Is there a zero-mass, vacuum Schwarzschild GR solution?

[fox26]

The Schwarzschild solution of the Einstein Equation of GR is said to be the only time-independent matter-free solution of that equation. In this usage, does “matter-free solution” mean without matter everywhere
except at the singularity of the solution? I thought that the only solutions of the Einstein Eq. for a totally
matter-free universe were either just empty Minkowski space, or space-times that had only gravitational
waves in them. The S. solution is neither one. The universe of the S. solution supposedly is not totally
matter-, or at least mass-free, there is matter in the infinite-density singularity at the center of the S.
solution, or at least non-zero mass attributed to the black hole represented by the solution, a
non-rotating, uncharged black hole having non-zero mass, with R = 2Gm/c^2, R representing the
Schwarzschild radius of the black hole and m its mass. However, Stephen Hawking and
others have said that such a singularity isn’t really in the space-time of the S. solution. Are they wrong?
Doesn’t matter have to be in a space-time to contribute to the stress-energy tensor of the Einstein
equation applicable to that space-time? The problem of whether singularities could exist in a correct
GR-quantum theory is separate from this question.

I have other questions about GR, black holes, and gravitational collapse, which I will ask in later posts.

 

[Nugatory]

In the Schwarzschild solution, $m$ is a parameter. What does the metric look like when you set it to 0?

 

[fox26]

In the Schwarzschild solution, $m$ is a parameter. What does the metric look like when you set it to 0?

True, it is the Minkowski, flat space-time, SR metric, which is a solution for a matter-free universe, but this is the trivial solution, while the statements about the S. solution being a matter-free solution were for every solution, not just the trivial one. The S. solution for m=0 has no singularity and no event horizon, doesn't represent a black hole, and isn't, I think, usually considered a Schwarzschild solution, all of which have m>0. You are quibbling.

 

[Dale]

The S. solution for m=0 has no singularity and no event horizon, doesn't represent a black hole, and isn't, I think, usually considered a Schwarzschild solution

I think that it is usually considered a Schwarzschild solution. Of course, it is the trivial solution, as you indicate, but trivial solutions are still solutions. It is simply spherical coordinates on Minkowski spacetime with the origin removed.

The Schwarzschild solution of the Einstein Equation of GR is said to be the only time-independent matter-free solution of that equation. In this usage, does “matter-free solution” mean without matter everywhere except at the singularity of the solution?

For technical reasons, the singularity itself is not part of spacetime. Basically, the mathematical framework of GR is pseudo Riemannian geometry, which treats spacetime as a manifold of signature (-+++). At the singularity itself the manifold structure breaks down, so it has to be removed from the spacetime, and then the Schwarzschild solution (even with M>0) is vacuum everywhere precisely because "everywhere" specifically excludes the singularity. It is kind of a mathematical "technicality" so I wouldn't read too much into it physically, but that is what your book was referring to.

 

[Ibix]

The Schwarzschild solution of the Einstein Equation of GR is said to be the only time-independent matter-free solution of that equation.

It's the only spherically symmetric one. Kerr-Newman is a counter example to your more general claim.

 

[PeterDonis]

The Schwarzschild solution of the Einstein Equation of GR is said to be the only time-independent matter-free solution of that equation.

No, it isn't. In addition to the obvious counterexample that @Ibix pointed out, there is at least one other:

https://en.wikipedia.org/wiki/Ozsváth–Schücking_metric

 

[fox26]

I think that it is usually considered a Schwarzschild solution. Of course, it is the trivial solution, as you indicate, but trivial solutions are still solutions. It is simply spherical coordinates on Minkowski spacetime with the origin removed.

For technical reasons, the singularity itself is not part of spacetime. Basically, the mathematical framework of GR is pseudo Riemannian geometry, which treats spacetime as a manifold of signature (-+++). At the singularity itself the manifold structure breaks down, so it has to be removed from the spacetime, and then the Schwarzschild solution (even with M>0) is vacuum everywhere precisely because "everywhere" specifically excludes the singularity. It is kind of a mathematical "technicality" so I wouldn't read too much into it physically, but that is what your book was referring to.

I understood that, for the non-trivial case with m>0, which is the interesting and problematic case, the origin of S. coordinates where the singularity is can't be included in a smooth manifold model, and in fact is not considered to be part of spacetime. That is the problem. In the S. model, all the matter is in the singularity, and so is considered not to be in the spacetime model. My major question, which you and all others so far have avoided answering or even considering, is how can matter which is not in spacetime lead, as supposedly shown by the Einstein equation, to a m>0, non-trivial S. solution with an infinite density singularity caused bv the matter? It doesn't work logically or mathematically to treat the matter and singularity as not in our model of spacetime or maybe even in spacetime itself because it doesn't fit the assumed axiomatic requirements of that model, yet consider it to be a factor in the Einstein equation which leads to the S. solution and singularity, but which supposedly includes only elements which are in the mathematical model representing the physical universe. Of course singularities are a problem in classical GR, and maybe some fudging is now necessary because of them, but I would want it to be honest fudging, which it appears not to be.

