How much is the Schwarzschild geometry a characteristic feature of GR

1. Nov 18, 2006

lalbatros

The Rindler geometry and its horizon can be obtained by a simple succession of Poincaré transformations to match the frame of an accelerated observer. By combining this SR result and the equivalence principle it follows that a uniform gravitational field is represented by the Rindler metric and inherits all its properties, specially the horizon.

Considering this, I have the feeling that:

- the Schwarzschild geometry does not need much more than the SR and the EP
- the additional need is merely related to the spherical symmetry

I would like to learn more about the additional physics to be understood in the Schwarzschild geometry as compared to the Rindler geometry.

I would also like to know how much it tells us about GR.

Thanks,

Michel

2. Nov 18, 2006

Los Bobos

The singularity of Rindler coordinates in flat spacetime comes from inappropriate choice of coordinates and is not physical. Only observers at rest in Rindler coordinates observe this surface of infinite redshift.

The Schwartzschild geometry comes as a solution of the field equations and all observers agree about the horizon of the geometry.

3. Nov 18, 2006

pervect

Staff Emeritus
It's fairly easy to see that the equivalence principle implies that the metic coefficient g_00 decreases as one gets deeper in a gravity well.

It may or may not be possible to deduce that g_00 goes to zero for a large enough mass, i.e. it may or may not be possible to deduce the existence of an event horizon.

I don't see any way of deducing the properties of the spatial components of the metric from the equivalence principle alone, i.e. g_rr, though. Which isn't quite the same thing as saying that it's impossible, but my intuition (which could be wrong) says that you'll need the full field equations for that.

4. Nov 18, 2006

robphy

This may be of interest. (I haven't read it myself yet.)

American Journal of Physics -- March 1988 -- Volume 56, Issue 3, pp. 265-269

The impossibility of a simple derivation of the Schwarzschild metric
Ronald P. Gruber/Richard H. Price/Stephen M. Matthews/William R. Cordwell/Lawrence F. Wagner

"The Schwarzschild metric, the general relativistic description of the space-time outside a spherical mass, has an extremely simple appearance. Because of this many attempts have been made to derive it by combining special relativity with concepts of Newtonian gravitation. It is shown here that such a derivation is impossible. A general discussion is given of the relationship of relativistic gravitation and its Newtonian limit with special emphasis on a particular non-Newtonian effect: spatial curvature."

5. Nov 19, 2006

Chris Hillman

Pretty essential

Hi, Michel,

I can only guess what you are trying to get at here. My guess is that you are trying to establish something like the following claims:

1. The "Rindler metric" can be obtained from the Minkowski metric by considering a sequence of (parallel) boosts applied to an observer in flat spacetime,

2. The "uniform gravitational field" in gtr is given by the Rindler metric, and this captures the content of "the" equivalence principle of gtr,

3. Therefore (?), assuming "the equivalence principle", the Schwarzschild metric can be obtained by elementary considerations in special relativity alone, without any reference to the vacuum field equations of gtr.

I don't know if this guess is accurate, I can offer some comments about these claims, irrespective of whether you were really trying to make them.

1. Indeed, the so-called "Rindler metric" is simply the usual Minkowski metric represented in another coordinate chart (note that changing coordinates is analogous to changing bases in linear algebra), and this Rindler coordinate chart (and the transformation from the Cartesian coordinate chart) can indeed be worked out from considering an observer in Minkowski vacuum (aka flat spacetime) who is accelerating with constant magnitude and direction of acceleration. This is done, for example, in the classic textbook by Misner, Thorne, & Wheeler, Gravitation (see section 6.6). This involves CONTINUOUS acceleration, but that can be approximated by a sequence of boosts. All you need know here is that acceleration corresponds to the path curvature of the world line of the accelerating test particle (observer), which is given by taking the exterior derivative of the tangent vector X to the curve, namely D_X X.

2. It is not VERY wrong to say this as a first approximation, but deeper examination shows that in some ways this claim is misleading. For one thing, there is more than one "equivalence principle" used in gtr (most textbooks discuss both strong and weak versions). More important, one important and valuable way to think about "the" EP is in terms of how the gravitational field varies over a larger and larger region of spacetime. In a sufficiently small region, a curved spacetime can always be approximated by a flat spacetime. This is simply the Lorentzian analog of a key insight of surface theory: curved two dimensional manifolds immersed in euclidean spaces have a tangent plane at each point. If you look at a larger region (or demand more accuracy), however, you must take account of quadratic curvature effects.

