Relative energy of a black hole.

In summary: However, the black hole's energy comes from the mass of all the matter that has fallen into it, and that mass contributes to the black hole's gravitational field. So if you're not in the black hole's gravitational field, then it doesn't have any energy.
  • #36
Q-reeus said:
Splendidly diplomatic. Consider an overseas embassy posting. :tongue:
Thanks!

Q-reeus said:
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.
 
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  • #37
DaleSpam said:
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.
 
  • #38
Q-reeus said:
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.

Q-reeus said:
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.

Q-reeus said:
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.

Q-reeus said:
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. :wink:
 
  • #39
Following up from my previous post:

Q-reeus said:
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.
 
  • #40
PeterDonis said:
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. :wink:
Not that I had any real fear, but nice to be reassured! :smile:
 
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  • #41
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.
 
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  • #42
Naty1 said:
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.
 
  • #43
Q-reeus said:
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.
 
  • #44
Naty1 said:
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.
 
  • #45
PeterDonis said:
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.
 
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  • #46
Q-reeus said:
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.

Q-reeus said:
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:

Q-reeus said:
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:

Q-reeus said:
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:

Q-reeus said:
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.

Q-reeus said:
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.

Q-reeus said:
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.
 
  • #47
PeterDonis said:
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?





PeterDonis said:
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.
 
  • #48
PeterDonis said:
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.
 
  • #49
Q-reeus said:
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.

Q-reeus said:
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.

Q-reeus said:
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.

Q-reeus said:
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.

Q-reeus said:
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.
 
  • #50
PeterDonis said:
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.
 
  • #51
Q-reeus said:
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.

Q-reeus said:
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.

Q-reeus said:
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.

Q-reeus said:
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.

Q-reeus said:
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.

Q-reeus said:
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.
 
  • #52
PeterDonis said:
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.
 
  • #53
Q-reeus said:
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.

Q-reeus said:
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.

Q-reeus said:
The total energy of this notionally closed system is constant.

Agreed.

Q-reeus said:
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.

Q-reeus said:
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.

Q-reeus said:
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.
 
  • #54
PeterDonis said:
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? :rolleyes:
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).
[STRIKE]So far so good.[/STRIKE] 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).
[STRIKE]And still good.[/STRIKE] 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.
 
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  • #55
Q-reeus said:
GR excludes gravitational field energy as source term.

In the precise sense of "source", meaning nonzero SET, yes.

Q-reeus said:
At the same time we must as general feature have conversion from non-gravitational to gravitational energy in any collapse scenario.

I would phrase this somewhat differently: the externally measured mass M of an isolated system, from one point of view, will, in general, contain a contribution that can be described as "energy stored in the gravitational field". Put another way, if we take a "snapshot" of the system at some instant of time (i.e., on a particular spacelike slice), and try to "count up" the contributions to the total mass M from individual parts of the system, we will find that we have to include a contribution from "gravitational potential energy" to make the final answer come out right.

However, this is only "from one point of view", and there is nothing in the physics that *requires* you to take that point of view, nor do you need to take it to figure out what actually happens. See below.

Q-reeus said:
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.

As I've said before, I believe this is only an issue for you because you are focusing on asking questions that your a priori conceptual scheme wants you to ask, such as "does gravity gravitate?", instead of first looking at the actual physics and then deriving your conceptual scheme from what the actual physics says. If you do the latter, there is no issue; the theory is well-defined and gives definite answers to all the questions you can ask about actual physical observables.

The actual physics, as I've said before, is simple: to figure out what the "observed field" is at a given event (meaning the metric, and therefore all quantities derivable from the metric, which includes the mass M, the "acceleration due to gravity", the "gravitational potential energy", tidal gravity, etc., etc.), it suffices to look in the past light cone of that event, figure out where the ultimate "sources" are (regions of nonzero SET), and then look at the (vacuum--zero SET) spacetime in between the sources and the event of interest to determine how the field generated by the sources "propagates" to the event of interest.

The above can be done without ever having to ask the questions you are asking. You don't need to know whether "gravity gravitates". You don't need to know how to "count up" individual parts of the system on a spacelike slice and add them up to get the externally measured mass M, or whether you need to include "gravitational potential energy" in the total. Those are simply not necessary questions to ask; they aren't needed to figure out what happens (what the observed field is); and "what happens" includes what the externally measured mass M of the system will be at a particular event. For a system whose mass M appears, from one point of view, to contain a contribution from "energy stored in the gravitational field", that same mass M can always be accounted for in the way I have described, without ever having to consider "energy stored in the gravitational field".

Another way of looking at this is to ask why you are so insistent on interpreting the mass M in terms of "adding up sources" on a spacelike slice, instead of doing it the way I have described (looking in the past light cone)--and therefore finding that, to make things "add up" correctly in this way, you need to include "energy stored in the field". I think the reason this way of looking at it is intuitively appealing is that we are used to looking at stationary, or nearly stationary, systems, for which two things are true: (1) a meaningful definition of "energy stored in the field" can be given that corresponds, intuitively, to "gravitational potential energy", which is familiar from Newtonian physics; (2) because the system is stationary, there is a very simple relationship between what's there on a spacelike slice and what's there in the past light cone of any particular event. The conceptual issues you are having are basically due to trying to extend the simple viewpoint that works reasonably well for stationary systems to a more general domain, non-stationary systems (systems that collapse, and systems that radiate energy) where items (1) and (2) no longer hold.
 
  • #56
Q-reeus said:
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.

This is probably just an issue of definition of terms. For "m" in my post, instead of reading "isolated rest mass at infinity", read "rest mass as-is, in the given initial separation", which, in your terminology, would be (m - e), where e is the portion of the "rest mass at infinity" m that was lost during the process of moving the two objects from infinity to a finite separation.

