# Schwarzschild's Metric 1916

1. Aug 31, 2013

### 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

2. Aug 31, 2013

### WannabeNewton

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

3. Aug 31, 2013

### StateOfTheEqn

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.

4. Aug 31, 2013

### WannabeNewton

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

5. Aug 31, 2013

### StateOfTheEqn

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.

6. Aug 31, 2013

### WannabeNewton

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 $r = 0$.

7. Aug 31, 2013

### Staff: Mentor

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.

8. Aug 31, 2013

### StateOfTheEqn

What does it say about the event horizon? I thought that was what we were discussing. I said in my first post
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).

Last edited: Sep 1, 2013
9. Sep 1, 2013

### StateOfTheEqn

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.

Last edited: Sep 1, 2013
10. Sep 1, 2013

### Staff: Mentor

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

11. Sep 1, 2013

### StateOfTheEqn

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.

12. Sep 1, 2013

### 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.

13. Sep 1, 2013

### Mentz114

$ds^2=(1-r_s/R)dt^2-(1-r_s/R)^{-1}dR^2-R^2d\Omega^2$
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.

14. Sep 2, 2013

### Staff: Mentor

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.

15. Sep 2, 2013

### Russell E

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."

16. Sep 3, 2013

### TrickyDicky

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.

17. Sep 3, 2013

### Mentz114

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.

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.

18. Sep 3, 2013

### atyy

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?

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

Last edited: Sep 3, 2013
19. Sep 3, 2013

### TrickyDicky

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.

20. Sep 3, 2013

### Mentz114

Yes, it does. I botched the transformation so my result is wrong. Apologies to all for my misleading ( but obviously wrong) assertion.

21. Sep 4, 2013

### Russell E

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.

22. Sep 4, 2013

### TrickyDicky

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.

23. Sep 4, 2013

### Staff: Mentor

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.)

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.

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]

Last edited by a moderator: May 6, 2017
24. Sep 4, 2013

### TrickyDicky

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.

See above qualifier.
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'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)
Sure, sorry about that, it doesn't add anything to my point or to the discussion anyway.

25. Sep 4, 2013

### Staff: Mentor

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.

This is true.

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