Are Finkelstein/Kruskal Black Hole Solutions Compatible with Einstein's GR?

[stevendaryl]

Of course, but when he says "the solution" he means the metric, which is an object independent of coordinates.

Yes, but to demonstrate mathematically that the metric is a solution to the field equations, you have to express the metric as a function of some coordinates, and show that the differential equations are satisfied. At least, I don't know how to specify a metric and prove that it satisfies the EFE without using coordinates.

 

[martinbn]

Yes, but to demonstrate mathematically that the metric is a solution to the field equations, you have to express the metric as a function of some coordinates, and show that the differential equations are satisfied. At least, I don't know how to specify a metric and prove that it satisfies the EFE without using coordinates.

Yes, of course, no one disagrees with this.

 

[Dale]

There is a little bit more to it than that. In the general case, you may need to split spacetime up into pieces, find metrics for each piece that solves the EFE as a differential equation, and then show that the pieces are compatible (in the boundary between neighboring regions, the metric in the two regions are compatible).

Sure, but even if it doesn't cover the entire manifold each piece individually is still a valid solution of the EFE and hence compatible with GR. So the extra work you mentioned is nice, but not necessary for the purpose of this thread.

 

[pervect]

There is a little bit more to it than that. In the general case, you may need to split spacetime up into pieces, find metrics for each piece that solves the EFE as a differential equation, and then show that the pieces are compatible (in the boundary between neighboring regions, the metric in the two regions are compatible).

That's an interesting observation, though it's not necessary in this case since the Kruskal coordinates cover the whole space-time. But you could use it to show that the Kruskal coordinates "glue together" the interior and exterior Schwarzschid coordinates.

The only problem is that I very strongly suspect all the people (2, I think) who are still arguing (against the position of the 3-4 Science Advisors which seems pretty uniform, so it's fair to divide it up into sides) in this thread don't follow the math even for the simpler case of showing that the Kruskal coordinates satisfy the EFE. (The main reason for this suspicion is that I don't see how they could continue to argue if they did follow the math.)

So I doubt demonstrating how Kruskal glues together Schwarzschild would actually accomplish anything as fa as the argument goes. But it might generate an interesting disussion (the neverending argument here begins to pall for me).

 

[Mentz114]

The only problem is that I very strongly suspect all the people (2, I think) who are still arguing (against the position of the 3-4 Science Advisors which seems pretty uniform, so it's fair to divide it up into sides) in this thread don't follow the math even for the simpler case of showing that the Kruskal coordinates satisfy the EFE. (The main reason for this suspicion is that I don't see how they could continue to argue if they did follow the math.)

I endorse this wholeheartedly. Not only do the 2 seem to not follow the maths, but they don't seem to know what 'field equations' are, never mind solutions to same. People who cannot refer to GR (or GTR if you prefer) without tacking 'Einsteins' in front of it also warrant suspicion.

The OP's original question should have been answered with one word - 'Yes' and the thread closed.

 

[harrylin]

Then it seems to me that the question is poorly-defined. In my opinion the foundation of GR is the EFE, so demonstrating that a metric is a solution to the EFE is the same as demonstrating that it is consistent with the foundations of GR.

In other words, I feel that the question has been taken "by the horns" and answered clearly and unambiguously: KS is consistent with GR. To me there seems absolutely no room whatsoever for doubt on the matter, so I do not understand exactly what you think remains.

OK, I'm reading it now and the room for doubt immediately strikes the eye: Kruskal includes the white hole solution, which is held to be not realistic and according to PeterDonis not consistent with GR. And I must say, while I have great difficulty understanding anything of that article, the Wikipedia article on so-called(?!) Finkelstein coordinates looks much clearer and much more useful. Moreover, that article mentions a disadvantage of Kruskal coordinates that sounds very weird: "in those coordinates the metric depends on both the time and space coordinates."
- http://en.wikipedia.org/wiki/Eddington-Finkelstein_coordinates

I am thus going to read that one instead, and will ask more about it next. The whole point of this topic is to do a reality check by means of a worked out example such as the one pervect gave but for the Earth falling towards a black hole.