The key point is that the externally measured total mass M of the system as a whole, in the initial state (objects separated by some distance and, at least momentarily, at rest relative to each other) *cannot* be simply the sum of the "masses" of the two objects individually, if you are trying to compute it the way you are trying to compute it; there *has* to be an additional contribution from "energy stored in the field", because that extra energy will appear as "tangible" energy when the two objects fall towards each other, collide, and form a single object with a positive temperature. You appear to agree with this:

Q-reeus said:
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.
 
  • #57
PeterDonis said:
This is probably just an issue of definition of terms. For "m" in my post, instead of reading "isolated rest mass at infinity", read "rest mass as-is, in the given initial separation", which, in your terminology, would be (m - e), where e is the portion of the "rest mass at infinity" m that was lost during the process of moving the two objects from infinity to a finite separation.

The key point is that the externally measured total mass M of the system as a whole, in the initial state (objects separated by some distance and, at least momentarily, at rest relative to each other) *cannot* be simply the sum of the "masses" of the two objects individually, if you are trying to compute it the way you are trying to compute it; there *has* to be an additional contribution from "energy stored in the field", because that extra energy will appear as "tangible" energy when the two objects fall towards each other, collide, and form a single object with a positive temperature. You appear to agree with this:
Peter, thanks for your clarification and with that I agree with the above. On the broader picture, while I respect you are an accomplished master of GR maths and it's application, sad to say there is no final consensus. Bravo though for putting in a lot of effort in trying to evaporate my scepticism. At the least it has given me a clearer understanding on how this issue is seen by the GR community. Have a nice day. :smile:
 
  • #58
Q-reeus said:
On the broader picture, while I respect you are an accomplished master of GR maths and it's application, sad to say there is no final consensus. Bravo though for putting in a lot of effort in trying to evaporate my scepticism. At the least it has given me a clearer understanding on how this issue is seen by the GR community. Have a nice day. :smile:

No problem, we can't always reach consensus. I do have one final question, though, about the precise nature of your disagreement. I'm still not entirely clear whether:

(1) You disagree with my contention that the observed field at a given event can always be explained (calculated) entirely in terms of "sources" (regions of nonzero SET) in the past light cone of that event; or

(2) You agree that the observed field at a given event can be explained (calculated) as above, but you don't think this is enough--that something more is needed for a proper physical understanding of what's going on.
 
  • #59
PeterDonis said:
No problem, we can't always reach consensus. I do have one final question, though, about the precise nature of your disagreement. I'm still not entirely clear whether:

(1) You disagree with my contention that the observed field at a given event can always be explained (calculated) entirely in terms of "sources" (regions of nonzero SET) in the past light cone of that event; or

(2) You agree that the observed field at a given event can be explained (calculated) as above, but you don't think this is enough--that something more is needed for a proper physical understanding of what's going on.
Neither of the above really. (1) is fine in principle, except for the specific contention in GR that SET never includes gravitational energy density Wg - however one wishes to precisely define the latter. And my suspicion is the reason gets down to ambiguities re non-localizability of Wg from the GR geometric perspective (free-fall and it's gone). That I find almost amusing. One could create the same vanishing trick in standard EM. A rather brief 'world' consisting of charged particles all having the same sign and charge-to-mass ratio, allowed to suddenly 'free-fall' apart, will have no charged observer detecting an E field, apart from gradients ('tidal forces'). The particular analogy to gravitation is obvious (of course excludes non-linear features present in GR). Physicists in such a world might justifiably conclude EM field energy was ill-defined and non-localizable, but that would be their mistaken perspective. What I am saying here is it seems natural to make coordinate measure the proper perspective for working out a clear working definition of gravitational energy, with free-fall the improper frame ('inaccessibility' and all that).

Another thing that to me screams 'gravity gravitates' not brought up earlier is the implications I see of a zero Nordtvedt effect: http://relativity.livingreviews.org/Articles/lrr-2006-3/index.html [Broken]. All three of active mass ma, passive mass mp, and inertial mass ma, are implied exactly equal. Saying gravitational energy does not gravitate (ma = 0) is one thing, but zero Nordtvedt requires it also have no inertial contribution either. Anyway, why go on - this is just my layman's reasoning. Running late. :zzz:
 
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  • #60
Q-reeus said:
Neither of the above really.

Then I'm still confused. But see below.

Q-reeus said:
(1) is fine in principle, except for the specific contention in GR that SET never includes gravitational energy density Wg - however one wishes to precisely define the latter.

Why is that a problem? The SET has a clear physical meaning, based on being on the RHS of the Einstein Field Equation; and that equation does not work if you try to add "gravitational energy density" into the RHS of the EFE. That was the point of one of the quotes we discussed early on, which AFAIK you agreed with.

Another way of putting this would be to ask: why do you need to even define "gravitational energy" in the first place? See further comments below.

Q-reeus said:
One could create the same vanishing trick in standard EM. A rather brief 'world' consisting of charged particles all having the same sign and charge-to-mass ratio...

But there is a big difference here: we know that this "world" does not match reality; we know that, in reality, there are particles with varying charge/mass ratios (and signs). We have *no* evidence of anything in reality with varying "energy/mass ratio" (or different "active gravitational mass", "passive gravitational mass", and "inertial mass", using the terminology you introduce below). If we did have such evidence, obviously we would have to change our model of gravity; but we don't. GR predicts that we never will, because all three of those "masses" are really the same thing, so they must all be the same as a matter of physical law.

If we ever found such evidence, GR would be falsified. But if it is actually true that all three "masses" are the same as a matter of physical law, then what GR is basically telling us is that we are asking the wrong question: we are using a conceptual scheme that doesn't quite match reality, because it leads us to ask a question ("why are active m_g, passive m_g, and inertial mass m_a all the same?") that, from a proper conceptual scheme, would never even be asked, because it would be "obvious" that there was only one kind of mass-energy to begin with.

(It's possible, btw, that such a conceptual scheme already exists: I believe there are some versions of quantum gravity in which there is no room for more than one kind of "mass", so to speak. But I'm not very up to date on developments in that area.)

