Black Holes - the two points of view.

[stevendaryl]

It is easy to fall into a Black Hole in a finite time, but not according to MY clock.
Mike

This may be a trivial observation, but I'll make it anyway: In GR, you can't really talk about what time something happens according to your clock unless it is an event that takes place at your clock. You can talk about what time something happens according to this or that coordinate system, but not according to this or that clock.

 

[PeterDonis]

Peter, as I ubnderstand it, that is what Painleve did with his "raindrop" coordinates - provide a cioordinate frame that covers the object right through the falling process and beyond.

Yes.

But I am at a loss to understand why there is an argument here.

I was responding to this question of yours:

If you think something can fall through the event horizon in a finite time, how does it avoid the time dilation?

The object falls into the hole in finite proper time by its own clock; there is no "time dilation" for it. Time dilation is relative. You keep on talking as if it isn't. Or, rather, you keep switching between talking as if it is, and talking as if it isn't. The question I quoted just above is talking as if it isn't; but when you say something like this...

It is easy to fall into a Black Hole in a finite time, but not according to MY clock.

...you are talking as if it is. You need to make up your mind which position you are taking: if you think that time dilation *is* relative, then the infalling object doesn't have to "avoid" it; it simply isn't there for the infalling object, period. If you think that time dilation *isn't* relative, then you are either misunderstanding GR, or trying to argue that GR is false.

 

[PeterDonis]

Yes, and it's a Killing vector field too, as long as Birkhoff's theorem holds.

No, it isn't. A Killing vector field is a vector field along whose integral curves the local metric is unchanged. The local metric is not unchanged along the worldline of an infalling observer, since the metric coefficients are a function of r; only a curve with constant r can be the integral curve of a Killing vector field in Schwarzschild spacetime.

According to Birkhoff's theorem a spherically symmetric vacuum has a timelike ∂t Killing vector field, in this context, the integral curves of this KVF are timelike geodesics.

No, this is false. The integral curves of ∂t are timelike *curves*, but they are *not* timelike *geodesics*. The proper acceleration along such curves is nonzero.

Your example is of an ingoing geodesic, so I don't understand what you mean by "are not geodesics".

The example I gave was for the worldline of a radial freely falling observer, which is indeed a geodesic, but it is *not* the integral curve of a Killing vector field. Conversely, the integral curves of ∂t *are* integral curves of a Killing vector field, but they are not geodesics. See above.

You said in the other thread " the event horizon is the boundary of the region in which the Killing vector field of the "time translation" isometry of Schwarzschild spacetime is timelike--i.e., the EH is the Killing horizon associated with the "time translation" isometry."

Yes, that's true.

But then how is this compatible with the fact that spherically symmetric vacuums have timelike Killing vector fields (or to use Carroll's words in his Notes on GR: "We have therefore proven a crucial result: any spherically symmetric vacuum metric possesses a timelike Killing vector"), the logical question here would be then, is the region inside the EH (region II) a vacuum or not? If it is it must have ∂t KVF and therefore your EH definition would not be valid.

Ah, perhaps this is the source of the confusion. The vector field ∂t is a KVF everywhere in the spacetime; this can be shown easily from Killing's equation. But ∂t is only timelike outside the EH; it is null on the EH and spacelike inside it. And the region inside the EH is certainly a vacuum everywhere (at least, it is in the maximally extended Schwarzschild spacetime). So I think Carroll left out a qualifier in his statement of the "crucial result"; he should have said that any spherically symmetric vacuum metric possesses a Killing vector field that is timelike *within the region where the spacetime is static*.

I haven't got time to dig into Carroll's notes right now, and I can't remember how much detail he gives about the derivation of Birkhoff's theorem. But as I understand it, spherical symmetry plus vacuum is enough to show that (1) there must be a static region of the spacetime; (2) there must be a Killing vector field on the spacetime which is timelike within that static region; and (3) the static region must be asymptotically flat (i.e., the metric goes to Minkowski as r -> infinity, so the static region must include r -> infinity). But those three things together do not require that the static region cover the entire manifold: and if there is a Killing horizon associated with the Killing vector field, then the static region will *not* cover the entire manifold.

 

[TrickyDicky]

No, it isn't. A Killing vector field is a vector field along whose integral curves the local metric is unchanged. The local metric is not unchanged along the worldline of an infalling observer, since the metric coefficients are a function of r; only a curve with constant r can be the integral curve of a Killing vector field in Schwarzschild spacetime.