 

[fox26]

It's the only spherically symmetric one. Kerr-Newman is a counter example to your more general claim.

I didn't claim that the S. solution was the only time-independent matter-free solution, only that it was said to be so. You are right in that the Kerr-Newman solution, for black holes with non-zero charge and angular momentum, is also said by some to be a matter-free solution. I didn't think of this as an obvious claimed counterexample to the matter-free claim for the S. solution when I read that claim. However, my original question was mainly to question how the matter-free claim for the S. solution could be true, as I restated in my reply to Dale above. My objections to that apply also to the Kerr-Newman solution being a matter-free solution. Do you have any solution to the main problem I indicated above and in my reply to Dale?

 

[fox26]

No, it isn't. In addition to the obvious counterexample that @Ibix pointed out, there is at least one other:

https://en.wikipedia.org/wiki/Ozsváth–Schücking_metric

Your counterexample is maybe (I haven't followed the link yet) a counterexample to the matter-free claim about the S. solution that I said was said, not to what I said. See my reply to ibix above. The main problem of my question, restated in my reply to Dale above, remains so far unanswered. Do you have such an answer?

 

[Ibix]

I didn't claim that the S. solution was the only time-independent matter-free solution, only that it was said to be so.

You went on to make several statements based on this, though (e.g. just Minkowski, or only gravitational waves). So I thought it worth correcting.

However, my original question was mainly to question how the matter-free claim for the S. solution could be true, as I restated in my reply to Dale above.

What you said in your post to Dale was:

It doesn't work logically or mathematically to treat the matter and singularity as not in our model of spacetime

But this doesn't make sense, since it clearly does work mathematically. That's what the Schwarzschild solution is - the maths of that exact situation, working.

Do note, however, that the Schwarzschild solution describes an eternal black hole in an otherwise empty universe. This is not a realistic model of a black hole. A real black hole has a history when it was not a black hole, and all the gravitational effects one sees outside the hole can be traced back to matter outside the hole, either before the horizon formed or before it fell into the horizon. This must be so - the past light cone of an event outside the horizon cannot contain anything inside the horizon. So whether all the mass inside the horizon is in spacetime or not is irrelevant to our experience outside the event horizon.

Inside the event horizon, the details of the interior are of much more importance. But it's not a place we can test - even if we had a black hole handy, we couldn't get any data out of probes we sent in. For what it's worth, I don't think anyone thinks the singularities are anything other than a product of the maths departing from reality. A successor theory for relativity that hopefully won't have singularities is a topic of active research. It's way above my pay grade, but I think the major problem is that we've never caught reality doing anything significantly different from what GR says. That means everyone's shooting blind.

Of course singularities are a problem in classical GR, and maybe some fudging is now necessary because of them, but I would want it to be honest fudging, which it appears not to be.

I don't think there's any fudging going on, except in the sense of us applying slightly idealised solutions to reality because the maths is tractable and the errors are small. The maths works, whether you like it or not, and keeps passing experimental tests to the best of our ability to measure. One day it won't, and that's when there'll be progress.

One other point to make about the Schwarzschild spacetime. The uniqueness of it means that it's the spacetime outside any spherically symmetric non-rotating eternal mass. So it's (approximately) applicable to spacetime outside all sorts of things like planets and stars that are nearly spherically symmetric and only slowly rotating - not just to black holes.

 

[PeterDonis]

I didn't claim that the S. solution was the only time-independent matter-free solution, only that it was said to be so.

And you didn't give a reference for where it was "said" to be so. Can you give one?

You are right in that the Kerr-Newman solution, for black holes with non-zero charge and angular momentum, is also said by some to be a matter-free solution.

It isn't "said by some" to be a matter-free solution, it is a matter-free solution, as any GR textbook will tell you.

You need to stop this "said to be" stuff and give specific references.

 

[PeterDonis]

how can matter which is not in spacetime lead, as supposedly shown by the Einstein equation, to a m>0, non-trivial S. solution with an infinite density singularity caused bv the matter?

The "mass" $m$ in the idealized Schwarzschild solution (and in other solutions such as the Kerr solution) is not a sign that there is "matter" somewhere. It is a property of the spacetime geometry. It is called "mass" for historical reasons, not because it being greater than zero requires there to be "matter" somewhere. It's simply an idealized solution which is vacuum everywhere and has a singularity (actually two of them, since maximally extended Schwarzschild spacetime has a "white hole" as well as a black hole), with a geometric property that is called "mass" for historical reasons.

