Relative energy of a black hole.

[cragar]

Does kinetic energy of a body contribute to the stress-energy tensor of a black hole.
If I move with respect to a black hole I would perceive it to have more energy. So would it have a stronger G field in my frame. Would it have a larger Schwarzschild radius. My assumptions are probably wrong. I don't know that much about GR but would I use the
Schwarzschild metric to try to solve this. How would I factor in the relative velocity. Dont make your response to complicated because I don't know that much about GR. Any help will be much appreciated.

 

[Bill_K]

cragar, It's true that if you are in a moving frame you will perceive the black hole to have greater energy. The key word here is "perceive". It's easy enough to write the Schwarzschild solution in a moving frame. However the attributes of the hole including its Schwarzschild radius are intrinsic and won't be affected. You will of course perceive them differently.

 

[Naty1]

If I move with respect to a black hole would I perceive it to have more energy. So would it have a stronger G field in my frame.

Here is one 'trick' to help resolve such questions:

Any situation where you ask about a rapidly moving massive body's gravitational effect and a 'stationary' observer can be transformed to an equivalent question about the interaction between a rapidly moving observer and a 'stationary' massive body. So all observations relating to a rapidly moving massive body can be answered as if the body is stationary...as if all measures are local. Local measures trump distant measures.

And a further perspective:

For a single electron, or black hole] as an example, the rest energy density of the electron is the only thing that causes spacetime curvature. The kinetic energy is frame-dependent, just as the velocity is; in the electron's rest frame it is zero, and we can predict all physical observables, like whether the electron forms a black hole, by solving the EFE in the electron's rest frame.

This means that no matter how fast you see a particle whizzing by, it will never become a black hole. It turns out, however, that even though such speed does not add to gravitational curvature, you generally PERCEIVE the spacetime as being curved ...but it is not gravitational curvature.

Does kinetic energy of a body contribute to the stress-energy tensor of a black hole.

Only if the KE exists in the rest frame of the body:

Say you have an atom, [or a black hole] and you HEAT that atom, now the constituent particles [or degrees of freedom]gain energy, move faster, and so even in the rest frame of the atom there is additional kinetic energy: that kind of kinetic energy, in the rest frame of the center of mass, DOES contribute to additional gravitational curvature. Another example, would be the energy required to compress a spring: such energy exists in the rest frame of the spring and so it's equivalent 'mass' increases.


Another, equivalent way to picture this:

In GR, spacetime [gravitational] curvature is a property INTRINSIC to a mass. GR is background independent...The gravitational field 'g is' frame-independent.

And docAl provided me this explanation several years ago:[paraphrased]

Assume you have a flat sheet of graph paper to represent two dimensional space without gravity. [You can think of time as a third dimension if you like.] When we introduce gravitation, the paper itself becomes curved. (Curvature that cannot be "flattened" without distortion. Gravitational "spacetime curvature" refers to this curvature of the graph paper, regardless of observer, whereas visible/perceived curvature in space is related to distorted, non-square grid lines drawn on the graph paper, and depends on the frame choice of the observer..."

Found another related discussion:

https://www.physicsforums.com/showthread.php?p=3661242&posted=1#post3661242

 

[Dale]

Does kinetic energy of a body contribute to the stress-energy tensor of a black hole.

Yes. However, kinetic energy does not only contribute to the time-time component of the stress-energy tensor, but it also contributes to the spatial components. So the overall effect is not straightforward.

 

[Agerhell]

Does kinetic energy of a body contribute to the stress-energy tensor of a black hole.
If I move with respect to a black hole I would perceive it to have more energy. So would it have a stronger G field in my frame. Would it have a larger Schwarzschild radius. My assumptions are probably wrong. I don't know that much about GR but would I use the
Schwarzschild metric to try to solve this. How would I factor in the relative velocity. Dont make your response to complicated because I don't know that much about GR. Any help will be much appreciated.

When you approach a black hole your own clock would start to tick slower due to what is known as gravitational time dilation. This could make you think that the black hole has gained energy(mass) and is thus able to accelerate you faster. (When your clock starts ticking slower you will think that you are speeding up)

When you are speeding up, accelerated by the black hole, your clock would start ticking slower for that reason contributing to your possible feeling that the black hole has gained mass.

At the same time light travels slower close to massive bodies (Shapiro effect) so if you are measuring your velocity somehow in relation to the speed of light, you may come to other conclusions.

A black hole is per definition black, but if there is some heated gas close to the black hole I think you will be perceiving that gas to be hotter the closer you get to the black hole, there will be lesser gravitational redshift. Of course if you travel faster and faster towards the black hole there will be more and more doppler shift contributing that you will be perceiving the gas as being more energetic.

Those are the physical effects I can think of... Regarding G and the Schwarzshild radius I do not know, maybe it depends upon how you are attempting to determine those...

 

[cragar]

Naty 1 said if I heat the atom it will contribute to the gravitational curvature.
So If I have a rotating black hole it will contribute to the curvature. What If I was going around in a circle outside the black hole using rocket power. But I guess I would know I was accelerating and I wouldn't think the black hole was rotating. And also if I had something orbiting a black hole that would contribute to its gravitational curvature.

 

[stevebd1]

Naty 1 said if I heat the atom it will contribute to the gravitational curvature.
So If I have a rotating black hole it will contribute to the curvature.

You might be interested to know that a black holes gravitational curvature is a combination of its irreducible mass, spin and charge where (in geometric units)-

M^2=\frac{J^2}{4M_{ir}^{2}}+\left(\frac{Q^2}{4M_{ir}}+M_{ir}\right)^2

M_{ir}=\frac{1}{2}\sqrt{\left(M+\sqrt{M^2-Q^2-a^2}\right)^2+a^2}

where M=Gm/c^2,\,Q=C\sqrt(Gk_e)/c^2 and J=aM where a=j/mc where j is angular momentum in SI units.

M_{ir} is the mass you would have left if all the charge and spin were extracted (i.e. a Schwarzschild BH).

Source- http://www.ece.uic.edu/~tsarkar/Goodies/Black Hole.pdf (page 12)

 

[cragar]

interesting thanks for that reply. So if a Black hole is charged It will have more gravitational curvature because of the Energy in the electric field. Does a charged rotating BH create a B field. But does the B field only exist inside the event horizon or can it exist outside.

 

[stevebd1]

interesting thanks for that reply. So if a Black hole is charged It will have more gravitational curvature because of the Energy in the electric field. Does a charged rotating BH create a B field. But does the B field only exist inside the event horizon or can it exist outside.

The charged field (and any consequential magnetic field) would reside outside the event horizon. Extract from 'How does the gravity get out of the black hole?'-

Purely in terms of general relativity, there is no problem here. The gravity doesn't have to get out of the black hole. General relativity is a local theory, which means that the field at a certain point in spacetime is determined entirely by things going on at places that can communicate with it at speeds less than or equal to c. If a star collapses into a black hole, the gravitational field outside the black hole may be calculated entirely from the properties of the star and its external gravitational field before it becomes a black hole. Just as the light registering late stages in my fall takes longer and longer to get out to you at a large distance, the gravitational consequences of events late in the star's collapse take longer and longer to ripple out to the world at large. In this sense the black hole is a kind of "frozen star": the gravitational field is a fossil field. The same is true of the electromagnetic field that a black hole may possess.

Source- http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html

 

[Q-reeus]

...Extract from 'How does the gravity get out of the black hole?'-
"...the gravitational field is a fossil field. The same is true of the electromagnetic field that a black hole may possess.
Source- http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html

And that same source goes on to say:

Often this question is phrased in terms of gravitons, the hypothetical quanta of spacetime distortion. If things like gravity correspond to the exchange of "particles" like gravitons, how can they get out of the event horizon to do their job?
Gravitons don't exist in general relativity, because GR is not a quantum theory. They might be part of a theory of quantum gravity when it is completely developed, but even then it might not be best to describe gravitational attraction as produced by virtual gravitons. See the physics FAQ on virtual particles for a discussion of this.

Nevertheless, the question in this form is still worth asking, because black holes can have static electric fields, and we know that these may be described in terms of virtual photons. So how do the virtual photons get out of the event horizon? Well, for one thing, they can come from the charged matter prior to collapse, just like classical effects. In addition, however, virtual particles aren't confined to the interiors of light cones: they can go faster than light! Consequently the event horizon, which is really just a surface that moves at the speed of light, presents no barrier.

Nice try, but presenting the problems and pretending to answer them doesn't cut it imo. So the 'fossil field', be it gravitational or electric, is a source unto itself? Perhaps someone can enlighten me here. It is well known that in GR the stress-energy tensor T as source of gravity contains zero contribution from the field itself. Yet when we get to a notional BH, seems some kind of magic takes over and the 'fossil field' of necessity becomes it's own source. Pray tell how is this not an act of sweeping under the rug an embarrassing contradiction? Carefully avoiding the words 'field as source term' and vaguely substituting 'fossil field' is not simply calling a rose by another name?

Further, just how can there be a necessarily *continuous* virtual particle exchange process giving rise to an exterior electric field? Seems especially problematic to me given that all temporal processes for any and all objects at or interior to the BH EH have come to a screeching halt wrt the BH exterior, where the E field supposedly effortlessly extends. It surely takes two to tango if an exchange process is to occur. You can have a genuine exchange if at one end of the telegrapher's line the telegrapher is in deep suspended animation?! In other words, if the exchange *rate* is necessarily zero, how is any exchange at all taking place between an external 'hovering' entity and charged matter at or interior to the EH? What deep principle am I missing here?

 

[Sam Gralla]

This means that no matter how fast you see a particle whizzing by, it will never become a black hole.

Nope, at least not for a boson star model of a particle. Choptuik and Pretorious showed a few years back that "center of mass frame energy" (including "kinetic energy") really does contribute to the total energy relevant for gravitational collapse.

http://arxiv.org/abs/0908.1780

 

[Q-reeus]

Nope, at least not for a boson star model of a particle. Choptuik and Pretorious showed a few years back that "center of mass frame energy" (including "kinetic energy") really does contribute to the total energy relevant for gravitational collapse.
http://arxiv.org/abs/0908.1780

I will dare to speak on Naty1's behalf here, and suggest he was referring to a single massive object 'in flight', which is totaly different to the situation of head-on collision of two such massive objects. It is obviously true that a moving observer cannot influence the physics in some other reference frame merely by having relative motion to it. And that surely was the gist of Naty1's point.
And btw folks, it would be nice if some GR buffs here actually responded to my #10! Or is it all too boring?

