Relative energy of a black hole.

Q-reeus

Splendidly diplomatic. Consider an overseas embassy posting.
OK then won't press any further, at least this squares the ledger in a way.

DaleSpam

Thanks!

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

Q-reeus

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

PeterDonis

I agree that that topic should be a separate thread.

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.

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.

I assure you that that is not an intended effect.

PeterDonis

Following up from my previous post:

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

Naty1

Q-reeus posts #18

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


Q-reeus: post #20 :
Dalespam: BRAVO!!!! much better than all the sniping that too often abounds. Had me LOL.

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


PeterDonis: post#28

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


from Q-reeus: post #31
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.

Q-reeus

Right, but occasionally beating a drum oft and loud gets results - just use that approach sparingly!
Yes to the very last bit but it seems evidently no to the first. From the first paragraph in that Wiki link:
Yet comments made in earlier posts here suggest (on my reading) otherwise - this is still being thrashed out. Stay tuned.

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.

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

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

That clarifies things.
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

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

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:

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:

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:

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.

No problem here.

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

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?





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

Q-reeus

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.
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?
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

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

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.

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.

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.

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

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?
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.
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.

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.

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.

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.

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.

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.

See my comments on the meaning of M, and how it is measured, above.