In a more realistic solution describing, for example, the collapse of a massive star to a black hole, there is matter present. That matter collapses to $r = 0$ and forms the singularity there. The vacuum region outside the matter has the Schwarzschild geometry, but now it's obvious where the "mass" in the vacuum region comes from: it's due to the matter that collapsed. But in the vacuum region, $m$ is still a geometric property of the spacetime, not something that is due to the "mass" at the singularity. The only difference is that in this more realistic solution, the geometry property has an obvious linkage to a region containing matter.

 

[Dale]

My major question, which you and all others so far have avoided answering or even considering, is how can matter which is not in spacetime lead, as supposedly shown by the Einstein equation, to a m>0, non-trivial S. solution with an infinite density singularity caused bv the matter?

It changes the boundary condition at the boundary surrounding the excised singularity.

Please don’t speak so accusatorily. I am doing the best I can. That aspect of your question was not clear to me.

It doesn't work logically or mathematically to treat the matter and singularity as not in our model of spacetime or maybe even in spacetime itself because it doesn't fit the assumed axiomatic requirements of that model, yet consider it to be a factor in the Einstein equation which leads to the S. solution and singularity, but which supposedly includes only elements which are in the mathematical model representing the physical universe.

Sure it works, it just doesn’t mean what you think it means. When writing down the spherically symmetric, static, vacuum solution you come up with not just a single solution, but rather a family of solutions with a single free parameter. That parameter changes the boundary condition on the inside of the manifold. That parameter has units of length and is called the Schwarzschild radius.

However, we are free to rewrite this length parameter in terms of mass or time or anything else we can think of, scales by appropriate universal constants. It was found that when the length parameter was rewritten in terms of mass, that the weak field limit recovered Newtonian gravity around a point mass. So even though it was a length parameter it became common to write that length in terms of mass.

It does not imply that there is mass anywhere in the solution since the parameter only changes the boundary condition and the vacuum condition holds everywhere in the solution. It is simply a different way to write the length scale for the boundary condition.

Of course singularities are a problem in classical GR, and maybe some fudging is now necessary because of them, but I would want it to be honest fudging, which it appears not to be.

This is a rather poor attitude. I am not sure what you think is “fudging” about boundary conditions. They are a standard part of solving any partial differential equation! Indeed, how could you think that a boundary condition could be avoided? To call a natural and unavoidable part of a solution “fudging” or “dishonest” is highly inappropriate.

 

[fox26]

You went on to make several statements based on this, though (e.g. just Minkowski, or only gravitational waves). So I thought it worth correcting.
What you said in your post to Dale was:
But this doesn't make sense, since it clearly does work mathematically. That's what the Schwarzschild solution is - the maths of that exact situation, working.

Do note, however, that the Schwarzschild solution describes an eternal black hole in an otherwise empty universe. This is not a realistic model of a black hole. A real black hole has a history when it was not a black hole, and all the gravitational effects one sees outside the hole can be traced back to matter outside the hole, either before the horizon formed or before it fell into the horizon. This must be so - the past light cone of an event outside the horizon cannot contain anything inside the horizon. So whether all the mass inside the horizon is in spacetime or not is irrelevant to our experience outside the event horizon.

Inside the event horizon, the details of the interior are of much more importance. But it's not a place we can test - even if we had a black hole handy, we couldn't get any data out of probes we sent in. For what it's worth, I don't think anyone thinks the singularities are anything other than a product of the maths departing from reality. A successor theory for relativity that hopefully won't have singularities is a topic of active research. It's way above my pay grade, but I think the major problem is that we've never caught reality doing anything significantly different from what GR says. That means everyone's shooting blind.
I don't think there's any fudging going on, except in the sense of us applying slightly idealised solutions to reality because the maths is tractable and the errors are small. The maths works, whether you like it or not, and keeps passing experimental tests to the best of our ability to measure. One day it won't, and that's when there'll be progress.

One other point to make about the Schwarzschild spacetime. The uniqueness of it means that it's the spacetime outside any spherically symmetric non-rotating eternal mass. So it's (approximately) applicable to spacetime outside all sorts of things like planets and stars that are nearly spherically symmetric and only slowly rotating - not just to black holes.

My statement that I thought the only solutions to the Einstein Eq. for a totally matter-free universe were empty Minkowski space, or (maybe "and" would have been better) spaces with only gravitational waves in them was not based in any way on the statement that you quoted and said it was based on. I think I read it (what I thought, not that I thought it) somewhere, but it seems plausible to me on its own merits, even though I don't have a proof. Do you know of a counterexample, that is, a GR spacetime containing no matter but which isn't either Minkowski Space or one containing only gravitational waves?