 

[Sam Gralla]

It is obviously true that a moving observer cannot influence the physics in some other reference frame merely by having relative motion to it.

Sure, if by "observer" you mean a limit where the mass of a real observer goes to zero. All I'm saying is that if you have any mass at all, then there will be a speed for which, when the electron comes whizzing by, you better get ready to become a black hole =).

 

[Q-reeus]

Sure, if by "observer" you mean a limit where the mass of a real observer goes to zero...

Agreed; and to be picky one should throw in 'at an arbitrarily large minimum separation distance'.

All I'm saying is that if you have any mass at all, then there will be a speed for which, when the electron comes whizzing by, you better get ready to become a black hole =).

That's an interesting thought. So for instance a sufficiently ultra-ultra relativistic particle, boring straight through the Earth say, would leave behind a black hole wake?! Yikes.

 

[Sam Gralla]

That's an interesting thought. So for instance a sufficiently ultra-ultra relativistic particle, boring straight through the Earth say, would leave behind a black hole wake?! Yikes.

Now that I think about it this may not be true. Intuitively, the energy has to stick around long enough in order for a black hole to actually form, and now that I think about it, I think this is what Choptuik and Prestorius found (and what Penrose originally argued, although I've never read that work). So the correct statement is probably that for a fixed size/mass of the Earth and fixed size/mass of the particle coming whizzing by, there is a (possibly vanishing) finite range of speeds for which a black hole forms.

 

[Q-reeus]

Now that I think about it this may not be true. Intuitively, the energy has to stick around long enough in order for a black hole to actually form, and now that I think about it, I think this is what Choptuik and Prestorius found (and what Penrose originally argued, although I've never read that work). So the correct statement is probably that for a fixed size/mass of the Earth and fixed size/mass of the particle coming whizzing by, there is a (possibly vanishing) finite range of speeds for which a black hole forms.

By speed I guess you mean energy, since for ultra-relativistic v = c-(an absolute whisker). In the case of glancing motion rather than direct head-on collision, wouldn't it tend to be a case of a minimum product of (tidal 'g')*(impulse duration) for crushing nearby matter sufficiently? By my crude reasoning, at ultra-relativistic energies the particle and field becomes a 2D pancake whose thickness is inversely proportional to particle KE E, but whose gravitational curvature (read tidal acceleration) at a given transverse radial distance grows quadratically with E. Again crudely, as we therefore have an impulse duration dt ~ E^-1, tidal accelerations a ~ E², then using good old S = 1/2a*dt² (S being some 'critical crushing displacement' for a given blob of stationary matter), it appears to be a stalemate with no advantage in going beyond a presumed minimum energy. This does not take into account gravito-magnetic interaction but I would think it relatively insignificant. The real complications might be the time dilational factor as crushed matter approaches it's Schwarzschild radius, and have little clue how that should be factored in.
But why am I contemplating the above, given general misgivings expressed in #10? Just for fun!

 

[pervect]

Nope, at least not for a boson star model of a particle. Choptuik and Pretorious showed a few years back that "center of mass frame energy" (including "kinetic energy") really does contribute to the total energy relevant for gravitational collapse.

http://arxiv.org/abs/0908.1780

There is nothing in this paper that says that a single particle moving at a high velocity will become a black hole.

It does say that a pair of particles colliding can become a black hole (which is true) - but it does not say that a single particle moving at high speed will become a black hole - because this statement, as previously mentioned, is false.

Relative to some observers, you are right now moving at 99.999999% of the speed of light. But you are not a black hole. Not to yourself, and not to the observer at which you are moving at such a high velocity - because being a black hole is frame indepenent.

The correct description of the gravitational field as seen by a rapidly moving observer is given by the Aichelberg-Sexl ultraboost. See for instance http://arxiv.org/abs/gr-qc/0110032

It may take a little bit of advanced knowledge to interpret the comonents of the field tensor, given by the Riemann, in semi-Newtonian terms. Basically, certain components of the Riemann tensor correspond to the gradient of the Newtonian gravitatioanl field, i.e. the Newtonian Tidal tensor. Other components of the Riemann include gravitomagnetic effects, and effects that have no direct Newtonian counterparts (the topogravitic part of the tensor).

The analogous case of the field of a moving charge is simpler to understand, and very helpful. Basically, the field of the moving charge becomes in the relativistic limit an impulsive plane wave, and something very similar happens to a moving mass.

 

[Q-reeus]

Too late to edit my #16, but a few extra considerations to round that off:
1: From considerations there it is now seems obvious there is no real possibility of glancing motion ever generating conditions favouring 'BH' formation in adjacent matter, even if the ultra-relativistic mass was a 'BH' in it's own frame. Certainly out of the question for a mere elementary particle such as an electron. Accretion onto an existent BH is the best that could be hoped for.
2: Even assuming it were possible for an ultra-relativistic elementary particle to collapse nearby matter into a BH state, by reciprocity of reference frame, it would of necessity be a Kamikaze affair ending in mutual 'BH' creation - not just one acting unilaterally on the other.

And this is the last plug here re entry #10. Why no takers? It only fuels suspicions there are no decent and believable answers. So come on, someone give it a shot please.

 

[Dale]

And this is the last plug here re entry #10. Why no takers? It only fuels suspicions there are no decent and believable answers. So come on, someone give it a shot please.

I don't see anything new in #10 that hasn't been asked and answered several times on this forum already. Try doing a brief search.

 

[Q-reeus]

I don't see anything new in #10 that hasn't been asked and answered several times on this forum already. Try doing a brief search.

Not quite the response I was hoping for, but then you and I did cross swords on those matters some time back. There was imo certainly no satisfactory resolution then, which is why I have raised it anew in the faint hope of a better outcome now. Strange to me that yourself and many others here can devote page after page of detailed response to such mundane issues as proving Gauss's law in EM, or magnetism in straight wires, or choo-choo trains with mirrors and flashing lights etc. etc. Yet when it comes to this apparently taboo topic (if not taboo, why the silence from the experts in GR?), I get a terse directive to 'just search'.

Don't want to appear sulky about that, but honestly, if you actually believe the issues in #10 have been well answered, how about your own detailed summary afresh? I have never been able to figure out effective searching using PF. As someone willing to endlessly discuss oft repeated SR/GR topics, this shouldn't be a hard ask surely. I could and if pressed am willing to reference back to that earlier thread, but prefer we leave that baggage behind. Oh, and I should thank you for at least stepping up and actually breaking the silence. Now if it seems like thread hijacking just say so and I will launch the issue(s) under my own banner.

 

[Dale]

I have never been able to figure out effective searching using PF.

I typically use a google search with the delimiter:

site:www.physicsforums.com

Regarding the rest, the fact that you are aware of the previous responses on this topic and consider them all unsatisfactory seems to indicate that further discussion will probably lead to the same result. As you point out, when available we are often willing to devote many fruitless pages to topics that we notice which we consider interesting at the time.

It appears that nobody with any current interest and availability has noticed your #10. For me, the lack of interest is simple exhaustion on the topic, which is a temporary condition but present now.

 

[Sam Gralla]

There is nothing in this paper that says that a single particle moving at a high velocity will become a black hole.

It does say that a pair of particles colliding can become a black hole (which is true) - but it does not say that a single particle moving at high speed will become a black hole - because this statement, as previously mentioned, is false.

If the "observer" is a real observer and has any finite amount of mass, then a particle moving very fast by it is in fact a particle collision, and a black hole may very well be formed. But I repeat myself.

 

[Q-reeus]

I typically use a google search with the delimiter:
site:www.physicsforums.com

Thanks for the tip.

Regarding the rest, the fact that you are aware of the previous responses on this topic and consider them all unsatisfactory seems to indicate that further discussion will probably lead to the same result.

Only active and willing participation this time round can decide that. I'm certainly open to serious feedback. And for the record, here's the link to that thread: https://www.physicsforums.com/showthread.php?t=508950 (begin at #3). Apart from a single oblique cautionary response, there was no participation from any real GR pro there. Disappointing to say the least. As I said in #18 here, that just tends to fuel suspicions.

It appears that nobody with any current interest and availability has noticed your #10. For me, the lack of interest is simple exhaustion on the topic, which is a temporary condition but present now.

A call for more patience I take it, but I suspect #10 has not slipped under more than a few of the resident experts radar. Just can't quite figure a reason for the extreme reticence. And what, you have no ready answers personally willing to offer here? (saw that cute 'no peaking' trick you inserted in another thread - think about applying it to yourself here and try answering maybe, without referencing the link I gave above). For sure I could try starting a new thread, but if nothing here, then most likely nothing there either, which would be so sad.

 

[PeterDonis]

Hi Q-reeus, I just came across this thread and read your post #10. It's been some time, but I know I've given the simple classical answer to this in at least one thread on PF. (Unfortunately it doesn't appear that that thread was one of the ones you've participated in). The simple classical answer is that the "field of the black hole" doesn't actually come from inside the hole; it comes from the past, from the object that originally collapsed to form the hole. That is actually what the Usenet Physics FAQ page you linked to means when it says that the hole's field is a "fossil field"; it's a stationary remnant left by the collapsing object.

IMO, the FAQ page's mention of virtual particles is a little misleading since it invites the interpretation that the BH field actually does come from inside the hole, but somehow virtual gravitons can escape because they can move faster than light. Even on a quantum view of gravity I don't think that's correct; I think the explanation I gave above still applies. (Which is not to say that there is not a sense in which virtual particles can travel faster than light; there is. I just don't think that sort of thing is needed to resolve the question of "how gravity gets out of the black hole"--it doesn't have to get out because that's not where it's actually coming from to begin with.)

 

[Q-reeus]

Hi Q-reeus, I just came across this thread and read your post #10. It's been some time, but I know I've given the simple classical answer to this in at least one thread on PF. (Unfortunately it doesn't appear that that thread was one of the ones you've participated in). The simple classical answer is that the "field of the black hole" doesn't actually come from inside the hole; it comes from the past, from the object that originally collapsed to form the hole. That is actually what the Usenet Physics FAQ page you linked to means when it says that the hole's field is a "fossil field"; it's a stationary remnant left by the collapsing object.

Hi Peter. Well you qualify as GR expert so good at least there is some useful feedback. I understand that by 'fossil field' it is meant that the static exterior spacetime curvature persists though causally cut off from the original source. But this begs the question - if the source matter has effectively vanished wrt the exterior region, how can this fossil field persist there? As stated in #10, the inescapable implication surely is that the field acts as it's own source. And that's what makes sense to me, but then it has to be explained how there is no contribution whatsoever from field curvature in the stress-energy tensor, which in GR is the sole source of such curvature!

