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

In summary: Adam-like object) on a Schwarzschild spacetime, then what she observes changes depending on her frame of reference. But that's not what happens on a proper acceleration spacetime.In summary, on a proper acceleration spacetime, what Eve observes changes depending on her frame of reference, but on a Schwarzschild spacetime, she remains in the same place and observes the same thing.
  • #71


PeterDonis said:
Can you be more specific about the boundary condition you have in mind? I assume it's at the horizon, because the boundary condition at spatial infinity is the same (asymptotic flatness). But I'm not sure I would characterize the presence of a coordinate singularity at the horizon in SC coordinates as a "boundary condition", and I'm not sure what else you could be referring to.

I'm referring simply to the fact that you require r>Rs, else you have coordinate patch of a TBD complete solution rather than a complete solution. This is a boundary condition.
 
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  • #72


PAllen said:
I'm referring simply to the fact that you require r>Rs, else you have coordinate patch of a TBD complete solution rather than a complete solution. This is a boundary condition.

As I said before, I would not call r > Rs a "boundary condition", because it's not imposed prior to deriving the solution. It's something you *discover* after you've already derived the solution: you solve the vacuum Einstein equation subject to the condition of spherical symmetry, with coordinate conditions that put the line element into a certain general form:

[tex]ds^2 = A(r) dt^2 + B(r) dr^2 + r^2 d\Omega^2[/tex]

Then you solve the vacuum EFE to find A(r) and B(r). Only after you've done that do you discover that there is a coordinate singularity at r = Rs, where A(r) = 0 and B(r) is undefined, meaning that the solution you derived, with the form of the line element you used, is only valid on a patch with r > Rs. That's not a boundary condition, because you didn't assume it, you derived it; it's a limitation of the solution when given in that form.

(Actually, strictly speaking the line element you derive is perfectly valid for 0 < r < Rs as well as for Rs < r < infinity. So what you've actually discovered is that there are *two* disconnected coordinate patches on which your line element is valid. But that's still not a boundary condition.)

Also, I would not characterize the exterior SC chart as a "complete solution", because it's geodesically incomplete at the horizon--infalling geodesics have a finite length when they reach the edge of the coordinate patch at r -> Rs--and also because all of the physical invariants that are defined at the horizon are finite there, indicating that there is no curvature singularity or any other reason why the manifold would not continue. So geometrically the coordinate patch with r > Rs cannot be the entire manifold.
 
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  • #73


PeterDonis said:
As I said before, I would not call r > Rs a "boundary condition", because it's not imposed prior to deriving the solution. It's something you *discover* after you've already derived the solution: you solve the vacuum Einstein equation subject to the condition of spherical symmetry, with coordinate conditions that put the line element into a certain general form:

[tex]ds^2 = A(r) dt^2 + B(r) dr^2 + r^2 d\Omega^2[/tex]

Then you solve the vacuum EFE to find A(r) and B(r). Only after you've done that do you discover that there is a coordinate singularity at r = Rs, where A(r) = 0 and B(r) is undefined, meaning that the solution you derived, with the form of the line element you used, is only valid on a patch with r > Rs. That's not a boundary condition, because you didn't assume it, you derived it; it's a limitation of the solution when given in that form.

(Actually, strictly speaking the line element you derive is perfectly valid for 0 < r < Rs as well as for Rs < r < infinity. So what you've actually discovered is that there are *two* disconnected coordinate patches on which your line element is valid. But that's still not a boundary condition.)
The way I look at this is that this is what you do as part of the search for the complete solution without any boundary conditions except of asymptotic flatness. A series of further analyses lead you to the complete solution. However, having got this far, you are entitled to say you are only interested in r>Rs, and posit this as a boundary condition.
PeterDonis said:
Also, I would not characterize the exterior SC chart as a "complete solution", because it's geodesically incomplete at the horizon--infalling geodesics have a finite length when they reach the edge of the coordinate patch at r -> Rs--and also because all of the physical invariants that are defined at the horizon are finite there, indicating that there is no curvature singularity or any other reason why the manifold would not continue. So geometrically the coordinate patch with r > Rs cannot be the entire manifold.
The definition of manifold has no requirement of geodesic completeness. A 2-sphere minus a pair of antipodal points is a riemanninan manifold, for example. There is a requirement that it be a collection of open sets, but there is no completeness requirement. As I said, if you don't impose some other defining condition for the manifold, you see the SC metric corresponds to two patches, and if you want the most complete solution you have to investigate further. However, there is nothing wrong (mathematically - physically is another matter) with declaring r> Rs as a boundary condition.

