The Schwarzschild (Exterior) Solution

[evanallmighty]

This might be a silly question to people who are have more practice in this subject, but how does the Schwarzschild Solution/metric have a singularity at r=0 if it is only valid outside the gravitational mass (black hole)? wouldn't that be in the interior solution, describing the inside of a Schwarzschild black hole, where the gravitational singularity is actually located? I thought the only "singularity" in the exterior was the coordinate singularity at the event horizon.

Thanks to anyone who can help!

 

[PAllen]

This might be a silly question to people who are have more practice in this subject, but how does the Schwarzschild Solution/metric have a singularity at r=0 if it is only valid outside the gravitational mass (black hole)? wouldn't that be in the interior solution, describing the inside of a Schwarzschild black hole, where the gravitational singularity is actually located? I thought the only "singularity" in the exterior was the coordinate singularity at the event horizon.

Thanks to anyone who can help!

You can extent the exterior solution with either fluid interior solutions (no singularity), or a vacuum interior solution (with non-removable singularity at r=0).

 

[evanallmighty]

Yes I know that, but what about the singularity in the exterior solution, if it is specifically an exterior solution "cut off from the interior" if it has to be, why does it have a singularity at r=0 EVEN if it doesn't even go "deep" enough to describe r=0, it just goes to the event horizon. "How does it even" know r=0 is there?

 

[evanallmighty]

I thought is can only describe to the event horizon, not where the gravitational singularity is.

 

[atyy]

It's not that the Schwarzschild solution isn't valid below the Schwarzschild radius. It's that we don't know from the Schwarzschild solution alone how to join it to the part above the Schwarzschild radius, because of the coordinate singularity. I think it's most straightforward to think of the solution in terms of Kruskal-Szekeres (KS) coordinates, which cover the whole spacetime without coordinate singularities. Then the different bits of the Schwarzschild solution are seen to correspond to different parts of the KS solution, and r=0 of Schwarzschild is the singularity in KS.

http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

 

[PAllen]

Yes I know that, but what about the singularity in the exterior solution, if it is specifically an exterior solution "cut off from the interior" if it has to be, why does it have a singularity at r=0 EVEN if it doesn't even go "deep" enough to describe r=0, it just goes to the event horizon. "How does it even" know r=0 is there?

You extend it. Say you have two series , one valid x < 1, the other x > 1. The both approach the same value at 1 and satisfy smoothness conditions. Then you can treat them as one function adding a point at 1.

That is really all that is going on. The vacuum interior solution can be shown to properly fit the the exterior solution. Easiest way is simply to use different coordinates (e.g. Kruskal) that don't have the horizon singularity at all. Then show that that you can transform the vacuum interor Schwarzschild metric to one part of the more complete map, same for the exterior, with only the measure zero horizon not covered.

 

[evanallmighty]

Thanks to both of you, I am beginning to understand more. Now, I don't know if you know this too, but say you wanted to create a non singular Schwarzschild black hole model.
If you somehow found a non singular interior solution, could you smoothly paste it to the exterior solution (ignoring the coordinate singularity for now) even though it has the ugly r=0 (in the exterior solution)/ Would it still be non singular as long as the interior solution is non singular?

 

[atyy]

By definition, a black hole has an event horizon. There are theorems that if a spacetime has an event horizon, then it will also have a singularity - I'm not sure of the exact technical conditions for that to hold - but an example is given in http://en.wikipedia.org/wiki/Penrose–Hawking_singularity_theorems

Edit: Actually that's wrong!

http://arxiv.org/abs/gr-qc/0508107
"Thus, trapped surfaces really are trapped. Finding a trapped surface in a spacetime is also sufficient to imply the existence of a singularity (in the form of an inextendible geodesic) somewhere in its causal future [1, 3, 9].We note that there is no corresponding theorem for event horizons—event horizons can exist that do not contain singularities."

 

[evanallmighty]

Yes, the Penrose Hawking theorem does say that naked singularities, which is what they call singularities with no event horizon to "cover" cannot exist, there is no such thing as naked singularities in that theorem. You can have a Schwarzschild black hole, with an event horizon, and infinite density at r=0 but with no singularity (I'm not sure how to say that technically) but the solutions for it admit that information can be transferred at r=0. I suppose a non singular interior solution would still have the ugly r^2 (r=0) but it just would not be a singularity, just a location. (I'm guessing). The only other question I have (for now) is can you paste a non singular interior solution (with matching boundary conditions to the exterior solutions) to an (exterior) Schwarzschild solution, even thought the exterior solution described a singularity at r=0? Would that cause it to not be a non singular model? Does the singularity described in the exterior solution matter in the grand scheme of things? Can you get rid of it too?