Q-reeus said:
What I am saying here is it seems natural to make coordinate measure the proper perspective for working out a clear working definition of gravitational energy, with free-fall the improper frame ('inaccessibility' and all that).

Once again, why are you trying to find a "proper perspective for working out a clear working definition of gravitational energy"? What physics does it capture that isn't captured in the method I have described (look in the past light cone for nonzero SET regions)? (My answer to this question, of course, is "none".)

Also, trying to use "coordinate measure" as a standard creates a problem: how do I tell *which* state of motion is the "standard" one? For the specific case I just described, the spacetime has a time translation symmetry which picks out the "hovering" observer--but what about, for example, an FRW spacetime, which doesn't have a time translation symmetry--let alone a generic spacetime where there is *no* symmetry? (You will note that these are also cases where it is much harder to come up with a definition of "gravitational energy".)

In other words, the "coordinate measure" criterion, while it is intuitively appealing in the simple cases that we have ordinary, everyday experience of, does not generalize well to more complicated cases. The beauty of the free-fall condition is that it always works: I don't have to assume *anything* about the spacetime. I can always test to see if an object is in free fall by direct physical observation: does the object feel any weight? So I can always use freely falling worldlines as "standard" worldlines to refer things to, no matter what kind of spacetime I am trying to analyze.

(Similar remarks apply, btw, to the prescription to look at the "standard" SET in the past light cone and then work through the vacuum region from there, "propagating" the field to the current event of interest. The definition of the "standard" SET is straightforward and unambiguous, so it can always be applied, regardless of the spacetime, and does not require any symmetry to be present.)

Q-reeus said:
Another thing that to me screams 'gravity gravitates'

And once again, why are you even asking this question to begin with? The actual physical observables, as I pointed out before, can be entirely explained and calculated without ever having to ask this question at all. (Although you still don't appear to entirely accept that this is true--but if it is false, then so is the Einstein Field Equation, since that, as I said above, only includes the "standard" SET, with no "gravitational energy" terms, otherwise it wouldn't work. So if you really want to dispute my #1, you'll need to show that the EFE, as it stands, with no "gravitational energy" terms, gives incorrect predictions.)
 
  • #61
I have some sympathy for scepticism expressed by Q-reeus about "fossil gravitational fields". For example let us say a star is just about to collapse to a black hole and it is orbiting a much larger gravitational object. When it finally collapses to a black hole its "frozen" gravitational field continues to orbit. Now if we assume the universe can no longer interact with the mass of the black hole (hidden behind the event horizon) then we have to conclude that the frozen gravitational field of the black hole has all the gravitational and inertial properties of the original mass without requiring the mass to be there. In other words it is just the field that is orbiting. This in turn implies that the momentum of a massive object that we normally associate with its mass, is actually a property of its gravitational field and not of the mass itself.

I also wonder where this leaves Hawking radiation. A very small black hole can evaporate in a matter of minutes, but go along with the idea that the gravitational field does not care if the mass is still there or not, we would be unaware that it had evaporated for a very long time (possibly infinite).

Also consider the merger of 2 or more black holes. The gravitational fields surrounding the merging objects changes in a complex, rapid dynamic way that seems inconsistent with the idea of frozen fossil gravitational fields.
 
  • #62
yuiop said:
Now if we assume the universe can no longer interact with the mass of the black hole (hidden behind the event horizon) then we have to conclude that the frozen gravitational field of the black hole has all the gravitational and inertial properties of the original mass without requiring the mass to be there. In other words it is just the field that is orbiting. This in turn implies that the momentum of a massive object that we normally associate with its mass, is actually a property of its gravitational field and not of the mass itself.

You are making the same conceptual error that Q-reeus is making: you are thinking of the BH as an "object" that has to "interact" with things, instead of thinking of it as spacetime curvature that was produced by some region of nonzero SET somewhere in the past.

For purposes of visualizing what's going on in a scenario like you describe, this is fine; if the BH's externally observed mass is much smaller than that of the object it is orbiting, you can treat the BH like a "test object" orbiting the other object, without having to worry about the BH's internal structure. For practical purposes this can work fine. But it is only an approximation; you are trying to extend the approximation beyond its domain of validity. If you want to think about the fundamentals of the BH, things like "where does its mass come from?", "where does its momentum come from?", etc., you simply can't use this approximation: you have to go back to the fundamentals, the Einstein Field Equation and the specific solution of it that produces the spacetime you are looking at--which is based ultimately on what regions of non-zero SET are present in the spacetime, and where. All of the dynamics of the BH, including how it orbits another body, are ultimately derived from this; there is no need to view the BH as an "object" that has to somehow carry mass and momentum independently of what is propagated to it from the regions of nonzero SET in the past.

yuiop said:
I also wonder where this leaves Hawking radiation. A very small black hole can evaporate in a matter of minutes, but go along with the idea that the gravitational field does not care if the mass is still there or not, we would be unaware that it had evaporated for a very long time (possibly infinite).

No, we wouldn't. If we include Hawking radiation, the BH is not stationary; its mass slowly decreases. That means the radius of the horizon also decreases, which means that the time for light rays emitted from "close to the horizon" to get out to far distant observers no longer diverges to infinity.

Also, it is incorrect to say that "the gravitational field does not care if the mass is still there or not". It does. As a given "packet" of Hawking radiation passes a given radius, observers at that radius will see a slightly smaller mass for the BH--as observed, for example, by a decrease in the rocket thrust it takes to hold station at a constant radius.

yuiop said:
Also consider the merger of 2 or more black holes. The gravitational fields surrounding the merging objects changes in a complex, rapid dynamic way that seems inconsistent with the idea of frozen fossil gravitational fields.

Once again, you are thinking of the BH's as "objects" instead of as curvature produced and propagated from nonzero SET regions in the past. For a spacetime where 2 BH's merge, there will be two such regions--the two bodies that originally collapsed to form the two BH's. Everything about the merger, including the rapid, dynamic changes in the field as they merge, followed by a "settling down" to a new quasi-stationary state with the final BH, is determined by those initial nonzero SET regions, including their positions relative to each other. There is no need to think about "fossil fields"--again, this is simply a scenario where that approximation breaks down if taken too seriously.
 