Your next claim, referring to my partial sentence "It doesn't work logically or mathematically to treat the matter and singularity as not in our model of spacetime" was that this doesn't make sense, since it clearly does work mathematically. This is a blatant example of quoting only part of someone's sentence and criticizing it as if it were an entire sentence. If Mr. Smith said "It isn't true that one equals two.", would it be correct to say that Mr. Smith was an idiot since he said "one equals two."? Look again at my entire sentence and try to understand it, in case you really didn't. If you did understand it, and just quoted part of it and then criticized that part knowing that the complete sentence said something quite different from what the fragment you quoted said, you are not worth bothering with any more.

I will assume, for the present, that you just failed to understand that the complete sentence meant something different from what your quote meant.

Black holes can be observed, even though their interiors cannot, from outside. There have been quite a few observations confirming the existence of black holes, even if not the singularities present non-quantum GR theory says exist inside them. The observations by LIGO in the past few years of gravitational waves from several pairs of coalescing black holes, confirming, as far as I know, all predictions of GR for such events which are observable by LIGO, are a good example. If you think that there is no problem with the predictions by GR of such black holes which are based on the existence in our spacetime of matter or just mass in singularities which are not in that spacetime, read and understand my complete reply to Dale.

 

[PeterDonis]

Do you know of a counterexample, that is, a GR spacetime containing no matter but which isn't either Minkowski Space or one containing only gravitational waves?

We have already given two counterexamples in this thread: Schwarzschild spacetime and Kerr spacetime. Neither of those are flat Minkowski spacetime, and neither of them contain any gravitational waves. The Ozvath-Schucking metric, referred to earlier, does contain only gravitational waves.

However, there are also other counterexamples, see the list here:

https://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)#Examples

Of these, in addition to the ones listed above, the Taub-NUT, Kerns-Wild, and Kasner metrics also contain no matter but aren't flat Minkowski space and do not contain gravitational waves. The double Kerr vacuum could be considered another counterexample, but the "unphysical zero active mass thread" makes it problematic.

 

[PeterDonis]

It doesn't work logically or mathematically to treat the matter and singularity as not in our model of spacetime or maybe even in spacetime itself because it doesn't fit the assumed axiomatic requirements of that model, yet consider it to be a factor in the Einstein equation which leads to the S. solution

There is no matter included in solving the Einstein equation to get the Schwarzschild solution. The stress-energy tensor is assumed to be identically zero, which means "no matter". This is true for all of the vacuum solutions referenced in the Wikipedia article I linked to in my previous post.

I explained in post #12 why the parameter $m$ that appears in the Schwarzschild solution, and its common interpretation as a "mass", does not mean there must be matter anywhere.

 

[fox26]

And you didn't give a reference for where it was "said" to be so. Can you give one?
It isn't "said by some" to be a matter-free solution, it is a matter-free solution, as any GR textbook will tell you.

You need to stop this "said to be" stuff and give specific references.

I read the 'matter-free" statement on the Internet recently, but I don't remember where. However, that statement isn't important, since it might plausibly be interpreted as being an unfortunately loose way of saying " matter free except at the singularity", as I suggested. However, your statement "it is a matter-free solution" will probably be acceptable to you as a reference, if you accept yourself as an adequate authority. What is important is Hawking's (and others) statement that the singularity really isn't in the spacetime of the S. solution. The singularity can't be in our models of spacetime, for mathematical reasons, although that doesn't insure that it can't be in this universe's actual spacetime. The statement by Hawking is somewhere in The Large Scale Structure of Spacetime, by Hawking and Ellis.

I just saw an alert about your most recent post, which seems to be getting to an answer to the main question my original post was asking. I will temporarily skip replying to any other replies and comment on that post.

 

[fox26]

There is no matter included in solving the Einstein equation to get the Schwarzschild solution. The stress-energy tensor is assumed to be identically zero, which means "no matter". This is true for all of the vacuum solutions referenced in the Wikipedia article I linked to in my previous post.

I explained in post #12 why the parameter $m$ that appears in the Schwarzschild solution, and its common interpretation as a "mass", does not mean there must be matter anywhere.