Now it has been stated that one must 'read between the lines'; meaning in effect that the non-linear character of gravity in GR is evidence 'gravity gravitates' there in some de facto sense. I'm not convinced of that reasoning. The mere fact that matter induces curvature demands non-linearity - rulers and clocks necessarily vary with distance from source, but this of itself does not imply any further contribution to curvature from that curvature. I don't particularly like citing authorities, but to back that point up, here's a quote from a Sascha Vongehr who at least styles himself as knowledgeable in GR:
"All you need to be clear about is that pressure is part of the pressure energy tensor, i.e. part of the physical source of what happens in general relativity, one side of the Einstein equation. Gravity is on the other side, and it is not a source. It does not somehow walk around to the other side and become pressure, it is plain curvature in space-time. It is non-linear and so on, so it looks to us like a force that self-interacts (= being its own source)." (last paragraph at: http://www.science20.com/comments/50735/add_gravitational)
Don't know whether that qualifies him as 'crackpot' but clearly he has studied the subject and came to that conclusion. Is there actually an overwhelming consensus position on 'gravity does/does not gravitate'?

IMO, the FAQ page's mention of virtual particles is a little misleading since it invites the interpretation that the BH field actually does come from inside the hole, but somehow virtual gravitons can escape because they can move faster than light. Even on a quantum view of gravity I don't think that's correct; I think the explanation I gave above still applies. (Which is not to say that there is not a sense in which virtual particles can travel faster than light; there is. I just don't think that sort of thing is needed to resolve the question of "how gravity gets out of the black hole"--it doesn't have to get out because that's not where it's actually coming from to begin with.)

I have stated elsewhere that if the virtual graviton picture is wrong as suggested, then it surely spells instant death to what is styled as the leading candidate for a TOE - string/superstring/M theory. Because in that theory gravity is described in terms of a spin-2 virtual graviton (and real gravitons for GW's). Unless I suppose vg's are their own source in that theory - I don't really know. That would place it in the 'gravity gravitates' camp. If so though, it gets back to why this is not correspondingly reflected in GR where 'officially' only non-gravitational energy contributes to curvature.

Regardless of the source of BH gravity itself, the issue of charged BH presents a headache for GR imo. Again as stated in #10, one either posit that a static electric (or magnetic) field can act as it's own source (and where would that leave Maxwell's linear eqn's and the definitions of EM fields wrt to potentials, which in turn are defined wrt source charge/current), or one posits something truly magical imo. Namely that infalling charged matter, already at or interior to the EH, can continue to act as a continual source of vp (virtual photon) exchange with the exterior regions - despite all temporal processes ceasing wrt the exterior. If I'm missing something obvious here, please enlighen.
Bet you wished already to have not bitten this one!

 

[PeterDonis]

I understand that by 'fossil field' it is meant that the static exterior spacetime curvature persists though causally cut off from the original source.

It appears that by "the original source" you mean the actual black hole region of the spacetime. But that is *not* the "original source" of the observed field in the exterior region. The "original source" is the collapsing object, while it was still collapsing, i.e., before the hole's event horizon was formed; in other words, it's matter with a nonzero stress-energy tensor in a region of spacetime that is *not* causally cut off from the event where the field is being observed.

if the source matter has effectively vanished wrt the exterior region, how can this fossil field persist there?

Because the "imprint" of the collapsing matter is propagated throughout the vacuum exterior region of the black hole spacetime.

As stated in #10, the inescapable implication surely is that the field acts as it's own source.

No, it doesn't have to. A "gravitational field", in the sense of a nonzero Riemann curvature tensor, can propagate through source-free space (space with a stress-energy tensor of zero), just as an electromagnetic field, in the sense of a nonzero Faraday tensor, can propagate through charge-free space (space with a charge-current density 4-vector of zero). Some people might say that this propagation means "the field acts as its own source", but that usage of the term "source" is imprecise; the precise meaning of "source" is the SET in GR, or the charge-current 4-vector in EM.

And that's what makes sense to me, but then it has to be explained how there is no contribution whatsoever from field curvature in the stress-energy tensor, which in GR is the sole source of such curvature!

No, it isn't. See above. The correct statement here is that the SET is the source of *Ricci* curvature; the Einstein tensor, which is what appears on the LHS of the Einstein Field Equation, is the Ricci tensor with a trace term added. But the "gravitational field" of a black hole, what you observe as the "field" in the exterior region, is entirely Weyl curvature; the Ricci tensor is zero (because it's a vacuum solution of the EFE), but the Weyl tensor is nonzero. The Weyl tensor doesn't appear at all in the EFE, so it can be nonzero in regions where the SET is zero.

Now it has been stated that one must 'read between the lines'; meaning in effect that the non-linear character of gravity in GR is evidence 'gravity gravitates' there in some de facto sense. I'm not convinced of that reasoning.

People can mean lots of different things by "gravity gravitates". The main thing this illustrates to me is that you have to be very careful reasoning from statements in ordinary English about GR (or indeed about any scientific theory). The true theory is the math, the equations and the precise predictions that they give rise to. Statements like "gravity gravitates" are attempts to capture some aspect of the math in ordinary English, but I would never want to hang my hat on an implication of such a statement without checking it against the actual math, to be sure my reasoning wasn't being led astray by a mismatch between the actual math and the approximation to it in English.

"All you need to be clear about is that pressure is part of the pressure energy tensor, i.e. part of the physical source of what happens in general relativity, one side of the Einstein equation. Gravity is on the other side, and it is not a source. It does not somehow walk around to the other side and become pressure, it is plain curvature in space-time.

This looks OK to me; I was saying the same sort of thing above when I talked about precisely defining the term "source".

It is non-linear and so on, so it looks to us like a force that self-interacts (= being its own source)." (last paragraph at: http://www.science20.com/comments/50735/add_gravitational)

People can also mean different things by "self-interacts". The precise meaning of that term really requires quantum theory; in quantum gravity, gravitons can interact with other gravitons directly, i.e., there are graviton-graviton vertices in the allowed set of Feynman diagrams. By contrast, quantum electrodynamics is not self-interacting: there are no direct photon-photon vertices.

I have stated elsewhere that if the virtual graviton picture is wrong as suggested, then it surely spells instant death to what is styled as the leading candidate for a TOE - string/superstring/M theory. Because in that theory gravity is described in terms of a spin-2 virtual graviton (and real gravitons for GW's). Unless I suppose vg's are their own source in that theory - I don't really know. That would place it in the 'gravity gravitates' camp. If so though, it gets back to why this is not correspondingly reflected in GR where 'officially' only non-gravitational energy contributes to curvature.

Just to be clear, I was not saying that the quantum picture of "virtual gravitons" was wrong per se; I was only saying that it was not necessary for understanding where the observed field around a black hole comes from. The general question of the validity of the "virtual graviton picture" is, IMO, a topic for a separate thread.

Regardless of the source of BH gravity itself, the issue of charged BH presents a headache for GR imo.

This question has the same answer as the answer for gravity: 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.

 

[Q-reeus]

Only if by "the original source" you mean the actual black hole region of the spacetime. But that is *not* the "original source" of the observed field in the exterior region. The "original source" is the collapsing object, while it was still collapsing, i.e., before the hole's event horizon was formed; in other words, it's matter with a nonzero stress-energy tensor in a region of spacetime that is *not* causally cut off from the event where the field is being observed.

Umm - first part above is past sense 'while it was still collapsing', but ends with a present sense 'where the field is being observed.' Still not seeing how the present - exterior field, is continuously sustained by the past - collapsing matter that is now at or behind an EH.

Q-reeus: "As stated in #10, the inescapable implication surely is that the field acts as it's own source."
No, it doesn't have to. A "gravitational field", in the sense of a nonzero Riemann curvature tensor, can propagate through source-free space (space with a stress-energy tensor of zero), just as an electromagnetic field, in the sense of a nonzero Faraday tensor, can propagate through charge-free space (space with a charge-current density 4-vector of zero).

I'm presuming that is referring primarily to static fields (radiation fields or time variation of fields in general are a non-issue here). If so then seems to me the ability to 'propagate' through source-free space means nothing more than the ability to exist at all in that spatial region, otherwise there is no meaning to the term field. What I'm having trouble with is the notion, unique to GR afaik, that a static field can exist apart from it's original source (and you have I think made it clear that source in GR cannot be curved spacetime itself).

...The correct statement here is that the SET is the source of *Ricci* curvature; the Einstein tensor, which is what appears on the LHS of the Einstein Field Equation, is the Ricci tensor with a trace term added. But the "gravitational field" of a black hole, what you observe as the "field" in the exterior region, is entirely Weyl curvature; the Ricci tensor is zero (because it's a vacuum solution of the EFE), but the Weyl tensor is nonzero. The Weyl tensor doesn't appear at all in the EFE, so it can be nonzero in regions where the SET is zero.

Now you have me really confused. From the wiki article on Weyl tensor at http://en.wikipedia.org/wiki/Weyl_tensor
"The Weyl tensor differs from the Riemann curvature tensor in that it does not convey information on how the volume of the body changes, but rather only how the shape of the body is distorted by the tidal force. The Ricci curvature, or trace component of the Riemann tensor contains precisely the information about how volumes change in the presence of tidal forces, so the Weyl tensor is the traceless component of the Riemann tensor."

That part is fine, and intuitively it's obvious that in a non-zero matter density region curvature will be different than in vacuum regions. Which has similarity to the case in EM where divergence is non-zero only where charge density is non-zero. But then:
"In general relativity, the Weyl curvature is the only part of the curvature that exists in free space — a solution of the vacuum Einstein equation — and it governs the propagation of gravitational radiation through regions of space devoid of matter. More generally, the Weyl curvature is the only component of curvature for Ricci-flat manifolds and always governs the characteristics of the field equations of an Einstein manifold."

What I get from that is that the exterior static field is a solution of the vacuum Einstein eq'n. So then, you are saying in effect that the latter is quite disconnected from the SET? But what then is the mass of a gravitating body other than an integration over the relevant SET for the matter region? And surely you cannot deny that for any static gravitating body, it's mass M determines the field.
Going back to the analogy with EM, while it's true divergence is zero exterior to a charged region, it's equally true the divergence-free field in that exterior region derives from the charge and nothing else! Yet it seems something totally different applies in GR. I need a clear statement: if the exterior, Weyl curvature field has no source in the SET (i.e., mass = integrated stress-energy-momnetum), what is it's source?