 

[PAllen]

Thanks to both of you, I am beginning to understand more. Now, I don't know if you know this too, but say you wanted to create a non singular Schwarzschild black hole model.
If you somehow found a non singular interior solution, could you smoothly paste it to the exterior solution (ignoring the coordinate singularity for now) even though it has the ugly r=0 (in the exterior solution)/ Would it still be non singular as long as the interior solution is non singular?

You can smoothly paste an interior non-singular, non-vaccuum, solution to the extertior solution at an r > event horizon. However, if you do it at the event horizon, various singularity theorems guarantee the interior will evolve to a singularity.

[Edit: this is not inconsitent with Atyy's addendum. While there can be event horizons without singularities, if you fit the Schwarzschild one at the horizon, you must have (in some sense) mass = M inside, which will force collapse. ]

 

[evanallmighty]

Evolve? What theorems state that? (I'm not questioning you, I'm just wondering)
You said "If you do it at the event horizon", is there another way to do it (place to do it) Besides the event horizon, or did I just interpret how you said it wrong?

 

[pervect]

The Schwarzschild interior solution is a perfectly valid solution of Einstein's equations. Certain of the assumptions it makes are unlikely to be true for an actual physical collapse, however.

One of the big issues is that the Schwarzschild solution is perfectly spherically symmetrical. A real dust cloud will never be symmetrical. Suppose it has clumps - the question is, will those clumps tend to spread out over time, through diffusion, or will they tend to grow, because of gravity attracting more material to the clump - and hence grow?

My understanding of the situation is that in the interior, gravity is so intense that the effect of gravity is likely overwhelm any mechanism that could make the clump to "spread itself out". Therefore, the assumption of perfect spherical symmetry is one of the suspect assumptions. The Schwarzschild solution is not "stable", a small departure from the perfect symmetry will tend to grow larger.

People who have looked at the issue (specifically, Kip Thorne) seem to prefer the BKL solution for the non-rotating case. http://en.wikipedia.org/w/index.php?title=BKL_singularity&oldid=431041130.

This is only one of the issues. The other issue is that any actual solution will probably have some angular momentum. While we do have a rotating spherical symmetrial solution, the Kerr solution, it again makes the assumption that the geometry is perfectly symmetrical. (It's also not the Schwarzschild solution which we started out discussing, so it's another reason the Schwarzschild solution won't likely be found in reality.). Again, this assumption of perfect symmetry is suspect. The seemingly innocuous assumption of spherical symmetry leads to some unpleasant and rather unphysical "artifacts" in the Kerr solution, such as infinite blueshifts at the "inner horizon". "Mass inflation" see http://arxiv.org/abs/0811.1926, is expected to eliminate such infinite blueshifts, and to very significantly alter the nature of the "inner horizon".

 

[PAllen]

Does the singularity described in the exterior solution matter in the grand scheme of things? Can you get rid of it too?

Not quite sure what you're asking here. The singularity at the event horizon is easy to remove by coordinate transform. The event horizon doesn't go awy, just the coordinate singularity. There is no way to remove the singularity from the vacuum interior solution.

 

[PAllen]

Evolve? What theorems state that? (I'm not questioning you, I'm just wondering)
You said "If you do it at the event horizon", is there another way to do it (place to do it) Besides the event horizon, or did I just interpret how you said it wrong?

You can match above the event horizon. That way you get a simplified model of a (fictitious) nonrotating star. Match a spherical perfect fluid solution to the Schwarzschild solution somewhere above the event horizon. Then you have no singularities anywhere, coordinate or otherwise. The fluid solution still must, in some sense, have a mass corresponding to M of the exterior solution, but the density will be non-critical.

By evolve, I mean the following. If you find some matter solution that some time t has finite density at r=0, and total mass-energy and density such that it can be smoothly fit below the event horizon to the vacuum interior Schwarzschild solution, it is guarnteed that for for some later t, there will be true singularity (given the preconditions of the singularity theorems).

 

[evanallmighty]

Not quite sure what you're asking here. The singularity at the event horizon is easy to remove by coordinate transform. The event horizon doesn't go awy, just the coordinate singularity. There is no way to remove the singularity from the vacuum interior solution.