  • #63
Peter, could you take a look at my #47?
 
  • #64
PeterDonis said:
You are making the same conceptual error that Q-reeus is making: you are thinking of the BH as an "object" that has to "interact" with things, instead of thinking of it as spacetime curvature that was produced by some region of nonzero SET somewhere in the past.
For purposes of visualizing what's going on in a scenario like you describe, this is fine; if the BH's externally observed mass is much smaller than that of the object it is orbiting, you can treat the BH like a "test object" orbiting the other object, without having to worry about the BH's internal structure. For practical purposes this can work fine. But it is only an approximation; you are trying to extend the approximation beyond its domain of validity. If you want to think about the fundamentals of the BH, things like "where does its mass come from?", "where does its momentum come from?", etc., you simply can't use this approximation: you have to go back to the fundamentals, the Einstein Field Equation and the specific solution of it that produces the spacetime you are looking at--which is based ultimately on what regions of non-zero SET are present in the spacetime, and where. All of the dynamics of the BH, including how it orbits another body, are ultimately derived from this; there is no need to view the BH as an "object" that has to somehow carry mass and momentum independently of what is propagated to it from the regions of nonzero SET in the past.
Sorry Peter, but I agree with yuiop here. How can a 'BH' not be modeled as an object that interacts with things when you have previously described it just that way - has a characteristic mass M according to Keplerian dynamics of an orbiting test mass. Let's not play with words here.
yuiop: "Also consider the merger of 2 or more black holes. The gravitational fields surrounding the merging objects changes in a complex, rapid dynamic way that seems inconsistent with the idea of frozen fossil gravitational fields."
Once again, you are thinking of the BH's as "objects" instead of as curvature produced and propagated from nonzero SET regions in the past.
A matter of definition surely - the two are synonymous by any reasonable score imo. A rose by any other name is still a rose.
For a spacetime where 2 BH's merge, there will be two such regions--the two bodies that originally collapsed to form the two BH's. Everything about the merger, including the rapid, dynamic changes in the field as they merge, followed by a "settling down" to a new quasi-stationary state with the final BH, is determined by those initial nonzero SET regions, including their positions relative to each other. There is no need to think about "fossil fields"--again, this is simply a scenario where that approximation breaks down if taken too seriously.
Again, I agree with yuiop. YouTube provides some nice CGI examples of 'BH merger events':http://www.youtube.com/watch?v=4m-ZVsLf070&feature=related http://www.youtube.com/watch?v=L478ZPy_2Ys&feature=related (and these from respected numerical GR groups)
Looks impressive, like lava lamp blobs fusing together, with electrostatics thrown into achieve that 'necking' effect. But is this real world or just some snippet from a sci-fi flick? Presumably the animations are from a coordinate or near enough to coordinate perspective. But from that perspective we know that Schwarzschild metric predicts infinite time dilation and radial length contraction at the EH of each 'pre-merging' BH. One can argue it's not a physical surface, but point is, logically to deform an infinitely curved region requires infinite coordinate time! So I'm having trouble seeing how an infinitely curved BH boundary is anything other than infinitely rigid in effect. So my idea of 'merger' would be roughly akin to say two basketballs, with a thin foam rubber sheet placed between, approaching each other and deforming slightly but never merging. The squashing bit allows that there is some finite mutual metric distortion of one on the other as seen in coordinate measure, but key word here is *finite*. (just to be clear here; my own interpretation is that it shows the inconsistency of 'BH' in the first place. I am not speaking on anyone else's behalf in saying that.)
 
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  • #65
PeterDonis said:
Q-reeus: "(1) is fine in principle, except for the specific contention in GR that SET never includes gravitational energy density Wg - however one wishes to precisely define the latter."
Why is that a problem? The SET has a clear physical meaning, based on being on the RHS of the Einstein Field Equation; and that equation does not work if you try to add "gravitational energy density" into the RHS of the EFE. That was the point of one of the quotes we discussed early on, which AFAIK you agreed with.
I agreed only that this was the official GR position - never mine as made abundantly clear in many posts. But I have a confession to make. My 'proof' of positive energy in a gravitational field neglected to fully account for one entity. In #45 pressure, which apart from rest mass/energy, forms the only other GR approved contribution to the SET in the scenario considered, was declared negligible, but there was no evidence given that it was negligible wrt gravitational energy as source. I had implicitly lumped pressure together with matter rest mass. My argument there and in #52 was not strictly correct (neither of us picked it up) - conversion, partial or fully, of non-gravitational energy to gravitational energy is not sufficient proof of itself that net system gravitating mass declines. One must account for pressure changes also. On two accounts it turns out my earlier claim holds true overall:

1: Scaling law. Take the thin shell stipulated in #45. Double it's assembled mass M'. Pressure has doubled, and in GR the pressure contribution to curvature is a linear function of that pressure. But the field energy density, in this weak gravity linear region, is quadratic wrt M'. Working out the specifics can be a little messy, but bottom line is, pressure cannot in general act to cancel gravitational contribution to M'. Phew.

