler shift and does not need to be accounted for by an additional "redshift" of the charge. In fact, if there were additionally a "redshift" in charge then you would get a double-redshift in the energy of the received gamma rays. Your conclusion only follows from the premise that the Doppler shift is illusory and that the observed decreased energy in the gamma rays therefore needs to be accounted for via some other mechanism besides the Doppler shift.

 

[Q-reeus]

Then your conclusion doesn't follow from your premise. If the Doppler shift is "very physically real" then the decreased energy from the two gamma rays is entirely accounted for by the Doppler shift and does not need to be accounted for by an additional "redshift" of the charge. In fact, if there were additionally a "redshift" in charge then you would get a double-redshift in the energy of the received gamma rays. Your conclusion only follows from the premise that the Doppler shift is illusory and that the observed decreased energy in the gamma rays therefore needs to be accounted for via some other mechanism besides the Doppler shift.

Stop smoking it DaleSpam - it's not doing you any good! :rofl:
Sorry, but you have everything screwy here. There's nothing 'Einstein' in seeing that redshift was in #5 *equated* to reduced energy of the original dipole! Where does this 'double reduction' logic you are finding come from? Not from me. Again - please re-read #5 *carefully*!

 

[Dale]

Where does this 'double reduction' logic you are finding come from? Not from me.

No, it comes from applying both your stated premise ("Doppler shift is 'very physically real'") and your stated conclusion ("charge is 'redshifted'"). If both are true then clearly we would measure a double Doppler shift. Since we do not measure a double Doppler shift then at least one must be false.

 

[Q-reeus]

No, it comes from applying both your stated premise ("Doppler shift is 'very physically real'") and your stated conclusion ("charge is 'redshifted'"). If both are true then clearly we would measure a double Doppler shift. Since we do not measure a double Doppler shift then at least one must be false.

Maybe there is in a sense fault on my part. Perhaps I was not crystal clear enough back in #5. So allow me to correct that possibility now.
Start with our dipole of moment ql far away from the NS. have it undergo annihilation as described in #5. We have converted,in entirety, the two original charges into gamma radiation. Rest energy tied up in the charges q+ & q- is now fully in the form of radiation hf1 + hf2 +... Net change in energy is zero - this is merely a conversion of one form into another.

Now for scenario 2, an identical dipole lies horizontally at rest on the surface of a NS. In order to be at rest it had to have done work in going from 'infinity' to the lowered gravitational potential extant at the NS surface, and in so doing has lost a measure of it's original rest energy applying far from the NS. And this reduction in rest energy is *precisely reflected* in the redshifted frequencies of released gamma rays when performing the same dipole charges annihilation process as before. All so far is standard fare I'm sure we both agree fully on - I hope anyway.

The confusion seems to come in next. In #5 as figure of speech I equated the gamma radiation redshift to an equivalent 'redshift' of charge strength (and used inverted commas there also) - meaning I would have thought obvious the gravitationally depressed dipole charges rest energy. No double accounting here whatsoever. It's the next step where my argument for GR failure of charge invariance comes in. Namely that as the dipole moment arm l has not been affected by the gravitational potential according to standard SC's, we must properly equate mutual collapse over that distance l and ensuing annihilation to reduction in charge rest energy only - as determined by a distant observer. In hindsight this is a mixed event and it would have been better to have looked at simply a partial contraction of the dipole by some fraction of original separation length with no annihilation - just generation of heat, which then escapes with the same redshift reduction factor. Same conclusion though. Charge rest mass is depressed via NS gravitational potential. That rest mass is tied up (presumably by the fixed 3/4 factor) in it's electric field. Well gravitationally depressed EM mass equates in my book to reduced field strength and thus reduced value of charge - all relative to 'infinity'. Are we now clear on the logical process involved here?

 

[dm4b]

Then if it utterly fails a la BH EH 'time freeze' argument

Well it doesn't necessairly.

