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Question about Black Holes (simple question)

by briduende
Tags: black, holes, simple
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briduende
#1
Jun22-11, 05:59 PM
P: 1
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!!
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Bill_K
#2
Jun22-11, 06:12 PM
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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
#3
Jun23-11, 02:44 AM
P: 1,115
Quote Quote by Bill_K View Post
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 http://www.physicsforums.com/showthr...=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.

DaleSpam
#4
Jun23-11, 03:02 AM
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Question about Black Holes (simple question)

Quote Quote by Q-reeus View Post
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
#5
Jun23-11, 03:59 AM
P: 1,115
Quote Quote by DaleSpam View Post
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.
DaleSpam
#6
Jun23-11, 04:44 AM
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Quote Quote by Q-reeus View Post
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
#7
Jun23-11, 05:16 AM
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Quote Quote by DaleSpam View Post
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.
DaleSpam
#8
Jun23-11, 02:43 PM
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Quote Quote by Q-reeus View Post
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.
DaleSpam
#9
Jun23-11, 02:47 PM
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Quote Quote by Q-reeus View Post
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
#10
Jun23-11, 04:09 PM
P: 292
Quote Quote by Q-reeus View Post
Back here http://www.physicsforums.com/showthr...=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
#11
Jun23-11, 04:16 PM
P: 317
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
#12
Jun24-11, 03:33 AM
P: 1,115
Quote Quote by DaleSpam View Post
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
#13
Jun24-11, 03:34 AM
P: 1,115
Quote Quote by DaleSpam View Post
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
#14
Jun24-11, 03:35 AM
P: 1,115
Quote Quote by cosmik debris View Post
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
#15
Jun24-11, 03:36 AM
P: 1,115
Quote Quote by dm4b View Post
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
#16
Jun24-11, 04:07 AM
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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
#17
Jun24-11, 04:48 AM
P: 1,115
Quote Quote by pervect View Post
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.
DaleSpam
#18
Jun24-11, 06:26 AM
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Quote Quote by Q-reeus View Post
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.


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