Black Holes - the two points of view.

[PeterDonis]

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

Yes, it does. Have you actually read the proof? It explicitly uses the *vacuum* Einstein Field Equation, and the conclusion of the theorem explicitly allows for a region inside the EH (where the $\partial / \partial t$ KVF is not timelike). So the theorem *does* say the interior region is a vacuum.

This is not in the proof of the theorem (either MTW's or Carroll's) so I take it is your own interpretation

You are correct that the proofs of Birkhoff's theorem do not talk about specific cases of "matching" a vacuum region to a non-vacuum region, as the Oppenheimer-Snyder solution does. But the proofs certainly *do* specifically use the *vacuum* EFE, as I said above, and so the conclusion of the theorem certainly does say that there can be a *vacuum* region inside the EH.

which I respect but implies that the surface of collapsing matter has recoiled to the point singularity at the origin and therefore there is no EH nor inside region.

It implies no such thing. The collapsing matter does eventually implode to a point singularity at r = 0, but that does not mean the EH or the region inside it do not exist. I suggest that you look up some references on the Oppenheimer-Snyder solution. MTW discusses it in section 32.4.

 

[TrickyDicky]

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

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

That's fair, here are the quotes: Chapter 7 of the Notes, pages 164, 165 and 166:

"we will sketch a proof of Birkhoff’s theorem, which states that the
Schwarzschild solution is the unique spherically symmetric solution to Einstein’s equations
in vacuum. The procedure will be to first present some non-rigorous arguments that any
spherically symmetric metric (whether or not it solves Einstein’s equations) must take on a
certain form, and then work from there to more carefully derive the actual solution in such
a case...
every point will be on exactly one of these spheres. (Actually, it’s almost every point — we will show below how it can fail to be absolutely every point.) Thus, we say that a spherically symmetric manifold can be foliated into spheres.
It is these spheres which foliate R3. Of course, they don’t really foliate all of the space, since the origin itself just stays put under rotations — it doesn’t move around on some two-sphere.
But it should be clear that almost all of the space is properly foliated, and this will turn out to be enough for us.

We can also have spherical symmetry without an “origin” to rotate things around. An
example is provided by a “wormhole”, with topology R × S2.

This foliated structure suggests that we put coordinates on our manifold in a way which
is adapted to the foliation.
...
We are now through with handwaving, and can commence some honest calculation. For
the case at hand, our submanifolds are two-spheres..."

If points in a graphic of the maximally extended Schwarzschild space are all 2-spheres I'd say we have homogeneity, since every point's 2-sphere has a center, Carroll makes cler that the proper foliation is the one with one only origin as depicted in the page 165 figure.
By allowing every point in the manifold to be a 2-sphere we certainly retain spherical symmetry but we introduce a homogeneity that seems more akin to a matter configuration, and it's certainly not what Birkhoff's theorem initially contemplated.

Sure there is still only one r = 0 for the exterior solution (the original Schwarzschild metric) but the 2-sphere foliation is what allows the inside to turn the timelike KVF to spacelike .

 

[PeterDonis]

If points in a graphic of the maximally extended Schwarzschild space are all 2-spheres I'd say we have homogeneity, since every point's 2-sphere has a center

That's not what "homogeneity" means. It doesn't mean that every point's 2-sphere *has* a center; that's just spherical symmetry. Homogeneity means that any point can *be* the center; that's *not* true for Schwarzschild spacetime. You appear to agree with this, since later on, you say:

Sure there is still only one r = 0 for the exterior solution (the original Schwarzschild metric)

Not just for the exterior solution; for the interior as well.

but the 2-sphere foliation is what allows the inside to turn the timelike KVF to spacelike .

The 2-sphere foliation by itself isn't enough. Minkowski spacetime can also be foliated by 2-spheres, and that doesn't allow any KVF on that spacetime to be timelike in some regions and spacelike in others. The fact that the $\partial / \partial t$ KVF is timelike in one region and spacelike in another is due to spherical symmetry + vacuum + the "M" parameter in the metric being nonzero; if M is zero (as it is for Minkowski spacetime), the $\partial / \partial t$ KVF is timelike everywhere. (And if M = 0 there are actually an infinite number of KVFs that are timelike everywhere, not just one, since there are an infinite number of global inertial frames.)

