# Schwarzschild's Metric 1916

No, we have an R2's worth of S2 surfaces, so to speak; that is, we have a 2-parameter family of S2 surfaces, each of which is an "intrinsic S2 manifold" in and of itself, and each one of which has the appropriate metric on it for spherical symmetry.
Neither you nor me are geometers so I would appreciate assistance on this point from an expert.

The way I understand the Schwarzschild spacetime topology S2XR2 is this, first of all one of the 4 dimensions is evidently time so we are left to study the spherical sy mmetry of the three dimensional hypersurface (the staticity of the manifold allows us to do this clea cut foliation). Agree so far?
We are left then with the direct product of the sphere with the real line(S2XR), unless you think it is R2XS1, but I don't think it is the case as it would lead to closed timelike curves (S1 time dimension).
Here is what I found online from a seminar of prof. Farkas on this:
"There are eight homogeneous simply connected geometries which give rise to compact three-manifolds. One of the simplest of the non-constant curvature ones is the space S2xR, which as its name suggests, is the direct product of the sphere with the real line. [....] in turn give rise to the 4 well-known compact manifolds admitting S2xR geometry. "
Now it is my understanding a constant curvature 3-manifold demands both homogeneity and isotropy, this space is homogeneous but doesn't have constant curvature.

The geodesic incompleteness doesn't have any effect on the spherical symmetry; it's a separate issue.
I could argue about this but let's leave this aside for a moment as we would only really need to get clear on the above.

WannabeNewton
See section 6.1 of Wald, section 5.2 of Carroll, and/or section 4.1 of Straumann; all three talk about what Peter mentioned. All we are doing is creating a foliation out of 2-spheres. In other words, we write the Schwarzschild space-time locally as ##\mathbb{R}\times \Sigma##, the metric as ##ds^2 = -\varphi^2 dt^2 + h##, and foliate ##(\Sigma,h)## by (invariant) 2-spheres.

All spherical symmetry means, for a space-time ##(M,g_{ab})##, is that there exists three space-like killing fields, call them ##L_1, L_2, L_3##, such that ##\mathcal{L}_{L_1}L_2 = L_3, \mathcal{L}_{L_2}L_3 = L_1, \mathcal{L}_{L_3}L_1 = L_2##. For Schwarzschild space-time we have the extra property that ##\mathcal{L}_{\xi}L_i = 0## where ##\xi## is the time-like killing field.

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PeterDonis
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For Schwarzschild space-time we have the extra property that ##\mathcal{L}_{\xi}L_i = 0## where ##\xi## is the time-like killing field.
Pedantic note: ##\xi## is timelike outside the horizon, null on the horizon, and spacelike inside the horizon; but it's a distinct KVF satisfying ##\mathcal{L}_{\xi}L_i = 0## everywhere. (Yes, that gets really weird inside the horizon, where all 4 KVFs are spacelike.)

PeterDonis
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The way I understand the Schwarzschild spacetime topology S2XR2 is this, first of all one of the 4 dimensions is evidently time
I think this is an optimistic use of the word "evidently". The dimension corresponding to the timelike coordinate in the K-S chart works ok for this; but the "dimension" corresponding to, say, the Schwarzschild t coordinate does *not*. (Nor does the "dimension" corresponding to the Painleve or Eddington-Finkelstein t coordinate.)

so we are left to study the spherical symmetry of the three dimensional hypersurface (the staticity of the manifold allows us to do this clea cut foliation). Agree so far?
Not really, because the manifold is only static outside the horizon. There is no way to foliate the entire spacetime by spacelike hypersurfaces that all have the same geometry. The best you can do is to foliate a portion of the spacetime that way. (For more on this, see the follow-up response to WannabeNewton's post that I'm about to post.)

We are left then with the direct product of the sphere with the real line(S2XR), unless you think it is R2XS1, but I don't think it is the case as it would lead to closed timelike curves (S1 time dimension).
I agree that the S2 portion of the topology is spacelike.

