# Relative energy of a black hole.

by cragar
Tags: black, energy, hole, relative
P: 1,115
 Quote by DaleSpam 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.
PF Patron
P: 4,770
 Quote by 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.
I agree that that topic should be a separate thread.

 Quote by Q-reeus 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.

 Quote by Q-reeus 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.

 Quote by 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.
PF Patron
P: 4,770
Following up from my previous post:

 Quote by Q-reeus 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.
P: 1,115
 Quote by PeterDonis 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!
PF Patron
P: 5,495
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-..._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.
P: 1,115
 Quote by Naty1 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-..._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.
PF Patron
P: 4,770
 Quote by Q-reeus 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.
PF Patron
P: 4,770
 Quote by Naty1 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.
P: 1,115
 Quote by PeterDonis 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.
PF Patron
P: 4,770
 Quote by Q-reeus 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.

 Quote by Q-reeus 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%...m_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:

 Quote by Q-reeus 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:

 Quote by Q-reeus 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:

 Quote by Q-reeus 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.

 Quote by Q-reeus 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.

 Quote by Q-reeus 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.
P: 2,757
 Quote by PeterDonis 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?

 Quote by PeterDonis 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.
P: 1,115
 Quote by PeterDonis 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 tLL 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.
PF Patron
P: 4,770
 Quote by Q-reeus 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.

 Quote by Q-reeus 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.

 Quote by Q-reeus 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.

 Quote by Q-reeus 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.

 Quote by Q-reeus 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.
P: 1,115
 Quote by PeterDonis 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 = Mt, and after merger we have M3 ~ 60% Mt (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 Mt. After collision/merger/ringdown, necesarily a portion of that original Mt 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.
PF Patron
P: 4,770
 Quote by Q-reeus 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.
I think you are misunderstanding the meaning of the term "mass"; or, rather, you are conflating two different possible meanings. The "M" that appears in the metric, for example the Schwarzschild metric, is "not" the same as the "mass" (actually "energy density", or "0-0 component") that appears in the stress-energy tensor as a "source" of curvature. It's critical to understand the distinction between these two concepts. See further comments below.

 Quote by Q-reeus Now, it is common to label a black hole with a certain *gravitating* mass M, right?
It's not just "labeling". The "M" that appears in the metric has a definite physical meaning: it's the mass you would measure if you put a test object in orbit around the black hole, measured its orbital parameters, and applied Kepler's Third Law. The same applies for any gravitating object--the Sun, for example. If we write down an expression for the metric in the vacuum region exterior to the Sun, it will have an "M" in it which is the Sun's mass measured the same way.

But measuring "M" this way tells us nothing about how it relates to the presence of a non-zero SET in the spacetime (except that there must be one *somewhere*). If we only went by the measured mass M, we would not know whether the Sun was a star or a black hole; either would give the same M. See below.

 Quote by Q-reeus 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.
Remember how we measure M: we put a test object in orbit. That orbit has to be at some radial coordinate r. To understand "where the M comes from", follow the prescription I gave earlier: pick an event somewhere on the worldline of the test object orbiting at that r; look in the past light cone of that event; and find a region with a nonzero SET. Suppose we have M = M(Sun), and we have used an orbit at r = r(Earth) to measure it. Then if the Sun is actually a star, we will find a region of nonzero SET pretty quickly--only eight light-minutes into the past light cone. But if the Sun is a black hole, we may have to look much further to find the nonzero SET region. It just so happens that, because of the particular symmetry of the situation (remember we are assuming perfect spherical symmetry, since that's a condition of the Schwarzschild solution), the "field" at a particular radial coordinate r in the exterior vacuum region is the same for all times t to the future of the nonzero SET region; so it doesn't matter whether that region is eight light-minutes or a billion light-years into the past light-cone, you get the same field--meaning the same metric, and therefore the same measured mass M--either way.

 Quote by Q-reeus Is it not still the case, in pre-merger we say start with M1 + M2 = Mt, and after merger we have M3 ~ 60% Mt (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?
Here we have violated the condition of spherical symmetry during the merger, so the relationship between the measured M and the nonzero SET regions in the past light cone can be more complicated. Before the merger, if we assume each BH was stationary, we can relate M1 and M2 to two nonzero SET regions in the past light cones as above. But after the merger, there is a region of spacetime where there are violent curvature fluctuations because of the violation of spherical symmetry; and those curvature fluctuations carry off energy in the form of gravitational waves. That changes the relationship between the final measured mass M3 and the nonzero SET regions in the past light-cone. It doesn't change anything about the SET regions themselves; no actual stress-energy escapes during the BH merger (it's all trapped behind the horizons of the BH's). But "nonzero SET" and "gravitating mass" in the sense of the value M appearing in the metric are, as I said above, not the same; and the relationship between them depends on the configuration of the spacetime in between the nonzero SET region and the event at which the metric, and thus M, is being measured.