But they are not a function of t, and I was referring to a timelike KVF, not a spacelike one that is what you are talking about here.

No, this is false. The integral curves of ∂t are timelike *curves*, but they are *not* timelike *geodesics*. The proper acceleration along such curves is nonzero.

In the general case that is usually true, but we are talking about free-falling towards the singularity thru the EH in a spherically symmetric vacuum, not about hovering observers.

The example I gave was for the worldline of a radial freely falling observer, which is indeed a geodesic, but it is *not* the integral curve of a Killing vector field.

It is indeed a timelike geodesic and its time-symmetry is derived by being an integral curve of a time-symmetric Killing vector field.

Ah, perhaps this is the source of the confusion. The vector field ∂t is a KVF everywhere in the spacetime; this can be shown easily from Killing's equation. But ∂t is only timelike outside the EH; it is null on the EH and spacelike inside it.

Indeed there is confusion about this point but I'm not so sure is on my part.
A killing vector field that is timelike:∂/∂t, generates a time-symmetry, if you say that such a KVF is everywhere in the spacetime you are agreeing with me. If in the next phrase you say that a timelike (∂/∂t) killing vector field is only timelike outside the EH you are contradicting yourself. What I mean is that if they are timelike they cannot be spacelike and viceversa. Probably what you want to say is that the Killing vector field(without qualifiers,not the timelike KVF) is spacelike inside the EH, null at the EH and timelike outside the EH. And this is what is IMO incompatible with the maximally extended Schwarzschild spacetime being a spherically vacuum solution.

And the region inside the EH is certainly a vacuum everywhere (at least, it is in the maximally extended Schwarzschild spacetime).
So I think Carroll left out a qualifier in his statement of the "crucial result"; he should have said that any spherically symmetric vacuum metric possesses a Killing vector field that is timelike *within the region where the spacetime is static*.

If you define the region inside the EH as a region free of timelike KVF then I don't see how that region can be a vacuum, we know the Birkhoff theorem states that a spherically symmetric vacuum must be static, and the region inside the EH is certainly not static.

I haven't got time to dig into Carroll's notes right now, and I can't remember how much detail he gives about the derivation of Birkhoff's theorem. But as I understand it, spherical symmetry plus vacuum is enough to show that (1) there must be a static region of the spacetime; (2) there must be a Killing vector field on the spacetime which is timelike within that static region; and (3) the static region must be asymptotically flat (i.e., the metric goes to Minkowski as r -> infinity, so the static region must include r -> infinity). But those three things together do not require that the static region cover the entire manifold: and if there is a Killing horizon associated with the Killing vector field, then the static region will *not* cover the entire manifold.

Thus my question. According to wikipedia Birkhoff's theorem states that any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat. I would say this implies that where it is not static it is no vacuum solution.

 

[TrickyDicky]

The Birkhoff theorem from a different source (scienceworld.wolfram):a spherically symmetric gravitational field in empty space must be static.

 

[PeterDonis]

It is indeed a timelike geodesic and its time-symmetry is derived by being an integral curve of a time-symmetric Killing vector field.

No, it is *not*. The worldlines of radial freely falling observers are *not* integral curves of any Killing vector field. Why do you think they are?

Just for explicitness, the only Killing vector fields in Schwarzschild spacetime are: ∂/∂t, ∂/∂theta, ∂/∂phi, and linear combinations of those with constant coefficients. The only one of those which is timelike in any portion of the spacetime is ∂/∂t.

A killing vector field that is timelike:∂/∂t, generates a time-symmetry

A KVF *always* generates a symmetry; that's the definition of a KVF. Whether the integral curves of that KVF are timelike, spacelike, or null can vary from one integral curve to another; being timelike, spacelike, or null is not an intrinsic property of the KVF that has to be the same everywhere. So a KVF only generates a "time-symmetry" in regions where it is timelike.

Probably what you want to say is that the Killing vector field(without qualifiers,not the timelike KVF) is spacelike inside the EH, null at the EH and timelike outside the EH.

I see that in one particular spot I did say the integral curves of ∂/∂t were timelike curves (but not timelike geodesics), but I only meant "timelike outside the EH". I should have said that explicitly, I suppose, but even from the context it should have been clear that that's what I meant (since I explicitly stated at the end of the same post that ∂/∂t was only timelike outside the horizon). In any case, I agree with the quoted statement just above (since it's the same thing I said at the end of that post).

Thus my question. According to wikipedia Birkhoff's theorem states that any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat. I would say this implies that where it is not static it is no vacuum solution.