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PeterDonis
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All we are doing is creating a foliation out of 2-spheres. In other words, we write the Schwarzschild space-time locally as ##\mathbb{R}\times \Sigma##, the metric as ##ds^2 = -\varphi^2 dt^2 + h##, and foliate ##(\Sigma,h)## by (invariant) 2-spheres.
I don't have the references handy to check, but is this foliation supposed to cover the entire maximally extended Schwarzschild spacetime? Or is it only supposed to cover the exterior and future interior regions? The two cases work differently.

For example: I can create a foliation of the entire maximally extended Schwarzschild spacetime using the K-S chart. The "vertical" dimension of the chart is the "time" (and is timelike everywhere), and I can find a family of spacelike hypersurfaces ##\Sigma## such that any vertical line on the chart intersects each surface exactly once. (Note that one of these hypersurfaces will be the horizontal axis of the chart, but it is the only such surface that will be exactly horizontal; the surfaces above it must "curve upward" so that they approach the future singularity hyperbola, and the surfaces below it must "curve downward" so that they approach the past singularity hyperbola).

But this (K-S) foliation has a different spatial geometry for each spacelike hypersurface ##\Sigma##; each hypersurface can be foliated by 2-spheres, but the range of areas of these 2-spheres will be different for different hypersurfaces. For example, the horizontal axis will be foliated by 2-spheres with areas ranging from infinity, down to ##16 \pi M^2##, and back up to infinity. Other hypersurfaces will be foliated by 2-spheres with areas ranging from infinity, down to some value less than ##16 \pi M^2## (a smaller value the closer the hypersurface is to the future or past singularity), and back up to infinity. So I don't think this can be described as foliating each ##\Sigma## by "invariant" 2-spheres.

Or, I can create a foliation of the exterior and future interior regions, only, by using the Painleve chart. This foliation will use hypersurfaces ##\Sigma## of constant Painleve ##t## coordinate, and all such hypersurfaces will be spacelike, and all will have the same geometry: each one can be foliated by an identical family of nested 2-spheres with areas ranging from zero to infinity. However, the curves orthogonal to this foliation--the integral curves of ##\partial_t##--are not all timelike, so it is a bit dodgy to think of the foliation as being composed of "surfaces of constant time". Also, of course, it doesn't cover the entire maximally extended spacetime, only the two regions I named.

I don't think any of the above affects spherical symmetry, since all that requires is that there is *some* way to foliate the spacetime (or each of a family of spacelike hypersurfaces ##\Sigma##) with 2-spheres; it doesn't require that the foliation satisfy additional properties like every hypersurface ##\Sigma## having exactly the same geometry. But I think it's important to realize that Schwarzschild spacetime does not work the way our intuitions think it ought to work.

See section 6.1 of Wald, section 5.2 of Carroll, and/or section 4.1 of Straumann; all three talk about what Peter mentioned. All we are doing is creating a foliation out of 2-spheres. In other words, we write the Schwarzschild space-time locally as ##\mathbb{R}\times \Sigma##, the metric as ##ds^2 = -\varphi^2 dt^2 + h##, and foliate ##(\Sigma,h)## by (invariant) 2-spheres. .
Hi, mate, good to see you!
The key word here is "locally", I guess one can do the sphere foliation locally and since in GR we are basically concerned with the local geometry there is a sense in wich physicists or relativists in particular can call this way of writing or representing the metric as 2-sphere points in a 2-dimensional plane of x versus t can say there is "local" spherical symetry", we would be back to terminology issues . But since we are talking about whole spacetimes I am still claiming that in the truthful global 4 dimensional representation global 3-dimensional isotropy of the spatial part of current Schwarzschild spacetime is lost.
Do you at least agree that S2XR is not an isotropic hypersurface?