(Physically, I stand by my argument that the boundary condition r> Rs leads to a violation of the equivalence principle for classical GR; but for mathematical GR, it's not a problem).
 
  • #74


PAllen said:
The definition of manifold has no requirement of geodesic completeness.

I agree as a matter of mathematics. As a matter of physics, I think it depends on what you're trying to do. See below.

PAllen said:
As I said, if you don't impose some other defining condition for the manifold, you see the SC metric corresponds to two patches, and if you want the most complete solution you have to investigate further. However, there is nothing wrong (mathematically - physically is another matter) with declaring r> Rs as a boundary condition.

I'm still not sure I would use the term "boundary condition" for this; I would prefer to describe it as restricting attention to only a portion of the complete manifold, for whatever reason (maybe you're only interested in static observers outside the horizon, so you don't care that infalling geodesics are incomplete at the horizon on your restricted patch). But that's really a matter of terminology, not physics; we appear to agree that physically, the geodesic incompleteness at the horizon is a sign that you need to look for a completion of the manifold if you're interested in a complete solution.
 
  • #75


I'm with PAllen on this one. Since the exterior Schwarzschild coordinates only include the portion of the manifold outside the horizon the horizon is a boundary and the Riemann curvature at the boundary is a boundary condition. This boundary condition can be characterized by a single parameter, usually denoted M.
 
  • #76


PeterDonis said:
I'm still not sure I would use the term "boundary condition" for this; I would prefer to describe it as restricting attention to only a portion of the complete manifold

Just as a matter of terminology, if you wanted to describe, e.g. a flat 2-manifold consisting of a plane missing a disc, how would you describe the 'missing a disc' condition? Given the definition of local flatness, I don't know what else to call this condition besides a boundary condition.
 
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  • #77


PAllen said:
Just as a matter of terminology, if you wanted to describe, e.g. a flat 2-manifold missing disc, how would you describe the 'missing a disc' condition? Given the definition of local flatness, I don't know what else to call this condition.

It would depend on why I was imposing the condition. If I were trying to model, say, an actual physical disk with a hole in it, then I would call the "missing a disk" condition a boundary condition. But if the actual physical object was a disk without a hole, but for some reason I was only interested in an annular section of it, I would not call the "missing a disk" condition a boundary condition; I would say that I was only interested in an incomplete portion of the complete physical object.
 
  • #78


I would call it a boundary condition because you need to know the value on that edge in order to solve the differential equation. That is what a boundary condition is. But, you need to know more than just that there is an edge, you need to know the value of your function at the edge.

Sorry, I am rambling, I think I will go sleep.
 
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  • #79


DaleSpam said:
I would call it a boundary condition because you need to know the value on that edge in order to solve the differential equation. That is what a boundary condition is. But, you need to know more than just that there is an edge, you need to know the value of your function at the edge.

If the actual, physical object is annular (a disk with an actual hole in it), then yes, the values of physical parameters at the edge are going to change because of the boundary condition there (something like "density goes to zero"), and that's going to affect the solution.

If the actual, physical object is a disk with no hole, but I restrict attention to an annular region, then no, that "boundary condition" does not affect the solution; I have to solve the equations for a disk with no hole, and then restrict attention to the portion of the solution that I'm interested in. If I solve the equations with the "boundary conditions" included, I will get the wrong solution, the one for a disk with a hole, not the one for a disk with no hole where I'm not looking at a region in the center.