 Quote by Q-reeus The consequence is that a dispersed system, whether neutron stars or BH's, carries there a maximal total energy/gravitating mass Mt. After collision/merger/ringdown, necesarily a portion of that original Mt 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?
There is a reduction in M, yes; as you say, there has to be by conservation of energy. There is no "reduction" in the "stress-energy content"--see above. M and "stress-energy content" are not identical.

 Quote by Q-reeus 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.
See my comments on the meaning of M, and how it is measured, above.
P: 1,115
 Quote by PeterDonis I think you are misunderstanding the meaning of the term "mass"; or, rather, you are conflating two different possible meanings. The "M" that appears in the metric, for example the Schwarzschild metric, is "not" the same as the "mass" (actually "energy density", or "0-0 component") that appears in the stress-energy tensor as a "source" of curvature. It's critical to understand the distinction between these two concepts. See further comments below.
Agreed entirely - and it would aid greatly if the two were given a specific differentiation/identification wrt some simple, static and non-BH system. More on that later.
 Remember how we measure M: we put a test object in orbit. That orbit has to be at some radial coordinate r. To understand "where the M comes from", follow the prescription I gave earlier: pick an event somewhere on the worldline of the test object orbiting at that r; look in the past light cone of that event; and find a region with a nonzero SET. Suppose we have M = M(Sun), and we have used an orbit at r = r(Earth) to measure it. Then if the Sun is actually a star, we will find a region of nonzero SET pretty quickly--only eight light-minutes into the past light cone. But if the Sun is a black hole, we may have to look much further to find the nonzero SET region. It just so happens that, because of the particular symmetry of the situation (remember we are assuming perfect spherical symmetry, since that's a condition of the Schwarzschild solution), the "field" at a particular radial coordinate r in the exterior vacuum region is the same for all times t to the future of the nonzero SET region; so it doesn't matter whether that region is eight light-minutes or a billion light-years into the past light-cone, you get the same field--meaning the same metric, and therefore the same measured mass M--either way.
Which is speaking to me that the distinction between M, and source of non-zero SET in the past light cone, is relevant only for a non-static system. A now stable and static planet that formed long ago from a collapsing gas/dust cloud easily qualifies. So I would say there M = volume integral of non-zero SET. The two are here synonymous, agreed? Anyway the following will attempt to clear all confusion about how factors relate and pan out.

Consider please the following scenario: A large bounding box of mass Mb with perfectly reflecting walls. Inside we have diffuse dust of mass Md >> Mb that over time gravitationally collapses symmetrically to form a stable, static planet of assembled mass Mp < Md. Heat radiated away during collapse is trapped inside the box. The total energy of this notionally closed system is constant. But the internal state has changed. Without question there has been a partial transfer from non-gravitational to gravitational energy. In GR the latter is 'dead weight' wrt acting as gravitational mass, the former is not. How can it be argued the net gravitating mass, as presented to a region exterior to the box, has not thereby diminished? No need to introduce GW's - whenever gravitational energy of any kind is created, a net reduction in overall system gravitational mass ensues (and note 'system' here means everything including radiation). Or so it seems bleeding obvious to me.
Past light cone is not an issue as this is a stable final system with all contributing sources (and non-sources) accessible in perfectly reasonable time to a region exterior to the box.
 See my comments on the meaning of M, and how it is measured, above.
But my 'labelling' convention for M in #45 referred to the dispersed matter prior to collapse/assembly. It was never to be confused with an M for the final gravitating system, which I designated as M' - the assembled mass.
PF Patron
P: 4,770
 Quote by Q-reeus Which is speaking to me that the distinction between M, and source of non-zero SET in the past light cone, is relevant only for a non-static system. A now stable and static planet that formed long ago from a collapsing gas/dust cloud easily qualifies. So I would say there M = volume integral of non-zero SET. The two are here synonymous, agreed?
Not quite. The *relationship* between non-zero SET in the past light cone and M, the quantity appearing in the metric, is simplest for the static case; but that still doesn't mean the two are identical.