All spherical symmetry means, for a space-time ##(M,g_{ab})##, is that there exists three space-like killing fields, call them ##L_1, L_2, L_3##, such that ##\mathcal{L}_{L_1}L_2 = L_3, \mathcal{L}_{L_2}L_3 = L_1, \mathcal{L}_{L_3}L_1 = L_2##. For Schwarzschild space-time we have the extra property that ##\mathcal{L}_{\xi}L_i = 0## where ##\xi## is the time-like killing field.
Pedantic note: ##\xi## is timelike outside the horizon, null on the horizon, and spacelike inside the horizon; but it's a distinct KVF satisfying ##\mathcal{L}_{\xi}L_i = 0## everywhere. (Yes, that gets really weird inside the horizon, where all 4 KVFs are spacelike.)
As Peter reminds there is a switch in the nature of the killing vector fields in the Schwarzschild spacetime, not only one of the turns from timelike to spacelike, losing staticity, but part of the rotational spacelike turn to translational spacetime, losing isotropy .

I think this is an optimistic use of the word "evidently". The dimension corresponding to the timelike coordinate in the K-S chart works ok for this; but the "dimension" corresponding to, say, the Schwarzschild t coordinate does *not*. (Nor does the "dimension" corresponding to the Painleve or Eddington-Finkelstein t coordinate.).
I should have referred to the killng vector fields instead that are of course coordinate independent.

Not really, because the manifold is only static outside the horizon. There is no way to foliate the entire spacetime by spacelike hypersurfaces that all have the same geometry. The best you can do is to foliate a portion of the spacetime that way. (For more on this, see the follow-up response to WannabeNewton's post that I'm about to post.).
Sure, staticity is lost, but the key here is hypersurface orthogonality(consider the FRW spacetime for instance wich is obviously not static but still has hypersurface orthogonality) and that is not lost in the switch from static to non-static.
EDIT: wrote this before reading the previous post and Edit by Peter

WannabeNewton
@Peter, yes Peter, I meant the latter of the two you mentioned (in particular the exterior region).

@Tricky, hi Tricky :)! Yes I was speaking strictly of local properties. Also, you probably already know this, but hypersurface orthogonality is a property of vector fields not of space-times. A static space-time has by definition a hypersurface orthogonal time-like killing field but a stationary space-time simply has a time-like killing field (you can check, for example, that the time-like killing field of Kerr space-time fails to be hypersurface orthogonal-simply compute its twist 4-vector and note that it is non-vanishing). In the FRW case, the 4-velocity field of the fundamental observers is hypersurface orthogonal but that isn't related to being static nor stationary, it's simply a property of said vector field.

I'll try to respond in more detail a while after because I'm stuck in a class at the moment :)

Bill_K
As Peter reminds there is a switch in the nature of the killing vector fields in the Schwarzschild spacetime, not only one of the turns from timelike to spacelike, losing staticity, but part of the rotational spacelike turn to translational spacetime, losing isotropy
No, that's false. What happens at the event horizon is that the time translation turns into a space translation. Just that. The sphere and its three Killing vectors remains a sphere.

PeterDonis
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part of the rotational spacelike turn to translational spacetime, losing isotropy.
This is not correct; the 3 rotational KVFs are the same everywhere in the spacetime. All that happens inside the horizon is that the 4th KVF is spacelike instead of timelike; but it's still a distinct KVF, and it's the one that is "translational".

Bill_K
I meant that one of the three spacelike rotations turns to timelike.
You are mistaken. It does not.

You are mistaken. It does not.
Sorry, wrote too fast.
Deleted

Before I got muddled up with the KVF switches at the event horizon(sorry about that), I was trying to achieve a better understanding of the Birkhoff's theorem and spherical symmetry in the Schwarzschild spacetime (this is not exactly the topic of the thread although highly related so I''ll try and keep it brief).
There is an element of confusion(at least for me) that I briefly mentioned in a previous post, namely the distinction between local isometries(local KVFs) versus global isometries(global KVFs) in spacetimes with somewhat complex 4D topologies like the Schwarzschild spacetime.
Certainly the foliation of spheres allows to have spherical symmetry(the 3 rotational KVFs) in this spacetime locally seemingly at any point. And since GR is only concerned with the local geometry I guess that is considered sufficient for practical matters. I admit I'm not at all certain that the spacetime has spherical symmetry as a global feature due to global geometrical-topological reasons but that is probably outside the scope of GR.