The real question here should be: why do quadratic effects suffice? Why does one not need to consider cubic or even more complicated forms? Why can Riemann and his heirs get away with the simplest thing beyond "bundled" linear forms, namely "bundled" quadratic forms, in order to define a suitable notion of "distance" along a curve on a smooth manifold? Answers, some of them good, have been offered in the literature. Be this as it may, in Riemannian (or Lorentzian) geometry, these quadratic deviations are precisely what is captured by the Riemann curvature tensor. In a vacuum solution in gtr, a kind of tensorial "average" of the Riemann curvature tensor (the Ricci curvature tensor) is identifiable as a relativistic analog of the tidal force tensor known from vacuum solutions in Newtonian gravitation. Thus slogans such as this: "the gravitational field is only observable as second order effects in the geometry". C.f. the Jacobi geodesic deviation formula, for example.

In both Newtonian gravitation nor gtr, in a sense, "all gravitational fields appear uniform on sufficiently small scales"; what is interesting is how they vary on larger scales. In particular, curved spacetimes such as the Schwarzschild vacuum are easily distinguished mathematically (e.g. using curvature invariants) from a Rindler scenario.

3. Like Pervect, I have, from time to time, encountered claims to the effect that the Schwarzschild geometry can be derived from "physical first principles", without appealing to the vacuum field equations of general relativity. Some of these are not entirely incorrect. For example, there are various other classical relativistic theories of gravitation which admit the Schwarzschild spacetime as a solution, even as a vacuum solution, so in that sense you don't need gtr to "derive" the Schwarzschild geometry (as a Lorentzian spacetime, i.e. a smooth manifold endowed with a specific metric tensor). But this probably isn't what you had in mind, and more ambitious claims of this nature tend to be far more dubious. Regarding the debunking eprint cited by Pervect, I note the Richard Price is well known for his work on how black holes respond (according to gtr) to perturbations (roughly speaking, outside the horizon, they radiate away imperfections in their prefered Kerr geometry as gravitational radiation) and is a coauthor of the problem book I mentioned in my first PF post (this being my second post), so he is certainly a leading expert on how gravitational fields behave (according to general relativity) under small perturbations from some reference spacetime, such as the Schwarzschild or Minkowski vacuum solutions of gtr.

Chris Hillman

6. Nov 19, 2006

Los Bobos

For this reason i consider this to be very wrong.

7. Nov 19, 2006

lalbatros

Dear Los Bobos,
Dear All,

When you consider the rindler geometry as the view an accelerated observer has on spacetime, you are are totally right.

"My" point is that by using the equivalence principle, the Rindler geometry is also what observers at rest in a uniform gravitational field will see. In this case, the Rindler metric is exactly the same thing for plane geometry as the Schwartzschild metric is in a central field.

Therefore, I would say that the Rindler metric is far from being an unphysical toy. And, precisely, what catched my interrest is that all features of the Schwartzschild metric seem to be there, while the embedding space is actually a flat space. Therefore my question: is there more physics in the Schwartzschild metric that in the Rindler metric.

I want to tell it still in another way.
Instead of considering a point mass as a the source of gravity, consider a slab mass distribution. Solve the Einstein equations, and you will find the Rindler geometry. The same metric as the one obtained by using SR in an accelerated motion and the equivalence principle to convert the result for a gravity field. This all is no surprise since GR is the extension of SR based on the equivamence principle. I would like to be sure that the difference with the Schwartzschild problem is only related to the geometry considered, and/or I would like to know what can be learned concerning the design of GR.

Thanks,

Michel

8. Nov 19, 2006

lalbatros

Pervect,

That's precisely why I would like to pinpoint some difference in the two geometries that could be used to illustrate the design of the field equations.

Michel

9. Nov 19, 2006

Tomaz Kristan

Quite off topic I guess, but Einstein personally dismissed Schwarzschild's work as "not smelling right". So it is at least wrong to credit only Einstein for GR. Schwarzschild is a contributor.

Doesn't matter much nowdays, but this is the history.

10. Nov 19, 2006

lalbatros

robphy,

This paper looks extremely interresting indeed (but I have no -free- access):

I precisely wrote this post because I believed the opposite.
Of course I don't believe a rigourous simple derivation could be done.
But I believed that SR+EP with some plausibility arguments could do it.
That's why the difference between Rindler and Schwartzschild is interresting.

Thanks,

Michel

11. Nov 19, 2006

Los Bobos

This was also my point :). Only observers at rest see this horizon, while everyone sees the Schwarzschild horizon. And this is the big point here, in my opinion. Geometry is more than choice of coordinates.