 Quote by Q-reeus Consider please the following scenario: A large bounding box of mass Mb with perfectly reflecting walls. Inside we have diffuse dust of mass Md >> Mb that over time gravitationally collapses symmetrically to form a stable, static planet of assembled mass Mp < Md. Heat radiated away during collapse is trapped inside the box.
But this changes the scenario from your original one, where the radiated heat can escape to infinity. If the heat is trapped by reflecting walls, then it will "fall" back into the planet, raising its temperature (and hence its energy). So the equilibrium state will be quite different than a "cold" planet with essentially zero temperature and radiation escaping to infinity.

 Quote by Q-reeus The total energy of this notionally closed system is constant.
Agreed.

 Quote by Q-reeus But the internal state has changed. Without question there has been a partial transfer from non-gravitational to gravitational energy.
Not necessarily. See my comments above. But in any case, this is a red herring. See below.

 Quote by Q-reeus In GR the latter is 'dead weight' wrt acting as gravitational mass, the former is not. How can it be argued the net gravitating mass, as presented to a region exterior to the box, has not thereby diminished?
It hasn't. Your specification has ensured that, as you said above and I agreed, the total energy inside the box is constant. That will mean that, if we put a test object in orbit about the box and measured its mass M externally, we would continue to get the same answer regardless of what happens inside the box.

Let's try a simpler system: a box with perfectly reflecting walls but zero mass (so it doesn't affect the curvature of the spacetime) enclosing two objects of equal rest mass m that start at mutual rest at some distance r apart. What will the externally measured mass of this system be? You might think it will simply be 2m, but think again. Suppose we let the system evolve for a while: the two objects fall towards each other, and at the instant right before they hit each other, they both still have rest mass m, but they also (in the center of mass frame, which is just the frame in which they were initially at rest) have each a considerable kinetic energy k. So at this point we would expect the externally measured mass to be 2(m + k).

Now let the two objects collide, and suppose the collision is perfectly inelastic; the two objects plop into each other and come to rest at the point where they collided, which is the center of mass of the combined system. Obviously, by conservation of energy, the final object must have total energy 2(m + k); but the kinetic energy portion is now converted into heat inside the combined object, which will be at some significant temperature (we assume both initial objects started out at zero temperature). If the enclosing box were not there, that heat could eventually be radiated away to infinity, so that we would end up with a final object of mass 2m. But the box prevents that from happening; the heat might be radiated, but it would then be reflected off the walls and converge on the central object again, until finally some thermal equilibrium was established with some portion of the "heat energy" 2k residing in the object and some portion residing in radiation bouncing around inside the box. In any case, the total energy of the system, as measured from outside the box, will continue to be 2(m + k).

What all this tells us is that, by conservation of energy, the *initial* externally observed mass M of the system must have been, not 2m, but 2(m + k). In the initial state, the energy that became the kinetic energy 2k of the objects, and then the heat inside the final combined object, was instead stored as "gravitational potential energy" in the mutual field created by the two objects combined. (At least, this is the usual way of putting it; but as should now be apparent, that way of putting it can lead to considerable conceptual difficulties.) Similar remarks would hold if we replaced the two initial objects by a spherical shell of dust and let it collapse. But notice that, on the "energy stored in the field" interpretation, the "energy stored in the field" is nonzero in the *initial* state, and is *zero* once the objects have collided! In other words, this scenario converts energy *from* "stored field energy" into "tangible" energy, not the other way around!

I'll follow up with more on this in another post when I have more time; but this should at least give some food for thought.