Is the Schwarzschild solution really a solution of the Einstein Equation with stress-energy tensor zero, in normal space, or does such a solution require some additional assumption about, say, boundary conditions on a subpart of total spacetime, which is conveniently excluded from that spacetime, such as is mentioned in a recent reply by Dale, to which I haven't yet replied? The solutions which Einstein considered to be physically possible, and those which in fact are possible, are surely those which have as matter (or just mass) inputs things which are actually in spacetime. In addition, the solutions which he would have considered physically possible are those in a realistic spacetime, not a spacetime with a spatial "inner boundary" chosen to be the boundary of either a singularity, or a complete black hole, which itself is excluded from the spacetime, which Dale's reply seems to be referring to. Such a spacetime is one with a hole in it, which hole is nonrealistic unless something like a singularity or black hole is assumed to potentially have existed in the space or spacetime, where the singularity or black hole doesn't exist but would have been if it hadn't been excluded (this is somewhat weird, but it is Dale's example, not mine). By suitably choosing the "inner boundary" conditions, the field at large distances from the hole is, according to Dale's post, I think, about equal to Newtonian gravity around a point mass. This says nothing about whether a singularity or a stationary black hole can be the solution of Einstein's Equation with zero mass or stress-energy tensor in a reasonable spacetime. If I understand Dale's example correctly, it gives only a (approximately) Newtonian gravity field at large distances, saying nothing about what is "inside" the hole omitted from space, which inside doesn't really exist in our universe.

In reply to Dale, whom I will refer to this reply in a reply I will make to his recent post: Boundary conditions are by no means always part of solving a partial differential equation. Initial conditions can be used instead for many problems, and are more suitable for the Cauchy problem in physical spacetime than assumed boundary conditions for an assumed space with a non-physical hole, since physical space doesn't have a hole with a boundary on which boundary conditions can be chosen to fit the solution (outside of the hole) desired. What I think you are suggesting is pure mathematics, not physics.

It is well-known that GR breaks down, is self-inconsistent, in situations where singularities caused by gravitational collapse of a massive body occur.
What I was asking in my question was mainly clarification of the situation and the seemingly inconsistent terminology used by and statements referring to such situations made by, physicists and others talking about such situations. In summary, the problem is that a singularity cannot be in our spacetime according to present theory, and a Schwarzschild solution without the mass at the singularity (in a physically realistic space that is simply connected except for the point where the omitted singularity would have been, and is without any other mass) isn't a solution of the Einstein Equation. If I am wrong in this, please tell me just how, without irrelevant diversions or non-physically relevant mathematical considerations.

 

[fox26]

It changes the boundary condition at the boundary surrounding the excised singularity.

Please don’t speak so accusatorily. I am doing the best I can. That aspect of your question was not clear to me.

Sure it works, it just doesn’t mean what you think it means. When writing down the spherically symmetric, static, vacuum solution you come up with not just a single solution, but rather a family of solutions with a single free parameter. That parameter changes the boundary condition on the inside of the manifold. That parameter has units of length and is called the Schwarzschild radius.

However, we are free to rewrite this length parameter in terms of mass or time or anything else we can think of, scales by appropriate universal constants. It was found that when the length parameter was rewritten in terms of mass, that the weak field limit recovered Newtonian gravity around a point mass. So even though it was a length parameter it became common to write that length in terms of mass.

It does not imply that there is mass anywhere in the solution since the parameter only changes the boundary condition and the vacuum condition holds everywhere in the solution. It is simply a different way to write the length scale for the boundary condition.

This is a rather poor attitude. I am not sure what you think is “fudging” about boundary conditions. They are a standard part of solving any partial differential equation! Indeed, how could you think that a boundary condition could be avoided? To call a natural and unavoidable part of a solution “fudging” or “dishonest” is highly inappropriate.

See my reply to PeterDonis's most recent post, which includes a reply to you.

 

[Ibix]

Do you know of a counterexample, that is, a GR spacetime containing no matter but which isn't either Minkowski Space or one containing only gravitational waves?

Multiple ones. Both PeterDonis and I have given such examples in posts 5 and 6. After the post I am quoting, Peter provided a link to a list of vacuum solutions and highlighted some specific ones.

I will assume, for the present, that you just failed to understand that the complete sentence meant something different from what your quote meant.

How very generous and forbearing of you. Dale has already commented on your language in post 13. We are here, in our free time, offering help and discussion solely for the fun of it. Assume a modicum of good faith, please, and lose the "martyred intellectual" attitude.

The full sentence you objected to my partial quotation of was (and linked to the right post this time):

It doesn't work logically or mathematically to treat the matter and singularity as not in our model of spacetime or maybe even in spacetime itself because it doesn't fit the assumed axiomatic requirements of that model, yet consider it to be a factor in the Einstein equation which leads to the S. solution and singularity, but which supposedly includes only elements which are in the mathematical model representing the physical universe.

