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Are Finkelstein/Kruskal interior black hole solution compatible with Einstein's GR?

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stevendaryl
#91
Dec5-12, 02:03 PM
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Quote Quote by Mentz114 View Post
There is no 't' in Schwarzshild.
...because the metric is independent of t.
stevendaryl
#92
Dec5-12, 02:10 PM
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Quote Quote by harrylin View Post
According to GR there is matter in the universe
I would say that it is an empirical question. Look around, you see matter, so the vacuum solutions are not relevant. But GR can describe both a universe with matter and a universe without matter.
PeterDonis
#93
Dec5-12, 02:27 PM
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Quote Quote by stevendaryl View Post
I would say that it is an empirical question. Look around, you see matter, so the vacuum solutions are not relevant.
There is matter, but not everywhere; there are certainly regions of the actual universe which are, at least to a very good approximation, vacuum. Vacuum solutions are certainly relevant for describing such regions. We use the Schwarzschild metric to describe spacetime around the Earth; we just don't use the entire global manifold, we use a portion of it. The EFE is local, so this is perfectly valid.
harrylin
#94
Dec5-12, 02:33 PM
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Quote Quote by harrylin View Post
[..] I do realise that the title of this thread is unclear, so I will continue this with a clearer title.
By chance (or perhaps not?) just now a new topic has been started that is very close to the topic title that I had in mind to continue with. In order not to duplicate threads I joined the discussion there:
http://www.physicsforums.com/showthread.php?p=4185579
harrylin
#95
Dec5-12, 02:38 PM
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Quote Quote by stevendaryl View Post
I would say that it is an empirical question. Look around, you see matter, so the vacuum solutions are not relevant. But GR can describe both a universe with matter and a universe without matter.
Concerning the second part, that's assuming that GR was intended for other universes than our own. I don't think so. Why would other universes have the same laws of nature as ours? I'm afraid that this really gets too philosophical and speculative...
Quote Quote by PeterDonis View Post
There is matter, but not everywhere; there are certainly regions of the actual universe which are, at least to a very good approximation, vacuum. [..]
Quite so; but wasn't the white hole solution intended to start near a black hole?
Mentz114
#96
Dec5-12, 02:47 PM
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Quote Quote by stevendaryl View Post
...because the metric is independent of t.
But there are 2 'c's, one of which I missed out. Irony.
harrylin
#97
Dec5-12, 02:49 PM
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Quote Quote by Mentz114 View Post
I apologise for my tetchy attitude. But when it comes to GR there is only one brand. The one invented by Einstein - so why put the soubriquet on. You seem to think there are many brands.

Anyhow, I'll stay out of this now.
In the literature and discussions I found different flavours of GR, and for me it is an unanswered question if that matters or not for the metaphysics. But thanks for your apology, such little things make PF a nice place to be in.
Quote Quote by PeterDonis View Post
It would depend on what the discussion was about. Yes, I noticed "mislabeling", in the sense that, as I said, the "Einstein Equivalence Principle" as it is currently used in GR (Pepsi) is not precisely the same principle that Einstein himself stated (Coke).

If the discussion is about what Einstein said, then yes, asking for Coke is perfectly reasonable. But if the discussion is about what's currently used in GR, then a client who keeps asking for Coke even after everybody has pointed out repeatedly that the discussion is really about Pepsi would seem a little weird.
This thread was intended to have the exact taste of Coke, in order to reduce the mutual misunderstandings that were experienced earlier (but it didn't work because I didn't explain the topic well enough).
harrylin
#98
Dec5-12, 03:00 PM
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Quote Quote by PeterDonis View Post
In other words, you meant "Adam' thinks he is in inertial motion, but then he discovers that he really isn't". But on what basis would Adam' even *think* he was in inertial motion? On the standard definition of "inertial motion", Adam' could measure directly that he was in inertial motion, by using an accelerometer, as I said. But on your definition, inertial motion doesn't mean free fall, it means motion in a straight line with respect to the gravitating body. On what basis would Adam' first think he is moving in a straight line, but then be forced to conclude otherwise?
He would only be forced to conclude otherwise if he could discern the existence of a gravitating body in the vicinity. Else, he simply wouldn't know. Similarly, a bee flying towards a window has no reason to expect the existence of the window - until hitting the glass.
stevendaryl
#99
Dec5-12, 03:08 PM
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Quote Quote by harrylin View Post
Concerning the second part, that's assuming that GR was intended for other universes than our own
I think you're being a little silly. Laws of physics typically describe more than what is actually the case. Newtonian physics will tell you what would happen in a perfectly elastic collision of two perfect spheres the size of the Earth. There are no perfect spheres the size of the Earth that are perfectly elastic.

