Black Holes - the two points of view.

[TrickyDicky]

I don't think so. The 4 Killing vector fields of the Schwarzschild exterior are:

$\partial_t$
$\partial_\phi$
$sin(\phi) \partial_\theta + cot(\theta) cos(\phi) \partial_\phi$
$cos(\phi) \partial_\theta - cot(\theta) sin(\phi) \partial_\phi$

In the interior, I think the same 4 are still Killing vector fields, because they still obey the Killing equation. But all 4 are now spacelike. That's what it seems to me, although I can't find a definitive statement one way or the other.

I agree.
What I'm saying is that I think the definition of isotropic vacuum solution (according to Birkhoff) demands at least one timelike KVF, so the interior wouldn't be a vacuum,(it would certainly be isotropic and have a time asymmetry quite lke the FRW metric).

 

[Dale]

I think the definition of isotropic vacuum solution (according to Birkhoff) demands at least one timelike KVF

I am not sure that is correct, but if it is then maybe Schwarzschild is not isotropic. It is definitely vacuum. But the insistence on a KVF which is everywhere timelike seems in doubt to me.

 

[Dale]

I don't think so. The 4 Killing vector fields of the Schwarzschild exterior are:

$\partial_t$
$\partial_\phi$
$sin(\phi) \partial_\theta + cot(\theta) cos(\phi) \partial_\phi$
$cos(\phi) \partial_\theta - cot(\theta) sin(\phi) \partial_\phi$

In the interior, I think the same 4 are still Killing vector fields, because they still obey the Killing equation. But all 4 are now spacelike. That's what it seems to me, although I can't find a definitive statement one way or the other.

Then perhaps a linear combination is timelike. It is possible to get a timelike vector from a linear combination of two spacelike vectors.

 

[TrickyDicky]

I am not sure that is correct, but if it is then maybe Schwarzschild is not isotropic. It is definitely vacuum.

Isotropic means spherically symmetric and that certainly is fulfilled by the Schwarzschild spacetime.

But the insistence on a KVF which is everywhere timelike seems in doubt to me.

In fact killing vector fields are usually defined globally in a space because they define isometries, that is they are geometric invariants. Regardless of the fact that there may be different type of geodesics

Then perhaps a linear combination is timelike. It is possible to get a timelike vector from a linear combination of two spacelike vectors.

But we are talking about killing vectors here, not vector fields in general.

 

[Dale]

Isotropic means spherically symmetric and that certainly is fulfilled by the Schwarzschild spacetime.

Well, since Schwarzschild is definitely symmetric and definitely vacuum and definitely a solution to the EFE, then it cannot be correct that every symmetric vacuum solution must have a KVF which is everywhere timelike.

 

[Austin0]

Going back to your earlier post, my answer is YES to all those points. Yes, the falling observer would take an infinite time to reach the event horizon - in OUR reference frame. Yes, he wouild fall through very quickly, in HIS reference frame. We "see" him trapped there for eternity, ever edging closer to the EH. But he does not then fall through quickly as our clocks tick over to infinity. That is meaningless. His clock is never stopped, in either reference frame, because our clocks never "reach" infinity.

you say the falling observers clock is never stopped in either frame because the distant observers clock never reaches infinity.
I agree. but you seem to ignore the fact that this is only true in the region where the faller has NOT reached the singularity.
you then want to magically have the faller PASS the horizon without ever having reached it.

It appears you interpret time dilation in a way that creates alternate contradictory realities.
If your premise that reaching the horizon requires infinite coordinate time for the distant observer is correct, that means that at all points in that interval the times at the two locations will be related by the SC metric. Both observers will agree on these relative elapsed times and both observers will agree that the faller has not reached the horizon.

the fact that the time subjectively passes normally for the faller does not affect this relationship.

An analogous scenario:
As system passes that is accelerating from the distant past that now has a gamma factor of 10²⁰ At this point we "observe" a passenger starting to walk from one end to the other. A stroll requiring 10 sec of ship time.
We jump ahead an interval on the order of the age of the universe 10¹⁰ Earth years. A future observer would see the passenger in virtually the same point in the walk with an elapsed time on his watch of 0.0018 secs.
Ahead another 10¹⁰ years etc.etc.