 

[TrickyDicky]

That's not what "homogeneity" means. It doesn't mean that every point's 2-sphere *has* a center; that's just spherical symmetry. Homogeneity means that any point can *be* the center; that's *not* true for Schwarzschild spacetime.

Homogeneity is space translation symmetry, that is, momentum conservation, that is the isometry generated by the spacelike KVF.
Without the 2-spheres foliation you don't get the spacelike KVF

The 2-sphere foliation by itself isn't enough. Minkowski spacetime can also be foliated by 2-spheres, and that doesn't allow any KVF on that spacetime to be timelike in some regions and spacelike in others. The fact that the $\partial / \partial t$ KVF is timelike in one region and spacelike in another is due to spherical symmetry + vacuum + the "M" parameter in the metric being nonzero; if M is zero (as it is for Minkowski spacetime), the $\partial / \partial t$ KVF is timelike everywhere. (And if M = 0 there are actually an infinite number of KVFs that are timelike everywhere, not just one, since there are an infinite number of global inertial frames.)

Yes, the 2-sphere foliation is necessary but not sufficient.
The key here is that this foliation as admitted by Carroll is basically non-rigorous, handwavy and was introduced long after the Birkhoff's theoren was first proved.

 

[TrickyDicky]

It implies no such thing. The collapsing matter does eventually implode to a point singularity at r = 0, but that does not mean the EH or the region inside it do not exist.

Theoretically is possible, check the super-extremal black hole of the Reissner-Nordstrom solution, I know is usually not considered physical but neither is the Schwarzschild black hole that
we are discussing.

 

[Dale]

Homogeneity is space translation symmetry, that is, momentum conservation, that is the isometry generated by the spacelike KVF.

Just because a KVF is spacelike does not mean that it represents space translation symmetry. The spacelike KVFs of the Schwarzschild spacetime represent rotations (isotropy), not translations (homogeneity).

 

[PeterDonis]

Homogeneity is space translation symmetry, that is, momentum conservation, that is the isometry generated by the spacelike KVF.

You are conflating different concepts here. "Homgeneity" in the sense the term is applied to FRW spacetime, which is the sense I thought we were talking about, is *not* the same as "space translation symmetry" in the sense of momentum conservation. Nor is it the same as an isometry generated by a spacelike KVF. Neither of those properties are sufficient to make it the case that the spacetime is spherically symmetric about *any* spatial point, which is what "homogeneity" means in the sense the term is used to describe FRW spacetime.

Without the 2-spheres foliation you don't get the spacelike KVF

Yes, you can. The vector field $\partial / \partial t$ is a KVF and is spacelike inside the EH.

The key here is that this foliation as admitted by Carroll is basically non-rigorous, handwavy and was introduced long after the Birkhoff's theoren was first proved.

I'm not really convinced this is true, since adopting such a foliation on any spherically symmetric spacetime is so obvious. But let's assume it's true for the sake of argument. So what?

 

[PeterDonis]

The spacelike KVFs of the Schwarzschild spacetime represent rotations (isotropy), not translations (homogeneity).

This is not quite true; as I noted in my post just now in response to TrickyDicky, the vector field $\partial / \partial t$ is a KVF and is spacelike inside the horizon, and the isometry it represents is a sort of "translation". However, that doesn't make that isometry a "homogeneity" in the sense that term is used to describe FRW spacetime.

 

[PeterDonis]

Theoretically is possible, check the super-extremal black hole of the Reissner-Nordstrom solution

Yes, R-N spacetime with Q > M has a naked singularity, with no horizon. So what? I wasn't saying that it's impossible for any spacetime to have a singularity without a horizon. The fact is that Schwarzschild spacetime *does* have a horizon and a region inside it--as does R-N spacetime with Q <= M, for that matter.

 

[Meselwulf]

When he reaches the Schwarzschild radius, along with all the other collapsing matter, he does not travel any further because space and time are distorted in such a way that the distance between him and the centre becomes a time dimension.

Actually, space doesn't become a time dimension, space becomes ''timelike'' which is different.

It is not that space becomes time or time becomes space, but rather it is a coordinate artefact. Space begins to behave as though it was a time dimension, it hasn't actually became a time dimension.