"To explain how gravitons escape from a black hole in order to cause the gravitational field we see at great distances, you have to accept the fact that the event horizon is not an infinitly high 'potential barrier' that would impede the wave function of a quantum system. Because general relativity shows that, in the appropriately selected coordinate system, the black hole only has ONE curvature singularity at its core, NOT at the event horizon as well, the event horizon is just a region of space-time which has a particular gravitational potential. This can now be translated into a problem in quantum mechanics where you are asking what the penetration or 'tunneling' probability is for quantum particles like gravitons, electrons etc. Outgoing quanta can therefore pass across the potential barrier at the spatial distance of the event horizon, and tunnel across it. The event horizon is poorly localized to quantum mechanical particles, and is just like any other potential barriers across which quantum mechanical particles can tunnel like a leaky membrane."

http://www.astronomycafe.net/qadir/q756.html

Another way to think about it, which also demonstrates why some folks hate virtual particles:

"The key point is that electromagnetic interactions (and gravity, if quantum gravity ends up looking like quantum electrodynamics) are mediated by the exchange of *virtual* particles. This allows a standard loophole: virtual particles can pretty much "do" whatever they like, including traveling faster than light, so long as they disappear before they violate the Heisenberg uncertainty principle.

The black hole event horizon is where normal matter (and forces) must exceed the speed of light in order to escape, and thus are trapped. The horizon is meaningless to a virtual particle with enough speed. In particular, a charged black hole is a source of virtual photons that can then do their usual virtual business with the rest of the universe. Once again, we don't know for sure that quantum gravity will have a description in terms of gravitons, but if it does, the same loophole will apply---gravitational attraction will be mediated by virtual gravitons, which are free to ignore a black hole event horizon."

http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/980601a.html

 

[Dale]

Now for scenario 2, an identical dipole lies horizontally at rest on the surface of a NS. In order to be at rest it had to have done work in going from 'infinity' to the lowered gravitational potential extant at the NS surface, and in so doing has lost a measure of it's original rest energy applying far from the NS.

If p is the four-momentum of the dipole then $g_{\mu\nu}p^{\mu}p^{\nu}$ is the same at infinity and on the surface. What may have lost energy is the NS-dipole system, not specifically the charge dipole, but even that depends on other things such as the kinetic energy of the dipole and the temperature of the NS.

You may want to read a little on binding energy and mass deficit:
http://en.wikipedia.org/wiki/Binding_energy

Starting with the NS-dipole system separated the system has some given ADM energy. As the dipole falls it exchanges potential energy for kinetic energy, and the ADM energy remains the same. At the surface assume that the kinetic energy is entirely converted thermal energy, then the temperature of the NS increases and the ADM energy again remains the same. As the NS cools back to the original temperature energy radiates away to infinity and the ADM energy decreases by the same amount. After the anhilation the gamma rays also radiate a "redshifted" amount of energy away to infinity and the ADM energy decreases the same amount. The sum of the redshifted energy and the thermal energy radiated is the same as the non-redshifted energy. There is nothing additional to account for via a reduction in charge.

 

[Q-reeus]

Well it doesn't necessairly...
...Another way to think about it, which also demonstrates why some folks hate virtual particles:..

Thanks for the links, which support what I suppose is the standard picture. Clear back in #3 though made the point that imo allowing infinite vp propagation velocity is just one factor here. The vp exchange orchestra has to be conducted at both ends, and that's where I really have a problem seeing it working consistently. One has to suppose that with SC's saying 'baton rate' is zero for charge maestro at the EH - as determined externally, a finite *rate* of return vp's is actually logically consistent. Slight of hand going on somewhere here methinks.

 

[Q-reeus]

If p is the four-momentum of the dipole then $g_{\mu\nu}p^{\mu}p^{\nu}$ is the same at infinity and on the surface. What may have lost energy is the NS-dipole system, not specifically the charge dipole, but even that depends on other things such as the kinetic energy of the dipole and the temperature of the NS.

You may want to read a little on binding energy and mass deficit:
http://en.wikipedia.org/wiki/Binding_energy

Starting with the NS-dipole system separated the system has some given ADM energy. As the dipole falls it exchanges potential energy for kinetic energy, and the ADM energy remains the same. At the surface assume that the kinetic energy is entirely converted thermal energy, then the temperature of the NS increases and the ADM energy again remains the same. As the NS cools back to the original temperature energy radiates away to infinity and the ADM energy decreases by the same amount. After the anhilation the gamma rays also radiate a "redshifted" amount of energy away to infinity and the ADM energy decreases the same amount. The sum of the redshifted energy and the thermal energy radiated is the same as the non-redshifted energy. There is nothing additional to account for via a reduction in charge.

Agreed about the net energy exchanges taking place as you describe it. But that last bit is a non-sequitur. The annihilation scenario was simply a round about way of showing depressed charge rest energy, when at rest in a grav potential. Keep the un-annihilated charges there! Their depressed rest mass equates imo to a depressed field - but see my following post re caveats which have since come to mind.