I mean the apparent singularity at r=0 described in the Schwarzschild exterior solution. Would it affect a non singular black hole model (consisting of some kind of non singular interior solution and the Schwarzschild exterior solution)

As in, could the model still be non singular if I take:
1) the schwarzschild exterior solution and:
2) some kind (I don't know exactly what) of non singular interior solution

 

[PAllen]

I mean the apparent singularity at r=0 described in the Schwarzschild exterior solution. Would it affect a non singular black hole model (consisting of some kind of non singular interior solution and the Schwarzschild exterior solution)

As in, could the model still be non singular if I take:
1) the schwarzschild exterior solution and:
2) some kind (I don't know exactly what) of non singular interior solution

The r=0 singularity is real not apparent.

What I've been trying to get across with respect to (2) is that any solution that smoothly fits at some r to Schwarzschild must contain the same mass as the M parameter of the Schwarzschild (with some vagueness about different mass defintions in GR). This means that if it is fit well outside of the horizon, there need not be any singularity at r=0, but if it is fit inside the event horizon, there must be a singularity (but not necessarily an eternal singularity as there is in the Schwarzschild interior vacuum solution). It is because the singularity of a non-vaccuum solution fit inside the horizon need not be an eternal singularity, that I speak of evolving toward the singularity. The existince of some form of singularity is what is guaranteed by the singularity theorems.

 

[evanallmighty]

I know it is not apparent I just referred to it as apparent because I was not sure about the singularity in the exterior solution (that was what my original question was about) but yes, I know it is a real physical entity there, not apparent.
Sorry to keep stretching the same question, but from what you and others have said:
I can (hypothetically) take the exterior Schwarzschild solution (even though it describes a singularity) and some kind of non singular interior solution and paste the two together at the boundary conditions. I am guessing this could possibly form a non singular black hole model, but your post about what you mean by is evolving into a singularity is disturbing me, although I THINK you are talking about a normal singular interior solution (vacuum) and an exterior solution.

 

[Matterwave]

The Schwarzschild solution is the solution for a non-rotating, non-charged, black hole. The metric that you solve for, doesn't necessarily have to be expressed in Schwarzschild coordinates. General relativity has the property of general coordinate covariance so that you are free to express the solution in any coordinates you want. There's no need to dwell on using Schwarzschild coodinates to describe this geometry. If you are getting near the event horizon, then it's best to switch to either Kruskal coordinates or Eddington-Finklestein coordinates. These coordinates do not have a singularity at the event horizon. But even these coordinates DO have a singularity at r=0, so r=0 is what we call an "essential singularity".

 

[atyy]

Apart form the Hawking-Penrose singularity theorems, the other relevant theorem known to me is Buchdal's theorem.

http://people.sissa.it/~rezzolla/lnotes/mondragone/collapse.pdf
http://www.roma1.infn.it/teongrav/VALERIA/TEACHING/ONDE_GRAV_STELLE_BUCHINERI/AA2010_2011/NSb.pdf

 

[PAllen]

I can (hypothetically) take the exterior Schwarzschild solution (even though it describes a singularity) and some kind of non singular interior solution and paste the two together at the boundary conditions. I am guessing this could possibly form a non singular black hole model, but your post about what you mean by is evolving into a singularity is disturbing me, although I THINK you are talking about a normal singular interior solution (vacuum) and an exterior solution.

I'll try again to say the same thing a different way.

1) Suppose you have a valid total solution of the Einstein Field equations (EFE) that matches Schwarzschild geometry beyond some r well beyond the horizon, but differs less than r. Then it is possible for this solution to have no singularity (or horizon, for that matter).

2) Suppose you have some total solution of the EFE that matches the Schwarzschild geometry beyond some r at or inside the horizon. Suppose further, that at some time t, this solution has finite density throughout all of its matter part. Then, it still must be true that there is an essential singularity within the event horizon at some later time t (and beyond).

If you need to ask again, can you try to describe exactly what part you don't understand?

 

[atyy]

Here are some things that I think fill in details of PAllen's statements.

A spherically symmetric vacuum solution outside a spherically symmetric star must be the Schwarzschild solution. (Birkhoff's theorem)
http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html

If a star is less than 9/8 Schwarzschild radius, it will collapse. (Buchdal's theorem)
http://people.sissa.it/~rezzolla/lnotes/mondragone/collapse.pdf
http://www.roma1.infn.it/teongrav/VALERIA/TEACHING/ONDE_GRAV_STELLE_BUCHINERI/AA2010_2011/NSb.pdf

If it is collapses to less than the Schwarzschild radius, there will be a trapped surface. (Penrose handwave)
If there is a trapped surface, there will be a singularity. (Hawking-Penrose theorem)
http://homepages.physik.uni-muenchen.de/~winitzki/T7/ , section 4.1.7

Lots of fun pictures: http://arxiv.org/abs/gr-qc/0508107

 

[evanallmighty]

2) Suppose you have some total solution of the EFE that matches the Schwarzschild geometry beyond some r at or inside the horizon. Suppose further, that at some time t, this solution has finite density throughout all of its matter part. Then, it still must be true that there is an essential singularity within the event horizon at some later time t (and beyond).