Typically, theories of physics tell you what follows from hypothesized initial conditions. Usually, the theories don't tell you what the initial conditions are, you have to find those out empirically.

It would be pretty weird if GR only applied to our universe.
PeterDonis
#100
Dec5-12, 03:20 PM
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Quote Quote by harrylin View Post
Quite so; but wasn't the white hole solution intended to start near a black hole?
The short answer is "no", but perhaps it's worth expanding on this.

(First, a quick note: "near the black hole" is still vacuum. The black hole region is vacuum at the horizon, and all the way down to r = 0. But I think that's a minor point compared to what I'm going to say below.)

Suppose we want to solve the Einstein Field Equation subject to the following conditions:

(1) The spacetime is spherically symmetric.
(2) The spacetime is vacuum everywhere--i.e., there is no matter *anywhere*, ever.

The complete solution to the EFE under these conditions includes an exterior region (which I'll call region I), a black hole region (region II), a second exterior region (region III), and a white hole region (region IV). The solution doesn't "start near a black hole"; it doesn't "start near" anywhere. It's just the complete solution we get when we impose those conditions ("complete" meaning "including all possible regions which are indicated by the math, whether they are physically reasonable or not").

Suppose we want to solve the Einstein Field Equation subject to the following somewhat different conditions:

(1') The spacetime is spherically symmetric.
(2') On some spacelike slice, the spacetime is vacuum for radius > R_0 (where R_0 is some positive value), but is *not* vacuum for radius <= R_0; instead, the region r <= R_0 on this spacelike slice is filled with dust (where "dust" means "a perfect fluid with positive energy density and zero pressure") which is momentarily at rest.
(3') We are only interested in the spacetime to the future of the spacelike slice given in #3.

The complete solution we get when we impose these conditions is what I'll call the "modernized Oppenheimer-Snyder model" ("modernized" to avoid any concerns about whether or not it was the model O-S originally proposed; this model is described, for example, in MTW). This spacetime has three regions: an exterior vacuum region (which I'll call region I'), a black hole interior vacuum region (region II'), and a non-vacuum collapsing region (region C'). There is no white hole region, and no second exterior region, in this spacetime.

Now, in the vacuum regions I' and II', the solution of the EFE is the vacuum solution: that is, it is *exactly the same* as the solution in the corresponding portions of regions I and II. Another way of saying this: if I describe regions I and II in a suitable coordinate chart, and regions I' and II' in a suitable coordinate chart, I can identify an open set of coordinate values in regions I and II that meet the following conditions:

(A) The coordinate values are exactly the same as the ones in regions I' and II'; and
(B) The invariant quantities at each corresponding set of coordinate values (I <-> I', and II <-> II') are identical.

A fairly common shorthand, I believe, for what I've said above is that region I' is isometric to a portion of region I, and region II' is isometric to a portion of region II. Or, speaking loosely, regions I' and II' can be thought of as "pieces" of regions I and II that have been "cut and glued" to region C'.

Hopefully all this makes somewhat clearer how the term "solution" is being used, and what it means to say that "the same solution" appears in different models.
PeterDonis
#101
Dec5-12, 03:24 PM
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Quote Quote by harrylin View Post
This thread was intended to have the exact taste of Coke
But in so far as Coke is different from Pepsi here, nobody actually uses Coke as a physical theory today. Everybody uses Pepsi (i.e., "modern GR", not "Einstein's GR", to whatever extent they are different, which I'm not even taking a position on right now). So if you're really interested in whether the Finkelstein or Kruskal metrics are consistent with Einstein's GR, as opposed to the GR that is actually used as a scientific theory today, you're interested in a question that only matters for history, not physics. If that's really your intent, you should make it crystal clear in the OP of a new thread that you're interested in the history, not the physics.
PeterDonis
#102
Dec5-12, 03:35 PM
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Quote Quote by harrylin View Post
He would only be forced to conclude otherwise if he could discern the existence of a gravitating body in the vicinity.
We've already stipulated that he can, because he can detect tidal gravity (as can Eve'). But given that, why would he ever assume he was moving in a straight line in the first place?

Maybe I should expound a bit more on what I'm looking for here. The standard view of this scenario is that the two cases are exactly parallel: in both cases, the accelerated observer (Eve, Eve'), because of her proper acceleration, is unable to observe or explore a region of spacetime that the free-falling observer (Adam, Adam') can. The physical criterion that distinguishes them is clear, and is the same in both cases (zero vs. nonzero proper acceleration).