Again, I think there should be a qualifier: instead of "static" it should say "asymptotically static" or something like that. As I said at the end of my last post, I believe that strictly speaking, Birkhoff's theorem only shows that there must be a static *region* in a spherically symmetric vacuum spacetime, and that region must be asymptotically flat; I do not believe Birkhoff's theorem shows that that static region must cover the entire manifold. However, I'm not able to look up specific references right now that state the actual assumptions and steps of the proof.

 

[TrickyDicky]

No, it is *not*. The worldlines of radial freely falling observers are *not* integral curves of any Killing vector field. Why do you think they are?

Because Lorentzian symmetric spaces (look up symmetric spaces) have geodesics symmetries that are isometries, on a general pseudoRiemannian manifold, they need not be isometric.
Remember Schwarzschild spacetime is equivalent to Minkowski spacetime with an straight line(the axial singularity) removed.

Again, I think there should be a qualifier: instead of "static" it should say "asymptotically static" or something like that.

No qualifiers AFAIK.

 

[Dale]

Just for explicitness, the only Killing vector fields in Schwarzschild spacetime are: ∂/∂t, ∂/∂theta, ∂/∂phi, and linear combinations of those with constant coefficients. The only one of those which is timelike in any portion of the spacetime is ∂/∂t.

And none of the integral curves of those Killing vector fields are geodesics except at spatial infinity. But I thought that there were 4 Killing vector fields.

 

[PeterDonis]

Because Lorentzian symmetric spaces (look up symmetric spaces) have geodesics symmetries that are isometries, on a general pseudoRiemannian manifold, they need not be isometric.

On a quick Google, a Lorentzian symmetric space appears to be a space in which the covariant derivative of the curvature tensor with respect to the Levi-Civita connection vanishes. How does that make the worldline of an infalling observer an integral curve of a Killing vector field? I am not familiar with the general subject of symmetric spaces, but I don't see that the details about them matter here; whether or not a given vector field satisfies Killing's equation is simple to check. Have you checked it for the vector field I gave, whose integral curves are the worldlines of infalling observers?

Also, I gave a simple enumeration of *all* the Killing vector fields in Schwarzschild spacetime in my last post*; are you disputing what I said? If so, please show your explicit proof of how the worldline of an infalling observer satisfies Killing's equation. If you are not disputing what I said, then it's obvious that the worldline of an infalling observer is *not* the integral curve of a KVF, regardless of anything else that may or may not be true about symmetric spaces.

(* - I should note that, strictly speaking, I didn't; there is a third KVF on the 2-sphere that I did not list; Carroll's lecture notes go into this, so I won't belabor it, since it's not really germane to the point under discussion.)

No qualifiers AFAIK.

Well, I was able to take a look at Carroll's lecture notes just now, and at the point where he first makes the claim about the spacetime being static, he is actually pulling a bit of a fast one. Here's what he says, on p. 169:

All of the metric components are independent of the coordinate t. We have therefore proven a crucial result: any spherically symmetric vacuum metric possesses a timelike Killing vector.

But actually, he hasn't really proven that. What he's proven is that the spherically symmetric vacuum metric is independent of the t coordinate, so d/dt is a KVF; but he has *not* proven that the t coordinate is timelike everywhere. He doesn't make this very clear, but there's a hint, earlier, on p. 168, where he's talking about how to rewrite his equation 7.12 for the metric in a form that's more suitable for what he's going to do with it:

We know that the spacetime under consideration is Lorentzian, so either m or n will have to be negative. Let us choose m, the coefficient of dt^2, to be negative. This is not a choice we are simply allowed to make, and in fact we will see later that it can go wrong, but we will assume it for now.

Assuming that m is negative is equivalent to assuming that the t coordinate is timelike; and as this quote makes clear, he is *assuming* it, not proving it--and as he also says, later on it will become evident that that assumption is not valid everywhere in the spacetime. In other words, he has *not* actually proven that a spherically symmetric, vacuum solution to the EFE must be static everywhere. He *has* proven that it must have a static region and that that region must be asymptotically flat; that's obvious from the fact that the metric he finally comes up with (the standard Schwarzschild metric) approaches the Minkowski metric in the limit as r -> infinity. But he has *not* proven that the entire manifold is static.

So even if the qualifiers aren't explicitly stated, they're there, at least in the case of Carroll's notes. AFAIK his notes are a valid version of the proof of Birkhoff's theorem, so it would seem to me that the qualifiers are there, period, whether various other sources explicitly state them or not.