I understand most of what you are saying now. But... (I'm sorry to keep asking questions) WHY does there have to be an essential singularity within the event horizon. You said nothing that shows one is required, or am I missing something? You might of said it I'm just missing it. Thanks

 

[PAllen]

I understand most of what you are saying now. But... (I'm sorry to keep asking questions) WHY does there have to be an essential singularity within the event horizon. You said nothing that shows one is required, or am I missing something? You might of said it I'm just missing it. Thanks

It's a well known mathematical theorem GR. See Atyy's post for some references. Previously, it seemed your issue was understanding what I was claiming. Only now are you clarifying that want justification for it. For justification, read Atyy's references.

 

[evanallmighty]

Yes... I read that right after I asked the previous question. I'm sorry. But this is a statement in 4.2 (page 98) that I don't understand:

"Statement 1: An event horizon is a null surface whose generators
are null geodesics without future endpoints, except if
a generator hits a singularity. (Generators cannot leave H and
enter the interior of B.)"

I'm just not quite sure what this is saying. I assume it is saying the geodesic never end.
Could you have a solution with infinite density (I've seen one before but forgot the name) but still be able to transfer information at r=0?

EDIT---Here it is: (sadly you can't read it unless you want to spend $31...)

A non singular solution for spherical configuration with infinite central density
Fuloria, Pratibha; Durgapal, M. C.
Astrophysics and Space Science, Volume 314, Issue 4, pp.249-250
A non-singular exact solution with an infinite central density is obtained for the interior of spherically symmetric and static structures. Both the energy density and the pressure are infinite at the center but we have e λ(0)=1 and e ν(0)≠0. The solution admits the possibility of receiving signals from the region of infinite pressure.
Keywords: General relativity, Exact solution, Astrophysics

 

[PAllen]

Yes... I read that right after I asked the previous question. I'm sorry. But this is a statement in 4.2 (page 98) that I don't understand:

"Statement 1: An event horizon is a null surface whose generators
are null geodesics without future endpoints, except if
a generator hits a singularity. (Generators cannot leave H and
enter the interior of B.)"

I'm just not quite sure what this is saying.
Could you have a solution with infinite density (I've seen one before but forgot the name) but still be able to transfer information at r=0?

EDIT---Here it is: (sadly you can't read it unless you want to spend $31...)

A non singular solution for spherical configuration with infinite central density
Fuloria, Pratibha; Durgapal, M. C.
Astrophysics and Space Science, Volume 314, Issue 4, pp.249-250
A non-singular exact solution with an infinite central density is obtained for the interior of spherically symmetric and static structures. Both the energy density and the pressure are infinite at the center but we have e λ(0)=1 and e ν(0)≠0. The solution admits the possibility of receiving signals from the region of infinite pressure.
Keywords: General relativity, Exact solution, Astrophysics

Without seeing the paper, I can't tell, but since it can't be inconsistent with theorems, one guesses it escapes there assumptions. The obvious ways (all consistent with the abstract as given) are:

1) The radius is greater than 9/8 the Schwarzschild radius.
2) The positive energy constraints are violated.


As for your first question, I don't have time now to look at it. Perhaps someone else will.

 

[evanallmighty]

You can match above the event horizon. That way you get a simplified model of a (fictitious) nonrotating star. Match a spherical perfect fluid solution to the Schwarzschild solution somewhere above the event horizon. Then you have no singularities anywhere, coordinate or otherwise. The fluid solution still must, in some sense, have a mass corresponding to M of the exterior solution, but the density will be non-critical.

I know you posted this a while back but I was just reading through again. So what your saying, is if you paste them together above the event horizon, it is describing a star. I want to describe a black hole. A nonsingular black hole more specifically. From what I have gathered, I can paste a perfect fluid interior (which can always be pasted to a static, spherical exterior solution, and the Schwarzschild exterior solution, and create a non singular black hole model, from what your first posts said.

 

[atyy]

Yes... I read that right after I asked the previous question. I'm sorry. But this is a statement in 4.2 (page 98) that I don't understand:

"Statement 1: An event horizon is a null surface whose generators
are null geodesics without future endpoints, except if
a generator hits a singularity. (Generators cannot leave H and
enter the interior of B.)"

I'm just not quite sure what this is saying.

It is saying that an event horizon is a surface made of spacetime paths on which light rays can travel forever. This is the boundary that light rays from inside the event horizon can never go past.