2: There is good reason to doubt pressure makes *any* contribution to gravitational mass. Consider the case of two 'G'-clamps welded back-to-back. Tighten both screws evenly. This generates positive stress in the screw sides, and negative stress in the opposite sides. There is also bending and shear stresses present elsewhere, but they are self-cancelling wrt net positive or negative pressure. The pressure distribution by inspection will have a quadrupole character. Hence if the screws are periodically tightened then loosened, we have a harmonic source of quadrupole pressure. It follows this arrangement generates GW's. GW amplitude is linear wrt pressure. But material displacement of the twin clamps under pressure is inversely proportional to the material elastic constant. Plastic clamps will flex far more for a given generated pressure than for say steel clamps. the kicker then is this: any metric back reaction from generating GW's must induce far greater power drain in the plastic clamps case than for the steel ones. There cannot be in general a conservative power balance. Unless that is, pressure is nor a source term in fact! We have not included GW contribution owing to just motion of the clamp material, but that's ok since that contribution will be proportional to material density, which need have no relation to elastic constant. The two contributions are thus decoupled.
Another way of putting this would be to ask: why do you need to even define "gravitational energy" in the first place?...
For all the reasons given before in numerous entries! Only by denying the very existence of gravitational energy can the problem seemingly go away. But then e.g. Hulse-Tayler-binary-pulsar-data-as-proof-of-GW's issue, as before discussed, becomes somewhat problematic.
Q-reeus: "One could create the same vanishing trick in standard EM. A rather brief 'world' consisting of charged particles all having the same sign and charge-to-mass ratio..."
But there is a big difference here: we know that this "world" does not match reality; we know that, in reality, there are particles with varying charge/mass ratios (and signs)...
Which misses the point; this is a valid gedanken experiment. It is possible for such a situation to exist and it implies certain things, which I have stated.
Once again, why are you trying to find a "proper perspective for working out a clear working definition of gravitational energy"? What physics does it capture that isn't captured in the method I have described (look in the past light cone for nonzero SET regions)? (My answer to this question, of course, is "none".)
I'll repeat. With pressure now taken care of, I have shown that there is necessarily such a beast as positive gravitational energy density in a static field. Conversion at least partially from non-gravitational to gravitational energy accompanies any collapse scenario. We now know from the forgoing this logically requires a net reduction in net system gravitating mass. Bingo - monopole GW's etc. Your argument is to just stick with finding the SET in the event past light cone, but what's missing here is crucial. The *recipe* for what constitutes part of the SET. if gravitational energy is missing from that recipe (as GR insists), my last umpteen entries here have been trying to drive home the inconsistencies that then invariably result. Take it or leave it.
 
  • #66
Peter: Great explanations, bravo for your patience...I, and perhaps others who have read and not posted, learned a lot from your explanations and while perhaps frustrating to you at times, subsequent explanations with slightly different perspectives added further clarity.

Here are a few summary points I really liked:

#43:
...the planet's mass density and pressure are the only things that contribute to the SET...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.


#46
..In other words, the "energy in the gravitational field" is *not* "stress-energy...

EM waves have zero charge, and gravitational waves have zero stress-energy...

the gravitational waves... *are* curvature, propagated from one region of spacetime to another... without any "source" present.



Yuiop:
I have some sympathy for scepticism expressed by Q-reeus about "fossil gravitational fields". ... When it finally collapses to a black hole its "frozen" gravitational field continues to orbit. ...In other words it is just the field that is orbiting.

Me too...A key point for me in reconciling this 'approximate' view with the Einstein formalism was this explanation from Peter:
#51
...remember we are assuming perfect spherical symmetry...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...

That really clarified for me your earlier explanation of the 'past light cone' perspective.

Further, and it's a minor point, one doesn't even need to 'assume' perfect spherical symmetry for a real black hole somebody proved [Was it Hawking?] spherical symmetry for a Schwarzschild type black hole...any initial irregularities would be smoothed out...


Peter:
...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... 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...

This business about different M's between the SET and metric is something I need to think about further...it hasn't clicked yet:

edit:
"This is what I am puzzling over...what's the physical background/explanation:
"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..."


from #51:
...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. ...But measuring "M" this way tells us nothing about how it relates to the presence of a non-zero SET in the spacetime...

[ok on that piece]

and this example helps...
...But after the merger [of two black holes] 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...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...

I just know I don't have an intuitive grasp yet:
[Is this too naive?: Look dopey (me)!, they are measures at different spacetime points so of course there will be different values.]

In your explanation of SET, a planet's mass density and pressure contribute to the SET...[ok, I get that] In your definition of M [in the metric] they would also be included in that entity, right...How might these differ between the two...what conditions??

#49:
...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...

So here is an example of a zero SET and a finite metric M, outside the horizon, right?

I also wondered about trickydicky question. (#47)"
"What about dark energy."
 
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  • #67
TrickyDicky said:
Peter, could you take a look at my #47?

Just did; good questions, sorry I hadn't responded before.

TrickyDicky said:
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?

The word "energy" can have different meanings. That's why I've tried not to use it in my explanations (and when I have, I've later clarified and qualified what I've said to make clear what specific thing I was using the word "energy" to refer to). Thinking that "energy is energy" is another example of an approximation that works well in the range of our ordinary experience, but breaks down when we try to extend it too far.

Once again, the only way to make sure we're being precise and are properly capturing the physics is to go back to the fundamentals: the Einstein Field Equation and its solutions. The RHS of that equation, the SET, has a precise physical meaning, and the "rules" it follows are also precise (as captured in that equation). That equation is sufficient, as I've said a number of times, to explain and calculate *all* classical gravitational phenomena, including those that are sometimes referred to as "gravitational energy". The latter is *not* a fundamental concept; it is just an approximate way of looking at the physics in a limited domain that works reasonably well in that domain.

TrickyDicky said:
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?

The precise definition of "dark energy" is a nonzero SET that is proportional to the metric. A nonzero cosmological constant is one possible form of dark energy, but not the only one. (The fact that the SET of dark energy is proportional to the metric is the key difference from an "ordinary" SET derived from matter or EM radiation.) Since dark energy, precisely defined, has a nonzero SET, it appears on the RHS of the EFE, and that is how it affects the physics. Whether or not this kind of energy "counts" as "gravitational field energy" depends on how you define the latter term; but as I've noted already, you don't have to define that term to figure out the physics, so answering the question of whether dark energy counts as gravitational energy is not necessary for the physics.