In any case, let's say the premise is fulfilled in what concerns to the Birkhoff's theorem, if the theorem states that spherical symmetry implies staticity (and asymptotic flatness), I don't know how to interpret the theorem in the regions where there is spherical symmetry but not staticity (inside the event horizon).

I'm still left with the unchallenged to this date claim(but I would like to hear the possible mathematical obstructions to this) that I can fabricate mathematically a curved Lorentzian manifold(unrelated to the current Schwarzschild spacetime) with a naked singularity at the origin to which I can assign as boundary condition a constant mass, that is a solution of the EFE in vacuum and has of course spherical symmetry around the origin and with a line element apparently similar to the one Schwarzschild found in his original paper.

PeterDonis
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if the theorem states that spherical symmetry implies staticity
It doesn't. It only states that spherical symmetry + vacuum implies a 4th KVF whose integral curves are orthogonal to the integral curves of 3 KVFs arising from spherical symmetry. It does *not* say the 4th KVF has to be timelike. (That was part of the point of the blog post I mentioned--I think--in an earlier post in this thread: to do the proof of the theorem in a way that makes it absolutely obvious that the 4th KVF does not have to be timelike.)

PAllen
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.. I was trying to achieve a better understanding of the Birkhoff's theorem and spherical symmetry in the Schwarzschild spacetime (this is not exactly the topic of the thread although highly related so I''ll try and keep it brief).
There is an element of confusion(at least for me) that I briefly mentioned in a previous post, namely the distinction between local isometries(local KVFs) versus global isometries(global KVFs) in spacetimes with somewhat complex 4D topologies like the Schwarzschild spacetime.
Certainly the foliation of spheres allows to have spherical symmetry(the 3 rotational KVFs) in this spacetime locally seemingly at any point. And since GR is only concerned with the local geometry I guess that is considered sufficient for practical matters. I admit I'm not at all certain that the spacetime has spherical symmetry as a global feature due to global geometrical-topological reasons but that is probably outside the scope of GR.

In any case, let's say the premise is fulfilled in what concerns to the Birkhoff's theorem, if the theorem states that spherical symmetry implies staticity (and asymptotic flatness), I don't know how to interpret the theorem in the regions where there is spherical symmetry but not staticity (inside the event horizon).
This has been explained before. The correct statement of the theorem is more like (from a paper I link later):

"Birkhoff 's theorem shows that any spherically symmetric solution of the vacuum Einstein equations is locally
isometric to a neighborhood in Schwarzschild spacetime. Hence it is a local uniqueness theorem whose corollary is
that locally spherically symmetric solutions exhibit an additional local Killing vector field; however this fi eld is not
necessarily timelike."

It is only when the extra local killing field is timelike that you have staticity.

I'm still left with the unchallenged to this date claim(but I would like to hear the possible mathematical obstructions to this) that I can fabricate mathematically a curved Lorentzian manifold(unrelated to the current Schwarzschild spacetime) with a naked singularity at the origin to which I can assign as boundary condition a constant mass, that is a solution of the EFE in vacuum and has of course spherical symmetry around the origin and with a line element apparently similar to the one Schwarzschild found in his original paper.
If one admits unusual topologies, as described in the linked paper, you can sort of do this. You can think of these as sections of Kruskal, glued in various ways, or with various identifications (analogous to how you get an every where metrically flat torus by cutting and gluing the flat plane; or the non orientable Mobius strip or Klein bottle by similar operations). This must be done very carefully to ensure that the resulting manifold is everywhere spherically symmetric. However, you can get non-orientable manifolds consistent with Birkhoff by these techniques. At least one of their examples (see page 6) has singularities with no horizons (but is a closed universe). It is nowhere static, but is still consistent with Birkhoff - it has an extra translational killing vector field.