 

[atyy]

I know you posted this a while back but I was just reading through again. So what your saying, is if you paste them together above the event horizon, it is describing a star. I want to describe a black hole. A nonsingular black hole more specifically. From what I have gathered, I can paste a perfect fluid interior (which can always be pasted to a static, spherical exterior solution, and the Schwarzschild exterior solution, and create a non singular black hole model, from what your first posts said.

Perhaps there is no proof that you cannot do that. But you must evade all the theorems in post #21.

 

[evanallmighty]

Perhaps there is no proof that you cannot do that. But you must evade all the theorems in post #21.

Well, there might not be, but unless I am mistaken, I have read some pretty convincing scientific papers trying to do what I want to do.

 

[PAllen]

I know you posted this a while back but I was just reading through again. So what your saying, is if you paste them together above the event horizon, it is describing a star. I want to describe a black hole. A nonsingular black hole more specifically. From what I have gathered, I can paste a perfect fluid interior (which can always be pasted to a static, spherical exterior solution, and the Schwarzschild exterior solution, and create a non singular black hole model, from what your first posts said.

No, you misread them. In every case I've said if you match the Schwarzschild geometry well above the event horizon, you can avoid a singularity. If you match at or below (or very close) to the event horizon you cannot avoid a singularity. I glossed over technical requirements of realism, e.g. energy conditions. I have never seen a reputable claim otherwise (and I've read numerous discussions by top GR experts verifying variants of what I've said).

 

[PAllen]

Well, there might not be, but unless I am mistaken, I have read some pretty convincing scientific papers trying to do what I want to do.

My guess is that you mis-understood some part of them. This area has been well established since the early 1970s.

 

[evanallmighty]

Ok PAllen... I think I have it! (And yes, the scientific papers I have read are very vague and I probably did miss something)

If I find a perfect fluid interior solution and the Schwarzschild exterior solution, I can paste them smoothly together above the event horizon in order to develop a non singular static spherical black hole model. The event horizon is still there and everything... right? Still a black hole?

Now... It doesn't seem as straight forward to paste them ABOVE the event horizon as AT the event horizon... hmmm I hate to keep asking things, but do you have any more information on that?

 

[atyy]

So I googled "nonsingular black hole" and via Hossenfelder et al's http://arxiv.org/abs/0912.1823 came across

http://arxiv.org/abs/gr-qc/9612057
Regular Black Holes and Topology Change
Arvind Borde
(Submitted on 20 Dec 1996)
Abstract: The conditions are clarified under which regular (i.e., singularity-free) black holes can exist. It is shown that in a large class of spacetimes that satisfy the weak energy condition the existence of a regular black hole requires topology change.


http://arxiv.org/abs/gr-qc/9911046
Regular Black Hole in General Relativity Coupled to Nonlinear Electrodynamics
Eloy Ayón-Beato, Alberto García (CINVESTAV-IPN, Mexico)
(Submitted on 11 Nov 1999)
The first regular exact black hole solution in General Relativity is presented. The source is a nonlinear electrodynamic field satisfying the weak energy condition, which in the limit of weak field becomes the Maxwell field.

Let me post again some references on the singularity theorems, so we can see how they are evaded.
Winitzki http://sites.google.com/site/winitzki/index/topics-in-general-relativity, section 4.1
Senovilla http://arxiv.org/abs/physics/0605007

Edit: They evade the theorem in Winitzki's section 4.1.7 by not having a noncompact Cauchy surface.

 

[pervect]

Well, there might not be, but unless I am mistaken, I have read some pretty convincing scientific papers trying to do what I want to do.

If I'm following the thread correctly, the Hawking Penrose theorems have been mentioned as obstacles. They don't say that you can't do it, they DO say you need "exotic matter" that violates one of the energy conditions to do it. (the Penrose theorem requires the weak energy condition to be violated, I'm not as familar with the other theorems mentioned).

Perhaps you don't care, but a little disclaimer that you know you are violating these energy conditions your statements will go over much better, even more so if you give some motivation for why these energy conditions would be violated other than "I want to do this and they're getting in the way".

 

[evanallmighty]

Perhaps you don't care, but a little disclaimer that you know you are violating these energy conditions your statements will go over much better, even more so if you give some motivation for why these energy conditions would be violated other than "I want to do this and they're getting in the way".

Yes, I know about the energy conditions. I'm just asking about this and learning more about it because I am very interested in learning about non singular black holes. Here's a cool article that got me interested in asking PF this:
http://koberlein.main.ad.rit.edu/journal/2009/12/non-singular-black-holes.php

It's pretty cool.