"All you need to be clear about is that pressure is part of the pressure energy tensor, i.e. part of the physical source of what happens in general relativity, one side of the Einstein equation. Gravity is on the other side, and it is not a source. It does not somehow walk around to the other side and become pressure, it is plain curvature in space-time."
This looks OK to me; I was saying the same sort of thing above when I talked about precisely defining the term "source".

Right but as per above I am still very unclear on what does constitute source. And I can't find it now but pretty sure Clifford Will is on record as stating that gravity is a source of further gravity. Not really arguing from authority, but it does seem there are diverging opinions in the GR community.

People can also mean different things by "self-interacts". The precise meaning of that term really requires quantum theory; in quantum gravity, gravitons can interact with other gravitons directly, i.e., there are graviton-graviton vertices in the allowed set of Feynman diagrams. By contrast, quantum electrodynamics is not self-interacting: there are no direct photon-photon vertices.

Interesting but mutual interaction suggests to me that 'gravitons gravitate', but know way too little here to press on that one.

Just to be clear, I was not saying that the quantum picture of "virtual gravitons" was wrong per se; I was only saying that it was not necessary for understanding where the observed field around a black hole comes from. The general question of the validity of the "virtual graviton picture" is, IMO, a topic for a separate thread.

Fair enough on that last point, but it has to be conceeded surely that without a 'gravitons gravitate' picture, field as vg's runs into real trouble in a BH scenario. As does EM field as vp exchange. And I'm not particularly arguing against vg's or vp's - I suspect the problem is with 'BH'.

Q-reeus: "Regardless of the source of BH gravity itself, the issue of charged BH presents a headache for GR imo."
This question has the same answer as the answer for gravity: 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.

That is a position statement, but as per my opening paragraph here I'm not seeing how it addresses the specific points raised in #10. If the EM field is not made somehow to be it's own source (but Maxwell's eqn's?!), how is it continuosly sustained? It just hangs there by fiat? Where does that leave the vp exchange picture popular in QFT? The only alternative I see is to posit a connection to the source charge/current. And that seems impossible given the temporal problem mentioned there.

So far this is how I see it. Your position is that both gravitational and electric fields exterior to a BH can be indefinitely sustained, based on past connection to mass/charge. But is that more than just bare postulate, albeit a standard one in GR? Without a position such as 'gravity gravitates', or an attempt to explain connection between exterior and interior 'original' source matter/charge, seems to me such fields are held up by mathematical fiat lacking any real underlying principle. My layman's prejudice perhaps.

 

[PeterDonis]

Umm - first part above is past sense 'while it was still collapsing', but ends with a present sense 'where the field is being observed.' Still not seeing how the present - exterior field, is continuously sustained by the past - collapsing matter that is now at or behind an EH.

You have to be careful using words like "present" in a BH spacetime; "present" according to whose time? But anyway, if you are thinking that the field somehow has to be "sustained" by the collapsed matter behind the EH, you are thinking of it wrong. Consider the case of an ordinary gravitating body that is *not* a BH, like the Sun. The field that the Earth experiences "now" due to the Sun is *not* "sustained" by the Sun "now"; it is determined by the way the Sun was eight minutes ago. In other words, it is determined by the way the Sun was in the Earth's past light cone.

The BH case is the same thing, except that, because the gravitating body has collapsed behind an EH, the past light cone of any event in the exterior region only includes the collapsing body *before* it fell behind the EH. That collapsing matter in the past, before it fell beneath the EH, is what corresponds to the Sun eight minutes ago for us on Earth. The collapsed matter inside the EH is irrelevant to the field observed in the exterior, just as the Sun's state "now" is irrelevant to the field we experience on Earth "now". Of course, in the case of the Sun, we will soon know what the Sun's state is "now", by observing the field eight minutes from now, whereas an observer in the exterior region will *never* know the state of the collapsing matter inside the EH, because he will never get an "update" from that region; but again, that is simply because the matter in the BH case has collapsed behind an EH, whereas the Sun has not.

I'm presuming that is referring primarily to static fields (radiation fields or time variation of fields in general are a non-issue here). If so then seems to me the ability to 'propagate' through source-free space means nothing more than the ability to exist at all in that spatial region, otherwise there is no meaning to the term field. What I'm having trouble with is the notion, unique to GR afaik, that a static field can exist apart from it's original source (and you have I think made it clear that source in GR cannot be curved spacetime itself).

Again, you are interpreting things incorrectly if you think the static field can exist apart from its source. The source is the collapsing matter; you just have to be careful about specifying what parts of the collapsing matter's worldline act as "source" for the field in the exterior region of the BH. There is no claim, in GR or anywhere else, that the field can exist with no "sources" anywhere. (Technically, an "eternal" BH spacetime does have vacuum everywhere, but it also has a white hole singularity, which is in the past light cone of any observer in the exterior region of the BH; that singularity effectively becomes the "source" of the exterior field. However, the "eternal" BH spacetime is unphysical, and I think it's much better for this discussion to consider the physically realistic case of a BH formed by collapsing matter.)

What I get from that is that the exterior static field is a solution of the vacuum Einstein eq'n. So then, you are saying in effect that the latter is quite disconnected from the SET?

No, I'm saying that the exterior static field will only exist in the exterior vacuum region if, somewhere in the past light cone of that region, there is a region of collapsing matter with a nonzero SET. The Weyl tensor at a given event in the exterior vacuum region is then determined by "propagation" (which may not be the best word here, but I don't know of any better one) of the field from the nonzero SET region in the past light cone of that event.

But what then is the mass of a gravitating body other than an integration over the relevant SET for the matter region?

That's one *component* of the externally observed mass, yes. But it's not the only one; if you just naively do the integral you have described, you will get the wrong answer. For example, if you just integrate the Sun's SET over its volume you will not get the Sun's actual observed mass; you will get a number that, roughly speaking, corresponds to the Sun's actual observed mass plus its gravitational binding energy, the energy it would take to "disassemble" the Sun and move all of its parts to spatial infinity, so they were no longer gravitationally bound to each other. See, for example, this Wiki page:

http://en.wikipedia.org/wiki/Mass_in_general_relativity

(See the section on the Newtonian limit for nearly flat spacetimes.)

For a BH spacetime, you can't assume that the spacetime is "nearly flat", so the Newtonian limit doesn't apply. But the more general methods discussed in the Wiki article (ADM mass, Komar mass, etc.) still work, because they only require the spacetime to be stationary and/or asymptotically flat, and a BH spacetime meets both of those requirements. But you'll notice that none of those methods require a nonzero SET! Basically, in a stationary and/or asymptotically flat spacetime, you can come up with a workable definition of "energy stored in the gravitational field", which cannot be done in a generic spacetime without those special properties. It then turns out that a BH's mass is *entirely* composed of "energy stored in the gravitational field".

Right but as per above I am still very unclear on what does constitute source. And I can't find it now but pretty sure Clifford Will is on record as stating that gravity is a source of further gravity. Not really arguing from authority, but it does seem there are diverging opinions in the GR community.

Generally, I would expect statements like the one you refer to from Will to be talking about the fact that gravitational waves carry energy. (The graviton-graviton interactions I talked about before are the quantum version of this.) Since they carry energy, they can also gravitate. But since gravitational waves can carry energy through a region of zero SET, figuring out *how* they gravitate is not straightforward. You end up having to work backwards to the original source of the waves, which must be a body or a system of bodies with nonzero SET that is vibrating or oscillating in some way. (For example, a binary pulsar system.)

Interesting but mutual interaction suggests to me that 'gravitons gravitate'

They do. See just above.

Fair enough on that last point, but it has to be conceeded surely that without a 'gravitons gravitate' picture, field as vg's runs into real trouble in a BH scenario. As does EM field as vp exchange. And I'm not particularly arguing against vg's or vp's - I suspect the problem is with 'BH'.

No, the problem is your apparent assumption that, at the quantum level, viewing a static field as due to virtual particle exchange is the only option. It isn't. But again, I think that's a topic for a separate thread.

If the EM field is not made somehow to be it's own source (but Maxwell's eqn's?!), how is it continuosly sustained?

Same answer as for gravity above, as I said. The EM field at a given event is ultimately due to a source--a region of nonzero charge-current density--somewhere in the past light cone of that event. The field "propagates" (again, not the best word IMO but I don't have a better one) from that source to the event where the field is observed.

Your position is that both gravitational and electric fields exterior to a BH can be indefinitely sustained, based on past connection to mass/charge.

Your use of the word "sustained" is an indication of the conceptual problem you are having. You are thinking of the field as a two-way interaction between "source" and "observer". It isn't. The field is just "propagated" in one direction--from the source in the observer's past light cone, to the observer (more precisely, to a specific event at which the observer measures the field). That can be explained entirely in terms of the model I have given. Nothing has to propagate back from the observer to the source.

 

[Q-reeus]

...But anyway, if you are thinking that the field somehow has to be "sustained" by the collapsed matter behind the EH, you are thinking of it wrong.

Thought I made my position clear enough in #25: "As stated in #10, the inescapable implication surely is that the field acts as it's own source. And that's what makes sense to me,..." All along have pointed out what I see as the impossibility of connection between exterior and interior regions.

Consider the case of an ordinary gravitating body that is *not* a BH, like the Sun. The field that the Earth experiences "now" due to the Sun is *not* "sustained" by the Sun "now"; it is determined by the way the Sun was eight minutes ago. In other words, it is determined by the way the Sun was in the Earth's past light cone.

Sure, but once established, there is in that situation a continuous exchange process occurring based on a QFT (charge) or string (gravity also) model. One that can't sensibly apply if source is behind a BH EH.

Q-reeus: "...What I'm having trouble with is the notion, unique to GR afaik, that a static field can exist apart from it's original source (and you have I think made it clear that source in GR cannot be curved spacetime itself)."
Again, you are interpreting things incorrectly if you think the static field can exist apart from its source.

No, as per earlier comment, I've argued the opposite (particularly in the case of charge), but believe that for consistency gravity should be part of it's own source.

Q-reeus: "But what then is the mass of a gravitating body other than an integration over the relevant SET for the matter region?"
That's one *component* of the externally observed mass, yes. But it's not the only one; if you just naively do the integral you have described, you will get the wrong answer.

Agreed - but then I was arguing there assuming a position based on certain of your statements made earlier - and I will quote you on them later here!

