# Schwarzschild's Metric 1916

StateOfTheEqn
Our current version of the Schwarzschild metric is

$$d\tau^2=(1-r_s/r)dt^2-(1-r_s/r)^{-1}dr^2-r^2d\Omega^2$$

where c is set to 1, r is the scalar distance, $$r_s$$ is the 'event horizon' radius, and $$d\Omega^2=d\theta^2+sin^2\theta d\phi^2$$.

In Schwarzschild's original paper from 1916 he does not use r the same way. His equation is equivalent to:
$$ds^2=(1-r_s/R)dt^2-(1-r_s/R)^{-1}dR^2-R^2d\Omega^2$$ where $$R=(r^3+r_s^3)^{1/3}$$ with (+,-,-,-) as the Minkowski signature. See equation (14) in reference linked below. BTW, In his paper he uses the notation $$\alpha=r_s$$
Note: $$ds^2=d\tau^2$$ for a free falling test particle.

One implication of his original formulation is that the coordinate singularity at $$r_s$$ gets removed which has implications for the theory of black holes. So, the two metrics obviously cannot be equivalent.

Another important question is: how can two contradictory metrics for a central gravitational field both be consistent with the GR field equations?

Reference in German but the mathematics is quite clear: http://de.wikisource.org/wiki/Über_..._Massenpunktes_nach_der_Einsteinschen_Theorie

## Answers and Replies

Coordinate singularities can be removed by coordinate transformations. There's no contradiction here, it's something that is done routinely in GR texts.

StateOfTheEqn
Coordinate singularities can be removed by coordinate transformations. There's no contradiction here, it's something that is done routinely in GR texts.
What is the criterion by which we can tell if the coordinate singularity is physically real or not? It seems to me that if it can be removed by a coordinate transformation then it may not be real. I apologize that my past education in GR was mostly coordinate free.

The term "coordinate singularity" actually refers to a singularity which isn't "real". It connotes a singularity which can be removed by means of a coordinate transformation. In order to tell if a singularity is a coordinate singularity or not, you could for example compute a scalar field such as ##R^{abcd}R_{abcd}## (this is called the Kretschmann scalar: http://en.wikipedia.org/wiki/Kretschmann_scalar) and see if the singularity persists. Since this is a scalar field, such a singularity cannot be removed by means of a coordinate transformation.

The definition of a "physical" singularity is a bit involved. It makes use of what is called geodesic incompleteness. Essentially, if a null or time-like geodesic terminates at some event we would expect that there is a physical singularity there. This isn't a perfect definition because there are pathological examples of geodesic incompleteness wherein problems don't persist but as far as realistic examples go, such as the Schwarzschild space-time, geodesic incompleteness provides an adequate definition of a physical singularity. See Hawking and Ellis ch8 and Wald ch9.

http://en.wikipedia.org/wiki/Singularity_theorems

StateOfTheEqn
you could for example compute a scalar field such as ##R^{abcd}R_{abcd}## (this is called the Kretschmann scalar: http://en.wikipedia.org/wiki/Kretschmann_scalar) and see if the singularity persists. Since this is a scalar field, such a singularity cannot be removed by means of a coordinate transformation.

Given a metric we can calculate ##K=R^{abcd}R_{abcd}## and for our current version of the Schwarzschild metric
$$d\tau^2=(1-r_s/r)dt^2-(1-r_s/r)^{-1}dr^2-r^2d\Omega^2$$ we have ##K=48G^2M^2/c^4r^6## (according to http://en.wikipedia.org/wiki/Kretschmann_scalar) which at ##r=r_s## is non-zero indicating a physically real singularity.

A solution to the GR field equations is a metric. Such a metric's representation ##g_{ij}## depends on the coordinate system. Two representations ##g_{ij}## and ##\overline{g}_{kl}## can be considered equivalent if they represent the same metric (but in diifferent coordinate systems). The above defined ##K=R^{abcd}R_{abcd}## would depend on the metric but not its representation. g(current) and g(1916) are obviously different representations but I think they also represent different metrics. Certainly, I would need proof that they represent the same one.

Anyway, my present thinking is g(1916) has no physically real singularity at ##r_s## but g(current) does.

The Kretschmann scalar being non-zero at ##r_s## means that it is not a physical singularity. Clearly the physical singularity is given by the space-like hypersurface ##r = 0##. For example a radially freely falling observer's worldline does not terminate at ##r_s## but it does terminate at ##r = 0##.

Mentor
What is the criterion by which we can tell if the coordinate singularity is physically real or not? It seems to me that if it can be removed by a coordinate transformation then it may not be real.