http://arxiv.org/abs/0910.5194

It doesn't. It only states that spherical symmetry + vacuum implies a 4th KVF whose integral curves are orthogonal to the integral curves of 3 KVFs arising from spherical symmetry. It does *not* say the 4th KVF has to be timelike. (That was part of the point of the blog post I mentioned--I think--in an earlier post in this thread: to do the proof of the theorem in a way that makes it absolutely obvious that the 4th KVF does not have to be timelike.)
Thanks for the input. I would appreciate a reference to a published paper that shows the wording you use and that includes a proof. (Not to understimate the proof on your blog of course, just to have as reference). Pallen's reference in the above post just mentions in passing that the field is not necessarily timelike.
Yet I am sure the way you state the theorem is an "a posteriori" reinterpretation of the original by Birkhoff to support the current Schwarzschild spacetime as the "unique spherically symmetric solution of the vacuum equations" view.
I say it because when Birkhoff in 1923(and apparently another nordic guy independently 2 years earlier) proposed and proved the theorem he couldn't have imagined the contrived extension found 40 years later by Kruskal et al. or any timelike to spacelike transition(not even the Finkelstein-Eddington coordinates were available at that moment). So the original theorem surely was stated using the technical word "static" for the spacetime that was implied by the premises as do most of the usual versions of the theorem.
Unfortunately I haven't been able to find the original paper by Birkhoff online.

PAllen
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Thanks for the input. I would appreciate a reference to a published paper that shows the wording you use and that includes a proof. (Not to understimate the proof on your blog of course, just to have as reference). Pallen's reference in the above post just mentions in passing that the field is not necessarily timelike.
Yet I am sure the way you state the theorem is an "a posteriori" reinterpretation of the original by Birkhoff to support the current Schwarzschild spacetime as the "unique spherically symmetric solution of the vacuum equations" view.
I say it because when Birkhoff in 1923(and apparently another nordic guy independently 2 years earlier) proposed and proved the theorem he couldn't have imagined the contrived extension found 40 years later by Kruskal et al. or any timelike to spacelike transition(not even the Finkelstein-Eddington coordinates were available at that moment). So the original theorem surely was stated using the technical word "static" for the spacetime that was implied by the premises as do most of the usual versions of the theorem.
Unfortunately I haven't been able to find the original paper by Birkhoff online.
Actually, by mathematical criteria (not physical), the Kruskal extension is the most natural extension, while those in the paper I linked are contrived. In exactly the sense that if you have a flat open disc, the natural extension is the Euclidean plane, while extension to a torus or Klein bottle is more contrived. Also true is that just as with the open flat disc, you know that there opportunity for extension, otherwise you have removable geodesic incompleteness.

I don't think it is interesting to worry about Birkhoff's paper. Many generalizations and new proof methods have been done over the years. It is accepted terminology to lump them all together and call the Birkhoff (who was preceded by two years anyway, in deriving 'his' theorem).

Here is another reference:

http://arxiv.org/abs/0908.4110

Note the statement:

"Although Birkhoff's theorem is a classic result, many current textbooks and review articles on general relativity
no longer provide a proof or even a careful statement of the theorem. Frequently it is cited as proving that the
spherically symmetric vacuum solution is static. This is clearly not the case..."

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PeterDonis
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I would appreciate a reference to a published paper that shows the wording you use and that includes a proof.
As I noted in the blog post, MTW's discussion makes it clear that the KVF does not have to be timelike, and they clearly state the wording of the theorem without using the word "static" or implying staticity. So far, they're the only source I'm aware of that never slips up about this (except for the minor wart in the details of their proof, which is why I wrote the post).

Yet I am sure the way you state the theorem is an "a posteriori" reinterpretation of the original by Birkhoff to support the current Schwarzschild spacetime as the "unique spherically symmetric solution of the vacuum equations" view.
I believe MTW's statement of the theorem is the "local" version, which basically says (my wording, not theirs) that any open neighborhood of a spacetime which is spherically symmetric and vacuum is isometric to an open neighborhood of Schwarzschild spacetime. That's not the same as the stronger claim that the entire spacetime must be the maximally extended Schwarzschild spacetime; I think you're correct that the latter claim requires specifying the global topology as well.