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

Just to clarify, I was only using the EM case as a analogy, and I wasn't trying to say that EM fields have zero stress-energy; you are correct that even "source-free" EM fields have a nonzero SET. The analogy I was making was simply that EM radiation can propagate through regions of spacetime that have a zero charge-current 4-vector, which is the "source" in the EM field equations. But the analogy is limited, and I don't insist on it if it doesn't help with understanding the gravity case.
 
  • #68
Q-reeus said:
Sorry Peter, but I agree with yuiop here. How can a 'BH' not be modeled as an object that interacts with things when you have previously described it just that way - has a characteristic mass M according to Keplerian dynamics of an orbiting test mass. Let's not play with words here.

A matter of definition surely - the two are synonymous by any reasonable score imo. A rose by any other name is still a rose.

I am not playing with words; I am pointing out what the words do and do not refer to, and what can and cannot be derived from them according to the actual, precise physics.

Take your statement about the mass M: it is measured by "Keplerian dynamics of an orbiting test mass". Very true. My point is that that does *not* imply that anything inside the BH horizon is interacting with anything outside. It simply doesn't. That's all. We are used to thinking of gravitating bodies as "interacting" with other bodies (like the Sun and the Earth), without bothering to always remind ourselves that the "interaction" does not occur instantaneously--the Earth is not interacting with the Sun "right now", it is interacting (if that's even the right word) with the Sun eight minutes ago. But the latter is in fact the case. A BH is just a much more extreme case, where we might have to go back billions of years to find the nonzero SET region in the past light cone--but that's still the correct precise description of the physics. Thinking of the BH itself as "interacting" with orbiting bodies is *not*; it's an approximation with a limited domain of validity. You are trying to stretch it beyond its domain of validity, and it is breaking down.

Q-reeus said:
One can argue it's not a physical surface, but point is, logically to deform an infinitely curved region requires infinite coordinate time!

First of all, the event horizon is not "infinitely curved", and the infinite Schwarzschild coordinate time is irrelevant; it's an artifact of the coordinate singularity at r = 2M in Schwarzschild coordinates. Do we need to have a separate discussion on that point, or can I just refer to all the hundreds of previous threads where that subject has been beaten to death?

Next, the horizons do not get "deformed" or "merged"; rather, we have a single spacetime whose horizon (there is only one horizon) happens to be shaped like a pair of trousers, so to speak, rather than a simple cylinder. And this shape of the horizon, once again, is entirely explained by the original configuration of nonzero SET regions that collapsed to form the two BH's that then "merged".

Once again, I'm not saying it's "wrong" to think of BH's as "objects" instead of curvature; I'm just saying that thinking of them as "objects" is an approximation with a limited domain of validity. You are trying to stretch that approximation beyond its domain of validity, and it is breaking down. If you go back to the fundamentals, the actual precise physics based on the EFE, there is no problem.

Q-reeus said:
(just to be clear here; my own interpretation is that it shows the inconsistency of 'BH' in the first place. I am not speaking on anyone else's behalf in saying that.)

Q-reeus said:
I agreed only that this was the official GR position - never mine as made abundantly clear in many posts.

In other words, you basically do not accept that standard GR, based on solutions to the EFE, is valid. If you don't accept that, then we really don't have a good basis for discussion at all, because everything I've said is based on the premise that the EFE is valid. If you don't accept the EFE, then of course you're not going to accept the rest of what I'm saying. But I very much doubt you'll be able to convince me that the EFE is not valid in the domain we have been discussing (though you're welcome to try).

Q-reeus said:
In #45 pressure, which apart from rest mass/energy, forms the only other GR approved contribution to the SET in the scenario considered, was declared negligible...I had implicitly lumped pressure together with matter rest mass.

For the purposes of the discussion we were having, I don't see a problem with this. If we wanted to be precise, we could say that where we were talking about "rest mass", we should instead read "all significant components of the SET".

Q-reeus said:
My argument there and in #52 was not strictly correct (neither of us picked it up) - conversion, partial or fully, of non-gravitational energy to gravitational energy is not sufficient proof of itself that net system gravitating mass declines. One must account for pressure changes also.

Strictly speaking, yes, this is true. But I don't think it affects the general points either of us were making.

Q-reeus said:
1: Scaling law. Take the thin shell stipulated in #45. Double it's assembled mass M'. Pressure has doubled, and in GR the pressure contribution to curvature is a linear function of that pressure.

For this idealized case, yes, the pressure "contribution to curvature" (meaning through the EFE) is linear in the pressure. But the pressure itself is not necessarily linear in the assembled mass (i.e,. doubling the assembled mass does not necessarily double the pressure). You have to actually look at the appropriate solution of the EFE to see how the pressure depends on the assembled mass.

Q-reeus said:
2: There is good reason to doubt pressure makes *any* contribution to gravitational mass.

If by "gravitational mass" you mean the "assembled mass" M of a spherically symmetric gravitating body, you are simply wrong here. Solutions describing, for example, static spherically symmetric stars have been well known for decades, and pressure most certainly does contribute to the "assembled mass" of the star.

Q-reeus said:
GW amplitude is linear wrt pressure.

Why do you think this?

Q-reeus said:
But material displacement of the twin clamps under pressure is inversely proportional to the material elastic constant.

Within a certain range of pressures and displacements (until the material's elastic limit is reached), yes.

Q-reeus said:
Plastic clamps will flex far more for a given generated pressure than for say steel clamps.

But they also have less energy density. See below.

Q-reeus said:
any metric back reaction from generating GW's must induce far greater power drain in the plastic clamps case than for the steel ones...We have not included GW contribution owing to just motion of the clamp material, but that's ok since that contribution will be proportional to material density, which need have no relation to elastic constant.

Really? I agree there is not a straight linear relationship, but there is still some relationship.

Q-reeus said:
Only by denying the very existence of gravitational energy can the problem seemingly go away. But then e.g. Hulse-Tayler-binary-pulsar-data-as-proof-of-GW's issue, as before discussed, becomes somewhat problematic.