BTW Perhaps I should have given this thread the title "Finkelstein vs Schwartzschild in the light of Einstein's GR; if that is clearer this discussion can continue with that title. It is the logical continuation of the Oppenheimer thread.

 

[PeterDonis]

Kruskal includes the white hole solution, which is held to be not realistic and according to PeterDonis not consistent with GR.

Please don't misquote me. Here's what I said (not direct quotes since I've said this in multiple places): The full K-S solution, including all four regions, is a consistent mathematical solution of the EFE. Nobody believes it is physically reasonable because of the presence of the white hole and the second exterior region (and the fact that it is vacuum everywhere, which is why the white hole and the second exterior region are there); but that doesn't change the fact that it's a consistent mathematical solution of the EFE. That makes it "consistent with GR".

Any physical theory is going to contain "unphysical" solutions--solutions which are consistent with the math of the theory but which make predictions that aren't physically reasonable. That doesn't mean those solutions are "incompatible" with the theory. It means that the theory isn't a mindless machine that cranks out physical predictions; it's a tool for humans to use in making physical predictions, and like any sophisticated tool, it requires care in its use.

If you are going to insist that only "physically reasonable" solutions be considered "consistent with GR" or "compatible with GR", then this needs to be a completely different discussion. It needs to be about what makes a solution physically reasonable or unreasonable. But we can't even have that discussion until we have agreement on what the space of solutions is, so we can sort them into the "reasonable" and "unreasonable" buckets. Everybody else appears to agree (I think--but I'm sure I'll be corrected if I'm wrong ) that the "space of solutions" with which to start sorting is the set of mathematical solutions to the EFE. The K-S solution is indisputably one such mathematical solution; DaleSpam even explicitly showed that in this thread.

So, as with previous threads on similar topics, I'm confused about what your position is.

 

[harrylin]

[..] If we do that, then Adam' is not in "inertial motion" in the first place; by the definition you give, it is Eve' who is in "inertial motion", because she is moving in a "straight line" in the coordinate system that you consider privileged (the one which is fixed with reference to the gravitating body). Your statement about Adam' having to conclude that his "inertial motion" is an illusion only makes sense on the standard definition of "inertial motion", the one I was using. If you're not going to stick to Einstein's definitions, how do you expect the rest of us to do so?

I checked and could not find an inconsistency in my description as based on the definitions... Please cite the "guilty" phrase and point it out exactly, thanks!

[..] If anything, this adds to the things that Adam' *cannot* conclude are "illusions". He can [..] directly measure tidal gravity in his vicinity, so he can conclude that he is freely falling towards a BH; so that's not an "illusion" either. And given that he is falling towards the black hole, he has no reason to think he is in "inertial" motion by your definition in the first place; as I said above, it's Eve' who is in "inertial" motion by that definition, not Adam'. What, exactly, is Adam' supposed to conclude is an "illusion"?

In out 1D example, it is not immediately clear for them (especially if only using accelerometers and not looking too far around) what physical reality is; and as we discussed earlier, the way they reckon such things as distant time and light propagation depends on their assessment of physical reality.

How does this work out with for example Finkelstein's coordinate system? (this reminds me a bit on the FAQ on the Twin paradox, with Finkelstein and Kruskal competing with Schwarzschild there are almost "too many solutions"!).

 

[Mentz114]

OK, I'm reading it now and the room for doubt immediately strikes the eye: Kruskal includes the white hole solution, which is held to be not realistic and according to PeterDonis not consistent with GR. And I must say, while I have great difficulty understanding anything of that article, the Wikipedia article on so-called(?!) Finkelstein coordinates looks much clearer and much more useful. Moreover, that article mentions a disadvantage of Kruskal coordinates that sounds very weird: "in those coordinates the metric depends on both the time and space coordinates."
- http://en.wikipedia.org/wiki/Eddington-Finkelstein_coordinates

Why is it weird if " the metric depends on both the time and space coordinates." What else could it depend on ?