In fact the 10 seconds on the ship would equate to approx 5.5 x 10¹³yrs.
even without factoring in the increased gamma from the acceleration over this time.

SO for the next 3,500 ages of the universe both frames will agree the passenger has not reached the far end of the ship. The fact that time appears to be passing normally for the passenger does not mean that he will ever complete his trip in the real universe.

Which is what you are suggesting here . One universe where the passenger never completes his walk (reaches the horizon) and another where he finishes his walk and moves on (reaches the horizon and moves past it)

It seems that either the inference of infinite time is incorrect and/or the Sc metric is not accurate approaching the singularity, in which case the faller reaches the horizon and beyond
Or the first two are correct and the faller does not pass the horizon.

You cannot have it both ways and be logically consistent as far as I can see.

It is like the old quandary "Do parallel lines never meet, or do they meet at infinity?". I'll know the answer when I get to infinity.

Mike

In this analogy I am saying that two lines cannot intersect and not intersect.

SO two observers moving along parallel lines will never reach a point of intersection if that point is at infinity no matter how long or far they travel. The fact that they can never be sure that the lines don't intersect at some more distant point is irrelevant.

You are saying that they both remain parallel AND intersect.
That for one observer with a fast ticking clock the lines never intersect no matter how long he travels.
But for the other observer, the lines do intersect at some finite point, simply because his clock is running slower and he is subjectively taking less time to reach infinity.
?

 

[Dale]

Then perhaps a linear combination is timelike. It is possible to get a timelike vector from a linear combination of two spacelike vectors.

According to my calculations in Mathematica there is no linear combination of the KVFs which is timelike inside the horizon. I am not super-confident, but it seems somewhat unlikely.

 

[PeterDonis]

But we know (tell me if you disagree) that a spherically symmetric vacuum region has 4 KVF: 3 spacelike (corresponding to those of the SO(3) group that any spherically symmetric space has and one timelike KVF.

I disagree. We know that a spherically symmetric vacuum spacetime has 4 KVFs. We know that 3 are spacelike because they are the KVFs corresponding to SO(3). We do *not* know that the 4th is timelike. There is nothing in the isometries of a spherically symmetric vacuum that requires the 4th KVF to be timelike everywhere.

TrickyDicky and PeterDonis, I am not sure that this is correct in general, but since the $\partial_t$ KVF becomes spacelike inside the EH perhaps one of the other three becomes timelike. Or perhaps there is a linear combination that becomes timelike.

I don't think this is correct; I think the the 3 KVFs corresponding to the SO(3) symmetries have to be spacelike everywhere. AFAIK all 4 KVFs in Schwarzschild spacetime are spacelike inside the EH. [Edit: it looks like DaleSpam and stevendaryl have confirmed this.] The timelike vector fields there are things like the frame field of infalling observers that I wrote down before, which is not a KVF.

In fact, that observation gives a good argument for why there can't be a timelike KVF inside the EH: *all* timelike vectors inside the EH have to have a nonzero (in fact negative) $\partial / \partial r$ component, and no such vector can be a KVF in Schwarzschild spacetime.

What I'm saying is that I think the definition of isotropic vacuum solution (according to Birkhoff) demands at least one timelike KVF

Isotropic means spherically symmetric and that certainly is fulfilled by the Schwarzschild spacetime.

I see two mistakes here. First of all, as I have now shown in detail, Birkhoff's theorem does *not* show that a spherically symmetric vacuum spacetime has a KVF that is timelike everywhere; it only shows that it has a KVF which is timelike in an asymptotically flat region, not that that region has to cover the entire spacetime. The proof in MTW that I posted about confirms this.

Second, regarding the comparison with FRW spacetime: Schwarzschild spacetime is isotropic about one "center" only (r = 0). FRW spacetime is isotropic about *every* point; you can choose any spatial point you like as the "center" and isotropy still holds. In other words, it is isotropic *and* homogeneous. That's a big difference in symmetry properties.