 

[Austin0]

If we are considering an EM emission from a free falling frame at a particular potential altitude to a receiver at infinity does it still hold? Or does it require calculation of the g dilation and the relativistic Doppler due to velocity?
I guess in a way I am just asking if there are two separate effects or only one, and if not two ; why not? Thanks

It still holds exactly as Dalespam described it, in all cases he listed (SR pure doppler, GR any case, any space time, any world lines). One calculation, one core phenomenon. Any possibility of separation into 'gravitational redshift' versus doppler in GR depends on the special case of a sufficiently static metric.

If I am understanding you correctly then in the case I outlined there is only one effect.
It sounds like the resulting shift is purely dependent on the relative velocity and in this case the received frequency would be the same whether the gravitating mass was there or not. Is this right??

You can factor it for a static metric because there is an identifiable class of static observers. Then you define redshift relations between these observers (computed e.g. with either Dalespam's approach or Synge's) as 'gravitational'. Then for, other observers, you figure total redshift, compare to instantly co-located static observers and call the difference kinematic. But for non-static metric, there is no class of static observers to perform this separation.

It is not true that mass makes no difference under this scheme. Stress-energy and geometry are interlinked, and parallel transport is affected by geometry (as are the way null paths connect world lines in Dalespam's approach). It is just that there is no need to factor it into separate effects, and in the general case, you can't.

.

But the time dilation you refer to is not a function of position, but a function of path. If, instead of comparing a distant observer with minimal proper acceleration with one who experiences enormous proper acceleration, we compare them with one in free fall toward the EH, starting from the distant observer, there is no significant time dilation between the near horizon free faller and the distant stationary observer. This near horizon observer sees their clock and the distant clock going at essentially the same rate.

Could you explain this in more detail? In earlier posts you seemed to say that Doppler shift of a freefalling source was affected by an additional factor beyond the motion effect ,due to location as measured by a distant receiver. I.e. motion Doppler and a dilation factor due to the potential at that location. This potential factor equivalent to the dilation factor of a momentarily co-located static source relative to the distant observer.

Here you seem to be saying there is no effect on the free faller from the geometry it is passing through.
How is it possible that the faller near the horizon has the same clock rate as a static observer at infinity??

 

[Dale]

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

It can be proven simply by taking the metric of the inside region and calculating the Einstein tensor. It is vacuum.

 

[Mentz114]

Is there any merit in this publication ?

Is there life inside black holes?, Vyacheslav I. Dokuchaeva.

arXiv:1103.6140v2 [gr-qc] 9 Apr 2011

Abstract. Inside the rotating or charged black holes there are bound periodic planetary
orbits, which not coming out nor terminated at the central singularity. The stable periodic
orbits inside black holes exist even for photons. We call these bound orbits by the orbits of
the third kind, following to Chandrasekhar classification for particle orbits in the black hole
gravitational field. It is shown that an existence domain for the third kind orbits is a rather
spacious, and so there is a place for life inside the supermassive black holes in the galactic
nuclei. The advanced civilizations of the third kind (according to Kardashev classification)
may inhabit the interiors of supermassive black holes, being invisible from the outside and
basking as in the light of the central singularity and the orbital photons.

 

[PAllen]

How is it possible that the faller near the horizon has the same clock rate as a static observer at infinity??

This all get's at the view I take that:

- time dilation is a convention; a coordinate dependent convenience for calculating physical observables. This includes so called gravitational time dilation.

- frequency shift or rate of signal reception from some O1 to O2 is an observable

- differential aging is an observable dependent on paths

As to the specific case you refer to, call:

O1 stationary observer far from horizon
O2 stationary observer near horizon
O3 observer who has free fallen from O1 to O2

Note, O3 is moving very fast (near c) relative to adjacent O2 (noting relative velocities in GR are observables only for nearby bodies; for distant bodies, they are in-determinate, and you must pick some convention if you want to assign distant relative velocities).

In discussing frequency shift observations between O2 and O3, since they would changing extremely rapidly, let's suppose we discuss each at moment it sees the other orthogonal to motion, so pure transverse doppler.