That should read "certainly is not real". If you can make a singularity go away with a coordinate transformation, then it's just an artifact of the choice of coordinates, with no more physical significance than the lack of a valid ##\theta## coordinate at the origin when you're using polar instead of cartesian coordinates in a two-dimensional plane, or the dubious longitude at the earth's north and south poles.

The other case is more interesting: If no one has found a coordinate transformation that makes the singularity go away, does that mean that we're dealing with a real singularity, or that we should be looking harder for a transformation that makes the singularity go away?

There's some history here. When the Schwarzchild solution was first discovered, it was generally accepted that the black hole solution (vacuum all the way down to and past ##r=R_S##) was unphysical because of the singularity at the Schwarzchild radius. The discovery of coordinate transformations (KS, GP, ...) that eliminated this singularity was a necessary condition for accepting the possibility that black holes really exist.

StateOfTheEqn
The Kretschmann scalar being non-zero at ##r_s## means that it is not a physical singularity. Clearly the physical singularity is given by the space-like hypersurface ##r = 0##. For example a radially freely falling observer's worldline does not terminate at ##r_s## but it does terminate at ##r = 0##.
What does it say about the event horizon? I thought that was what we were discussing. I said in my first post
One implication of his original formulation is that the coordinate singularity at ##r_s## gets removed which has implications for the theory of black holes. So, the two metrics obviously cannot be equivalent.

Certainly both metrics give a singularity at r=0. I thought that was obvious. Is the singularity at r=0 a "naked" singularity or not? Is transit through ##r=r_s## two-way or one-way? g(1916) says yes (two-way) but g(current) says no (one-way).

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StateOfTheEqn
That should read "certainly is not real". If you can make a singularity go away with a coordinate transformation, then it's just an artifact of the choice of coordinates, with no more physical significance than the lack of a valid ##\theta## coordinate at the origin when you're using polar instead of cartesian coordinates in a two-dimensional plane, or the dubious longitude at the earth's north and south poles.

The other case is more interesting: If no one has found a coordinate transformation that makes the singularity go away, does that mean that we're dealing with a real singularity, or that we should be looking harder for a transformation that makes the singularity go away?

There's some history here. When the Schwarzchild solution was first discovered, it was generally accepted that the black hole solution (vacuum all the way down to and past ##r=R_S##) was unphysical because of the singularity at the Schwarzchild radius. The discovery of coordinate transformations (KS, GP, ...) that eliminated this singularity was a necessary condition for accepting the possibility that black holes really exist.

One part of my question is whether the two metric representations (that I am calling g(1916) and g(current)) represent the same metric or different metrics. A transformation that makes a singularity go away is only valid if it is a transformation from one representation to another of the same metric (i.e. an isometry). That would not apply to different metrics as opposed to different representations of the same metric. g(1916) and g(current) treat the event horizon differently. It seems that with g(1916) it does not even exist. My other question was how, if they represent different metrics, we can get both as solutions of the GR field equations.

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Mentor
Anyway, my present thinking is g(1916) has no physically real singularity at ##r_s## but g(current) does.

If I write in what you're calling g(current) in KS or PG coordinates, is the singularity still there?

StateOfTheEqn
If I write in what you're calling g(current) in KS or PG coordinates, is the singularity still there?

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.

My answer to your question is this: if KS and PG preserve the metric represented by g(current) then the singularity at r=0 would still be there and they would predict that crossing the event horizon is a one-way trip.

StateOfTheEqn
Here is another thing some might find interesting.

If we set $d\sigma^2=R^2d\Omega^2=R^2(d\theta^2+sin^2\theta d\phi^2)$ we get a metric on the 2-sphere of radius $R=(r^3+r_s^3)^{1/3}$ where $r$, as before, is the scalar radius. From this we can compute that the circumference $C=2\pi R=2\pi (r^3+r_s^3)^{1/3}>2\pi r$ and the surface area $A=4\pi R^2=4\pi (r^3+r_s^3)^{2/3}>4\pi r^2$. This can only happen if space itself is negatively curved. Not only is space-time curved but space itself is slightly curved negatively and that curvature increases as $r$ approaches zero. Physically, the negative spatial curvature means that light rays initially on outbound radial paths diverge somewhat in excess of their radial divergence.

I have consulted a couple of modern derivations of what I have been calling g(current) and they both assume without much discussion that space (as opposed to space-time) is Euclidean around the gravitating body. Schwarzschild, in his derivation of g(1916), did not make that assumption but $R=(r^3+r_s^3)^{1/3}$ arose from his derivation.