 

[PeterDonis]

In fact killing vector fields are usually defined globally in a space because they define isometries, that is they are geometric invariants. Regardless of the fact that there may be different type of geodesics

This doesn't have anything to do with the causal nature of a KVF (timelike, spacelike, or null) or whether or not that nature can be different in different regions of the spacetime. An isometry doesn't have to be timelike everywhere (or spacelike everywhere, or null everywhere). Also, orbits of isometries don't have to be geodesics.

Regarding the bit about "geometric invariants": that term just means that, when you calculate an invariant *at a specific event in spacetime*, you must get the same answer no matter which coordinate chart you use. It does *not* mean that, once you get an answer about an invariant at one event (e.g., that a certain vector is timelike), that same answer must apply at all other events. A vector field (including an isometry) that is timelike at one event does not have to be timelike everywhere.

 

[TrickyDicky]

We do *not* know that the 4th is timelike. There is nothing in the isometries of a spherically symmetric vacuum that requires the 4th KVF to be timelike everywhere.

It seems that the issue here is with the word "everywhere" because I hope you at least admit that per Birkhoff's theorem a spherically symmetric vacuum requires the 4th KVF to be timelike locally (at the points where it is a solution of the vacuum EFE).
No theorem can claim that something is a determinate way "everywhere", it is only that way at the regions where the conditions of the theorem hold.
Some of the versions of the theorem (not the original, that I haven't had the chance to find yet) say that the theorem only holds in the exterior region. That would help clarify some things.
The version that can be found at the Einstein online site states it this way:
" The spherically symmetric spacetime around any spherically symmetric matter configuration has the same properties as space-time around a Schwarzschild black hole of the appropriate mass."

This suggests also that the theorem is only valid outside the Black hole, that is, outside the Event horizon, and that the interior is not required to be a vacuum.

I see two mistakes here. First of all, as I have now shown in detail, Birkhoff's theorem does *not* show that a spherically symmetric vacuum spacetime has a KVF that is timelike everywhere; it only shows that it has a KVF which is timelike in an asymptotically flat region, not that that region has to cover the entire spacetime. The proof in MTW that I posted about confirms this.

Yes, see above.

Second, regarding the comparison with FRW spacetime: Schwarzschild spacetime is isotropic about one "center" only (r = 0). FRW spacetime is isotropic about *every* point; you can choose any spatial point you like as the "center" and isotropy still holds. In other words, it is isotropic *and* homogeneous. That's a big difference in symmetry properties.

Yes, is the difference between vacuum and matter solutions. A vacuum cannot be expanding or contracting, there is nothing to contract or expand, and all vacuums are homogeneous, in case that property can be applied to vacuums at all, usually is referred to matter configurations.

 

[TrickyDicky]

This doesn't have anything to do with the causal nature of a KVF (timelike, spacelike, or null) or whether or not that nature can be different in different regions of the spacetime. An isometry doesn't have to be timelike everywhere (or spacelike everywhere, or null everywhere). Also, orbits of isometries don't have to be geodesics.

Regarding the bit about "geometric invariants": that term just means that, when you calculate an invariant *at a specific event in spacetime*, you must get the same answer no matter which coordinate chart you use. It does *not* mean that, once you get an answer about an invariant at one event (e.g., that a certain vector is timelike), that same answer must apply at all other events. A vector field (including an isometry) that is timelike at one event does not have to be timelike everywhere.

Isometries usually refer to spacetime global symmetries, I think it makes little sense to say a spacetime has a symmetry but just a little of it, or only in a region. A spacetime as a whole either has a global symmetry or it doesn't.
KVF's happen to generate global isometries.

 

[TrickyDicky]

Another version of the theorem:
"Theorem. Any rotationally symmetric solution of the VEE(vacuum Einstein equation) is static,
and is isometric to an open subset of the Schwarzschild exterior of the Schwarzschild black hole."

