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Relative energy of a black hole. |
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| Feb27-12, 02:39 PM | #52 |
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Relative energy of a black hole.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. |
| Feb27-12, 04:16 PM | #53 |
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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. |
| Feb28-12, 07:19 AM | #54 |
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![]() 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. |
| Feb28-12, 12:21 PM | #55 |
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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. 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. |
| Feb28-12, 12:27 PM | #56 |
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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: |
| Feb28-12, 01:43 PM | #57 |
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| Feb28-12, 02:36 PM | #58 |
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(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. |
| Feb28-12, 03:29 PM | #59 |
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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/...6-3/index.html. 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.
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| Feb28-12, 04:18 PM | #60 |
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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. 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.) 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.) |
| Feb28-12, 04:51 PM | #61 |
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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. |
| Feb28-12, 07:02 PM | #62 |
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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. 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. |
| Feb29-12, 09:46 AM | #63 |
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Peter, could you take a look at my #47?
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| Feb29-12, 10:13 AM | #64 |
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Looks impressive, like lava lamp blobs fusing together, with electrostatics thrown in to 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.) |
| Feb29-12, 10:17 AM | #65 |
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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. |
| Feb29-12, 10:29 AM | #66 |
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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: #46 Yuiop: #51 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: 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: and this example helps... [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: I also wondered about trickydicky question. (#47)" "What about dark energy." |
| Feb29-12, 10:46 AM | #67 |
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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. |
| Feb29-12, 11:30 AM | #68 |
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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. 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. 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? |
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