Mentz114
...
I do not know whether g(current) and g(1916) represent the same or different metrics. I strongly suspect the latter is true.
Your suspicion is justified.
##ds^2=(1-r_s/R)dt^2-(1-r_s/R)^{-1}dR^2-R^2d\Omega^2##
With ##R=r## and ##R=(r^3+r_s^3)^{1/3}## the line element represents different spacetimes. They have different K-invariants and Ricci tensors. The current metric has Rαβ=0 but the 1916 Rαβ is not zero.

Mentor
I have consulted a couple of modern derivations of what I have been calling g(current) and they both assume without much discussion that space (as opposed to space-time) is Euclidean around the gravitating body.

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.

Russell E
In Schwarzschild's original paper from 1916 he does not use r the same way... One implication of his original formulation is that the coordinate singularity at r_s gets removed which has implications for the theory of black holes. So, the two metrics obviously cannot be equivalent.

Not true. The two actually are equivalent. This is well known. You can find a discussion of this here

http://www.mathpages.com/rr/s8-07/8-07.htm

starting about half-way through the article, beginning with the words "Interestingly, the solution in Schwarzschild's 1916 paper was not presented in terms of what we today call Schwarzschild coordinates. Those were introduced a year later by Droste..." and concluding with the words "So Schwarzschild's version of the solution is not physically distinct from the usual interpretation introduced by Droste in 1917."

1 person
TrickyDicky
Your suspicion is justified.
##ds^2=(1-r_s/R)dt^2-(1-r_s/R)^{-1}dR^2-R^2d\Omega^2##
With ##R=r## and ##R=(r^3+r_s^3)^{1/3}## the line element represents different spacetimes. They have different K-invariants and Ricci tensors. The current metric has Rαβ=0 but the 1916 Rαβ is not zero.

I can't see what you mean by "the 1916 Ricci tensor is not zero". It was obtained by Schwarzschild as a solution to the equation Rab=0, wasn't it?
I would agree they have different Kretschman scalar at r=r^s, being non-zero at the current metric (meaning in that metric a singularity there is not physical but merely a coordinate artifact as commented by WN) and zero in the 1916 metric confirming we would be talking about different spacetimes(both solutions of the EFE), and making one wonder why all those endless and (often) boring discussions about the physics at the event horizon and the nature of the evident coordinate singularity at r=r^s(of the current and better or mostly only known current metric) ever come up other than the 1916 metric is rarely mentioned.

Mentz114
I can't see what you mean by "the 1916 Ricci tensor is not zero". It was obtained by Schwarzschild as a solution to the equation Rab=0, wasn't it?
I don't know how it was derived. But I calculated Gab= κTab and got a static fluid with anisotropic pressure. It is possible that I made a mistake in setting up the calculation, but I can't see it.

I would agree they have different Kretschman scalar at r=r^s, being non-zero at the current metric (meaning in that metric a singularity there is not physical but merely a coordinate artifact as commented by WN) and zero in the 1916 metric confirming we would be talking about different spacetimes(both solutions of the EFE), and making one wonder why all those endless and (often) boring discussions about the physics at the event horizon and the nature of the evident coordinate singularity at r=r^s(of the current and better or mostly only known current metric) ever come up other than the 1916 metric is rarely mentioned.
Well, boring to some. There are other good ways ( like KS and GP charts) to study the event horizon in any case. I don't feel deprived because I never saw this metric before this thread.

Your suspicion is justified.
##ds^2=(1-r_s/R)dt^2-(1-r_s/R)^{-1}dR^2-R^2d\Omega^2##
With ##R=r## and ##R=(r^3+r_s^3)^{1/3}## the line element represents different spacetimes. They have different K-invariants and Ricci tensors. The current metric has Rαβ=0 but the 1916 Rαβ is not zero.

But before the coordinate change ##R=(r^3+r_s^3)^{1/3}##, doesn't ##ds^2=(1-r_s/R)dt^2-(1-r_s/R)^{-1}dR^2-R^2d\Omega^2## have the same form as what we now usually call the Schwarzschild metric?

Not true. The two actually are equivalent. This is well known. You can find a discussion of this here

http://www.mathpages.com/rr/s8-07/8-07.htm

starting about half-way through the article, beginning with the words "Interestingly, the solution in Schwarzschild's 1916 paper was not presented in terms of what we today call Schwarzschild coordinates. Those were introduced a year later by Droste..." and concluding with the words "So Schwarzschild's version of the solution is not physically distinct from the usual interpretation introduced by Droste in 1917."