I say it because when Birkhoff in 1923(and apparently another nordic guy independently 2 years earlier) proposed and proved the theorem he couldn't have imagined the contrived extension found 40 years later by Kruskal et al. or any timelike to spacelike transition(not even the Finkelstein-Eddington coordinates were available at that moment).
I agree this is likely, but I'm not sure it's impossible: the Painleve chart was discovered in 1921 (by Painleve, and independently in 1922 by Gullstrand, according to Wikipedia), and AFAIK Painleve did include the extension of the chart to the future interior region (which was simple in his chart since it's nonsingular at the horizon).

Unfortunately I haven't been able to find the original paper by Birkhoff online.
Neither have I; I also haven't been able to find either Painleve's or Gullstrand's original papers online. It would be really interesting to see just what they were thinking at the time.

PAllen
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Thanks for the input. I would appreciate a reference to a published paper that shows the wording you use and that includes a proof. (Not to understimate the proof on your blog of course, just to have as reference). Pallen's reference in the above post just mentions in passing that the field is not necessarily timelike.
Yet I am sure the way you state the theorem is an "a posteriori" reinterpretation of the original by Birkhoff to support the current Schwarzschild spacetime as the "unique spherically symmetric solution of the vacuum equations" view.
I say it because when Birkhoff in 1923(and apparently another nordic guy independently 2 years earlier) proposed and proved the theorem he couldn't have imagined the contrived extension found 40 years later by Kruskal et al. or any timelike to spacelike transition(not even the Finkelstein-Eddington coordinates were available at that moment). So the original theorem surely was stated using the technical word "static" for the spacetime that was implied by the premises as do most of the usual versions of the theorem.
Unfortunately I haven't been able to find the original paper by Birkhoff online.
From http://arxiv.org/abs/0908.4110, the following is a rigorous statement of the theorem with cosmological constant aspects removed:

Theorem 1. The only locally spherically symmetric solutions to the vaccuum EFE are locally isometric to the Schwarzschild family of solutions. Furthermore, these solutions are
real analytic in each local coordinate chart.

Corollaries established are that the M parameter of each local coordinate chart must match. Also, it is clear from prior context that in the theorem statement, they mean Kruskal geometry, in that a local chart may be isometric to any part of Kruskal: interior, exterior, some of both, white hole, black hole. This statement also clarifies confusion over analyticity. Birkhoff does imply something about analyticity, but only for each chart, not globally. Thus (as shown especially in the other paper I linked) there are global solutions that cannot be geodesically completed further, that are analytic in each chart, but not globally analytic, and that have completely different topologies than Kruskal, including ones with naked singularities and also non-orientable topologies.

stevendaryl
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I say it because when Birkhoff in 1923(and apparently another nordic guy independently 2 years earlier) proposed and proved the theorem he couldn't have imagined the contrived extension found 40 years later by Kruskal et al. or any timelike to spacelike transition(not even the Finkelstein-Eddington coordinates were available at that moment).
By the "timelike to spacelike" transition, I assume you mean the character of the r and t coordinates of the Schwarzschild solution at the event horizon? In that case, that transition is blatantly obvious from the Schwarzschild metric.

What's interesting about the Kruskal coordinates is that they DON'T have such a transition. You have one coordinate that is timelike everywhere, and another coordinate that is spacelike everywhere.

By the "timelike to spacelike" transition, I assume you mean the character of the r and t coordinates of the Schwarzschild solution at the event horizon? In that case, that transition is blatantly obvious from the Schwarzschild metric.

What's interesting about the Kruskal coordinates is that they DON'T have such a transition. You have one coordinate that is timelike everywhere, and another coordinate that is spacelike everywhere.
I was actually thinking about the local Killing vector fields that as you know are coordinate independent.

@PAllen: Thanks for the interesting references.

