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

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PAllen
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Dec4-12, 09:55 PM
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Quote Quote by PeterDonis View Post
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.
Quote Quote by PeterDonis View Post

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).
PeterDonis
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Dec4-12, 10:04 PM
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Quote Quote by PAllen View Post
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.

Quote Quote by PAllen View Post
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.
DaleSpam
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Dec4-12, 10:10 PM
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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.
PAllen
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Dec4-12, 10:14 PM
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Quote Quote by PeterDonis View Post
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.
PeterDonis
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Dec4-12, 10:18 PM
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Quote Quote by PAllen View Post
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.
DaleSpam
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Dec4-12, 10:31 PM
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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.
PeterDonis
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Dec4-12, 10:45 PM
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Quote Quote by DaleSpam View Post
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.
PAllen
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Dec4-12, 11:01 PM
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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.
PeterDonis
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Dec4-12, 11:08 PM
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Quote Quote by PAllen View Post
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).
harrylin
#82
Dec5-12, 06:00 AM
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Quote Quote by PeterDonis View Post
[..] 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 realise 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).
DaleSpam
#83
Dec5-12, 06:13 AM
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Quote Quote by PeterDonis View Post
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.

Quote Quote by PeterDonis View Post
(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.

Quote Quote by PeterDonis View Post
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.
harrylin
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Dec5-12, 06:48 AM
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Quote Quote by Mentz114 View Post
[..] 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?
PAllen
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Dec5-12, 09:09 AM
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Quote Quote by DaleSpam View Post

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.]
PeterDonis
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Dec5-12, 09:27 AM
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Quote Quote by DaleSpam View Post
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.

Quote Quote by DaleSpam View Post
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.)

Quote Quote by DaleSpam View Post
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".
PeterDonis
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Dec5-12, 09:30 AM
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Quote Quote by PAllen View Post
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.
PeterDonis
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Dec5-12, 09:44 AM
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Quote Quote by harrylin View Post
Instead, I meant it in the sense of "the child will soon realise 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?
PeterDonis
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Dec5-12, 10:00 AM
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Quote Quote by harrylin View Post
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.
Mentz114
#90
Dec5-12, 10:35 AM
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Quote Quote by harrylin View Post
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.


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