 

[Dale]

The Schwarzschild spacetime is a solution to the vacuum EFE everywhere. The Schwarzschild spacetime is static. The Schwarzschild spacetime has a KVF which is timelike outside the EH, but not inside. Therefore, being a static solution to the vacuum EFE does not imply that there is a KVF which is everywhere timelike.

 

[TrickyDicky]

The Schwarzschild spacetime is a solution to the vacuum EFE everywhere. The Schwarzschild spacetime is static. The Schwarzschild spacetime has a KVF which is timelike outside the EH, but not inside. Therefore, being a static solution to the vacuum EFE does not imply that there is a KVF which is everywhere timelike.

No. The region inside the EH is not static, that is a well known fact, PeterDonis is also aware of it.

 

[Dale]

No. The region inside the EH is not static, that is a well known fact, PeterDonis is also aware of it.

Then Birkhoff's theorem, correctly stated, must refer to a spacetime which is asymptotically static or static at infinity or something similar. Something has to give, you are holding a set of definitions and theorems that are contradicted, as a whole, by the example of a Schwarzschild spacetime.

Your propositions:
1) Any spherically symmetric solution to the vacuum EFE is static (your statement of Birkhoff's theorem)
2) Schwarzschild spacetime is a spherically symmetric solution to the vacuum EFE
3) Schwarzschild spacetime is not static

You cannot have all 3 true. 2) can be easily proven. So either 1) or 3) must be false.

 

[TrickyDicky]

Then Birkhoff's theorem, correctly stated, must refer to a spacetime which is asymptotically static or static at infinity or something similar. Something has to give, you are holding a set of definitions and theorems that are contradicted, as a whole, by the example of a Schwarzschild spacetime.

Your propositions:
1) Any spherically symmetric solution to the vacuum EFE is static (your statement of Birkhoff's theorem)
2) Schwarzschild spacetime is a spherically symmetric solution to the vacuum EFE
3) Schwarzschild spacetime is not static

You cannot have all 3 true. 2) can be easily proven. So either 1) or 3) must be false.

I guess one just have to add to 1) outside an spherically symmetric matter configuration or a Schwarzschild black hole, and to 3) inside the EH of a Schwarzschild black hole.
That makes the 3 of them compatible.

 

[PAllen]

Here is a paper indicating we are not the only ones bothered by the variety of formulations of Birkhoff's theorem, and attempts to classify such statements into four(!) categories(!):

http://arxiv.org/abs/1208.5237

 

[TrickyDicky]

Here is a paper indicating we are not the only ones bothered by the variety of formulations of Birkhoff's theorem, and attempts to classify such statements into four(!) categories(!):

http://arxiv.org/abs/1208.5237

Interesting paper. Thanks

 

[Dale]

I guess one just have to add to 1) outside an spherically symmetric matter configuration or a Schwarzschild black hole, and to 3) inside the EH of a Schwarzschild black hole.
That makes the 3 of them compatible.

OK (assuming by "black hole" you refer to the EH rather than the singularity).

So, it should be clear that there is no conflict between those restated propositions and the idea that an object free-falling across the EH is represented by a geodesic which is everywhere timelike.

 

[TrickyDicky]

2) Schwarzschild spacetime is a spherically symmetric solution to the vacuum EFE
...

Actually there is an additional source of confusion, at the time the theorem was first formulated, the only Schwarzschild geometry was the original Schwarzschild metric, it took almost 40 years to come up with the extended space, and the added region was not static, so the theorem only applied outside that region.

 

[PAllen]

Actually there is an additional source of confusion, at the time the theorem was first formulated, the only Schwarzschild geometry was the original Schwarzschild metric, it took almost 40 years to come up with the extended space, and the added region was not static, so the theorem only applied outside that region.

No, this is not true. The complete manifold (maximal extension of every geodesic) was worked out independently by Synge (1960) and Kruskal later. However, the extension across the horizon (2 of the 4 regions), and recognition that the horizon was not singular in any way, was considered well established enough to be presented in a college textbook from 1942 (I have a copy; it is Bergmann's book, endorsed via lengthy forward by Einstein).

