
#37
Feb2512, 07:30 AM

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#38
Feb2512, 03:21 PM

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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 nonzero stressenergy 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 followup 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. 



#39
Feb2512, 03:27 PM

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Following up from my previous post:
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
Feb2612, 03:02 AM

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#41
Feb2612, 06:56 AM

P: 5,634

Qreeus posts #18
from my post #3: pervect #17 Qreeus: post #20 : Sam Gralla: PeterDonis: post#28 from Qreeus: post #31 Doesn’t a portion of the Einstein stress energy tensor capture the effects of an EM field, like T00 here: http://en.wikipedia.org/wiki/Stress..._of_the_tensor If so, the effects of 'self interaction' would likely lie within the remained of the Einstein stress energy tensor.....anybody know what I am trying to describe??? edit: the analogy I thought to myself at the time was that maybe gravitons self ineract in a way photons don'ts....but the original description was a classical one, not quantum. 



#42
Feb2612, 09:15 AM

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#43
Feb2612, 11:35 AM

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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 zerothe 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 "worldtube", 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" propagatingthe 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
Feb2612, 11:45 AM

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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
Feb2612, 02:05 PM

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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 massenergy is given off purely as heat that radiates away totally. If subsequently one constituent matter particle selfannihilates 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 N1 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 stressenergy contribute to curvature. Doesn't seem right. 



#46
Feb2612, 10:00 PM

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One example of a definition of "energy in the gravitational field" is described here: http://en.wikipedia.org/wiki/Stress%...m_pseudotensor However, as I said above, I don't believe this is the only definition in the literature. (Other GR experts here may know more about this.) The key point, though, is that "energy in the gravitational field" does *not* act as a "source" of gravity according to the Einstein Field Equation (because only a nonzero SET does that). In other words, the "energy in the gravitational field" is *not* "stressenergy". So: This may seem like playing with words, but let's consider the (very good) examples you bring up: 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 stressenergy. 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 curvaturethey *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: (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
Feb2712, 05:15 AM

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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 matterenergy SET). Why one gravitational field energy is "stressenergy" in one case but not in the other? 



#48
Feb2712, 06:54 AM

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#49
Feb2712, 08:34 AM

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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 stressenergy (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 energyatinfinity of the system). But the stressenergy 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. 



#50
Feb2712, 10:20 AM

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#51
Feb2712, 12:19 PM

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But measuring "M" this way tells us nothing about how it relates to the presence of a nonzero 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. 



#52
Feb2712, 02:39 PM

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Consider please the following scenario: A large bounding box of mass M_{b} with perfectly reflecting walls. Inside we have diffuse dust of mass M_{d} >> M_{b} that over time gravitationally collapses symmetrically to form a stable, static planet of assembled mass M_{p} < M_{d}. 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 nongravitational 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 nonsources) accessible in perfectly reasonable time to a region exterior to the box. 



#53
Feb2712, 04:16 PM

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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. 



#54
Feb2812, 07:19 AM

P: 1,115

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 nongravitational 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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