For example, if you just integrate the Sun's SET over its volume you will not get the Sun's actual observed mass; you will get a number that, roughly speaking, corresponds to the Sun's actual observed mass plus its gravitational binding energy, the energy it would take to "disassemble" the Sun and move all of its parts to spatial infinity, so they were no longer gravitationally bound to each other.

Warming up, not quite there yet. But wait!:

...Basically, in a stationary and/or asymptotically flat spacetime, you can come up with a workable definition of "energy stored in the gravitational field", which cannot be done in a generic spacetime without those special properties. It then turns out that a BH's mass is *entirely* composed of "energy stored in the gravitational field".

Oh wow! This looks awfully like at least a substantial convergence of viewpoint, but it leaves me baffled. Said above I would quote you on earlier statement(s). Here they are, both from #26:

Q-reeus: "As stated in #10, the inescapable implication surely is that the field acts as it's own source." (referred specifically to a BH scenario)
No, it doesn't have to. A "gravitational field", in the sense of a nonzero Riemann curvature tensor, can propagate through source-free space (space with a stress-energy tensor of zero), just as an electromagnetic field, in the sense of a nonzero Faraday tensor, can propagate through charge-free space (space with a charge-current density 4-vector of zero). Some people might say that this propagation means "the field acts as its own source", but that usage of the term "source" is imprecise; the precise meaning of "source" is the SET in GR, or the charge-current 4-vector in EM.

Q-reeus: "All you need to be clear about is that pressure is part of the pressure energy tensor, i.e. part of the physical source of what happens in general relativity, one side of the Einstein equation. Gravity is on the other side, and it is not a source. It does not somehow walk around to the other side and become pressure, it is plain curvature in space-time." (I was quoting someone else there)
This looks OK to me; I was saying the same sort of thing above when I talked about precisely defining the term "source".

Pardon my confusion Peter, but am having difficulty reconciling those quotes with what you now say: "It then turns out that a BH's mass is *entirely* composed of "energy stored in the gravitational field"." So surely to goodness gracious this is saying 'gravity gravitates' after all?

Q-reeus: "Interesting but mutual interaction suggests to me that 'gravitons gravitate'"
They do. See just above.

Fine - but - oh well - fine.

Q-reeus: ...but it has to be conceeded surely that without a 'gravitons gravitate' picture, field as vg's runs into real trouble in a BH scenario. As does EM field as vp exchange. And I'm not particularly arguing against vg's or vp's - I suspect the problem is with 'BH'.
No, the problem is your apparent assumption that, at the quantum level, viewing a static field as due to virtual particle exchange is the only option. It isn't. But again, I think that's a topic for a separate thread.

But if one accepts that QFT is fundamentally where it's at viz a viz EM, it has to be the only option! Realize many dismiss vp's as a perturbative mathematical trick, but is there any QFT expert who can explain EM in purely classical terms? Exchange processes are afaik absolutely integral to interactions in general in any quantum theory. If that can be made to work for a BH fossil E field, this would be real news to me. We seem to agree that the 'explanation' given in the link quoted in #10 just isn't. But what is?

Q-reeus: "If the EM field is not made somehow to be it's own source (but Maxwell's eqn's?!), how is it continuously sustained?"
Same answer as for gravity above, as I said. The EM field at a given event is ultimately due to a source--a region of nonzero charge-current density--somewhere in the past light cone of that event. The field "propagates" (again, not the best word IMO but I don't have a better one) from that source to the event where the field is observed.

As per previous remark, there is imo a particularly severe problem explaining a 'fossil E field' using a vp exchange (or equivalent in QFT). Unlike with 'gravity gravitates', for sure charge is not generated by an E field, classically or in QFT. If you insist the past light cone as source does the job, say goodbye to QFT!

Your use of the word "sustained" is an indication of the conceptual problem you are having. You are thinking of the field as a two-way interaction between "source" and "observer". It isn't. The field is just "propagated" in one direction--from the source in the observer's past light cone, to the observer (more precisely, to a specific event at which the observer measures the field). That can be explained entirely in terms of the model I have given. Nothing has to propagate back from the observer to the source.

If only QFT agreed with that, but I think not. Certainly not in the case of 'static' E/B fields. Still I cautiously think we have come to a common understanding on the matter of whether a gravitational field can be a further source of gravity. Yes?

 

[PeterDonis]

"As stated in #10, the inescapable implication surely is that the field acts as it's own source.

No, as I have said several times already, this apparent "implication" is wrong, and I've explained why: the "source" is in the past light cone of the event where the "field" is being measured. See below.

Sure, but once established, there is in that situation a continuous exchange process occurring based on a QFT (charge) or string (gravity also) model.

And I have said that this apparent "continuous exchange" is also wrong. There is no continuous exchange; more precisely, no "continuous exchange" model is required to explain the observed field. Models of virtual particle exchange may be useful, but that doesn't mean they are required; you can't take what they appear to be telling you about "reality" as absolute truth. See further comments below.

No, as per earlier comment, I've argued the opposite (particularly in the case of charge), but believe that for consistency gravity should be part of it's own source.

Not if "source" is defined in the precise way I defined it.

Pardon my confusion Peter, but am having difficulty reconciling those quotes with what you now say: "It then turns out that a BH's mass is *entirely* composed of "energy stored in the gravitational field"." So surely to goodness gracious this is saying 'gravity gravitates' after all?

No, at least not the way you mean it. I did warn you about trying to reason about these things, and draw conclusions from statements about them, using English instead of math. However, I can point out one thing in English that may help to resolve your confusion: even if energy is "stored in the gravitational field", it still needs to have got there from somewhere. The "mass" of a BH ultimately comes from the stress-energy of the object that collapsed to form it; but once the BH is formed, its mass can be viewed as being "stored in the gravitational field" that is left behind once the collapsing object has hit the singularity at r = 0 and vanished.

But even the above English version has problems (one is that I have implicitly assumed a notion of time in several places in the above, without saying what notion of time it is). Ultimately, if you want to be sure you are understanding things correctly, you *have* to look at the math. The best summation of the math that I can give in English is this:

(1) The observed "field" at any given event is entirely determined, ultimately, by what "sources" are in the past light cone of that event;

(2) A "source", as used in #1 above, is a region with a non-zero stress-energy tensor.

If you find yourself asking questions, in English, that don't seem to be addressed by the above, then that's probably a sign that English is leading you astray. "Does gravity gravitate?" is, IMO, one such question; see further comments below.

But if one accepts that QFT is fundamentally where it's at viz a viz EM, it has to be the only option!

You are assuming that "QFT is the only option" is equivalent to "virtual particles are the only option". That's not the case. As I keep saying, this should be a separate thread, but briefly, "virtual particles" are an aspect of one particular way of treating certain problems in QFT, based on perturbation theory, using approximations appropriate for those types of problems; they are not "fundamental" parts of QFT that must be accounted for in an explanation of quantum "reality".

We seem to agree that the 'explanation' given in the link quoted in #10 just isn't. But what is?

As I said before, classically, the EM field of a charged BH is explained the same way as the gravity field: at any given event, the EM field observed at that event is ultimately due to the presence of EM sources in the past light cone of that event. An EM "source" here is a region with a nonzero charge-current density 4-vector.

The plain answer to how this picture is expanded to account for quantum phenomena is that nobody knows for sure. We don't have a theory of quantum gravity. We do know how to do QFT in a curved spacetime, where the spacetime itself is treated classically (but possibly with quantum "back reaction" terms added to the stress-energy tensor); in that picture, causality still holds and the explanation I gave above of fields being determined by sources in the past light cone, is still essentially correct (both for gravity itself and for the EM field of a charged BH). The only thing that QFT in curved spacetime really adds to the classical picture at this level is Hawking radiation: black holes can slowly evaporate. But that doesn't change anything I said above.

My personal opinion is that, even when we do have a good theory of quantum gravity, its classical limit will still be GR, so it will still look like the picture I described above. The only way a complete quantum gravity theory could change that picture, IMO, would be if we somehow discovered that BH's can't exist at all: that there is some large quantum correction to the classical behavior that prevents event horizons from ever forming in the first place. But that would take away the problem we are discussing here altogether.

If only QFT agreed with that, but I think not. Certainly not in the case of 'static' E/B fields.

No; see above. QFT, at least as we have developed it so far, does not contradict the picture of fields being determined by sources in the past light cone. See my comments above about virtual particles; virtual particle exchange is not the only way to think about static fields in a quantum context, and in some cases (such as this one), the virtual particle view can be misleading.

Still I cautiously think we have come to a common understanding on the matter of whether a gravitational field can be a further source of gravity. Yes?

It depends on what you mean. See my caution above about trying to use English to describe this stuff instead of math. I don't think the question "does gravity gravitate?" is a good one to ask, because it's not a good translation of the math into English in the first place. For the precise definition of the term "source", which I gave above, the plain answer to the question above as you asked it is "no": a "gravitational field" by itself can exist in a region with a zero stress-energy tensor, so it can't be a "source".

 

[Q-reeus]

Peter; on the matter of BH E field source. You at least have a consistent position, in the sense of not changing position from one entry to another. I fundamentally disagree with your viewpoint so there can be no headway and best we drop that part here. I have something in mind for a new thread attacking it all differently, but later. Now on the issue of whether gravity is a source of further gravity, I cannot see a consistent position, even though in any given response you come across as presenting one, when I check against other statements I'm getting a mixed picture. On the one hand, there is this from #28:
1:

Basically, in a stationary and/or asymptotically flat spacetime, you can come up with a workable definition of "energy stored in the gravitational field", which cannot be done in a generic spacetime without those special properties. It then turns out that a BH's mass is *entirely* composed of "energy stored in the gravitational field".

2:

Q-reeus: "Right but as per above I am still very unclear on what does constitute source. And I can't find it now but pretty sure Clifford Will is on record as stating that gravity is a source of further gravity. Not really arguing from authority, but it does seem there are diverging opinions in the GR community."
Generally, I would expect statements like the one you refer to from Will to be talking about the fact that gravitational waves carry energy. (The graviton-graviton interactions I talked about before are the quantum version of this.) Since they carry energy, they can also gravitate.

3:

Q-reeus: "Interesting but mutual interaction suggests to me that 'gravitons gravitate'"
They do. See just above.

Sure seems crystal clear that 1,2,3 here all say in essence the same thing. That a gravitational field, whether static ('virtual gravitons'), or radiative ('real gravitons'), carries energy, and *therefore* gravitates (acts as a source of further gravity). Plain english perfectly adequate at this level. And 1: is specific - "a BH's mass is *entirely* composed of "energy stored in the gravitational field"." A plain english statement that the field must here entirely be it's own source. And yet you will probably say no!
But then this from #30:

Q-reeus: "As stated in #10, the inescapable implication surely is that the field acts as it's own source."
No, as I have said several times already, this apparent "implication" is wrong, and I've explained why: the "source" is in the past light cone of the event where the "field" is being measured.