Can you give some specific references? In the "current" metric you give in your OP, spacelike slices of constant time are certainly not Euclidean; that's obvious just from looking at the line element you wrote down.
The specific reference is http://de.wikisource.org/wiki/Über_..._Massenpunktes_nach_der_Einsteinschen_Theorie .

The line element I wrote down is $d\sigma^2=R^2d\Omega^2=R^2(d\theta^2+sin^2\theta d\phi^2)$ where $R=(r^3+r_s^3)^{1/3}$. Indeed that is not Euclidean but neither is it a slice of constant time. Would that not have the metric $d\sigma^2=dR^2/(1-r_s/R)+R^2d\Omega^2=dR^2/(1-r_s/R)+R^2(d\theta^2+sin^2\theta d\phi^2)$

PeterDonis
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That's Schwarzschild's 1916 paper. I was asking about the modern derivations of g(current) that you referred to, which you said assumed space was Euclidean. I've never seen a modern derivation that makes that assumption, and indeed, as I said, it's obvious from the g(current) you wrote down in the OP of this thread that slices of constant time in that metric are not Euclidean.

The line element I wrote down is $d\sigma^2=R^2d\Omega^2=R^2(d\theta^2+sin^2\theta d\phi^2)$ where $R=(r^3+r_s^3)^{1/3}$.
That's not what you have been calling g(current), unless I'm completely misunderstanding your terminology; it's part of what you have been calling g(1916).

Indeed that is not Euclidean
As a metric on the 2-sphere, no, of course not, because the 2-sphere is not flat. But that is not what makes the metric of a 3-dimensional slice of constant time non-Euclidean. See below.

but neither is it a slice of constant time.
It's a 2-sphere which is a portion of a slice of constant time, in the g(1916) metric.

Would that not have the metric $d\sigma^2=dR^2/(1-r_s/R)+R^2d\Omega^2=dR^2/(1-r_s/R)+R^2(d\theta^2+sin^2\theta d\phi^2)$
That would be a complete slice of constant time in what you are calling g(1916), yes. And as a metric on a 3-dimensional space, it is also not Euclidean. But the reason it is not Euclidean is not that ##R^2(d\theta^2+sin^2\theta d\phi^2)## is not Euclidean; the reason is that the ##dR^2## term has a coefficient that depends on ##R##, instead of just being 1. If the 3-metric were

$$d\sigma^2 = dR^2 + R^2(d\theta^2+sin^2\theta d\phi^2)$$

then that *would* be a Euclidean 3-metric; it would just be the standard Euclidean metric on 3-space expressed in spherical polar coordinates. But the factor of ##\left( 1 - r_s / R \right)^{-1}## in front of the ##dR^2## makes the metric, as a metric on 3-space, non-Euclidean.

Just to be clear, my interest is in the event horizon. I accept that there is a real physical singularity at r=0. In g(current) there is at the event horizon what I and perhaps some others call a coordinate singularity given that time slows down and, from the point of view of an external observer, seems to stop. With respect to the metric g(1916) that does not happen. I do not know whether g(current) and g(1916) reopresent the same or different metrics. I strongly suspect the latter is true.
The two metric representations do not represent the same metric.

Let $(t,R,\theta,\phi)=\Phi (t,r,\theta,\phi) = (t,(r^3 + r_s^3)^{1/3},\theta,\phi)$ where $r_s$ is a constant > 0.

Then $$D\Phi=\left(\begin{array}{cccc} 1&0&0&0\\ 0&\partial_r R&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right)$$

The metric representations are:

$$g=\left(\begin{array}{cccc} 1-r_s/r&0&0&0\\ 0&(1-r_s/r)^{-1}&0&0\\ 0&0&r^2&0\\ 0&0&0&r^2sin^2\theta \end{array}\right)$$

and
$$\overline g=\left(\begin{array}{cccc} 1-r_s/R&0&0&0\\ 0&(1-r_s/R)^{-1}&0&0\\ 0&0&R^2&0\\ 0&0&0&R^2sin^2\theta \end{array}\right)$$