I think "the Schwarzschild solution does not make sense logically or mathematically because..." is a fair paraphrase. And therefore pointing out that it clearly does work mathematically is a perfectly valid counterpoint, regardless of what comes after the "because". If you disagree with my reading, provide your own paraphrase instead of whining about my supposed intellectual dishonesty. That way we might come to a mutual understanding of what it is you are talking about.

Black holes can be observed, even though their interiors cannot, from outside. There have been quite a few observations confirming the existence of black holes, even if not the singularities present non-quantum GR theory says exist inside them. The observations by LIGO in the past few years of gravitational waves from several pairs of coalescing black holes, confirming, as far as I know, all predictions of GR for such events which are observable by LIGO, are a good example. If you think that there is no problem with the predictions by GR of such black holes which are based on the existence in our spacetime of matter or just mass in singularities which are not in that spacetime, read and understand my complete reply to Dale.

I don't understand how this is a response to anything I wrote. I explicitly said that I don't think anyone believes the singularities are real and people are working on successor theories for this reason, among others. I also said (as Peter also did) that in a realistic model of a black hole, all effects observable from outside are traceable to matter (etc) that is or was outside the event horizon at the time it caused the effect. So our observation of black holes can only tell us that GR isn't noticeably incorrect up to the event horizon. It doesn't say "there must be singularities", nor the opposite. Hopefully one day our measurement precision will improve and we'll observe something that GR does not predict.

 

[Dale]

Boundary conditions are by no means always part of solving a partial differential equation. Initial conditions can be used instead for many problems,

Yes, you are right, although in this case it is an unimportant distinction. Whether it is boundary or initial conditions the salient point is that solving a differential equation leads to a family of functions and you need to specify the value of a set of parameters in order to select a specific solution from among the family. In this case that set of parameters is a single non-negative real number which represents the length scale. No matter is assumed, it is just a boundary condition for picking one solution out of the family.

The solutions which Einstein considered to be physically possible, and those which in fact are possible, are surely those which have as matter (or just mass) inputs things which are actually in spacetime.

That is correct. The Schearzschild solution is not a physically reasonable solution. It is a simple “toy model” solution that is a useful approximation in many cases, but it is not in its entirely itself a physically reasonable model. In particular, the static condition is not realistic in the actual universe where a black hope cannot have existed forever. Also the black hole spacetime includes a white hole region which is considered non-physical. I.e. it does work logically and mathematically, but it is well known to not be a physically reasonable solution in its entirety.

Probably the simplest physically reasonable black hole spacetime is the Oppenheimer Snyder spacetime. It still has some “unreasonable” facets such as perfect spherical symmetry and 0 pressure, but it has matter in the manifold, it is not static, and it does not have a white hole region. It also captures a lot of the features that are found in more complicated numerical solutions where the symmetry is not perfectly spherical and the equation of state includes pressure.

 

[PeterDonis]

Is the Schwarzschild solution really a solution of the Einstein Equation with stress-energy tensor zero, in normal space

Yes.

or does such a solution require some additional assumption about, say, boundary conditions on a subpart of total spacetime

No. You discover that some boundary condition is required after you have already found the solution; it's not an assumption that you have to put in beforehand.

In the case of Schwarzschild, you discover that there is a free constant in the solution, usually called $m$, which can take on any real value. If $m = 0$, you obtain flat Minkowski spacetime; if $m > 0$, you find that there are two singularities, the "black hole" and the "white hole" in the maximal analytic extension of the solution. As @Dale has said, this solution is actually not physically reasonable as it stands, but constructing a physically reasonable solution which has a region that is Schwarzschild with $m > 0$ (such as the Oppenheimer Snyder model @Dale mentioned) does not require any additional assumptions about boundary conditions except for the obvious point that the solution has to be continuous across the boundary between the vacuum region and the matter region, which has nothing to do with the singularity. (We haven't discussed the case $m < 0$ for the Schwarzschild solution here, but it is also a mathematically valid solution and has a different structure from either of the two just mentioned.)

It is well-known that GR breaks down, is self-inconsistent, in situations where singularities caused by gravitational collapse of a massive body occur.

No, that is not what is well known. GR is not self-inconsistent anywhere. The issue with singularities is that in some region near enough to them, the spacetime curvature will be so large that new physics is expected to appear, which GR does not describe. Most physicists expect the new physics to be some form of quantum gravity, and consequently the scale at which new physics is expected to appear is roughly the Planck scale, i.e., the spacetime curvature would be of the order of the inverse Planck length squared.

a Schwarzschild solution without the mass at the singularity (in a physically realistic space that is simply connected except for the point where the omitted singularity would have been, and is without any other mass) isn't a solution of the Einstein Equation.