 

[PeterDonis]

It seems that the issue here is with the word "everywhere" because I hope you at least admit that per Birkhoff's theorem a spherically symmetric vacuum requires the 4th KVF to be timelike locally (at the points where it is a solution of the vacuum EFE).

No, I don't admit that. Did you read my post giving details from MTW's proof of Birkhoff's theorem? Please read it carefully; the Wikipedia article does not state the theorem correctly. The theorem does *not* say that a spherically symmetric vacuum spacetime must be static; it says that a spherically symmetric vacuum region in a spacetime must be a piece of the Schwarzschild geometry. The difference between those two statements is crucial to this discussion, and my post quotes statements from MTW's details on the proof that highlight what the difference is.

Some of the versions of the theorem (not the original, that I haven't had the chance to find yet) say that the theorem only holds in the exterior region.

Please give references. MTW certainly does not say this. Again, please read my previous post on that carefully. Same comment for the other versions of the theorem that you quote.

A quick comment on quoting various statements of the theorem: given the topic of discussion, just quoting statements of the theorem is useless. We need to look at the actual proof; that is what I did in my post quoting from MTW's proof. It's the actual proof, and the assumptions it uses, that are crucial here.

Isometries usually refer to spacetime global symmetries, I think it makes little sense to say a spacetime has a symmetry but just a little of it, or only in a region.

I haven't said that. The KVF under discussion, $\partial / \partial t$ in Schwarzschild coordinates, is a global KVF, meaning the particular vector that is a member of the KVF at each event is a is a Killing vector. But that doesn't mean all of the properties of the particular vector that is a member of the KVF at one event must be identical to all of the properties of the (different) vector that is a member of the KVF at another event.

No. The region inside the EH is not static, that is a well known fact, PeterDonis is also aware of it.

Yes, that's correct. And it's OK, because Birkhoff's theorem does not require the entire spacetime to be static. See above and my previous posts.

Then Birkhoff's theorem, correctly stated, must refer to a spacetime which is asymptotically static or static at infinity or something similar.

It does. That's been the point of a number of my recent posts, in particular the one quoting in detail from MTW's proof of the theorem.

Here is a paper indicating we are not the only ones bothered by the variety of formulations of Birkhoff's theorem, and attempts to classify such statements into four(!) categories(!)

It looks to me like these refer to four different theorems that all go under the name "Birkhoff theorems". The third of the four is the one we have been calling "Birkhoff's theorem". The others are not different statements or formulations of that one; they are different theorems. (They are related, but not the same.) At least, that's how it looks to me.

Actually there is an additional source of confusion, at the time the theorem was first formulated, the only Schwarzschild geometry was the original Schwarzschild metric, it took almost 40 years to come up with the extended space, and the added region was not static, so the theorem only applied outside that region.

I don't know about the original 1923 statement of the theorem, but certainly the statement and proof given in MTW is not subject to this restriction. And MTW at least implies (to me) that the 1923 statement was similarly unrestricted (though I can't be sure since they only reference the 1923 paper, they don't appear to quote directly from it). See above and my MTW post.

Edit: Since this thread has been accumulating posts fast and furious, here's the link to my previous post on the MTW statement and proof (post #169):

https://www.physicsforums.com/showpost.php?p=4070743&postcount=169

 

[Mike Holland]

If your premise that reaching the horizon requires infinite coordinate time for the distant observer is correct, that means that at all points in that interval the times at the two locations will be related by the SC metric. Both observers will agree on these relative elapsed times and both observers will agree that the faller has not reached the horizon.
. You cannot have it both ways and be logically consistent as far as I can see.
?

Ah! But I can have it both ways. Let's say that he has 4 minutes (on his clock) before he hits the EH. Then after a millenium on our clock he has only two minutes left. After another millenium he has 1 minute left etc. Then when our clocks "reach" infinity, he will have traveled for 2 + 1 + 1/2 + 1/4 + 1/8 etc which adds up to 4 minutes. So the faller covers an infinite number of time increments in a finite time. Zeno's paradox rearing its ugly head. You need to read the Greek philosophers again :)

I am in full agreement with the rest of your post, but my reference to parallel lines only confused the issue.