My own plain english attempt to sensibly synthesize the above would be to say that 'the "source" is in the past light cone' has to be *synonymous* with "a BH's mass is *entirely* composed of "energy stored in the gravitational field"." And I note; that field is all exterior to the EH and accessible in the here and 'now'. And as far as your repeated comments that expressing this in english is leading me astray, i would respond that plain english staements regarding conceptual basis take precedence everytime over just learnig a mathematical framework that may have a suspect conceptual basis.

If I didn't know better, could swear you're out to drive me insane. Saw both movie versions of Nineteen Eighty Four a while ago. Recommended viewing. In the end, O'brien the interrogator breaks poor old Winston Smith, who, with tears of joy, truly believes that when four fingers are held up to his face, there are really five. I have no tears of joy, just a frown. A little melodramatic, but what I'm saying is, please explain what seems to me are irreconcilable statements.

 

[Dale]

Regarding the rest, the fact that you are aware of the previous responses on this topic and consider them all unsatisfactory seems to indicate that further discussion will probably lead to the same result.

Only active and willing participation this time round can decide that. I'm certainly open to serious feedback.

The preliminary data doesn't contradict my hypothesis.

 

[Q-reeus]

The preliminary data doesn't contradict my hypothesis.

Possibly, but it's not over yet. If you have some personal insights feel free to share. And just for your quip, I'm inspired to remind you of a long outstanding committment: https://www.physicsforums.com/showpost.php?p=3567285&postcount=42
Prepared to make good on that one - like 'soon'?

 

[Dale]

Possibly, but it's not over yet. If you have some personal insights feel free to share.

Correct, it is not over yet, which is why I characterized the data as "preliminary". Nevertheless, the preliminary data suggests that sharing my personal insights will not make any difference. I will continue to monitor the data and, should my hypothesis be falsified, I would be willing to make the effort in the future.

I'm inspired to remind you of a long outstanding committment: https://www.physicsforums.com/showpost.php?p=3567285&postcount=42
Prepared to make good on that one - like 'soon'?

Yeah, you are justified in that, I definitely overpromised. I worked on it a few days, got stuck, and dropped it. The difficulty was that the matter distribution involves some discontinuities, which makes things messy both analytically and numerically.

 

[Q-reeus]

Correct, it is not over yet, which is why I characterized the data as "preliminary". Nevertheless, the preliminary data suggests that sharing my personal insights will not make any difference. I will continue to monitor the data and, should my hypothesis be falsified, I would be willing to make the effort in the future.

Splendidly diplomatic. Consider an overseas embassy posting.

Yeah, you are justified in that, I definitely overpromised. I worked on it a few days, got stuck, and dropped it. The difficulty was that the matter distribution involves some discontinuities, which makes things messy both analytically and numerically.

OK then won't press any further, at least this squares the ledger in a way.

 

[Dale]

Splendidly diplomatic. Consider an overseas embassy posting.

Thanks!

OK then won't press any further, at least this squares the ledger in a way.

I put a similar comment in the original thread also, so it would be easier to find.

 

[Q-reeus]

Thanks!

You're welcome.

I put a similar comment in the original thread also, so it would be easier to find.

So I've noticed. A nice way to finish there.

 

[PeterDonis]

Peter; on the matter of BH E field source. You at least have a consistent position, in the sense of not changing position from one entry to another. I fundamentally disagree with your viewpoint so there can be no headway and best we drop that part here. I have something in mind for a new thread attacking it all differently, but later.

I agree that that topic should be a separate thread.

My own plain english attempt to sensibly synthesize the above would be to say that 'the "source" is in the past light cone' has to be *synonymous* with "a BH's mass is *entirely* composed of "energy stored in the gravitational field"."

No, it doesn't; but here again is an example of the pitfalls of trying to use English to talk about this stuff. The English word "mass" as used in the latter sentence (the BH's "mass" can be viewed as being entirely composed of energy stored in the gravitational field) does *not* imply that the BH "mass" is synonymous with "the source of the BH's gravitational field". The distinction would be a lot clearer if we were using math to discuss this.

And as far as your repeated comments that expressing this in english is leading me astray, i would respond that plain english staements regarding conceptual basis take precedence everytime over just learnig a mathematical framework that may have a suspect conceptual basis.

But how do you describe the conceptual basis? To do that without ambiguity requires math as well. Or at least it requires something besides plain, ordinary English: it requires English with precise definitions of words as technical terms, even if the meanings thereby become different from their ordinary meanings. Ordinary English is not a precise language, so you can't use it "as is" to talk precisely about concepts. You have to add the precision somehow.

Let's go back to the two statements that I said were the best I could do at summing up the math in English:

(1) The observed "field" at any given event is entirely determined, ultimately, by what "sources" are in the past light cone of that event;

(2) A "source", as used in #1 above, is a region with a non-zero stress-energy tensor.

To make this really precise, I would have to define what "the observed field" means. There are actually at least two potential candidates. One is the Riemann curvature tensor; the other is the metric. I'll use the metric because from it you can derive the Riemann curvature tensor, as well as all the other quantities that are sometimes talked about as being "the field" (for example, the acceleration required to "hover" at a constant radial coordinate r above the hole). So (1) and (2) together really say that the metric at any given event is determined entirely by what regions of nonzero SET are in the past light cone of that event.

You will notice that I nowhere mentioned the BH's "mass" in the above. It is true that there is a quantity called "M" in the metric, which turns out to be the externally observed "mass" of the hole, in the sense that it's the mass you would come up with if you put objects in orbit about the hole, measured their orbital parameters, and applied Kepler's third law. But doing that does not require making any statements about "where the mass is located", or "how the mass is stored", or anything like that. Ultimately you are measuring the metric, since the quantity "M" is part of the metric; and we've already seen how the metric is determined.

So the question "where is the BH's mass located?" or similar questions, are like the question "does gravity gravitate?" They're not actually questions about the physics; they're questions prompted by attempting to capture the physics in English, and being led astray by the imprecision of English in doing so. I'm vulnerable to this too, which is why I originally tried to actually give an answer to your question about how the "mass" of the BH is determined. But if you go back and read the follow-up discussion, you will see that I quickly added caveats; in particular, right before I gave the statements (1) and (2) which I gave again above, I explicitly said there were "problems" with the view of the BH's mass that I had just given. As I said then, the statements (1) and (2) (with the clarification I gave above) are the best I can do at summing up the actual physics in English; if you find yourself asking a question, in English, that can't be answered by looking at those two statements, it's probably a sign that English is leading you astray. All of what I've said about this should be taken in that light.

If I didn't know better, could swear you're out to drive me insane.

I assure you that that is not an intended effect.

 

[PeterDonis]

Following up from my previous post:

Sure seems crystal clear that 1,2,3 here all say in essence the same thing. That a gravitational field, whether static ('virtual gravitons'), or radiative ('real gravitons'), carries energy, and *therefore* gravitates (acts as a source of further gravity). Plain english perfectly adequate at this level. And 1: is specific - "a BH's mass is *entirely* composed of "energy stored in the gravitational field"." A plain english statement that the field must here entirely be it's own source. And yet you will probably say no!

You're right, I do say no. The above does not follow from the statements (1) and (2) that I said were the best I could do at summing up the physics in English. There's nothing in those statements about the field "carrying energy" or about whether it "gravitates". And I talked in my last post about the problems with asking things like "where the BH's mass is located" or "how the BH's mass is stored".

Basically, you are focusing on the parts of my posts that I have explicitly said were problematic because of the limitations of English, and you are not looking hard enough at the statements that I have explicitly said are the best ones to use if you are trying to sum up the physics in English. I would recommend reversing your approach.

 

[Q-reeus]

Let's go back to the two statements that I said were the best I could do at summing up the math in English:

(1) The observed "field" at any given event is entirely determined, ultimately, by what "sources" are in the past light cone of that event;

(2) A "source", as used in #1 above, is a region with a non-zero stress-energy tensor.

To make this really precise, I would have to define what "the observed field" means. There are actually at least two potential candidates. One is the Riemann curvature tensor; the other is the metric. I'll use the metric because from it you can derive the Riemann curvature tensor, as well as all the other quantities that are sometimes talked about as being "the field" (for example, the acceleration required to "hover" at a constant radial coordinate r above the hole). So (1) and (2) together really say that the metric at any given event is determined entirely by what regions of nonzero SET are in the past light cone of that event.

Let's just run with the above then. Step away from the problematic BH scenario and consider a typical situation of say a spherically symmetric non-rotating planet of mass M, static wrt some external hovering observer. We agree I hope that here 'past light cone' is a trivial consideration as there is no time variation and M as SET source, and field at observation point, are manifestly part of the same spacetime manifold, in the same ordinary sense that source charge and resultant Coulombic field in electrostatics are. So the base question is, what does or does not contribute to the SET here? I have seen repeated statements from various authorities (but not all) that it's everything *except* gravitational energy-momentum. Do you agree with that?

Q-reeus: "If I didn't know better, could swear you're out to drive me insane."
I assure you that that is not an intended effect.

Not that I had any real fear, but nice to be reassured!

 

[Naty1]

Q-reeus posts #18

...Even assuming it were possible for an ultra-relativistic elementary particle to collapse nearby matter into a BH state, by reciprocity of reference frame, it would of necessity be a Kamikaze affair ending in mutual 'BH' creation ...

exactly. that is why I keep posting this 'trick', already posted in this thread and introduced to me in another thread by someone else:

from my post #3:

Any situation where you ask about a rapidly moving massive body's gravitational effect and a 'stationary' observer can be transformed to an equivalent question about the interaction between a rapidly moving observer and a 'stationary' massive body. So all observations relating to a rapidly moving massive body can be answered as if the body is stationary...as if all measures are local. Local measures trump distant measures.

You can easily expand this concept to a local inertial frame of a group of particles where local energy and momentum would contribute to gravitational effects, say the components of an atom, or a hot gas or two colliding particles.

pervect #17

There is nothing in this paper that says that a single particle moving at a high velocity will become a black hole.

It does say that a pair of particles colliding can become a black hole (which is true) - but it does not say that a single particle moving at high speed will become a black hole - because this statement, as previously mentioned, is false.