Why do you think this? The binary pulsar data is perfectly consistent with standard GR and the EFE, including the fact that the system is emitting GW's and that, consequently, the energy remaining in the system (which would correspond to its externally measured mass, if for example we put a test object in a far orbit about the system and measured its orbital parameters) is decreasing. All of this is perfectly well explained by the configuration of nonzero SET regions in the past light cone.

Q-reeus said:
Which misses the point; this is a valid gedanken experiment. It is possible for such a situation to exist and it implies certain things, which I have stated.

You are basically saying, if the evidence were different than it is, we would draw different conclusions. So what?

Q-reeus said:
I'll repeat. With pressure now taken care of, I have shown that there is necessarily such a beast as positive gravitational energy density in a static field.

You have shown no such thing. You have only shown that you can assign a reasonable meaning to the term "gravitational energy density in a static field" such that that density is positive. Again, so what? This says nothing about the fundamental physics; it only says that you can make a certain approximation work in a certain limited domain. I have never disputed that the approximation works within its limited domain.

Q-reeus said:
Conversion at least partially from non-gravitational to gravitational energy accompanies any collapse scenario.

Again, this is an approximation that works in a limited domain. It is not the fundamental physics.

Q-reeus said:
We now know from the forgoing this logically requires a net reduction in net system gravitating mass.

If GW's are emitted, yes.

Q-reeus said:
Your argument is to just stick with finding the SET in the event past light cone, but what's missing here is crucial. The *recipe* for what constitutes part of the SET. f gravitational energy is missing from that recipe (as GR insists), my last umpteen entries here have been trying to drive home the inconsistencies that then invariably result. Take it or leave it.

You have shown no such inconsistencies. Nothing you have said has rebutted my repeated claim that *all* of the observed physics can be explained and calculated using the standard GR recipe--solve the EFE using the nonzero SET regions (with the standard definition of SET) as the sources on the RHS. Why? Because nothing you have said is actually *derived* from trying to apply the standard recipe. Instead, you keep on applying your own recipe, based on your own approximate version of the physics, and finding that it doesn't work. You're right: it doesn't work.

In other words, all you have illustrated is that other, approximate ways of capturing the physics break down when you try to extend them beyond a limited domain. So what?
 
  • #69
Thanks for answering.

PeterDonis said:
The precise definition of "dark energy" is a nonzero SET that is proportional to the metric. A nonzero cosmological constant is one possible form of dark energy, but not the only one. (The fact that the SET of dark energy is proportional to the metric is the key difference from an "ordinary" SET derived from matter or EM radiation.) Since dark energy, precisely defined, has a nonzero SET, it appears on the RHS of the EFE, and that is how it affects the physics. Whether or not this kind of energy "counts" as "gravitational field energy" depends on how you define the latter term; but as I've noted already, you don't have to define that term to figure out the physics, so answering the question of whether dark energy counts as gravitational energy is not necessary for the physics.
Well the thing is in many GR texts the non-linearity of the EFE is attributed precisely to the very thing you are dismissing here as unnecessary or irrelevant for the physics:The gravitational field energy behaviour and the "gravity gravitates" issue. So they must have a different idea , or at least broader of what the physics of GR is.
Actually your answer to my question amounts to say it is not a relevant question for your idea of the relevant physics.
Dark energy in its most accepted interpretation, that which is compatible with GR, is thought to be precisely a repulsive gravitational field, and as you admit it is a nonzero SET. But you insist that the usual attractive gravitational field doesn't count as SET source, while the standard view is that precisely the fact that gravity gravitates is what makes the EFE non-linear.
 