As I noted above, your statement here is incorrect. You would do well to look at some GR textbooks where the solution is explicitly derived. You could also look at my Insights article giving a short proof of Birkhoff's Theorem, which is a key part of the derivation (basically it's the part that comes after you have assumed spherical symmetry and vacuum):

https://www.physicsforums.com/insights/short-proof-birkhoffs-theorem/

Note that I explicitly assume that all of the Einstein tensor components are zero, which is what you get when you assume "no matter anywhere", and that no assumption at all is required about any singularity or any "mass" anywhere. The "mass" $m$ arises as a free real constant in the course of the proof.

 

[PeterDonis]

it is not static

Actually, the vacuum region of the Oppenheimer-Snyder model is static outside the horizon (since it is just Schwarzschild spacetime outside the horizon). But the matter region is not static anywhere.

 

[Dale]

Actually, the vacuum region of the Oppenheimer-Snyder model is static outside the horizon (since it is just Schwarzschild spacetime outside the horizon). But the matter region is not static anywhere.

Precisely, I was referring to the entire spacetime, both inside and outside the matter. But that is a good point to mention: even though the Schwarzschild spacetime as a whole is not physically reasonable, portions of the solution do match with portions of physically reasonable spacetimes. Of course, the boundaries must be properly matched.

 

[fox26]

Yes.
No. You discover that some boundary condition is required after you have already found the solution; it's not an assumption that you have to put in beforehand.

In the case of Schwarzschild, you discover that there is a free constant in the solution, usually called $m$, which can take on any real value. If $m = 0$, you obtain flat Minkowski spacetime; if $m > 0$, you find that there are two singularities, the "black hole" and the "white hole" in the maximal analytic extension of the solution. As @Dale has said, this solution is actually not physically reasonable as it stands, but constructing a physically reasonable solution which has a region that is Schwarzschild with $m > 0$ (such as the Oppenheimer Snyder model @Dale mentioned) does not require any additional assumptions about boundary conditions except for the obvious point that the solution has to be continuous across the boundary between the vacuum region and the matter region, which has nothing to do with the singularity. (We haven't discussed the case $m < 0$ for the Schwarzschild solution here, but it is also a mathematically valid solution and has a different structure from either of the two just mentioned.)
No, that is not what is well known. GR is not self-inconsistent anywhere. The issue with singularities is that in some region near enough to them, the spacetime curvature will be so large that new physics is expected to appear, which GR does not describe. Most physicists expect the new physics to be some form of quantum gravity, and consequently the scale at which new physics is expected to appear is roughly the Planck scale, i.e., the spacetime curvature would be of the order of the inverse Planck length squared.As I noted above, your statement here is incorrect. You would do well to look at some GR textbooks where the solution is explicitly derived. You could also look at my Insights article giving a short proof of Birkhoff's Theorem, which is a key part of the derivation (basically it's the part that comes after you have assumed spherical symmetry and vacuum):

https://www.physicsforums.com/insights/short-proof-birkhoffs-theorem/

Note that I explicitly assume that all of the Einstein tensor components are zero, which is what you get when you assume "no matter anywhere", and that no assumption at all is required about any singularity or any "mass" anywhere. The "mass" $m$ arises as a free real constant in the course of the proof.

GR assumes, that is, contains as an axiom, that spacetime is a pseudo Riemannian manifold, as you pointed out. It then derives, according to its laws and the assumption of physically reasonable conditions, which are observed to actually occur, that a singularity occurs in that spacetime, which makes that spacetime not a pseudo Riemannian manifold. This is an inconsistency in GR (+ assumptions of conditions with which it is supposed to deal properly), in the standard mathematical sense, with GR + those reasonable assumptions implying both that spacetime is a pseudo Riemannian manifold and that it isn't.

For problems with the Schwarzschild solution, see my recent addition #25 to the original question.

 

[PAllen]

GR assumes, that is, contains as an axiom, that spacetime is a pseudo Riemannian manifold, as you pointed out. It then derives, according to its laws and the assumption of physically reasonable conditions, which are observed to actually occur, that a singularity occurs in that spacetime, which makes that spacetime not a pseudo Riemannian manifold. This is an inconsistency in GR (+ assumptions of conditions with which it is supposed to deal properly), in the standard mathematical sense, with GR + those reasonable assumptions implying both that spacetime is a pseudo Riemannian manifold and that it isn't.

For problems with the Schwarzschild solution, see my recent addition #25 to the original question.

A singularity is does not make a mean a manifold is not a pseudo-Riemannian manifold. A singularity is a feature some manifolds have, and it is surprisingly difficult to define invariant criteria for a singularity (and there are different strengths of singularity). A highly over simplified statement is that a manifold has a singularity if there are some geodesics that are incomplete, and this incompletess cannot be resolved by a smooth extension. There are enormous technical difficulties in making this criterion both rigorous and general. But in no sense is a manifold with a singularity not a manifold.