I am thus going to read that one instead, and will ask more about it next. The whole point of this topic is to do a reality check by means of a worked out example such as the one pervect gave but for the Earth falling towards a black hole.

There is no 'worked out example' for the Earth falling towards a black hole.

BTW Perhaps I should have given this thread the title "Finkelstein vs Schwartzschild in the light of Einstein's GR; if that is clearer this discussion can continue with that title. It is the logical continuation of the Oppenheimer thread.

There is no 't' in Schwarzshild.

Kruskal includes the white hole solution, which is held to be not realistic and according to PeterDonis not consistent with GR.

I'll bet he said no such thing. He may have said that it is non-physical - but nearly all solutions of the EFE are unphysical.

 

[Dale]

OK, I'm reading it now and the room for doubt immediately strikes the eye: Kruskal includes the white hole solution, which is held to be not realistic

I agree that it isn't realistic, but I wasn't addressing a question about its realism. I was addressing the question of its consistency with GR, and it IS clearly and unambiguously consistent with GR. Do you have any doubt about the answer to that specific question?

BTW Perhaps I should have given this thread the title "Finkelstein vs Schwartzschild in the light of Einstein's GR; if that is clearer this discussion can continue with that title. It is the logical continuation of the Oppenheimer thread.

OK, so then why don't you make a new thread with that title and ask the question that you are actually interested in. This question seems resolved.

 

[harrylin]

Please don't misquote me.

You said:

' "Modern GR" does not consider white holes to be physically reasonable. They are valid mathematical solutions of the EFE only if the spacetime is vacuum everywhere. Nobody believes that this mathematical solution describes any actual, physical spacetime. Any actual, physical spacetime contains matter somewhere;" '

I did not quote you but indicated the essence of your stated opinion. In case you tried to say that GR allows for white holes because according to GR there is vacuum everywhere, then my paraphrase was wrong.
[..]

So, as with previous threads on similar topics, I'm confused about what your position is.

My position is that I must make up for myself if Finkelstein's solution is better and more realistic (firstly according to theory and secondly according to my own philosophy) than that of Schwarzschild. And for that I gladly accept the help of you and others to come to an understanding of that solution.

 

[Dale]

Everybody else appears to agree (I think--but I'm sure I'll be corrected if I'm wrong ) that the "space of solutions" with which to start sorting is the set of mathematical solutions to the EFE. The K-S solution is indisputably one such mathematical solution; DaleSpam even explicitly showed that in this thread.

I don't know about everyone else, but I agree.

 

[Dale]

How does this work out with for example Finkelstein's coordinate system? (this reminds me a bit on the FAQ on the Twin paradox, with Finkelstein and Kruskal competing with Schwarzschild there are almost "too many solutions"!).

There are, in fact, an infinite number of spherically symmetric vacuum solutions, all related to each other via some coordinate transform.

 

[harrylin]

I agree that it isn't realistic, but I wasn't addressing a question about its realism. I was addressing the question of its consistency with GR, and it IS clearly and unambiguously consistent with GR. Do you have any doubt about the answer to that specific question?

OK, so then why don't you make a new thread with that title and ask the question that you are actually interested in. This question seems resolved.

According to GR there is matter in the universe - the difference between a theory of physics and mathematics can hardly be better clarified.
However, I do realize that the title of this thread is unclear, so I will continue this with a clearer title.

 

[Dale]

According to GR there is matter in the universe

This is false, GR does not assert that there is matter in the universe. In fact, many spacetimes which are clearly part of GR and widely discussed in the GR literature are vacuum solutions, including Schwarzschild.

In the sense in which there is no matter in KS there is also no matter in Schwarzschild (and Finkelstein). If you wish to exclude KS on that ground then you must also exclude Schwarzschild and all other vacuum solutions.

 

[Mentz114]

According to GR there is matter in the universe ...

No. There are plenty of solutions of the EFE where no matter is present. According to our experience, there is matter in the universe and this enables us to declare as unphysical such solutions of the EFE.