You are claiming that, contrary to the above, the cases are different: Adam is "privileged" in the first case, but Eve' is in the second. So I'm looking for some criterion that picks out Adam in the first case, but picks out Eve' in the second; in other words, something that applies to Adam but not Eve, and applies to Eve' but not Adam'. The only criterion I have so far is "moves in a straight line according to my chosen coordinates", but that only pushes the problem back a step: what is it that applies to the coordinates of Adam but not Eve, *and* to those of Eve' but not Adam'? I haven't seen an answer yet.
DrGreg
#103
Dec5-12, 03:39 PM
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Quote Quote by Mentz114 View Post
But there are 2 'c's, one of which I missed out. Irony.
OK if you assume c=1.
harrylin
#104
Dec5-12, 04:45 PM
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Quote Quote by PeterDonis View Post
But in so far as Coke is different from Pepsi here, nobody actually uses Coke as a physical theory today. Everybody uses Pepsi (i.e., "modern GR", not "Einstein's GR", to whatever extent they are different, which I'm not even taking a position on right now). So if you're really interested in whether the Finkelstein or Kruskal metrics are consistent with Einstein's GR, as opposed to the GR that is actually used as a scientific theory today, you're interested in a question that only matters for history, not physics. If that's really your intent, you should make it crystal clear in the OP of a new thread that you're interested in the history, not the physics.
A number of people who participated in these threads hold that the GR that is actually used is effectively that theory; I don't know, perhaps it only sounds different. But physics is concerned with predictions based on established theory that has not been invalidated by experiment - else it would be religion. Thus the question concerns not just history but correct current presentation of physics theory.
PeterDonis
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Dec5-12, 04:52 PM
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Quote Quote by harrylin View Post
A number of people who participated in these threads hold that the GR that is actually used is effectively that theory
Yes, I think I'm one of them. But it does depend on what you consider to be "effectively that theory", and that, to me, is a matter of history (and perhaps terminology), not physics.

Quote Quote by harrylin View Post
physics is concerned with predictions based on theory that has not been invalidated by experiment - else it would be religion.
I agree. My point about history vs. physics is simply that if you're interested in our best current theory that hasn't been invalidated by experiment, whether or not it's "the same theory that Einstein used" is irrelevant. You're not going to read Einstein to learn it anyway; you're going to read the most up to date textbooks and literature you can find.

Quote Quote by harrylin View Post
Thus the question concerns not just history but correct current presentation of physics theory.
To me these are two different questions, and I'm trying to figure out which one we should be talking about: the history question or the current physics question. I don't see how "correct presentation" of the current theory has to even take any position on the historical question. Of course the historical question is interesting, but the current theory stands or falls on its own merits regardless of how, historically, it has gotten to this point.
DaleSpam
#106
Dec6-12, 06:24 AM
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Quote Quote by PeterDonis View Post
No argument with that; I just don't like using the term "boundary condition" to refer to this, since it's not something you impose before you derive the solution, it's something you discover in the course of doing the solution. But as I said, that's a matter of terminology, not physics or mathematics.
Agreed. To me a condition on the boundary is a boundary condition regardless of whether you found it by solving the differential equation and then specifying the resulting constants or if you inserted in the condition before solving the differential equation. The math doesn't care about the order, but as you say, this is just terminology.

If you want to distinguish the two then I would suggest "constant of integration" for the post-hoc constants and "boundary condition" for the a-priori constants. Under that categorization (which I wouldn't use) I would agree that the curvature at the horizon arises from a constant of integration rather than a boundary condition.

You can always change a constant of integration into a boundary condition by changing the order of operations.
PeterDonis
#107
Dec6-12, 08:52 AM
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Quote Quote by DrGreg View Post
OK if you assume c=1.
Of course, in quantum gravity we would set c = h = 1, to get...Swarzsild?
harrylin
#108
Dec6-12, 09:15 AM
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Quote Quote by PeterDonis View Post
[..] My point about history vs. physics is simply that if you're interested in our best current theory that hasn't been invalidated by experiment, whether or not it's "the same theory that Einstein used" is irrelevant. You're not going to read Einstein to learn it anyway; you're going to read the most up to date textbooks and literature you can find. [..]
To my knowledge Einstein's GR as I defined here is our best current theory that hasn't been invalidated by experiment. It is always possible to reformulate a theory in such a way that the interpretation changes but the verifiable predictions remain the same. And I agree with the mentors that differing philosophies should not be debated on this forum, as that is useless. Tempting as it is to continue with discussing philosophy (which would deteriorate into debating it), I will insist on discussing numbers - as I also tried (but without insisting on it) in this thread.


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