Wilfred Owen and chemical weapons - how sadly relevant still. But "they shall not pass" - is that where Gandalf stole his phrase?

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TrickyDicky
Not true. The two actually are equivalent. This is well known. You can find a discussion of this here

http://www.mathpages.com/rr/s8-07/8-07.htm

The discussion is quite interesting, it shows that arriving to the conclusion of their equivalence is simple: first of all one must question most of Schwarzschild's original concerns and requirements( like "imposing continuity of metric coefficients") wich basically amounts to interpret his metric the way it is understood nowadays(and admittedly as it has basically been undertood since 1917's Droste and Hilbert take on the solution). And indeed it can be interpreted this way just by then assuming before hand analyticity (as it is also discussed in the mathpages article in the bit about analytic continuation) and finally assuming naturally that they are simply different coordinates for the same metric, taking advantage of the ambiguity that presents us the fact that one can never be completely sure if a metric tensor representation in coordinates is what we think it is apparently. I mean it is the case that one can represent two different geometries with apparenty the same formal appearance, just ike it is obviously possible to represent the same spacetime with completely different coordinates that at first sight would make very difficult to conclude they are describing the same geometry.
In these cases one must follow what the author of the metric solution intends to define with the specific line element that he presents, it is quite clear that Schwarzschild had in mind a different spacetime than the one we usually think of as the "Schwarzschild spacetime", whether he did it for wrong reasons as it is hinted at in the mathpages article is irrelevant in this respect from he moment this "other spacetime" is certainly a solution of the EFE too, and that was the goal of Schwarzschild while writing this solution in the trenches in 1915, giving Einstein an exact solution. It is also the case that his alternative spacetime with a physical singularity at r^s(wich let's recall it is actualy the origin of this spacetime) gives the same physical prediction for all the empirical tests of GR in the Solar system as the current Schwarzschild spacetime, it is only on more speculative and not directly verified to date predictions that the solutions differ.

Mentz114
But before the coordinate change ##R=(r^3+r_s^3)^{1/3}##, doesn't ##ds^2=(1-r_s/R)dt^2-(1-r_s/R)^{-1}dR^2-R^2d\Omega^2## have the same form as what we now usually call the Schwarzschild metric?
Yes, it does. I botched the transformation so my result is wrong. Apologies to all for my misleading ( but obviously wrong) assertion.

Russell E
...it is quite clear that Schwarzschild had in mind a different spacetime than the one we usually think of as the "Schwarzschild spacetime"... it is only on more speculative and not directly verified to date predictions that the solutions differ.

There is only one unique spherically symmetrical solution to the vacuum field equations (see Birkhoff's theorem), and it's the one Schwarzschild described. His description differs only in appearance from how the Schwarzschild solution is commonly presented, due to the use of an unusual radial coordinate. Changing coordinates doesn't change any of the physical properties, metrical relations, or predictions of the solution.

TrickyDicky
There is only one unique spherically symmetrical solution to the vacuum field equations (see Birkhoff's theorem), and it's the one Schwarzschild described. His description differs only in appearance from how the Schwarzschild solution is commonly presented, due to the use of an unusual radial coordinate. Changing coordinates doesn't change any of the physical properties, metrical relations, or predictions of the solution.

All these sentences are correct taken at face value, but I suspect you didn't understand what I explained if you are using them to counter my argument.

Take your last statement for instance :"Changing coordinates doesn't change any of the physical properties, metrical relations, or predictions of the solution." Totally true by itself, but it seems to miss the fact that I was considering two different solutions of the EFE in vacuum.
You can try and argue that you know for sure Schwarzschild was actually referring to the current Schwarzschild spacetime, fine it is useless to debate about what a death person thought, we only have what he wrote (and I ignore if you have read the original paper).
The truth is that as I said in a previous post if you take simply the original line element of Schwarzschild without any other consideration one can of course interpret it simply by assuming analyticity (and of course being willing to consider them the same solution helps ;) ) that it is referring to the same spacetime than the current extended one, one simply need to introduce a (rather contrived but admissible nevertheless) coordinate transformation.
One coment about the innocence (or lack of) of assuming analyticity, it is certainly a trivial assumption in geodesically complete Riemannian spaces like Euclidean space(which is the one more commonly used in undergraduate physics and mathematics) and even more automatic in holomorphic spaces like complex manifolds(which are by definition analytic). It is by no means trivial in Lorentzian manifolds (indefinite metric) that are not geodesically complete like it is the case here.