Mike

 

[TrickyDicky]

But notice, now, that there is nothing stopping the dt^2 or dr^2 terms from being of either sign, since they are no longer exponentials. (A mathematically stricter derivation would not have put the exponentials in in the first place, but would have left the line element in a form that explicitly allows either sign for the terms; the derivation still goes through just fine if you do it that way.) So there is nothing in this proof that requires the region of spacetime under consideration to be static, since there is nothing that requires the t coordinate to be timelike.

It seems to me you are deriving geometrical conclusions from a simple matter about signature conventions , the fact that the time coordinate can be positive or negative depending on the choice of signature bears no evident connection for me with what we are discussing here and in any case it is a purely coordinate thing.
Besides, I don't have any problem with the Birkhoff theorem as you quote it from MTW, it is perfectly compatible with the interior region not being a vacuum, since it only demands staticity of "a piece of the Schwarzschild's geometry", I take that piece to correspond with the outside region of the geometry.

 

[TrickyDicky]

No, this is not true. The complete manifold (maximal extension of every geodesic) was worked out independently by Synge (1960) and Kruskal later. However, the extension across the horizon (2 of the 4 regions), and recognition that the horizon was not singular in any way, was considered well established enough to be presented in a college textbook from 1942 (I have a copy; it is Bergmann's book, endorsed via lengthy forward by Einstein).

This seems completely irrelevant to the point I was making, so Ok let's say 20 years instead of 40 years,doesn't make any difference.

 

[TrickyDicky]

So, it should be clear that there is no conflict between those restated propositions and the idea that an object free-falling across the EH is represented by a geodesic which is everywhere timelike.

Well, it is tricky to say, it depends on the coordinates you use, in Schwarzschild coordinates it is a spacelike geodesic, but that seems to be a problem of the coordinates because they are swapped at the EH. In fact a test particle must always be timlike. As you said the norm of its tangent vector can't change sign.
This is taken from wikipedia page:"Geodesics in GR", I would like for anybody disagreeing with any of the quoted (especially the bolded ones) statements to speak up and explain why."With a metric signature of (-+++) being used,

timelike geodesics have a tangent vector whose norm is negative;
null geodesics have a tangent vector whose norm is zero;
spacelike geodesics have a tangent vector whose norm is positive.

Note that a geodesic cannot be spacelike at one point and timelike at another.

An ideal particle (ones whose gravitational field and size are ignored) not subject to electromagnetic forces (or any other non-gravitational force) will always follow timelike geodesics. Note that not all particles follow geodesics, as they may experience external forces, for example, a charged particle may experience an electric field — in such cases, the worldline of the particle will still be timelike, as the tangent vector at any point of a particle's worldline will always be timelike."

 

[TrickyDicky]

The free-falling observer does indeed follow a geodesic, but that geodesic is NOT an integral curve of the Killing vector field.

The curve corresponding to the time-like Killing vector is the worldline of someone hovering at a constant r, theta, phi in Schwarzschild coordinates, not someone freefalling.

I missed this key post.

Isn't a test particle orbiting a BH at a constant r (circular orbit, in a spherically symmetric vacuum setting this is possible unlike the real universe case) outside the EH also free-falling, that is, describing a timelike geodesic?

 

[Dale]

Well, it is tricky to say, it depends on the coordinates you use, in Schwarzschild coordinates it is a spacelike geodesic, but that seems to be a problem of the coordinates because they are swapped at the EH.

That isn't actually correct, but I can see why you might think so. The confusion stems from the fact that the Schwarzschild coordinates inside and outside the EH are two separate coordinate charts that don't overlap. If you solve the geodesic equation in Schwarzschild coordinates you will get multiple roots, one will be timelike for the r<2M chart and another will be timelike for the r>2M chart. Obviously, if you use the same root for both coordinate charts then you will get a mistake, but such a pair of paths does not represent a free-falling object.