Let $\overline{\textbf{x}}$ be a column vector w.r.t. the basis $(dt,dR,d\theta,d\phi)$ and $\textbf{x}$ be the same vector w.r.t the basis $(dt,dr,d\theta,d\phi)$. Then $\overline{\textbf{x}} = D\Phi \textbf{x}$. The vectors are assumed to be carrying a Minkowski signature. For example,

$$\textbf{x}=\left(\begin{array}{c} dt\\ idr\\ id\theta\\ id\phi \end{array}\right)$$

Let $<.,.>_g$ be the metric w.r.t. $g$ and $<.,.>_{\overline{g}}$ be the metric w.r.t. $\overline{g}$.

Then $$<\overline{\textbf{x}},\overline{\textbf{x}}>_{\overline{g}}=(\overline{\textbf{x}})^T(\overline{g}) (\overline{\textbf{x}})=(D\Phi\textbf{x})^T(\overline{g})(D\Phi\textbf{x})=\textbf{x}^T(D\Phi)^T(\overline{g})(D\Phi)\textbf{x} = <\textbf{x},\textbf{x}>_{(D\Phi)^T(\overline{g})(D\Phi)}$$

But a simple calculation shows $g \neq (D\Phi)^T(\overline{g})(D\Phi)$ unless $r=R$ but they are unequal by assumption. Therefore $g$ and $\overline g$ do not represent the same metric.

Calculating $r_s$:

In his 1916 paper Schwarzschild derives the formula ${\dot \theta}^2=r_s/2(r^3+r_s^3)=r_s/2R^3$. The equation $2{\dot \theta}^2R^3=r_s$ imposes a constraint on $\dot \theta$ given $R$ (and implicitly $r$) similar to Kepler's 3rd law: $T^2=Kr^3$. For a circular orbit, $K=4\pi^2/GM$

We can rewrite Kepler's 3rd law as $(2\pi/\dot\theta)^2=Kr^3$ and then $8\pi^2/K=2\dot\theta^2r^3$

So, we have
$$\begin{eqnarray} 2{\dot \theta}^2r^3&=&8\pi^2/K\\ 2{\dot \theta}^2R^3&=&r_s \end{eqnarray}$$
Since $r_s$ is a constant we have $R/r \rightarrow 1$ as $r \rightarrow \infty$. Then $r_sK/8\pi^2=1$
and so $r_s=8\pi^2/K=2GM$

No event horizon at $r_s$:

Setting $r=r_s$ we have:

$R=(r^3+r_s^3)^{1/3}=(2r_s^3)^{1/3}=\sqrt{2}r_s$ and
$ds^2=(1-1/\sqrt{2})dt^2-(1-1/\sqrt{2})^{-1}dR^2-R^2d\Omega^2$ with $(1-1/\sqrt{2}) >0$

In fact, $(1-r_s/R)>0$ for all $r>0$

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But the reason it is not Euclidean is not that ##R^2(d\theta^2+sin^2\theta d\phi^2)## is not Euclidean; the reason is that the ##dR^2## term has a coefficient that depends on ##R##, instead of just being 1. If the 3-metric were

$$d\sigma^2 = dR^2 + R^2(d\theta^2+sin^2\theta d\phi^2)$$

then that *would* be a Euclidean 3-metric; it would just be the standard Euclidean metric on 3-space expressed in spherical polar coordinates. But the factor of ##\left( 1 - r_s / R \right)^{-1}## in front of the ##dR^2## makes the metric, as a metric on 3-space, non-Euclidean.
Yes, you are right. It is the metric $d\sigma^2=dR^2/(1-r_s/R)+R^2d\Omega^2=dR^2/(1-r_s/R)+R^2(d\theta^2+sin^2\theta d\phi^2)$ that makes $\mathbb{R}^+ \times S^2$ non-Euclidean, not the previous one. But within that metric a surface of constant $R$ would have surface area $4\pi R^2>4\pi r^2$ would it not? That would indicate $\mathbb{R}^+ \times S^2$ has negative curvature and not be Euclidean as in g(current).

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