So to avoid useless debates let's just use our mathematical abstraction capacities, and let's check that it is doesn't take a great strain of our geometrical imagination to conceive and define a a vacuum spherically symmetric spacetime geodesically incomplete at the origin that is also a solution of the EFE, having a naked singularity is not considered very physical just like it is the case (not being quite physical I mean) with the Schwarzschild spacetime as discussed in the mathpages article, and like it happens with many solutions of the EFE like say, Godel spacetime with its closed timelike curves, etc..., but we are now only considering the math model, and it is perfectly definable mathematically.

I'm glad you bring up the Birkhoff theoerem because I think it is worth recalling that one of the premises of the theorem is spherical symmetry, that is rotational symmetry in three dimensions.
It is geometrically evident that "Schwarzschild spacetime"(wich is R2XS2) doesn't meet this premise except for its outside the event horizon region(also it also only meets the staticity premise in that region). The whole solution only enjoys cilindrical symmetry (axisymmetry), that is rotational symmetry in 2 dimensions.
There is typically a couple of obstacles to realize this, first the representation of the spacetime is usually done in 2 dimensional graphs, and most importantly in 4D rotational symmetry about a plane corresponds to corresponding 2D rotational symmetry in every perpendicular plane, about the point of intersection.

Mentor
I'm glad you bring up the Birkhoff theoerem because I think it is worth recalling that one of the premises of the theorem is spherical symmetry, that is rotational symmetry in three dimensions.

The more precise way of stating this is that spherical symmetry means 3 Killing vector fields corresponding to 3 spatial rotations. But S2 with the appropriate metric on it has that symmetry. (I think we've had a discussion similar to this before.)

It is geometrically evident that "Schwarzschild spacetime"(wich is R2XS2) doesn't meet this premise except for its outside the event horizon region

This is not correct; the S2 part of the topology, with the appropriate metric on it, is there throughout the entire spacetime, so the spherical symmetry is as well.

What *is* true is that, counterintuitively, there is no "center" to the spherical symmetry in the spacetime itself; i.e., there is no portion of the spacetime corresponding to the degenerate "sphere" with zero radius. The surface r = 0, the singularity, is not actually part of the spacetime. However, that doesn't take away the spherical symmetry anywhere else.

(also it also only meets the staticity premise in that region).

Staticity is not a premise of Birkhoff's Theorem. We've had this discussion before as well; I summarized the upshot of that discussion in a post on my PF blog:

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

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TrickyDicky
The more precise way of stating this is that spherical symmetry means 3 Killing vector fields corresponding to 3 spatial rotations. But S2 with the appropriate metric on it has that symmetry.
This is correct for the case of an intrinsic S2 manifold, that is, one that is not embedded in euclidean or other space. Here we have a spherical surface and a Euclidean surface.

This is not correct; the S2 part of the topology, with the appropriate metric on it, is there throughout the entire spacetime, so the spherical symmetry is as well.
See above qualifier.
What *is* true is that, counterintuitively, there is no "center" to the spherical symmetry in the spacetime itself; i.e., there is no portion of the spacetime corresponding to the degenerate "sphere" with zero radius. The surface r = 0, the singularity, is not actually part of the spacetime. However, that doesn't take away the spherical symmetry anywhere else.
I'll concede this is not such a clear cut case as I maybe made it appear, as there are certain complexities in 4 dimensional space that are hard to visualize.
But certainly for all physical purposes that I've seen the Scwarzschild solution applied to there is a "center" where the geodesic incompleteness lies that breaks spherical symmetry.
(I think we've had a discussion similar to this before.)
I'd appreciate if you stopped repeating this unless you point to the specific discussion with a link. the last discussion I recall with you you had to retract a couple of things (so did I for that matter)
Staticity is not a premise of Birkhoff's Theorem.
Sure, sorry about that, it doesn't add anything to my point or to the discussion anyway.

Mentor
This is correct for the case of an intrinsic S2 manifold, that is, one that is not embedded in euclidean or other space. Here we have a spherical surface and a Euclidean surface.

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.

But certainly for all physical purposes that I've seen the Scwarzschild solution applied to there is a "center" where the geodesic incompleteness lies

This is true.

that breaks spherical symmetry.

But this is false. The geodesic incompleteness doesn't have any effect on the spherical symmetry; it's a separate issue.

TrickyDicky
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.

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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Mentor
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.)

Mentor
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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Mentor
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.

TrickyDicky
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

@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 :)

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.

Mentor
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".