Or let's consider this case: I'm modeling a capacitor. Consider two different possibilities:

(1) The actual, physical capacitor has finite-sized plates, and I'm interested in their entire area. My solution will then have edge effects because of the finite plate size; in other words, the boundary condition affects the solution.

(2) The actual, physical capacitor has extremely large plates, but I'm only interested in a small area near the center. I impose a "boundary condition" of a small plate area; but if I then solve the equation using that condition, I will get the wrong answer: my solution will have edge effects, and there aren't any in the small area I'm considering--the edge effects are way out in a different place that I'm not modeling. To get the right answer, I have to ignore the "boundary condition" and instead solve the equations with a much larger "real" plate size, and then restrict my solution to the small area I'm interested in. (What I'll really do, of course, is solve the equations assuming infinitely large plates, but that's just a shortcut to make the math easier since I know it will give the same answer as the more complicated procedure that's strictly correct.) The "boundary condition" here does *not* affect the solution, which is why I would prefer *not* to call it a "boundary condition".

The parallel with the case of Schwarzschild spacetime should be clear. If I impose a "boundary condition" r > Rs, that doesn't change any of the actual geometric invariants at Rs or as r -> Rs. Those invariants are all given by the solution with no boundary condition at the horizon, the one I derive purely from the EFE with the assumptions of spherical symmetry and vacuum plus the coordinate conditions on the line element. The boundary condition adds nothing to the solution; it's just a way of restricting attention to a portion of the complete manifold.

Contrast this with, for example, a solution for a static, spherically symmetric star. Here there is a boundary condition that does affect the solution: there will be some radial coordinate r at which the spacetime is no longer vacuum. That does affect the geometric invariants.
 
  • #80


I'm looking at it this way:

1) If I want SC coordinates as one or two patches of a more complete manifold, there are only boundary conditions at infinity; further, I need to do further analysis to find the complete manifold consistent with no boundary conditions except at infinity.

2) If I want SC exterior coordinates to represent a complete GR solution (not a patch of a more complete solution), then I need additional boundary conditions. I must posit that r > Rs, and that the metric remains vacuum and spherically symmetric at this boundary. (How do I know the value of Rs beforehand? It doesn't matter. I can try (1), and let this suggest (2).)

For mathematical GR, (1) leads to the unique complete Kruskal manifold; (2) leads to one exterior region considered as a solution (manifold) unto itself. In either case, any of dozens of popular coordinates can be used (along with infinite others).

Physically, I can argue that (2) is absurd even on local grounds, and (1) is also absurd for different (global) reasons. To get an idealized, physically plausible model of something, the simplest is O-S collapse, incorporating part of one interior and one exterior sheet of the complete Kruskal manifold in its 'late' stage.
 
  • #81


PAllen said:
I'm looking at it this way:

As I said before, mathematically and physically I agree with this; my only quibble is with the term "boundary condition" in (2), but that's a matter of terminology, not physics (or math).
 
  • #82


PeterDonis said:
[..] You said that Adam' must conclude that his inertial motion is an illusion. That claim makes no sense unless you believe that Adam' is in inertial motion. [..]
Instead, I meant it in the sense of "the child will soon realize that his Santaclaus is an illusion" (which does not mean that I believe that Santaclaus exists). Thanks for pointing out that that sentence was unclear for you.
Finkelstein coordinates can be used to continuously describe the motion of Adam' at and below the horizon, while Schwarzschild coordinates cannot. This is exactly parallel to the way that Minkowski coordinates can be used to describe the motion of Adam at and beyond the Rindler horizon, while Rindler coordinates cannot.

I'm beginning to wonder if you understand what a coordinate chart is and what two charts both covering the same region of a spacetime means.
That was clear; we were discussing the physical interpretations (or, since nothing of the issue is verifiable to us: the metaphysical interpretation according to theory). The problem is that while I understand most of your answers, you misunderstand most of my questions. I noticed that the most difficult exercise with this type of discussions is not to explain the answers clearly enough but to explain the questions clearly enough. I'll try to better - and I may have come up with title question that is unlikely to lead to misunderstanding (to start after reading up on Finkelstein).
 