Thus, the error is in your knowledge. There is no inconsistency in GR.

 

[PeterDonis]

G = 0 everywhere except at r = 0, which isn't in the space, whereas before the singularity formed, G ≠\neq 0 at some points other than r = 0. Can such a G still give the same metric as before the singularity formed? I think not.

You are wrong. You are also operating with several misconceptions.

First, the singularity in Schwarzschild spacetime is not a point in space. It's a moment of time. And that moment is to the future of every event inside the black hole. So comparing what things are like "after" the singularity forms with "before" the singularity forms makes no sense; there is no "after". Similar remarks apply to the Oppenheimer-Snyder model that @Dale and I referred to: the singularity doesn't get "formed" by the collapsing matter and then stay in some particular place. It is the future boundary of the region inside the black hole; the collapsing matter disappears from the universe when it hits that future boundary (at the "left edge" of it, heuristically speaking).

A spacetime diagram of the vacuum portion of the Oppenheimer-Snyder model appears in this Insights article:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-4/#toggle-id-1

Second, you are mis-describing how a model like the Oppenheimer-Snyder model is constructed. The Einstein Field Equation is a differential equation; that means it describes the metric and its derivatives and relates them to the stress-energy at a particular point in spacetime. Figuring out a global solution requires integrating the local solution over a manifold. However, the stress-energy tensor might be different at different points on the manifold. For example, there might be a boundary between a region which is vacuum ($T = 0$) and a region which contains matter ($T \neq 0$). To construct a global solution containing two such regions with a boundary between them, you have to match the metric and appropriate derivatives across the boundary. Then you solve the Einstein Field Equation within each region subject to the boundary conditions at the boundary between the regions, plus any other boundaries you find as you extend the solution over the manifold (for example, the singularity at $r = 0$ and asymptotic flatness at infinity for the model under discussion).

So a correct description of the Oppenheimer-Snyder model would be that $G = 0$ in the vacuum region, $G \neq 0$ in the matter region, and appropriate boundary conditions are enforced between them. The metric in the vacuum region is therefore Schwarzschild; the metric in the matter region is not (it turns out to be a portion of a closed collapsing FRW geometry). That is all there is to it.

Third, a spacetime geometry is a spacetime geometry. There is no "before" or "after". There are one or more regions with (if applicable) boundaries between them. Those boundaries don't have to be "before" or "after" boundaries. In the Oppenheimer-Snyder model, the boundary between the two regions is timelike (it is just the set of worldlines of particles at the surface of the collapsing matter); so it marks out a "point" that is moving "in space", not a time. The boundary's past endpoint is in the infinite past (since this an idealized model), and its future endpoint is at the singularity (since that is the future boundary of the region inside the black hole). There is no "before" or "after" the boundary; the only concept that makes sense is "inside" and "outside" the boundary. Nor is there any question of "the same G giving the same metric" before and after the singularity forms. "The same G" applies everywhere in the vacuum region, so the metric is Schwarzschild everywhere in the vacuum region; that's all there is to it. There is no having to "match" what the G does "before" vs. "after" anything; the vacuum region is one contiguous region of the spacetime.

 

[PeterDonis]

GR assumes, that is, contains as an axiom, that spacetime is a pseudo Riemannian manifold, as you pointed out.

Correct.

It then derives, according to its laws and the assumption of physically reasonable conditions, which are observed to actually occur, that a singularity occurs in that spacetime, which makes that spacetime not a pseudo Riemannian manifold.

Wrong. What is derived from solving the EFE for the Schwarzschild case is that there are geodesics in the manifold that cannot be extended to arbitrary values of their affine parameters (these are all the geodesics that have an endpoint at $r = 0$). But the manifold is a perfectly good pseudo-Riemannian manifold; the existence of such geodesics does not change that.

This is an inconsistency in GR

No, it is not. See above and post #27 from @PAllen .

For problems with the Schwarzschild solution, see my recent addition #25 to the original question.

Those aren't problems with the Schwarzschild solution. They are just your misunderstandings.

I am closing the thread at this point because the discussion is going nowhere and you are getting close to a warning. You keep repeating incorrect claims even though you have been given the corrections right here in this thread. You need to read more carefully and think more carefully before you post. You also would do well to spend some time working through a GR textbook and studying various well-known solutions of the Einstein Field Equation and how they are derived. (The Insights article with the spacetime diagram that I linked to is a four part series on the Schwarzschild geometry that might be helpful. I also linked earlier to the article on Birkhoff's theorem which gives an example of deriving a solution of the EFE.)