Black Holes - the two points of view.

[TrickyDicky]

OK, so it sounds like you and I agree that an object which free falls across the EH must have a timelike worldline both inside and outside the EH.

No, I agreed with what I quoted. I'm still trying to decipher how your statement here that "an object which free falls across the EH must have a timelike worldline both inside and outside the EH" does not contradict the intrinsic geometry of the K-S space.

 

[PeterDonis]

No, I agreed with what I quoted. I'm still trying to decipher how your statement here that "an object which free falls across the EH must have a timelike worldline both inside and outside the EH" does not contradict the intrinsic geometry of the K-S space.

Look at a Kruskal diagram, such as the one on the Wikipedia page:

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

Any curve which is inclined to the vertical by less than 45 degrees on this diagram is timelike. There are many such curves that go from Region I to Region II; any such curve is a possible worldline for an infalling object. And of those curves, there are many that are geodesics.

If you want a mathematical description of such an infalling geodesic, it can be written in K-S coordinates, but it's kind of messy; it's easier to write it in ingoing Painleve coordinates for an ingoing timelike curve, or ingoing Eddington-Finkelstein coordinates for an ingoing null curve. For the timelike case, in Painleve coordinates, the ingoing geodesics are integral curves of the vector field $\partial_T - \sqrt{2M / r} \partial_r$, where T is the Painleve time coordinate. It is easy to show that this vector field is timelike everywhere.

 

[PeterDonis]

I think it is true of any spherically symmetric pseudoriemannian manifold. But let me find a good reference.

Please do, because Schwarzschild spacetime is a spherically symmetric pseudoriemannian manifold, and I gave an explicit example of a Killing vector field in it whose integral curves are not geodesics. So I'd be extremely surprised to find any reputable reference that claimed that the integral curves of a Killing vector field always *are* geodesics.

So when DrGreg said that worldlines must obey the intrinsic geoemetry I uderstand that if the geoemetry says a worldline (geodesic or not) must be spacelike that means it cannot be followed by a massive observer, right?

I would say "curve" instead of "worldline" for that precise reason; yes, a spacelike (or null) curve cannot be the worldline of any massive observer.

 

[Dale]

No, I agreed with what I quoted. I'm still trying to decipher how your statement here that "an object which free falls across the EH must have a timelike worldline both inside and outside the EH" does not contradict the intrinsic geometry of the K-S space.

What do you mean by "intrinsic geometry of the KS space"? I assume by KS you mean Kruskal–Szekeres coordinates, but that is a coordinate chart for a Schwarzschild spacetime and doesn't have any intrinsic geometry. The intrinsic geometry comes from the pseudo-Riemannian manifold representing the spacetime, not the coordinates.

In GR spacetime is represented by a pseudo-Riemannian manifold. The path of a massive particle is represented by a timelike curve in that manifold. So the worldline of any massive particle is always timelike regardless of whether it is geodesic or not and regardless of if it is inside or outside of an event horizon.

 

[Mike Holland]

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.

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.

But I am at a loss to understand why there is an argument here. All along I have claimed that a falling observer will pass through the event horizon in his proper time, but that he won't in my proper time. This is what I understand from the calculations based on Schwarzschlild coordinates. So as long a I sit here in my armchair, I will see him hovering near the Evenyt Horizon forever edging closer and closer. As far as I am concerned, he hasn't fallen through it. But from his point of view in his spaceship, he just flies through and into the singularity. If the Black Hole is large enough, he won't even notice that he has gone through.

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

Mike

 

[Mike Holland]

Mike Holland, I don't know if you are avoiding my post 117 or simply have not had time, but I would like a response, in particular, to the question I posed there:
Do you believe that changing coordinate systems can make something start or stop existing?

My apologies, DaleSpam, I have been very busy. Also, your posts are usually the most difficult ones to answer. I need time to think.

If two coordinate systems do not overlap, then things that exist in one will not exist in the other. Similarly, if two systems have common space coordinates but no overlap in time coordinate, then you cannot convert x,y,z,t in one system to x,y,z,t in the other, so you cannot have the same event in both systems. But there can be cases of partial overlap to complicate things.

In the case of an Event Horizon, we can have a common x,y,z, but t does not overlap between my coordinates and the local ones. Any time coordinate at the event horizon corresponds to t = infinity in my coords, so no event there can be mapped into an x,y,z,t in my system. Events there do not exist for me. I can change coordinate system by jumping into the forming Black Hole, and when I get there it will come into existence for me.