Relative to some observers, you are right now moving at 99.999999% of the speed of light. But you are not a black hole. Not to yourself, and not to the observer at which you are moving at such a high velocity - because being a black hole is frame indepenent.

Nice summary! [that last paragraph makes it into my notes!]


Q-reeus: post #20 :

Don't want to appear sulky about that, but honestly,...

Dalespam:

...the lack of interest is simple exhaustion on the topic, which is a temporary condition but present now.

BRAVO! much better than all the sniping that too often abounds. Had me LOL.

Sam Gralla:

If the "observer" is a real observer and has any finite amount of mass, then a particle moving very fast by it is in fact a particle collision, and a black hole may very well be formed. But I repeat myself.

Good luck with that hypothesis! Never going to happen. Good thing it can't happen that easily.


PeterDonis: post#28

...Technically, an "eternal" BH spacetime does have vacuum everywhere, but it also has a white hole singularity, which is in the past light cone of any observer in the exterior region of the BH; that singularity effectively becomes the "source" of the exterior field... Basically, in a stationary and/or asymptotically flat spacetime, you can come up with a workable definition of "energy stored in the gravitational field", ... It then turns out that a BH's mass is *entirely* composed of "energy stored in the gravitational field"…

Great explanation... why do you consider the white hole version 'unphysical'...relative to the second part?


from Q-reeus: post #31

... Now on the issue of whether gravity is a source of further gravity,

I can't find it now but pretty sure Clifford Will is on record as stating that gravity is a source of further gravity...

I have read something similar, have been unable to locate the source and the exact wording...and never understood what was intended. I think the gist of it was that gravity interacts with itself in a way that the EM field doesn’t….and the nature of that gravitational ‘self ineraction’ is captured within the Einstein mathematics.
Doesn’t a portion of the Einstein stress energy tensor capture the effects of an EM field, like T00 here:

http://en.wikipedia.org/wiki/Stress-energy_tensor#Identifying_the_components_of_the_tensor

If so, the effects of 'self interaction' would likely lie within the remained of the Einstein stress
energy tensor...anybody know what I am trying to describe?

edit: the analogy I thought to myself at the time was that maybe gravitons self ineract in a way photons don'ts...but the original description was a classical one, not quantum.

 

[Q-reeus]

Q-reeus: post #20 : Don't want to appear sulky about that, but honestly,...
Dalespam:...the lack of interest is simple exhaustion on the topic, which is a temporary condition but present now.

BRAVO! much better than all the sniping that too often abounds. Had me LOL.

Right, but occasionally beating a drum oft and loud gets results - just use that approach sparingly!

Q-reeus: #31 ...pretty sure Clifford Will is on record as stating that gravity is a source of further gravity...

I have read something similar, have been unable to locate the source and the exact wording...and never understood what was intended. I think the gist of it was that gravity interacts with itself in a way that the EM field doesn’t….and the nature of that gravitational ‘self ineraction’ is captured within the Einstein mathematics.
Doesn’t a portion of the Einstein stress energy tensor capture the effects of an EM field, like T00 here:
http://en.wikipedia.org/wiki/Stress-energy_tensor#Identifying_the_components_of_the_tensor
If so, the effects of 'self interaction' would likely lie within the remained of the Einstein stress-energy tensor...anybody know what I am trying to describe?

Yes to the very last bit but it seems evidently no to the first. From the first paragraph in that Wiki link:

The stress–energy tensor...is an attribute of matter, radiation, and non-gravitational force fields. The stress-energy tensor is the source of the gravitational field in the Einstein field equations of general relativity, just as mass is the source of such a field in Newtonian gravity.

Yet comments made in earlier posts here suggest (on my reading) otherwise - this is still being thrashed out. Stay tuned.

 

[PeterDonis]

So the base question is, what does or does not contribute to the SET here? I have seen repeated statements from various authorities (but not all) that it's everything *except* gravitational energy-momentum. Do you agree with that?

Yes. For the region of spacetime occupied by the planet, the SET is determined by the matter of the planet. In the simplest case where we can model the planet as a perfect fluid in hydrostatic equilibrium, the planet's mass density and pressure are the only things that contribute to the SET.

However, bear in mind that the SET is only nonzero in the region of spacetime occupied by the planet. In the region exterior to the planet, including the point where the "field" is being measured, the SET is zero--the exterior region is a vacuum. So to determine the "field" at an event in the exterior region, you have to do two things, as I've been saying:

First, determine what "sources" (regions of nonzero SET) are in the past light cone of that event (in this case, that would be the intersection of the planet's "world-tube", the region of spacetime occupied by the planet, and the event's past light cone). I described this above.

Second, determine how the spacetime curvature produced by those sources "propagates" through the vacuum to the event at which the field is being measured. (I put "propagates" in quotes because there are no actual gravitational waves or other measurable "signals" propagating--the spacetime is stationary; actually static in the simplest case where the planet has no net electric charge). You can derive this by solving the vacuum Einstein Field Equation in the exterior region and deriving whatever "field" quantities you are interested in from the resulting metric.

 

[PeterDonis]

PeterDonis: post#28

Great explanation... why do you consider the white hole version 'unphysical'...relative to the second part?

In a spacetime where a black hole is formed by gravitational collapse, the white hole region does not exist: the only vacuum regions of the spacetime are region I, the exterior, and region II, the interior of the black hole (behind the future horizon). The rest of the spacetime is the non-vacuum region occupied by the collapsing matter.

Since the above is the only known physical mechanism for forming a black hole, the white hole would appear to be unphysical. I know there are speculations about how the white hole solution might be physically useful; see, for example, the Wiki page:

http://en.wikipedia.org/wiki/White_hole

However, these are just speculations; we'll have to wait and see if any of them pan out.

 

[Q-reeus]

Yes. For the region of spacetime occupied by the planet, the SET is determined by the matter of the planet. In the simplest case where we can model the planet as a perfect fluid in hydrostatic equilibrium, the planet's mass density and pressure are the only things that contribute to the SET.

That clarifies things.

However, bear in mind that the SET is only nonzero in the region of spacetime occupied by the planet.

By definition from above. But doesn't this seem strange in principle? As has been acknowledged earlier there is energy in the curvature/field (both within and outside the matter region), so that position inevitably breaks that all forms of stress-energy should contribute to curvature. Or, if one holds the latter is in fact observed, there must be zero energy density in a gravitational field. Where would the latter leave e.g. Hulse-Taylor binary pulsar orbital decay data as proof of GW's? But it's easy to show energy must be in the field.

Consider the case of dispersed matter of total mass M brought 'from infinity' and assembled as a spherical thin shell of mean radius R. Let the Newtonian potential V = -M/R (with G=c=1) be small so 'linear gravity' applies, and assumes pressure is negligible. Let the original matter consisting of a large number N of identical particles conserve N during assembly so overall mass-energy is given off purely as heat that radiates away totally. If subsequently one constituent matter particle self-annihilates somehow and radiates to infinity, to a very good approximation that radiation has been frequency redshifted by a factor f = (1+2V)^1/2. The shell now of N-1 particles has lost an overall mass of essentially fM/N. The depressed mass of each particle (before that single particle annihilation and exit) was thus fM/N. This must be fractionally considerably smaller though than given by dividing the assembled mass M' by N, since if we were to continue annihilating particles until all have gone, redshift factor f progressively grows, finishing at essentially unity. Which coincides when worked out, with the assembled mass M' being M' = M(1+f)/2 (approx), which exceeds fM. We must conclude the gravitational field contributes a positive amount that balances the ledger (or abandon conservation of energy!). There must at least be energy in a gravitational field.

So I conclude that GR posits a fundamental break - gravitational energy is exempt from the less than universal requirement that all forms of stress-energy contribute to curvature. Doesn't seem right.

 

[PeterDonis]

But doesn't this seem strange in principle?

Not to me; but one does have to be precise about defining terms like "energy"--that English vs. math thing again.

As has been acknowledged earlier there is energy in the curvature/field

In a sense, yes, there is; but it's not "energy" in the sense of "non-zero stress-energy tensor". There is a lot of literature on the issue of "energy in the gravitational field", how it can be defined (there is no one unique way of defining it), how it figures into "energy conservation" (see further comments below on that), the fact that it can't be localized the way "energy" in the sense of a non-zero SET can, etc.

One example of a definition of "energy in the gravitational field" is described here:

http://en.wikipedia.org/wiki/Stress–energy–momentum_pseudotensor

However, as I said above, I don't believe this is the only definition in the literature. (Other GR experts here may know more about this.)

The key point, though, is that "energy in the gravitational field" does *not* act as a "source" of gravity according to the Einstein Field Equation (because only a nonzero SET does that). In other words, the "energy in the gravitational field" is *not* "stress-energy". So:

that position inevitably breaks that all forms of stress-energy should contribute to curvature.

This is wrong--all forms of *stress-energy* do contribute to curvature (by acting as a source in the EFE), but "energy in the gravitational field" is *not* "stress-energy" in this sense.

This may seem like playing with words, but let's consider the (very good) examples you bring up:

Where would the latter leave e.g. Hulse-Taylor binary pulsar orbital decay data as proof of GW's? But it's easy to show energy must be in the field.

The binary pulsar is indeed a good example of a system which is losing energy that apparently can only be carried by "the field"--specifically, gravitational waves. The fact that the system is losing energy is well documented by the observed change in orbital parameters; the fact that the energy lost can only be carried by gravitational waves is shown by the absence of any other observed energy coming out of the system of the right order of magnitude (the system of course radiates EM waves as well, but AFAIK their intensity is nowhere near large enough to explain the change in orbital parameters).

However, the gravitational waves emitted by the binary pulsar are *not* a "source" of gravity, for the same reason that EM waves are not sources of electromagnetism. EM waves have zero charge, and gravitational waves have zero stress-energy. The waves can carry energy from a "source" (the binary pulsar) to a "sink" (a gravitational wave detector, for example, if we ever succeed in detecting them), but they themselves do not produce any curvature--they *are* curvature, propagated from one region of spacetime to another purely by the properties of spacetime itself, without any "source" present.

Now let's look at your second example:

Consider the case of dispersed matter of total mass M brought 'from infinity' and assembled as a spherical thin shell of mean radius R. Let the Newtonian potential V = -M/R (with G=c=1) be small so 'linear gravity' applies, and assumes pressure is negligible.

Thin shells can be somewhat difficult to handle (I believe we had a thread about this some time back...) To make the scenario simpler, I would "assemble" the dispersed matter into a sphere in hydrostatic equilibrium, such as a planet; properly chosen values for the number and rest mass of the particles can ensure that the equilibrium is stable with negligible pressure compared to the energy density, and that the object will not collapse to a black hole.