This isn't so much a problem of the coordinates as a problem with the boundary conditions. Since neither chart contains r=2M you cannot match up the boundary conditions using only those two coordinate charts.

 

[TrickyDicky]

That isn't actually correct, but I can see why you might think so. The confusion stems from the fact that the Schwarzschild coordinates inside and outside the EH are two separate coordinate charts that don't overlap. If you solve the geodesic equation in Schwarzschild coordinates you will get multiple roots, one will be timelike for the r<2M chart and another will be timelike for the r>2M chart. Obviously, if you use the same root for both coordinate charts then you will get a mistake, but such a pair of paths does not represent a free-falling object.

This isn't so much a problem of the coordinates as a problem with the boundary conditions. Since neither chart contains r=2M you cannot match up the boundary conditions using only those two coordinate charts.

Well explained.

 

[TrickyDicky]

... regarding the comparison with FRW spacetime: Schwarzschild spacetime is isotropic about one "center" only (r = 0). FRW spacetime is isotropic about *every* point; you can choose any spatial point you like as the "center" and isotropy still holds. In other words, it is isotropic *and* homogeneous. That's a big difference in symmetry properties.

I want to go back to this point because if you read the first pages of Carroll's Notes chapter on Black holes, he clearly shows that at least in the region inside the EH the difference in symmetry you mentioned is nullified; he explicitly says that using non-rigorous arguments, it is deliberately chosen to extend isotropy to all points, meaning using a manifold completely foliated by two spheres, justifying it because although the origin shouldn't be included, almost all space validity is good enough to claim all space validity.
By doing this it demands not only spherical symmetry but homogeneity, at least in the region inside the EH. The problem is that in the abscence of cosmological constant, any isotropic and homogeneous solution of the EFE is an expanding or contracting, FRW-like metric with no timelike KVF

 

[Dale]

Well explained.

Thanks! I am glad.

 

[PeterDonis]

It seems to me you are deriving geometrical conclusions from a simple matter about signature conventions , the fact that the time coordinate can be positive or negative depending on the choice of signature bears no evident connection for me with what we are discussing here

It's not the t *coordinate* that can be positive or negative; it's the *metric coefficient* g_tt (and g_rr as well). It's true that we are making this observation in a particular coordinate chart; but only because it has already been shown that a spherically symmetric metric can *always* be expressed in such a chart without loss of generality. (MTW do that in Chapter 23, and the proof is referenced in the section I quoted from.) So the observation is not coordinate-dependent.

If you insist on having the observation re-stated in coordinate-free terms, it would be this: Birkhoff's theorem shows that a spherically symmetric vacuum metric has one extra KVF, in addition to the three that arise from the spherical symmetry; but there is nothing in the assumptions or the proof of Birkhoff's theorem that requires that KVF to be timelike everywhere. It is only required to be timelike in an asymptotically flat portion of the region of spacetime under consideration; but the region of spacetime under consideration, the region that MTW call "a piece of the Schwarzschild geometry", can include a portion inside the EH which is not static (because the extra KVF is not timelike there), but which is still vacuum.

Besides, I don't have any problem with the Birkhoff theorem as you quote it from MTW, it is perfectly compatible with the interior region not being a vacuum

No, it isn't. You're misunderstanding the theorem. The theorem says that any spherically symmetric vacuum region of spacetime must be a piece of the Schwarzschild geometry; but the "piece" in question can include the vacuum Schwarzschild geometry inside the EH. The entire "piece", including both inside and outside the EH, is vacuum.

since it only demands staticity of "a piece of the Schwarzschild's geometry", I take that piece to correspond with the outside region of the geometry.

Again, you're misunderstanding. The "piece" terminology is there in order to allow the theorem to apply to a model like the Oppenheimer-Snyder spherically symmetric collapse, where the vacuum region exterior to the collapsing matter is the "piece" of the Schwarzschild geometry that starts at r = the radius of the surface of the collapsing matter, and goes out to r -> infinity. That remains true even after the surface of the collapsing matter has fallen inside the EH, so the "piece" of the Schwarzschild geometry outside its surface, which is vacuum, includes both a vacuum region inside the EH and a vacuum region outside the EH.