 

[PeterDonis]

And none of the integral curves of those Killing vector fields are geodesics except at spatial infinity. But I thought that there were 4 Killing vector fields.

There are; I forgot to list one of the three that are symmetries of the 2-sphere. See the asterisk in my response to TrickyDicky just now.

 

[stevendaryl]

In the general case that is usually true, but we are talking about free-falling towards the singularity thru the EH in a spherically symmetric vacuum, not about hovering observers.

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.

 

[stevendaryl]

Yes, and it's a Killing vector field too, as long as Birkhoff's theorem holds.

Birkhoff's theorem doesn't imply any such thing. Birkhoff's theorem isn't about geodesics, it doesn't say anything about whether there is a timelike geodesic that is an integral of a Killing vector field.

 

[Austin0]

Since you're so insistent on doing calculations in Schwarzschild coordinates, try this one: write down the equation defining the proper time of an object freely falling radially inward from a finite radius r = R > 2M, to radius r = 2M. Write it so that the proper time is a function of r only (this is straightforward because it's easy to derive an equation relating r and the Schwarzschild coordinate time t, so you can eliminate t from the equation). This equation will be a definite integral of some function of r, from r = R to r = 2M. Evaluate the integral; you will see that it gives a finite answer. Therefore, the proper time elapsed for an infalling object is finite, even according to Schwarzschild coordinates.

Correct me if I am wrong but it appears to me that the integration of proper falling time does not have a finite value. It asymptotically approaches a finite limit but can never have a definite value until reaching the horizon, which of course it is also asymptotically approaching and so will never reach in any finite external coordinate time according to the application of the Sc metric.On the assumption that it is accurate up to and including the singularity at the horizon.

 

[PeterDonis]

Correct me if I am wrong but it appears to me that the integration of proper falling time does not have a finite value.

Yes, it appears that way, if you just try to intuitively guess the answer without deriving it. But when you actually derive it, you find that it *does* give a finite answer, despite your intuition. Intuition is not always a reliable guide. The actual calculation is in most standard GR textbooks (I know it's in MTW).

It asymptotically approaches a finite limit

This is equivalent to saying the proper time integral *does* have a finite value. If you try to evaluate the integral in the most "naively obvious" way in Schwarzschild coordinates, you have to take a limit as r -> 2m, since the metric is singular at r = 2m; but the limit, when you take it, is finite. However, even if you insist on doing the integral in Schwarzschild coordinates, you can still write it in a way that doesn't even require taking a limit; as I said in the previous post you quoted, you can eliminate the t coordinate altogether and obtain an integrand that is solely a function of r and is nonsingular at r = 2m, so you can evaluate the integral directly. Or, of course, you could do it in better coordinates, such as Painleve, where there is no coordinate singularity at r = 2m to begin with. The proper time elapsed along a given curve is an invariant, so you can calculate it in any coordinate system you like and get the same answer.

 

[TrickyDicky]

On a quick Google, a Lorentzian symmetric space appears to be a space in which the covariant derivative of the curvature tensor with respect to the Levi-Civita connection vanishes. How does that make the worldline of an infalling observer an integral curve of a Killing vector field? I am not familiar with the general subject of symmetric spaces, but I don't see that the details about them matter here; whether or not a given vector field satisfies Killing's equation is simple to check. Have you checked it for the vector field I gave, whose integral curves are the worldlines of infalling observers?

Also, I gave a simple enumeration of *all* the Killing vector fields in Schwarzschild spacetime in my last post*; are you disputing what I said? If so, please show your explicit proof of how the worldline of an infalling observer satisfies Killing's equation. If you are not disputing what I said, then it's obvious that the worldline of an infalling observer is *not* the integral curve of a KVF, regardless of anything else that may or may not be true about symmetric spaces.

(* - I should note that, strictly speaking, I didn't; there is a third KVF on the 2-sphere that I did not list; Carroll's lecture notes go into this, so I won't belabor it, since it's not really germane to the point under discussion.)