Then each one's observations are follows:

O1 sees redshift from O2 and greater redshift from O3
O2 sees redshift from O3 and blue shift from O1
O3 sees redshift from O2 and no significant shift from O1

Note the asymmetry between what O1 observes about O3 and vice versa.The explanation is seen by comparing O3 and O2. O2 sees blue shift from O1. O3 is moving away from the O1 as seen by O2. Thus O3 will experience red shift of the blue shift seen by O2. By stating that O3 has free fallen from O1, I have arranged that these particular red and blue shifts cancel.

 

[PAllen]

Is there any merit in this publication ?

Is there life inside black holes?, Vyacheslav I. Dokuchaeva.

arXiv:1103.6140v2 [gr-qc] 9 Apr 2011

I can offer only the following:

- I noticed this paper once and, at a quick perusal, it seemed plausible
- It is published in a reputable peer reviewed journal (Class. Quantum Grav)
- It is referenced as valid source in at least one later peer reviewed paper (Phys.Rev.D)

 

[Mentz114]

Thanks for that, PAllen. This paper is more interesting. From the conclusion

... we have shown that the solution of the Einstein equations (5)
for a point mass immersed in the universe with the positive cosmological constant
has very special properties: the metric is everywhere smooth, light can propagate
outward through the horizon, there is an antitrapped surface enclosing the point
mass and there is necessarily an initial singularity...

Horizons and the cosmological constant
Krzysztof A. Meissner

arXiv:0901.0640v1 [gr-qc] 6 Jan 2009

The coordinates he's introduced cover the whole spacetime, which is geodesically complete. I'll have to spend more time with this.

 

[TrickyDicky]

Is there any merit in this publication ?

Is there life inside black holes?, Vyacheslav I. Dokuchaeva.


Abstract. Inside the rotating or charged black holes there are bound periodic planetary
orbits, which not coming out nor terminated at the central singularity. The stable periodic
orbits inside black holes exist even for photons. We call these bound orbits by the orbits of
the third kind, following to Chandrasekhar classification for particle orbits in the black hole
gravitational field. It is shown that an existence domain for the third kind orbits is a rather
spacious, and so there is a place for life inside the supermassive black holes in the galactic
nuclei. The advanced civilizations of the third kind (according to Kardashev classification)
may inhabit the interiors of supermassive black holes, being invisible from the outside and
basking as in the light of the central singularity and the orbital photons.

T 11:29 AM

arXiv:1103.6140v2 [gr-qc] 9 Apr 2011

I can offer only the following:

- I noticed this paper once and, at a quick perusal, it seemed plausible
- It is published in a reputable peer reviewed journal (Class. Quantum Grav)
- It is referenced as valid source in at least one later peer reviewed paper (Phys.Rev.D)

But it is a bit of a crowded vacuum, no?

 

[PAllen]

But it is a bit of a crowded vacuum, no?

The paper is clearly of the 'what if' type. It also does make care to mention supermassive black holes. However, I doubt the author really thinks it is plausible that there is life in the intense radiation conditions of supermassive black holes.

 

[TrickyDicky]

It can be proven simply by taking the metric of the inside region and calculating the Einstein tensor. It is vacuum.

Yes, that would be the most straight forward way to check it. What metric of the inside region would be more suitable to do that?
Maybe PAllen or someone knows about some reference where this computation has been carried out.

One of the things that concerns me about this is that for instance Weyl curvature is usually associated to static spacetimes, is there any example of non-static metrics with non-vanishing Weyl curvature?

 

[PeterDonis]

Yes, that would be the most straight forward way to check it. What metric of the inside region would be more suitable to do that?

Um, the Schwarzschild metric? Meaning this:

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

with $r < 2m$.

Maybe PAllen or someone knows about some reference where this computation has been carried out.

It's in the same reference I've already given you. MTW explicitly shows that the above metric satisfies the vacuum EFE (meaning the Einstein tensor is zero) for $0 < r < 2m$ and $2m < r < \infty$. That's what "proving Birkhoff's theorem" means. (The only reason $r = 2m$ is left out is that the metric in the form given above is singular there; as I noted before, they show in an exercise various ways of dealing with that in order to verify that Birkhoff's theorem is also proved for $r = 2m$.)