Does that answer your question?
Mike

Edit: I've been thinking a bit further. What I said about Event Horizons still applies, but my ideas don't cover an event that did occur in my past, or will occur in my future. Such events exist on the timeline, but don't exist NOW. Things come into existence as I travel along my timeline, and others cease to exist as they move into my past. But in another sense of the word "exist", they do have an existence in my timeframe which Black Holes don't.

 

[TrickyDicky]

What do you mean by "intrinsic geometry of the KS space"? I assume by KS you mean Kruskal–Szekeres coordinates, but that is a coordinate chart for a Schwarzschild spacetime and doesn't have any intrinsic geometry. The intrinsic geometry comes from the pseudo-Riemannian manifold representing the spacetime, not the coordinates.

I was referring to the maximally extended Schwarzschild vacuum spacetime solution. KS coordinates happen to cover the whole spacetime.

In GR spacetime is represented by a pseudo-Riemannian manifold. The path of a massive particle is represented by a timelike curve in that manifold. So the worldline of any massive particle is always timelike regardless of whether it is geodesic or not and regardless of if it is inside or outside of an event horizon.

This is my understanding too, but IMO this contradicts the geometrically intrinsic definition of EH that PeterDonis gave.
See my answer to his posts below.

 

[TrickyDicky]

Look at a Kruskal diagram, such as the one on the Wikipedia page:

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

Any curve which is inclined to the vertical by less than 45 degrees on this diagram is timelike. There are many such curves that go from Region I to Region II; any such curve is a possible worldline for an infalling object. And of those curves, there are many that are geodesics.

If you want a mathematical description of such an infalling geodesic, it can be written in K-S coordinates, but it's kind of messy; it's easier to write it in ingoing Painleve coordinates for an ingoing timelike curve, or ingoing Eddington-Finkelstein coordinates for an ingoing null curve. For the timelike case, in Painleve coordinates, the ingoing geodesics are integral curves of the vector field $\partial_T - \sqrt{2M / r} \partial_r$, where T is the Painleve time coordinate. It is easy to show that this vector field is timelike everywhere.

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

Please do, because Schwarzschild spacetime is a spherically symmetric pseudoriemannian manifold, and I gave an explicit example of a Killing vector field in it whose integral curves are not geodesics. So I'd be extremely surprised to find any reputable reference that claimed that the integral curves of a Killing vector field always *are* geodesics.

I didn't say always, we are talking about spherically symmetric vacuums only.
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.
Yor example is of an ingoing geodesic, so I don't understand what you mean by "are not geodesics".

Let me try and clarify where I see a problem here.
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."
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.

 

[Dale]

My apologies, DaleSpam, I have been very busy. Also, your posts are usually the most difficult ones to answer. I need time to think.

No problem, I understand that it can be difficult to keep up with a multiple-poster thread like this.

Does that answer your question?

Yes. The short answer is that you do believe that changing coordinate systems can make something start or stop existing.

Unfortunately, that position becomes ridiculous pretty quickly. In GR you are free to choose your coordinates in almost any way you like, you don't even need three for space and one for time, they don't have to be orthogonal or normal nor cover the whole spacetime. So, I could choose coordinates that do not cover anything to my right. Then, according to you, that light to my right, from which I am receiving a signal, does not exist. A change in coordinates and suddenly it exists.

Carried to the extreme, I do not even need to use coordinates which include me. So I can make myself not exist. Which begs the question, if I don't exist when I choose coordinates that don't include me, then how can something which doesn't exist choose coordinates?

Hopefully you can now understand why the idea of existence being determined by coordinate systems is repugnant to many people who understand the flexibility of coordinate systems in GR.

 

[Dale]

I was referring to the maximally extended Schwarzschild vacuum spacetime solution. KS coordinates happen to cover the whole spacetime.

OK, thanks for the clarification.

This is my understanding too, but IMO this contradicts the geometrically intrinsic definition of EH that PeterDonis gave.
See my answer to his posts below.

Since you and I agree then I will stop arguing . I didn't see any conflict with PeterDonis' comments (particularly since Killing vectors don't generally represent worldlines nor geodesics), but rather than try to explain what I think he meant I will let him explain directly. Perhaps you picked up on something I missed.

 

[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$\Lambda$r^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.