 Quote by Q-reeus But my 'labelling' convention for M in #45 referred to the dispersed matter prior to collapse/assembly. It was never to be confused with an M for the final gravitating system, which I designated as M' - the assembled mass.
Ah, ok; I was wondering a little about that but didn't read carefully enough. Then my comments were really referring to M', not M.
P: 1,115
 Quote by PeterDonis But this changes the scenario from your original one, where the radiated heat can escape to infinity. If the heat is trapped by reflecting walls, then it will "fall" back into the planet, raising its temperature (and hence its energy). So the equilibrium state will be quite different than a "cold" planet with essentially zero temperature and radiation escaping to infinity.
Fair point - I had not specifically addressed that. However it changes nothing in respect of key principle. The easiest counterargument is to allow the enclosing box to grow as large as desired. Equilibrium temperature grows correspondingly small. Secondly, assuming a configuration where concentration of thermal energy in the central mass is high merely slightly increases concentration of the total system energy there, to a generally small degree. It makes no appreciable difference to the existence of an all important partial conversion of non-gravitational mass/energy to gravitational energy via collapse.
 Not necessarily. See my comments above. But in any case, this is a red herring. See below.
Another red herring, apart from the one dealt with above?
 Let's try a simpler system: a box with perfectly reflecting walls but zero mass (so it doesn't affect the curvature of the spacetime)...
Purist response: a massless box cannot withstand the radiant pressure it has to contain. But enough of nitpickery.
 ...enclosing two objects of equal rest mass m that start at mutual rest at some distance r apart. What will the externally measured mass of this system be? You might think it will simply be 2m, but think again. Suppose we let the system evolve for a while: the two objects fall towards each other, and at the instant right before they hit each other, they both still have rest mass m, but they also (in the center of mass frame, which is just the frame in which they were initially at rest) have each a considerable kinetic energy k. So at this point we would expect the externally measured mass to be 2(m + k).
So far so good. Oh, too casual reading that. No, if the isolated rest mass, assuming infinite separation, is m each, then the combined mass must be less than 2m when separated by r and stationary. A portion of m+m was lost (as heat, or mechanical work supplied elsewhere) to arrive at the 'initial', partially separated configuration you give. Binding energy is negative, so total mass declines. Subsequent collapse, where all energy is now contained within the enclosure, conserves total energy yes.
 Now let the two objects collide, and suppose the collision is perfectly inelastic; the two objects plop into each other and come to rest at the point where they collided, which is the center of mass of the combined system. Obviously, by conservation of energy, the final object must have total energy 2(m + k); but the kinetic energy portion is now converted into heat inside the combined object, which will be at some significant temperature (we assume both initial objects started out at zero temperature). If the enclosing box were not there, that heat could eventually be radiated away to infinity, so that we would end up with a final object of mass 2m. But the box prevents that from happening; the heat might be radiated, but it would then be reflected off the walls and converge on the central object again, until finally some thermal equilibrium was established with some portion of the "heat energy" 2k residing in the object and some portion residing in radiation bouncing around inside the box. In any case, the total energy of the system, as measured from outside the box, will continue to be 2(m + k).
And still good. As per previous edit. If m is substituted with an m' that reflects the reduced system initial mass, the rest of the argument just here I agree with.
 What all this tells us is that, by conservation of energy, the *initial* externally observed mass M of the system must have been, not 2m, but 2(m + k). In the initial state, the energy that became the kinetic energy 2k of the objects, and then the heat inside the final combined object, was instead stored as "gravitational potential energy" in the mutual field created by the two objects combined. (At least, this is the usual way of putting it; but as should now be apparent, that way of putting it can lead to considerable conceptual difficulties.)...
Yes those conceptual difficulties come across below. Note here though that the conventional 'modern' Newtonian interpretation of negative gravitational binding energy is to ascribe a corresponding negative energy to the field. What I have shown in #45 is this is incorrect. The fact of redshift demands imo for consistency that field (or curvature, by geometric interpretation) energy is both present and has a positive sign - in keeping with analogous EM, mechanical systems.
 Similar remarks would hold if we replaced the two initial objects by a spherical shell of dust and let it collapse. But notice that, on the "energy stored in the field" interpretation, the "energy stored in the field" is nonzero in the *initial* state, and is *zero* once the objects have collided! In other words, this scenario converts energy *from* "stored field energy" into "tangible" energy, not the other way around!
And here imo is the crimson fish indeed. Disagree entirely. If you want to go back and argue a basic flaw in #45 be my guest - I stand by it. It is clear there the difference M'-fM = M(1-f)/2, identified as necessarily gravitational energy (not explicitly stated there but implied), grows monotonically from zero at 'infinite' separation to a positive values at any finite final R. Naturally that expression is just a good approximation for R >> rs, but in that regime any error amounts to a tiny higher order correction.

The bottom line to all this is stark and simple. To repeat: GR excludes gravitational field energy as source term. At the same time we must as general feature have conversion from non-gravitational to gravitational energy in any collapse scenario. Simple math follows. To avoid this prospect (monopole GW's etc.) while holding to gravity does not gravitate, as stated before one can say there is no energy in static field curvature, while presumably keeping it for GW's. Now it can be postulated that nature truly behaves like that, but I suspect a truly bizarre playground follows. And you well know my suggested cure.

 Related Discussions Special & General Relativity 9 Special & General Relativity 13 Special & General Relativity 8 Special & General Relativity 12 General Astronomy 2