  • #83


PeterDonis said:
I have to solve the equations for a disk with no hole, and then restrict attention to the portion of the solution that I'm interested in.
I think this is the point where I disagree with you. If the coordinate chart you are using doesn't cover the entire disk then you cannot solve the equations for the disk with no hole using those coordinates. So not only do you not "have to" do what you suggest, what you suggest is actually mathematically impossible.

PeterDonis said:
(2) The actual, physical capacitor has extremely large plates, but I'm only interested in a small area near the center. I impose a "boundary condition" of a small plate area; but if I then solve the equation using that condition, I will get the wrong answer: my solution will have edge effects, and there aren't any in the small area I'm considering
No, the boundary condition is not the size of the plate, but the behavior of the currents and fields at the edge of the region of interest. In this case you would use a boundary condition appropriate for the situation, one which represents the condition that currents can enter and leave the boundary freely. If you do that, then you will get the correct answer.

Obviously, if you use the "non-conductive" boundary condition in a situation where the boundary is "conductive" then you will get a wrong answer, but that is because you used the wrong boundary condition, not because what you used isn't a boundary condition nor because you needed to solve a bigger problem.

PeterDonis said:
The parallel with the case of Schwarzschild spacetime should be clear. If I impose a "boundary condition" r > Rs, that doesn't change any of the actual geometric invariants at Rs or as r -> Rs.
The boundary condition isn't r>Rs. It is the curvature at the boundary. This is imposed when you choose the parameter M. Setting M is what sets your boundary condition, regardless if your chart covers all r>Rs or even just some r>R0>Rs.
 
  • #84


Mentz114 said:
[..] People who cannot refer to GR (or GTR if you prefer) without tacking 'Einsteins' in front of it also warrant suspicion. [..]
We (Peter and I) noticed mislabeling in an earlier thread, as referred here. If you are in a restaurant that is known to have mislabeled bottles in the past and you hear one waiter say to another that a client "warrants suspicion" because he asked for Coca Cola and not Pepsi Cola, what would you think?
 
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  • #85


DaleSpam said:
The boundary condition isn't r>Rs. It is the curvature at the boundary. This is imposed when you choose the parameter M. Setting M is what sets your boundary condition, regardless if your chart covers all r>Rs or even just some r>R0>Rs.

This is the key point, expressed much better by Dalespam. You've got boundary condition at infinity (asymptotic flatness); you've got constraint for (e.g.) vacuum and spherical symmetry. That still doesn't determine a unique solution. Specifying curvature on any 2-shere of given area respecting the symmetry, that you take as your inner boundary, is exactly a boundary condition.

[Edit: being more specific, suppose I don't know the form of the solution, and don't know how much information needs to be specified on the boundary to get a unique solution. So, since I am assuming vaccuum, I know the Ricci scalar is identically zero. I make a guess - let's set an arbitrary constant value of the Kretschmann scalar for the inner boundary, and further require staticity. Then, I find a unique solution (as long as I don't pick K too large; if I do, I can't meet staticity). Or, don't impose staticity. Then, you find for given boundary r, if you pick K above a critical value, you have a horizon in your solution at some greater r than your boundary.

Without the extra boundary condition, you find a one parameter family of solutions, of more general 'shape' - the Kruskal geometries. You can then use Newtonian behavior at weak field as a condition to define the free parameter physically.]
 
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  • #86


DaleSpam said:
I think this is the point where I disagree with you. If the coordinate chart you are using doesn't cover the entire disk then you cannot solve the equations for the disk with no hole using those coordinates.

Ah, ok, I see where you're coming from. Yes, you're right, if I hit a coordinate singularity when I reach the boundary of some annular region, I can't continue the solution using that chart. 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.

DaleSpam said:
No, the boundary condition is not the size of the plate, but the behavior of the currents and fields at the edge of the region of interest. In this case you would use a boundary condition appropriate for the situation, one which represents the condition that currents can enter and leave the boundary freely. If you do that, then you will get the correct answer.