Let the original matter consisting of a large number N of identical particles conserve N during assembly so overall mass-energy is given off purely as heat that radiates away totally.

No problem here.

If subsequently one constituent matter particle self-annihilates somehow and radiates to infinity, to a very good approximation that radiation has been frequency redshifted by a factor f = (1+2V)^1/2. The shell now of N-1 particles has lost an overall mass of essentially fM/N.

...

Which coincides when worked out, with the assembled mass M' being M' = M(1+f)/2 (approx), which exceeds fM.

I don't quite understand where the final expression here is coming from. I haven't had time to try to work through this scenario in detail. As a general comment, though, I would make the following observations:

(1) The externally measured mass, M, of the system once it has collapsed and all excess heat has radiated away, is *less* than the original energy at infinity, Nm (i.e., the number of particles N times the rest mass per particle m), of the particles. The difference is, of course, the energy at infinity of the radiated heat itself.

(2) Since the externally measured mass is smaller, the energy at infinity that will be seen by annihilating the first particle will be less than m (i.e., less than the rest mass a particle would have at infinity). Since there are N particles total, the average energy at infinity released per particle must be M/N (N particles, M total energy released). However, the energy at infinity released by the *last* particle should be m (because at that point the potential is unity; there is no "gravitational field" left). But m is greater than M/N, the average, so the energy released by the first particle should, indeed, be *less* than M/N.

(3) Some of the energy from the annihilation of the first particle can't be radiated to infinity: it has to go instead into the rest of the particles remaining in the object, making each of them slighly less tightly bound, gravitationally, than they were before. (This effect may be what you are thinking of as the energy in the field "balancing the ledger".) As fewer and fewer particles remain, this effect will become smaller and smaller, and more and more of the energy released by each particle's annihilation would be captured at infinity (to the point that the last particle's annihilation radiates its full rest mass, m, to infinity).

Not sure if all this helps, but as I said above, there is a lot of literature on this topic.

 

[TrickyDicky]

the "energy in the gravitational field" is *not* "stress-energy".

Energy is energy, right? Do you mean there are two types of energy, the regular one accounted by the SET and the gravitational one that follows different rules?
Let's consider "Dark energy" for a moment, it is thought to have a gravitational origin (as cosmological constant) and yet everyone agrees it is the source of a SET (with some differences with the usual matter-energy SET). Why one gravitational field energy is "stress-energy" in one case but not in the other?

However, the gravitational waves emitted by the binary pulsar are *not* a "source" of gravity, for the same reason that EM waves are not sources of electromagnetism. EM waves have zero charge, and gravitational waves have zero stress-energy.

EM waves have no charge but still carry energy and have nonzero stress-energy so the example is not valid wrt energy.

 

[Q-reeus]

One example of a definition of "energy in the gravitational field" is described here:
http://en.wikipedia.org/wiki/Stress%...m_pseudotensor
However, as I said above, I don't believe this is the only definition in the literature...

Thanks for the link, but how to interpret. It appears t_LL just addresses conservation of total energy-momentum, but I can't decide there if it is in fact making gravitational energy-momentum a curvature source term, which from your previous comments would place it as 'supplementary' to GR proper.

However, the gravitational waves emitted by the binary pulsar are *not* a "source" of gravity, for the same reason that EM waves are not sources of electromagnetism. EM waves have zero charge, and gravitational waves have zero stress-energy. The waves can carry energy from a "source" (the binary pulsar) to a "sink" (a gravitational wave detector, for example, if we ever succeed in detecting them), but they themselves do not produce any curvature--they *are* curvature, propagated from one region of spacetime to another purely by the properties of spacetime itself, without any "source" present.

And this seems to imply a big problem. When it comes to inspiral and merger of two BH's, as I recall something like up to 40% of the combined pre-merger mass can be radiated away as GW's. So ok allow that the tLL provides a full accounting of energy-momentum in that time evolving system. What though about the total system gravitating mass? We clearly have a huge conversion from gravitating SET (pre merger BH's) to non-gravitating GW's. The total system mass 'charge' is clearly not conserved. This means that a rather weak monopole GW wave component should of necessity be generated. Yet is that not strictly prohibited in GR? How is this reconciled consistently?

Q-reeus: "...Which coincides when worked out, with the assembled mass M' being M' = M(1+f)/2 (approx), which exceeds fM."
I don't quite understand where the final expression here is coming from. I haven't had time to try to work through this scenario in detail...

Just a hint - situation is analogous to say capacitor discharge, where average potential is exactly half the peak. All comes out easy enough, and much easier to come by with my thin shell model than your own preferred solid sphere model. The rest of your commentary on the shell thing mirrors my own in all essentials. We agree there is gravitational energy there, but how it 'acts' is another matter. I see TrickyDicky has added some points.

 

[PeterDonis]

I can't decide there if it is in fact making gravitational energy-momentum a curvature source term, which from your previous comments would place it as 'supplementary' to GR proper.

It's not; t_LL does not appear in the Einstein Field Equation.

And this seems to imply a big problem. When it comes to inspiral and merger of two BH's, as I recall something like up to 40% of the combined pre-merger mass can be radiated away as GW's. So ok allow that the tLL provides a full accounting of energy-momentum in that time evolving system.

Actually, any system radiating GW's would do for raising this question, e.g., the binary pulsar. This makes some aspects easier to think about: see next comment.

What though about the total system gravitating mass? We clearly have a huge conversion from gravitating SET (pre merger BH's) to non-gravitating GW's.

This apparent "conversion" is actually not straightforward for BH's, since the BH is a vacuum solution; true, a real BH is formed from the collapse of a massive object with a nonzero SET, but once the singularity is formed the SET is zero everywhere. An ordinary system like the binary pulsar, that radiates appreciable GW's, doesn't raise this issue; the system clearly has a sizable region with nonzero SET, and it appears (but only appears--see below) that this nonzero SET must gradually be "converted" into zero-SET GW energy.

The total system mass 'charge' is clearly not conserved. This means that a rather weak monopole GW wave component should of necessity be generated. Yet is that not strictly prohibited in GR? How is this reconciled consistently?

You're correct that monopole GW radiation doesn't occur; in fact, neither does dipole--quadrupole is the lowest order GW radiation.

However, you are *not* correct that the "charge", in the sense of nonzero SET, is not conserved. Consider first the simpler analogy of EM radiation: it carries away energy from the radiating source, but no charge is carried away along with it. The radiation is produced by *oscillating* charges, and as the radiation is emitted, the amplitude of the oscillations decreases; the charges are still there, but they oscillate less and less.

The same thing occurs with GW radiation: for example, the binary pulsar is a system of two objects orbiting each other, in other words, a system where stress-energy (gravitational "charge") is oscillating. The oscillation causes GW radiation to be emitted, and as it is emitted, the amplitude of the oscillations decreases (the two pulsars in the binary system get closer together, along with other accompanying orbital parameter changes that decrease the total energy-at-infinity of the system). But the stress-energy itself is still there; it's just oscillating less and less.

(The same general answer holds for two BH's that merge: the final BH will start out oscillating, or perhaps vibrating would be a better word, and the vibrations emit GW's, which decreases the amplitude of the vibrations, until the final BH settles down to its final stationary state, in which no further GW's are emitted. But as I said above, it's harder to relate this to the presence of nonzero SET.)

It is true, btw, that the *total mass* of the binary pulsar system is not conserved; it is slowly decreasing as GW's are emitted. But that is something different from the "total charge" you would obtain by looking only at the regions of nonzero SET. The "total mass" includes the effect of the orbital parameters, not just the contribution from the nonzero SET of the pulsars themselves.

Just a hint - situation is analogous to say capacitor discharge, where average potential is exactly half the peak.

I see, you're just averaging f over all the particle annihiliations (this assumes, btw, that f varies linearly during the process, which may not be the case). But the "assembled mass" of the system, before any particles are annihilated, is M, not fM; the value of M already takes any "redshift factor" into account.

 

[Q-reeus]

This apparent "conversion" is actually not straightforward for BH's, since the BH is a vacuum solution; true, a real BH is formed from the collapse of a massive object with a nonzero SET, but once the singularity is formed the SET is zero everywhere...

Peter, you've got my head spinning here again. Thought had this much bedded down: in GR all gravitational fields - static or GW's, contribute nothing to what we would term M, the gravitating mass (inclusive of momentum and pressure) that is the origin of all curvature - Ricci and Weyl etc. Now, it is common to label a black hole with a certain *gravitating* mass M, right? Zero SET, and zero contribution from the field. Oh my. So this gets back to 'past light cone' presumably - there *was* a SET but now... Alright, let's just say a BH's mass M derives from a 'fossil' SET. Is it not still the case, in pre-merger we say start with M₁ + M₂ = M_t, and after merger we have M₃ ~ 60% M_t (the deficit in purely *energy* terms carried off by GW's). All those M's representing gravitating mass. A net reduction, regardless of what we call the source of each M. What am I missing here?

However, you are *not* correct that the "charge", in the sense of nonzero SET, is not conserved. Consider first the simpler analogy of EM radiation: it carries away energy from the radiating source, but no charge is carried away along with it.

Conservation of charge and zero field divergence in EM wave gaurantee that in EM, but as per remark in #47 I am not seeing the carry over to GR being apt. Seems to boil down to one simple consequence, of one simple postulate (the field does not form part of the SET). The consequence is that a dispersed system, whether neutron stars or BH's, carries there a maximal total energy/gravitating mass M_t. After collision/merger/ringdown, necesarily a portion of that original M_t has been lost to GW's - the remainder has to be less than before - how can there not be a reduction and maintain conservation of energy? If the loss was all to heat radiation (that we all agree is a source of SET) I would agree with your position, but we also agree GW's will carry off a good portion at least. This seems especially evident if we consider the direct head-on collision of two BH's. Energy balance demands conversion from gravitating mass to non-gravitating GW's has to be there, hence net gravitating mass loss. If of course it is true gravitational energy is non-gravitating.

I see, you're just averaging f over all the particle annihiliations (this assumes, btw, that f varies linearly during the process, which may not be the case). But the "assembled mass" of the system, before any particles are annihilated, is M, not fM; the value of M already takes any "redshift factor" into account.

First part is essentially correct. The problem is you adopted a different meaning to the terms I had originally used in #45. Your M is not the M I used there. If you go back and check carefully I think there will be no conflicting opinion on that issue.