 

[PeterDonis]

I want to go back to this point because if you read the first pages of Carroll's Notes chapter on Black holes, he clearly shows that at least in the region inside the EH the difference in symmetry you mentioned is nullified; he explicitly says that using non-rigorous arguments, it is deliberately chosen to extend isotropy to all points, meaning using a manifold completely foliated by two spheres, justifying it because although the origin shouldn't be included, almost all space validity is good enough to claim all space validity.
By doing this it demands not only spherical symmetry but homogeneity, at least in the region inside the EH. The problem is that in the abscence of cosmological constant, any isotropic and homogeneous solution of the EFE is an expanding or contracting, FRW-like metric with no timelike KVF

You're going to have to give direct quotes, because I don't see Caroll saying that Schwarzschild spacetime inside the EH is homogeneous, or anything like that. He does say that the spacetime can be foliated by 2-spheres except for the origin at r = 0, but that doesn't make the spacetime homogeneous; there is still only *one* r = 0. In a homogeneous spacetime, such as FRW, *any* point can be chosen as r = 0. You can't do that in Schwarzschild spacetime.

"Foliated by 2-spheres" just means every point lies on one and only one 2-sphere; but that says nothing about how many *different* possible foliations there are. In Schwarzschild spacetime, there is only one, because there is only one possible r = 0. In an FRW spacetime, there are an infinite number of possible foliations, because you can choose any spatial point to be r = 0 and there are an infinite number of them.

 

[PeterDonis]

Well, it is tricky to say, it depends on the coordinates you use, in Schwarzschild coordinates it is a spacelike geodesic, but that seems to be a problem of the coordinates because they are swapped at the EH.

The worldline of an infalling observer is *not* spacelike in Schwarzschild coordinates inside the EH; as you yourself recognize, a geodesic's tangent vector can't change the sign of its norm.

What happens in Schwarzschild coordinates inside the EH is that the vector $\partial / \partial t$ becomes spacelike. But $\partial / \partial t$ is *not* the tangent vector to an infalling object's worldline. I wrote down that tangent vector in a previous post in Schwarzschild coordinates; it's easy to show that *that* tangent vector is always timelike, regardless of whether r > 2M or r < 2M.

 

[TrickyDicky]

If you insist on having the observation re-stated in coordinate-free terms, it would be this: Birkhoff's theorem shows that a spherically symmetric vacuum metric has one extra KVF, in addition to the three that arise from the spherical symmetry; but there is nothing in the assumptions or the proof of Birkhoff's theorem that requires that KVF to be timelike everywhere.

I'm not saying that Birkhoff requires that KVF to be timelike everywhere from quite a few posts back.

It is only required to be timelike in an asymptotically flat portion of the region of spacetime under consideration; but the region of spacetime under consideration, the region that MTW call "a piece of the Schwarzschild geometry", can include a portion inside the EH which is not static (because the extra KVF is not timelike there), but which is still vacuum.

So you claim, but nothing in the proof nor in the stament of the theorem says that the inside region is a vacuum.

the "piece" in question can include the vacuum Schwarzschild geometry inside the EH. The entire "piece", including both inside and outside the EH, is vacuum.

I think you are confusing the part with the whole.

Again, you're misunderstanding. The "piece" terminology is there in order to allow the theorem to apply to a model like the Oppenheimer-Snyder spherically symmetric collapse, where the vacuum region exterior to the collapsing matter is the "piece" of the Schwarzschild geometry that starts at r = the radius of the surface of the collapsing matter, and goes out to r -> infinity. That remains true even after the surface of the collapsing matter has fallen inside the EH, so the "piece" of the Schwarzschild geometry outside its surface, which is vacuum, includes both a vacuum region inside the EH and a vacuum region outside the EH.

This is not in the proof of the theorem (either MTW's or Carroll's) so I take it is your own interpretation which I respect but implies that the surface of collapsing matter has recoiled to the point singularity at the origin and therefore there is no EH nor inside region.