But what if you don't know the currents and fields prior to actually deriving the solution? In the Schwarzschild case, you don't know the curvature components until you've actually solved the EFE, so how can you state a "boundary condition" in terms of those components prior to solving the EFE? (But see further comments below.)

DaleSpam said:
The boundary condition isn't r>Rs. It is the curvature at the boundary. This is imposed when you choose the parameter M. Setting M is what sets your boundary condition, regardless if your chart covers all r>Rs or even just some r>R0>Rs.

But until you've solved the EFE, you don't even know that there *is* a parameter M. That is something that comes out of the solution, not something you impose prior to the solution.

But again, this is really a matter of terminology, now with the term "solution"; I am using it to mean the Schwarzschild line element, without specifying M; you (and PAllen; I'll respond to his post separately) are using it to mean the Schwarzschild line element, *plus* the specification of M. With that usage of the term, I agree with what you're saying; the line element itself only gives a "partial solution".
 
  • #87


PAllen said:
This is the key point, expressed much better by Dalespam. You've got boundary condition at infinity (asymptotic flatness); you've got constraint for (e.g.) vacuum and spherical symmetry. That still doesn't determine a unique solution. Specifying curvature on any 2-shere of given area respecting the symmetry, that you take as your inner boundary, is exactly a boundary condition.

Yes, as I said in my response to DaleSpam just now, if the "solution" has to include the actual value of M, then I agree with what you're saying. With this terminology, what I was calling a "solution" is only a "partial solution"; it specifies a particular infinite family of geometries, but not which particular one within the family we are using, and the boundary condition you are giving is a way of specifying the particular solution within the infinite family.

The only possible issue here is that, as I said in my response to DaleSpam, until you've got the partial solution (the line element, which gives you the curvature components explicitly in terms of M), you don't know how to specify the boundary condition that picks out one particular geometry from the infinite family that the line element describes. But that seems like a minor issue and I don't have a handy term for it anyway, so I'll stop here. :wink:
 
  • #88


harrylin said:
Instead, I meant it in the sense of "the child will soon realize that his Santaclaus is an illusion" (which does not mean that I believe that Santaclaus exists).

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?
 
  • #89


harrylin said:
We (Peter and I) noticed mislabeling in an earlier thread, as referred here. If you are in a restaurant that is known to have mislabeled bottles in the past and you hear one waiter say to another that a client "warrants suspicion" because he asked for Coca Cola and not Pepsi Cola, what would you think?

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.
 
  • #90


harrylin said:
We (Peter and I) noticed mislabeling in an earlier thread, as referred here. If you are in a restaurant that is known to have mislabeled bottles in the past and you hear one waiter say to another that a client "warrants suspicion" because he asked for Coca Cola and not Pepsi Cola, what would you think?
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.
 
  • #91


Mentz114 said:
There is no 't' in Schwarzshild.

...because the metric is independent of t.
 
  • #92


harrylin said:
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.
 
  • #93


stevendaryl said:
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.
 
  • #94


harrylin said:
[..] I do realize 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:
https://www.physicsforums.com/showthread.php?p=4185579
 
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  • #95


stevendaryl said:
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... :rolleyes:
PeterDonis said:
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? :confused:
 
  • #96


stevendaryl said:
...because the metric is independent of t.
But there are 2 'c's, one of which I missed out. Irony.
 
  • #97


Mentz114 said:
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. :!)
PeterDonis said:
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 :-p (but it didn't work because I didn't explain the topic well enough).
 
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  • #98


PeterDonis said:
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.
 
  • #99


harrylin said:
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.
 
  • #100


harrylin said:
Quite so; but wasn't the white hole solution intended to start near a black hole? :confused:

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.
 
  • #101


harrylin said:
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.
 
  • #102


harrylin said:
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.
 
  • #103


Mentz114 said:
But there are 2 'c's, one of which I missed out. Irony.
OK if you assume c=1. :smile:
 
  • #104


PeterDonis said:
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.
 
  • #105


harrylin said:
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. :wink: 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.

harrylin said:
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.

harrylin said:
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.
 

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