  • #70
PeterDonis said:
Take your statement about the mass M: it is measured by "Keplerian dynamics of an orbiting test mass". Very true. My point is that that does *not* imply that anything inside the BH horizon is interacting with anything outside. It simply doesn't.
Assuming the existence of such a truly causally isolated region, sure and neither I or I assume yuiop are disagreeing with that. But what matters obviously is the entity 'BH' is interacting, as a mass M, with it's surrounds. Ergo, the external field is doing this - by logical reduction from your own statements. If the interior isn't interacting, hey, that just leaves the exterior, which is just the field! Call it the SET in the by and by, still boils down to: if not the interior, only one thing left.
That's all. We are used to thinking of gravitating bodies as "interacting" with other bodies (like the Sun and the Earth), without bothering to always remind ourselves that the "interaction" does not occur instantaneously--the Earth is not interacting with the Sun "right now", it is interacting (if that's even the right word) with the Sun eight minutes ago. But the latter is in fact the case. A BH is just a much more extreme case, where we might have to go back billions of years to find the nonzero SET region in the past light cone--but that's still the correct precise description of the physics. Thinking of the BH itself as "interacting" with orbiting bodies is *not*; it's an approximation with a limited domain of validity. You are trying to stretch it beyond its domain of validity, and it is breaking down.
Please define precisely the nature of this 'approximation' - and to what numerical extent is it an approximation.
...because it leads us to ask a question ("why are active m_g, passive m_g, and inertial mass m_a all the same?") that, from a proper conceptual scheme, would never even be asked, because it would be "obvious" that there was only one kind of mass-energy to begin with.
Go back and check the link I gave re Nordtvedt effect in #59. These researchers sure take the concept of gravitational binding energy seriously, and the fact all three masses applied to such are experimentally equal has the consequence I stated there - if there is no active gravitational energy mass, neither is there an inertial mass. Comfortable with that? And let's get one thing clear. Your repeated claim the standard EFE/SET setup explains everything is not really true. The differences between GR and 'gravity gravitates' theories in general are below the level of detection in all current tests. Baryshev link in another thread sets out some of the details.
First of all, the event horizon is not "infinitely curved", and the infinite Schwarzschild coordinate time is irrelevant; it's an artifact of the coordinate singularity at r = 2M in Schwarzschild coordinates. Do we need to have a separate discussion on that point, or can I just refer to all the hundreds of previous threads where that subject has been beaten to death?
Hopefully not needed. From a coordinate observer perspective, looking for this merger event to be over by breakfast, it is infinitely curved - else we say the SC's are lying/useless.
For this idealized case, yes, the pressure "contribution to curvature" (meaning through the EFE) is linear in the pressure. But the pressure itself is not necessarily linear in the assembled mass (i.e,. doubling the assembled mass does not necessarily double the pressure). You have to actually look at the appropriate solution of the EFE to see how the pressure depends on the assembled mass.
I specified weak gravity regime. Do you deny pressure will there double to all but a tiny and inconsequential corrective factor?
If by "gravitational mass" you mean the "assembled mass" M of a spherically symmetric gravitating body, you are simply wrong here. Solutions describing, for example, static spherically symmetric stars have been well known for decades, and pressure most certainly does contribute to the "assembled mass" of the star.
You commented before reading the rest!
Q-reeus: "GW amplitude is linear wrt pressure."
Why do you think this?
How could it be otherwise? What do the EFE's say on this any differently (weak gravity regime, remember!)? What do you propose instead?
Q-reeus: "Plastic clamps will flex far more for a given generated pressure than for say steel clamps."
But they also have less energy density...
My turn to say; so what? We are not talking about elastic energy here. One linear in displacement, one parametric. Chalk and cheese.
Q-reeus: "any metric back reaction from generating GW's must induce far greater power drain in the plastic clamps case than for the steel ones...We have not included GW contribution owing to just motion of the clamp material, but that's ok since that contribution will be proportional to material density, which need have no relation to elastic constant."
Really? I agree there is not a straight linear relationship, but there is still some relationship.
Irrelevant. What matters is they are independent parameters, and that is sufficient as part of the proof. Looks though like it deserves a separate thread - it is spelling death to a key concept in GR and you are saying 'ho hum'!
Why do you think this? The binary pulsar data is perfectly consistent with standard GR and the EFE, including the fact that the system is emitting GW's and that, consequently, the energy remaining in the system (which would correspond to its externally measured mass, if for example we put a test object in a far orbit about the system and measured its orbital parameters) is decreasing. All of this is perfectly well explained by the configuration of nonzero SET regions in the past light cone.
Again, missing the point, which is there is undeniably energy in the GW field. Yet re static field situations you are putting "energy" in inverted commas like that. Why? Do you actually see a fundamental distinction?
Q-reeus: "Which misses the point; this is a valid gedanken experiment. It is possible for such a situation to exist and it implies certain things, which I have stated."
You are basically saying, if the evidence were different than it is, we would draw different conclusions. So what?
No, I am saying with things being as they are we can draw different conclusions. The very fact that pseudo-tensors are needed to get any sort of decent energy definitions in GR should be making that evident.
...Nothing you have said has rebutted my repeated claim that *all* of the observed physics can be explained and calculated using the standard GR recipe--solve the EFE using the nonzero SET regions (with the standard definition of SET) as the sources on the RHS. Why? Because nothing you have said is actually *derived* from trying to apply the standard recipe. Instead, you keep on applying your own recipe, based on your own approximate version of the physics, and finding that it doesn't work...
This is a repeated theme I deny is accurate. You are on record as stating the SET specifically includes only matter contributions, and that means gravitational energy/"energy" is excluded. I have dealt with pressure and shown it cannot nullify the need of a gravitational field energy ("energy" if you wish). Kindly go back then to the scenario in #52 and show me, point by specific point, where you think I an the one getting things wrong. Point by point, not dismissive generalities. That should be real interesting.
 
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<h2>1. What is the relative energy of a black hole?</h2><p>The relative energy of a black hole refers to the amount of energy contained within the black hole. This energy is primarily in the form of gravitational potential energy, which is created by the immense mass of the black hole.</p><h2>2. How is the relative energy of a black hole measured?</h2><p>The relative energy of a black hole is measured using a unit called the Schwarzschild radius, which is the distance from the center of the black hole at which the escape velocity equals the speed of light. This radius is directly proportional to the mass of the black hole, so the larger the mass, the greater the relative energy of the black hole.</p><h2>3. Can the relative energy of a black hole change?</h2><p>Yes, the relative energy of a black hole can change over time due to a process called Hawking radiation. This is a slow emission of energy from the black hole, causing it to gradually lose mass and decrease in relative energy.</p><h2>4. How does the relative energy of a black hole affect its surroundings?</h2><p>The relative energy of a black hole has a significant impact on its surroundings. Objects that come too close to the black hole can be pulled in due to its strong gravitational pull, and the intense energy can also distort the fabric of space-time around it.</p><h2>5. Is the relative energy of a black hole the same as its mass?</h2><p>No, the relative energy of a black hole is not the same as its mass. While the mass of a black hole contributes to its relative energy, there are other factors such as its spin and charge that also play a role in determining the total energy of the black hole.</p>

1. What is the relative energy of a black hole?

The relative energy of a black hole refers to the amount of energy contained within the black hole. This energy is primarily in the form of gravitational potential energy, which is created by the immense mass of the black hole.

2. How is the relative energy of a black hole measured?

The relative energy of a black hole is measured using a unit called the Schwarzschild radius, which is the distance from the center of the black hole at which the escape velocity equals the speed of light. This radius is directly proportional to the mass of the black hole, so the larger the mass, the greater the relative energy of the black hole.

3. Can the relative energy of a black hole change?

Yes, the relative energy of a black hole can change over time due to a process called Hawking radiation. This is a slow emission of energy from the black hole, causing it to gradually lose mass and decrease in relative energy.

4. How does the relative energy of a black hole affect its surroundings?

The relative energy of a black hole has a significant impact on its surroundings. Objects that come too close to the black hole can be pulled in due to its strong gravitational pull, and the intense energy can also distort the fabric of space-time around it.

5. Is the relative energy of a black hole the same as its mass?

No, the relative energy of a black hole is not the same as its mass. While the mass of a black hole contributes to its relative energy, there are other factors such as its spin and charge that also play a role in determining the total energy of the black hole.

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