 

[Q-reeus]

OK at this point I should acknowledge there is a 'split' issue re my preceeding argument relating to charge invariance holding in GR or not. One has F = qE, and in reduced gravitational potential phi, it is F that is unambiguously reduced by the factor (1+phi) for small phi, where phi is the Newtonian potential (and has a negative sign!). That leaves the partitioning between reduced q and/or E undecided to this point. Naturally if one assigns all the reduction to q, paradoxically *effective* charge invariance holds re external observer - since E would then be invariant wrt phi for a given charge number. But is that a consistent choice? Reading up on Gravitation and Relativity - M.G.Bowler, pp 73-74, a fully self-consistent set of relations for EM in a gravitational potential is developed. What is shown is that consistency for eg light speed, EM energy density, etc, then permittivity and permeability both vary as (1-2phi) for small phi, with q taken as invariant. And formally applying E = q/(4*pi*r²*epsilon), E is indeed the one shown to be depressed by (1+phi), and similarly for magnetic induction B. Well I think so - the type-setting is so-so and there may have been one or two typos there to confuse things. If someone has an alternate source leading to differing results, please provide it here!

 

[Dale]

The annihilation scenario was simply a round about way of showing depressed charge rest energy, when at rest in a grav potential.

But the charge dipole rest energy is not depressed:

If p is the four-momentum of the dipole then $g_{\mu\nu}p^{\mu}p^{\nu}$ is the same at infinity and on the surface.

 

[Q-reeus]

But the charge dipole rest energy is not depressed:

A gravitating spherical shell composed of N particles undergoes partial collapse to some reduced radius. Those same N particles are now collectively in a reduced gravitational potential. Work was done during the collapse, released as heat, light, which escapes. The total final rest energy of the shell is undeniably less than before. There are still the same N particles present. By my reckoning that means each particle has a reduced rest energy, owing to the changed gravitational potential. Should I exempt dipoles from this basic logic for some strange reason? Are you sure the 4-momentum argument is sound? Are you sure it applies to the *final* rest state of the dipole after heat etc has dissipated to infinity, and not just up to the moment where in your scenario it's free-fall from infinity abruptly ends with a crash on to the NS surface? I would have preferred a gentler winching down procedure, but whatever.

 

[Q-reeus]

In #30 things weren't expressed all that well. By analogy with the notions of active and passive gravitational mass, I would now put it as the problem of a split between 'active charge' q_a, as source of E field, and 'passive charge' q_p, as measure of response to an applied E. It is then the product q_aq_p that drops by the factor (1+phi). According to Bowler, by this parlance q_p is the invariant wrt gravitational potential phi, and q_a is the one gravitationally depressed. That implies failure of charge invariance consistent with previous arguments re clock rate effect on virtual photon exchange. However it is not the consensus view as per Wikipedia entry given in #12. So again, any textbook or preferably online links showing from first principles a contradictory finding that static EM fields are gravitational invariants?

 

[Dale]

The total final rest energy of the shell is undeniably less than before.

Yes.

By my reckoning that means each particle has a reduced rest energy, owing to the changed gravitational potential.

No. Each particle has its original rest energy. Did you read the wikipedia link on binding energy that I had given earlier:
http://en.wikipedia.org/wiki/Binding_energy

"A bound system typically has a lower potential energy than its constituent parts". So the rest energy of a system of bound particles is less than the rest energy of the constituents.

Are you sure it applies to the *final* rest state of the dipole after heat etc has dissipated to infinity, and not just up to the moment where in your scenario it's free-fall from infinity abruptly ends with a crash on to the NS surface?

Yes, I am sure about it, but I am not the one making "contrarian" claims. The real question is if you are sure that it does not apply. When you are challenging mainstream science you had better be very sure.

 

[Q-reeus]

...No. Each particle has its original rest energy. Did you read the wikipedia link on binding energy that I had given earlier:
http://en.wikipedia.org/wiki/Binding_energy
"A bound system typically has a lower potential energy than its constituent parts". So the rest energy of a system of bound particles is less than the rest energy of the constituents...

Yes I read it. A matter of careful definition - no point comparing apples with oranges. By 'constituent parts' I take it to mean the original state of gravitationally unaffected N constituent parts - ie when each part is 'at infinite separation' and in zero gravitational potential. After collapse, the very same but now gravitationally bound N constituent parts have reduced net energy. One can argue the semantics but by that definition everything gone over in #25 is logically coherent.