 

[Dale]

Yes, that would be the most straight forward way to check it. What metric of the inside region would be more suitable to do that?

The standard metric is fine. It works outside the EH and inside the EH, just not on the EH itself.

Maybe PAllen or someone knows about some reference where this computation has been carried out.

See L.3.2:
http://onlinelibrary.wiley.com/doi/10.1002/9783527622061.app12/pdf
It is vacuum.

Or you can just plug it into pretty much any GR-related software program. It takes 1.6 ms to calculate with Mathematica.

 

[TrickyDicky]

Um, the Schwarzschild metric? Meaning this:

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

with $r < 2m$.



It's in the same reference I've already given you. MTW explicitly shows that the above metric satisfies the vacuum EFE (meaning the Einstein tensor is zero) for $0 < r < 2m$ and $2m < r < \infty$. That's what "proving Birkhoff's theorem" means.

That's a vacuum alright, the problem is that the Schwarzschild metric regardless the signature convention used is static.

 

[TrickyDicky]

The standard metric is fine. It works outside the EH and inside the EH, just not on the EH itself.

See L.3.2:
http://onlinelibrary.wiley.com/doi/10.1002/9783527622061.app12/pdf
It is vacuum.

Or you can just plug it into pretty much any GR-related software program. It takes 1.6 ms to calculate with Mathematica.

Thanks but I'm only saying that I don't understand how that metric can be non-static and a vacuum at the same time.

 

[Dale]

Thanks but I'm only saying that I don't understand how that metric can be non-static and a vacuum at the same time.

What does static-ness have to do with vacuum-ness?

The Minkowski metric is static and vacuum. The exterior Reissner–Nordström metric is static and non-vacuum. The FLRW metric is non-static and non-vacuum. The metric of a gravitational wave is non-static and vacuum.

 

[TrickyDicky]

What does static-ness have to do with vacuum-ness?

That is what the Birkhoff's theorem is about. It relates vacumness+spherical symmetry to staticness.

The Minkowski metric is static and vacuum. The exterior Reissner–Nordström metric is static and non-vacuum. The FLRW metric is non-static and non-vacuum. The metric of a gravitational wave is non-static and vacuum.

Gravitational waves don't have the spherical symmetry requirement, and the RN metric is not a vacuum.

 

[Dale]

That is what the Birkhoff's theorem is about. It relates vacumness+spherical symmetry to staticness.

I thought we had already covered that and agreed that was a mis-statement of Birkhoff's theorem.

 

[TrickyDicky]

I thought we had already covered that and agreed that was a mis-statement of Birkhoff's theorem.

I'm not referring to the statements, all the proofs I've read show precisely how a general isotropic metric which has two functions depending on time and space, when those function are solved for Rab=0 you get that those functions are no longer depending on time.

 

[DrGreg]

...you get that those functions are no longer depending on time.

I suspect that the proofs show that "those functions are no longer depending on the coordinate labelled t", which is not the same thing. (Inside the event horizon, t is spacelike.)

 

[Dale]

I'm not referring to the statements, all the proofs I've read show precisely how a general isotropic metric which has two functions depending on time and space, when those function are solved for Rab=0 you get that those functions are no longer depending on time.

By "time" and "space" don't you mean a coordinate t and a radial coordinate r.

 

[TrickyDicky]

I suspect that the proofs show that "those functions are no longer depending on the coordinate labelled t", which is not the same thing. (Inside the event horizon, t is spacelike.)

I have found no proof of the theorem that interprets t as spacelike.
That interpretation relies on comparing an inside metric with an outside one, but the proofs derive the metric as a unique and independent entity.

If the labels are changed and now t is spacelike then the logical thing is to put t instead of r in the functions, isn't it?

 

[Dale]

If the labels are changed and now t is spacelike then the logical thing is to put t instead of r in the functions, isn't it?

No, r is still the radial coordinate and still has the usual meaning in terms of the surface area of the sphere, even though it is timelike.

 

[DrGreg]

I have found no proof of the theorem that interprets t as spacelike.
That interpretation relies on comparing an inside metric with an outside one, but the proofs derive the metric as a unique and independent entity.