[Edit]I didn't see Dalespam's post ...

 

[PAllen]

A few thoughts of mine on some of the recent discussion:

- There is exactly one complete (maximal) spherically symmetric solution of GR with asymptotically flat boundary conditions and vacuum throughout.

- Any open subset of this geometry is also a solution of GR (albeit, with different boundary conditions). For example, the subset that is covered by exterior SC coordinates has a boundary condition not present in the maximal geometry.

- The unique maximal geometry can be covered with uncountably infinite different coordinate systems or collections of coordinate patches. This is also true of any subset of the maximal solution. These different coordinate coverings are not different solutions of GR, they are different descriptions of the same solution. Kruskal coordinates, SC coordinates, Eddington-Fikelstein, Panlieve, Lemaitre, etc. are all just relabellings of any parts they cover in common.

The way to relate this to the EFE is that the EFE are insufficient by themselves to specify coordinate expression of the metric. To get a specific metric expression, you need to impose coordinate conditions in one of several common ways (e.g. Harmonic; De Donder; or use a metric 'ansatz').

 

[PeterDonis]

I checked and could not find an inconsistency in my description as based on the definitions... Please cite the "guilty" phrase and point it out exactly, thanks!

I already have. Are you reading my posts?

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. But inertial motion, according to the definition you cited, means motion in a straight line in a coordinate system which is fixed with reference to the gravitating body. By that definition, Adam' is not in inertial motion; Eve' is.

In out 1D example, it is not immediately clear for them (especially if only using accelerometers and not looking too far around) what physical reality is

I'm not sure I understand what this means, and I'm not sure it's worth trying to disentangle it. The physical observations that Adam' and Eve' can make are perfectly clear.

How does this work out with for example Finkelstein's coordinate system?

Since that coordinate system is physically equivalent to Schwarzschild coordinates in the region outside the horizon--both have exactly the same geometric invariants--everything I said applies in Finkelstein coordinates just as it does in Schwarzschild coordinates. The only difference is that 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.

 

[PeterDonis]

You said:

' "Modern GR" does not consider white holes to be physically reasonable.

Yes. That is not the same as saying "the K-S solution is not consistent with GR". At least, not the way I would use words. If that's the way you want to use words, then you should at least have said something like "PeterDonis says the K-S solution isn't physically reasonable; IMO that means it's not consistent with GR." Then at least people would know that it was you who were using the word "consistent" in a totally unusual way, not me.

I did not quote you but indicated the essence of your stated opinion.

No, you didn't. You indicated the essence of your additional claim based on my stated opinion. There's a difference. See above.

My position is that I must make up for myself if Finkelstein's solution is better and more realistic (firstly according to theory and secondly according to my own philosophy) than that of Schwarzschild. And for that I gladly accept the help of you and others to come to an understanding of that solution.

First we need to come to a common understanding of what a "solution" is. See, for example, PAllen's post #67 and my #68.

 

[PeterDonis]

the subset that is covered by exterior SC coordinates has a boundary condition not present in the maximal geometry.

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.

 

[PAllen]

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.

 

[PeterDonis]

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:

ds^2 = A(r) dt^2 + B(r) dr^2 + r^2 d\Omega^2

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.

 

[PAllen]

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:

ds^2 = A(r) dt^2 + B(r) dr^2 + r^2 d\Omega^2

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.

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]

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.

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.

 

[Dale]

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]

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]

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.

 

[Dale]

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]

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]

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]

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]

[..] 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).

 

[Dale]

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.

(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.

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]

[..] 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]

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]

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.

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.)

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]

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]

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?

 

[PeterDonis]

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]

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.

 

[stevendaryl]

There is no 't' in Schwarzshild.

...because the metric is independent of t.

 

[stevendaryl]

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]

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]

[..] 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

 

[harrylin]

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...

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]

...because the metric is independent of t.

But there are 2 'c's, one of which I missed out. Irony.

 

[harrylin]

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. :!)

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]

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]

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]

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.