12. Nov 19, 2006

lalbatros

Chris Hillman,

I have no doubt about the need for the Einstein fields equations.
I accept it quite easily when reading Landau and starting from and invariant least action principle.
But I am missing intuition and pictorial representations.
I believe that for the Schwartzschild geometry, the additional machinery from the field equations are needed. However, I also think that the Schwartzschild geometry may be the example showing the least about GR. This is an opportunity to try to illustrate it precisely, before going further away from flat space.

Thanks,

Michel

13. Nov 19, 2006

lalbatros

Los Bobos,

My guess is that the difference in the two horizons is simply the result of differences between the 1/r and the f(r)=Cst fields from a point and a slab geometry respectively. Naïvely, the point brings a stronger field (singularity) and results in a "stronger" horizon.

I would appreciate link explaining you point.

Thanks,

Michel

Last edited: Nov 19, 2006
14. Nov 19, 2006

Los Bobos

I will try to explain this a bit better (it's a while time since i've worked with gr). The Rindler and Schwarzschild horizons are very similiar and actually for both of them the singularity at the horizon is just a pseudosingularity - for Rindler, as said before, it can be only seen by observers at rest in Rindler coordinates and is thus unphysical and for Schwarzschild you can also do a clever change of coordinates (Kruskal-Szekeres) to get the singularity out of the horizon.
But, for Schwarzschild the horizon still remains at the same place.

15. Nov 21, 2006

Chris Hillman

Role of Schwarzschild vacuum solution?

Hi, Michel,

For the field equation? Most textbook authors try hard to motivate the EFE, and some offer insights not readily available elsewhere. I don't know what books you've already tried or how you learn best, but you might try the discussion in Frankel, Gravitational Curvature (unfortunately out of print, but well worth the effort if you can track down a copy). And if you haven't already seen the exposition by Baez and Bunn, I also recommend http://www.math.ucr.edu/home/baez/einstein/ and the expository paper by Penrose, "The Geometry of the Universe", which appeared in Mathematics Today, ed. by Lynn Arthur Steen (not to be confused with a completely different book with the same title).

Sorry, I don't understand.

By the way, it seems to me that the obvious derivation of the Schwarzschild vacuum in gtr is pretty simple, so I am not quite sure I understand why anyone wants anything simpler. Perhaps you should substitute "elementary" for "simple"? As in, "deriving" this spacetime without appealing to say tensor calculus? Or perhaps you should substitute "general" for "simple"? As in, showing that this spacetime is a static spherically symmetric vacuum solution for some large class of relativistic gravitation theories? (Then, you could try to identify for which of these theories Birkhoff's unicity theorem holds true.)

Chris Hillman

16. Nov 21, 2006

lalbatros

Chris,

Indeed, I am not looking for something simpler.
Anyway, with Mathematica it is straightforward to derive the ad-hoc differential equation or to check that the Schwarzschild metric is a solution to the Einstein equations.

My point was that the Rindler metric can entirely derived without the Einstein equations: starting in the frame of an accelerated observer and converting to a "gravitating observer" using the quivalence principle. For the Schwarzschild metric this can not be done, as robphy pointed with a reference. But I am still interrested to understand why and how the Einstein equation becomes necessary then.

Thanks,

Michel

17. Nov 21, 2006

Chris Hillman

Hi again, Michel,

Sorry, I still don't understand. "Necessary" in what sense, and in order to do or achieve what?

Chris Hillman

Last edited: Nov 21, 2006
18. Nov 22, 2006

lalbatros

Chris,

By "Necessary", I mean that the "Schwarzschild" metric can only be derived from the Einstein equations. On the contrary, the Rindler metric can be derived from the Eintein equation as well as by combined SR and the EP in a simple way.

Michel

19. Nov 22, 2006

Los Bobos

lalbatros, as I pointed out earlier, Rindler horizons are observer dependent and thus unphysical. I am kind of losing your point here...

20. Nov 22, 2006

Chris Hillman

Huh?

Hi, Michel,

Sorry, I am -still- unsure of precisely what you are trying to say. If you mean to claim that the Schwarzschild solution might be a vacuum solution in some relativistic classical field theories of gravitation other than gtr, I've already pointed out that this is true. If you mean to claim that the EFE is not neccessary to derive the Schwarzschild vacuum in gtr, I guess that depends upon what you mean. You might be thinking of the Birkhoff unicity theorem, but of course proving this requires the EFE.

Chris Hillman