Once you've got the equation$$ds^2 = - \left( 1 - \frac{2m}{r} \right) dt^2 + \frac{dr^2}{1 - 2m / r} + r^2 \left( d\theta^2 + \sin^2 \theta \, d\phi^2 \right)$$then t is automatically spacelike whenever $r < 2m$, almost directly from the definition of "spacelike". No comparison required.

If the labels are changed and now t is spacelike then the logical thing is to put t instead of r in the functions, isn't it?

If you wanted to, you could rewrite the metric inside the horizon as$$ds^2 = - \left( 1 - \frac{2m}{T} \right) dR^2 + \frac{dT^2}{1 - 2m / T} + T^2 \left( d\theta^2 + \sin^2 \theta \, d\phi^2 \right)$$and that might remove the confusion that it seems to be causing for you, but it has the disadvantage that you then have to repeat the same argument twice instead of just using the same formula everywhere.

 

[Dale]

I'm not referring to the statements, all the proofs I've read show precisely how a general isotropic metric which has two functions depending on time and space, when those function are solved for Rab=0 you get that those functions are no longer depending on time.

Did you go back and look at the proof you are thinking of? Are you sure that "time" and "space" are not just "t" and "r" coordinates? Please confirm.

 

[Dale]

Once you've got the equation$$ds^2 = - \left( 1 - \frac{2m}{r} \right) dt^2 + \frac{dr^2}{1 - 2m / r} + r^2 \left( d\theta^2 + \sin^2 \theta \, d\phi^2 \right)$$then t is automatically spacelike whenever $r < 2m$, almost directly from the definition of "spacelike". No comparison required.

If you wanted to, you could rewrite the metric inside the horizon as$$ds^2 = - \left( 1 - \frac{2m}{T} \right) dR^2 + \frac{dT^2}{1 - 2m / T} + T^2 \left( d\theta^2 + \sin^2 \theta \, d\phi^2 \right)$$and that might remove the confusion that it seems to be causing for you, but it has the disadvantage that you then have to repeat the same argument twice instead of just using the same formula everywhere.

The problem with this is that then your space is foliated by spheres of radius T and spherical symmetry implies that everything is a function of T only. Both of those are generally associated with the coordinate r. That, and you will confuse everyone else. Best to just stick to the standard notation and learn the quirks.

 

[PeterDonis]

That's a vacuum alright, the problem is that the Schwarzschild metric regardless the signature convention used is static.

This is not correct. See below.

I have found no proof of the theorem that interprets t as spacelike.

Yes, you have. I've given it to you.

That interpretation relies on comparing an inside metric with an outside one, but the proofs derive the metric as a unique and independent entity.

There is no "interpretation" involved. The proofs show that the metric is independent of the t coordinate, for all values of r (except r = 0 where there's a curvature singularity, and requiring special handling at r = 2m). You don't have to make any comparisons to show that; the proof for any given value of r goes through regardless of what happens at any other value of r. And as others have already noted, if $r < 2m$ then t is spacelike. So the proof is valid for t spacelike.

If the labels are changed and now t is spacelike then the logical thing is to put t instead of r in the functions, isn't it?

You can't just arbitrarily "put t instead of r in the functions". The r coordinate is defined by the fact that the area of a 2-sphere at r is $4 \pi r^2$. That fixes how r occurs in the "functions" (I'm not sure what you mean by that term but I assume you're referring to things like the metric coefficients and Einstein tensor components). You can change the name of the r coordinate to something else, but that doesn't change its meaning.

Similarly, the "t" coordinate is the coordinate that, once you've completed the derivation of Birkhoff's theorem, the metric turns out to be independent of. You can change the label on it, but that doesn't change its meaning. And that meaning is the same for $r < 2m$ as for $r > 2m$; it's the *same* coordinate in both regions.

Have you actually read the proof in MTW? If you don't have access to MTW, I have posted a similar proof on my PF blog here:

https://www.physicsforums.com/blog.php?b=4211

My version doesn't use exponentials to write the metric coefficients, as MTW does; I commented on that in a previous post. So my version makes it explicit that there is no restriction on the signs of $g_{tt}$ and $g_{rr}$, and therefore there's no assumption that either the t or the r coordinate is timelike or spacelike; the proof is valid for t both timelike and spacelike, and for r both spacelike and timelike.