Well, I was able to take a look at Carroll's lecture notes just now, and at the point where he first makes the claim about the spacetime being static, he is actually pulling a bit of a fast one. Here's what he says, on p. 169:



But actually, he hasn't really proven that. What he's proven is that the spherically symmetric vacuum metric is independent of the t coordinate, so d/dt is a KVF; but he has *not* proven that the t coordinate is timelike everywhere. He doesn't make this very clear, but there's a hint, earlier, on p. 168, where he's talking about how to rewrite his equation 7.12 for the metric in a form that's more suitable for what he's going to do with it:



Assuming that m is negative is equivalent to assuming that the t coordinate is timelike; and as this quote makes clear, he is *assuming* it, not proving it--and as he also says, later on it will become evident that that assumption is not valid everywhere in the spacetime. In other words, he has *not* actually proven that a spherically symmetric, vacuum solution to the EFE must be static everywhere. He *has* proven that it must have a static region and that that region must be asymptotically flat; that's obvious from the fact that the metric he finally comes up with (the standard Schwarzschild metric) approaches the Minkowski metric in the limit as r -> infinity. But he has *not* proven that the entire manifold is static.

So even if the qualifiers aren't explicitly stated, they're there, at least in the case of Carroll's notes. AFAIK his notes are a valid version of the proof of Birkhoff's theorem, so it would seem to me that the qualifiers are there, period, whether various other sources explicitly state them or not.

Honestly I find all these arguments not very convincing. As if you were talking about something remotely related to what I am saying.
You haven't answered my simple question, is the region inside of the EH a vacuum or not?

Carroll doesn't need to prove anything because Birkhoff's theorem was proved almost a century ago and without any of the qualifiers you didn't make explicit.

 

[TrickyDicky]

Have you checked it for the vector field I gave, whose integral curves are the worldlines of infalling observers?

Wasn't that vector field Killing?
And wasn't the infaller worldline a geodesic (free falling observer)?

 

[Dale]

You haven't answered my simple question, is the region inside of the EH a vacuum or not?

Yes. For the usual Schwarzschild metric R_{\mu \nu}=0 so R=0 and therefore R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=0. So it is definitely vacuum.

 

[PeterDonis]

You haven't answered my simple question, is the region inside of the EH a vacuum or not?

Yes, it is. But if that is your simple question, why are you going off on this long tangent based on your incorrect claim that the worldline of an infalling observer is an integral curve of a Killing vector field?

Wasn't that vector field Killing?
And wasn't the infaller worldline a geodesic (free falling observer)?

I guess I shouldn't be surprised that we've lost track of which vector field is which. Here's the quick summary. I'll give both vector fields in Schwarzschild coordinates this time (I gave the one for the infalling observer in Painleve coordinates before).

(1) The vector field \partial / \partial t is a Killing vector field. Its integral curves are *not* geodesics.

(2) The vector field \left[ 1 / \left( 1 - 2m / r \right) \right] \partial / \partial t - \left[ \sqrt{2m / r} \right] \partial / \partial r is the frame field for infalling observers (strictly speaking, for observers freely falling "from rest at infinity"); i.e., its integral curves are the worldlines of such observers. This vector field is *not* a Killing vector field, and its integral curves *are* geodesics.

Carroll doesn't need to prove anything because Birkhoff's theorem was proved almost a century ago and without any of the qualifiers you didn't make explicit.

Can you give a reference? So far the only one you've given that outlines an actual proof of the theorem is Carroll's notes, and they have the qualifier. Unless you can show me a proof that does not have the qualifier, I'm standing by what I said.

 

[PeterDonis]

Unless you can show me a proof that does not have the qualifier, I'm standing by what I said.

I have now looked at the proof of Birkhoff's theorem given in MTW (Section 32.2, pp. 843-844 in my edition). Based on what I see there, I'm still standing by what I said, but several points are worth noting. First, here is their statement of the theorem:

Let the geometry of a given region of spacetime (1) be spherically symmetric, and (2) be a solution to the Einstein field equations in vacuum. Then that geometry is necessarily a piece of the Schwarzschild geometry.

They reference Birkhoff (1923); I don't know if they are directly quoting his statement of the theorem or paraphrasing (I think the latter).

Notice that this statement does *not* say the given region of spacetime is static; it only says it is "a piece of the Schwarzschild geometry". See further comments below.

Second, the proof of the theorem, starting near the bottom of the same page (p. 843), writes the metric of the given region of spacetime in Schwarzschild coordinates:

ds^2 = - e^{2 \Phi} dt^2 + e^{2 \Lambda} dr^2 + r^2 \left( d\theta^2 + sin^2 \theta d\phi^2 \right)

But immediately following that equation is this:

...notice that: (1) for generality one must allow g_{tt} = - e^{2 \Phi} and g_{rr} = e^{2 \Lambda} to be positive or negative (no constraint on sign!)

Strictly speaking, allowing the dt^2 and dr^2 terms to be of either sign means you can't write the metric in the form given above, because the exponentials must be positive for real exponents (and all the functions in the metric are supposed to be real-valued). But it appears to be habit for MTW to write the metric this way, since they do it throughout the book, even in places where they are discussing the interior region where the signs of the terms switch.

In any case, MTW are clearly saying here that "for generality" one cannot assume that the t coordinate is timelike (since that assumption is equivalent to assuming that e^{2 \Phi}is positive); and therefore one cannot assume that the given region of spacetime is static.

In the same paragraph, they go on to say:

(2) at events where the gradient of the "circumference function" r is zero or null, Schwarzschild coordinates cannot be used.

In other words, to properly derive the conclusion of Birkhoff's theorem on the horizon, you have to use some other method that doesn't require Schwarzschild coordinates. (Some examples of other methods are given in exercise 32.1; they amount to finding alternate coordinate charts that are nonsingular on the horizon.)

The rest of the proof is straightforward: compute the Einstein tensor for the metric written above and set each component equal to zero (i.e., impose the vacuum Einstein field equation). Then solve for the unknown functions of r. The result is the standard Schwarzschild line element (with the caveat that the result thus derived can't be used on the horizon, but that's a minor technical point):

ds^2 = - \left(1 - \frac{2M}{r} \right) dt^2 + \frac{dr^2}{1 - 2M / r} + r^2 \left( d\theta^2 + sin^2 \theta d\phi^2 \right)

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; the line element is perfectly valid for r < 2M, where the signs of the dt^2 and dr^2 terms are switched and t is spacelike.

I don't have my copy of Wald handy so I can't check his discussion of Birkhoff's theorem. The Wikipedia page references d'Inverno's textbook, which I don't have, but if anyone does and can check it, I'd be interested to see how it's discussed there.

 

[PAllen]

I'll add, referencing one of my favorite GR authors, J.L. Synge(1960):

(Note: Synge is the only major textbook I know of that properly credits Jebsen(1921) and Alexandrov (1923) as well as Birkhoff(1923) for this theorem)

Synge derives this from scratch, including nonzero cosmological constant, and notes:

1) Static character applies only while: 1 - A/r - 1/3\Lambdar^2 >0
thus noting that for positive cosmological constant, the static feature breaks down both
for r too small or r too big(!). Note: Synge uses A to subsume all the generic constants of
the solution.

2) Applies for spherically symmetric spacetimes with matter, including arbitrary pulsations, as
long as you have vacuum beyond some r (also meeting the restrictions of (1)).

 

[TrickyDicky]

Yes, it is.

Ok, thanks for your opinion.
So maybe you can help me find the wrong link in my thought chain.
You say that in the region inside the EH there is no timelike Killing vector field, right?
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. These seem to be by definition the isometries of a vacuum spacetime (spherically symmetric and with vanishing cosmological constant).
You claim that everything that surrounds the singularity is a vacuum (according to your answer "yes, it is").
So how can a region of a spacetime that is defined by not having a timelike KVF be a spherically symmetric vacuum region?
Please, explain.

But if that is your simple question, why are you going off on this long tangent based on your incorrect claim that the worldline of an infalling observer is an integral curve of a Killing vector field?

Good question, to be honest I lost track of my initial thought process to bring in this point myself in relation with the other question. If I remember I'll tell. But anyway I am not now 100% sure we are dealing with a symmetric space here, the wikipage name as examples Minkowski, deSitter and anti deSitter spacetimes and none of them have singularities, and the last two vacuums have nonzero cosmological constant unlike the Schwarzschild vacuum.

 

[TrickyDicky]

It is also worth noting that the region inside the event horizon in the maximally extended spacetime seems to have the same 6 isometries as the FRW metric (3 spatial translations and three rotations), only in this case with the singularity future-oriented instead of in the past, that is contracting towards it instead of expanding away from it. And FRW metrics surely aren't vacuum solutions of the EFE as they have non-vanishing Ricci tensor.
I think we are demanding that Rab=0 locally at every point of the spacetime solution, correct?

 

[Dale]

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.

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.

 

[stevendaryl]

Ok, thanks for your opinion.
So maybe you can help me find the wrong link in my thought chain.
You say that in the region inside the EH there is no timelike Killing vector field, right?
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'm not very knowledgeable about these topics, but where are you getting that information from? I don't think that it is correct. I think all 4 are spacelike in the interior.

 

[stevendaryl]

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 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.

 

[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.