# Is a Pseudo-Riemann Metric Intrinsic to General Relativity?

1. Jan 5, 2015

### CSnowden

In considering special relativity as a limiting case of the general theory (without matter or curvature) the question arose as to whether the pseudo-riemann nature of the SR metric is actually an independant (essentially experimentally determined) assumption/property or derivable from the structure of the general theory. Can someone clarify if a positive-definite metric GR theory can be constructed which would still accurately describe the basic properties of matter/gravity via space-time curvature? Obviously the observed properties of nature (eqivalent reference frames, constancy of speed of light) dictate a pesudo-riemann metric, I'm just wondering if this is an independent assumption or 'baked into' the mathematical structure of the general theory. Many thanks!

2. Jan 5, 2015

### Matterwave

That space time is locally Minkowski in nature is a postulate of the general theory. This is usually given the name of "Einstein Equivalence principle".

3. Jan 5, 2015

### CSnowden

So by posulate, do you mean that it is a separate (independant) assumption which is not logically (mathematically) required to formulate a curved spacetime description of gravity?

4. Jan 5, 2015

### Matterwave

It is a postulate of GR, not all theories of gravity. I am not; however, familiar with attempts to describe gravity using a Riemannian metric. It seems that such a metric would be counter-productive, since it won't encompass SR as a natural limit, but I can't say for sure that it couldn't ever work...

5. Jan 5, 2015

### Staff: Mentor

Since there is an inherent physical difference between timelike, spacelike, and null quantities, it's hard to see how a Riemannian, as opposed to pseudo-Riemannian, metric could describe spacetime. It's certainly possible to construct a theory of curved Riemannian manifolds; but a Riemannian manifold would be locally Euclidean, not locally Minkowskian, so it could not describe the difference between timelike, spacelike, and null quantities.

6. Jan 5, 2015

### bcrowell

Staff Emeritus
You can't derive from the structure of GR the fact that the signature is semi-Riemannian. The standard formulation of GR is simply that you have a manifold and a metric, and you impose the Einstein field equations. The field equations are tensorial equations that don't say anything about the dimensionality or signature. They are equally meaningful in a Euclidean signature.

If you take the standard formulation of GR and make the signature Euclidean, you can't describe anything like the observed properties of matter and gravity.

However, the standard formulation of GR is not the only one out there. Some others are:

- spin-2 gravity in flat spacetime (Deser, 1970, http://arxiv.org/abs/gr-qc/0411023 )

- shape dynamics (Julian Barbour, Tim Koslowski, http://arxiv.org/abs/1305.1487 )

- Ashtekar equations

I don't pretend to know a lot about any of these approaches. Basically all of them are equivalent to the standard formulation GR under a pretty broad range of conditions, but they may be of interest for other reasons, e.g., Barbour hopes that shape dynamics will be easier to quantize than the standard formulation. The Ashtekar equations may be of particular interest here because, unlike the standard formulation, they don't break down if the metric becomes degenerate. If your metric evolves continuously from semi-Riemannian to Riemannian, it has to pass through a point where it's degenerate. Speaking loosely, I think the Ashtekar equations allow us to postulate a universe that is semi-Riemannian here and now, with all the observed properties of matter and gravity, but that evolves into a timeless Riemannian universe beyond some boundary in spacetime.

7. Jan 6, 2015

### ChrisVer

I am not sure about what I will write, so someone could try to have a look and correct me
Suppose that you have a metric that is in general positive... Then you diagonalize it, since it's positive definite, it will have positive eigenvalues, and thus it will be brought to the form:
$g_{ab} = \begin{pmatrix} g_{00} & 0 & 0 & 0 \\ 0 & g_{11}& 0 & 0 \\ 0 & 0 & g_{22}& 0 \\ 0&0&0&g_{33}\end{pmatrix} ,~~ g_{00 , 11,22,33} >0$
If you make a coordinate transformation (I think going to the comoving coordinates) to this metric, then I'm not sure if you can obtain the Minkowski metric, because you will have to change the sign of one of the eigenvalues...and so you are bringing down the equivalence principle...

8. Jan 6, 2015

### bcrowell

Staff Emeritus
Yes, this is all true.

9. Jan 6, 2015

### PAllen

A caveat is the meaning of the stress energy tensor. That is, the Einstein tensor certainly is certainly defined from the curvature tensor irrespective of dimensionality or signature, but then does it have any meaning interpreted as a stress energy tensor? It comes down to what does meaningful mean? (Clinton would be proud).

10. Jan 6, 2015

### bcrowell

Staff Emeritus
Right. In a Euclidean signature, the theory can't be related to the established properties of matter and gravity.

In a formulation like the Ashtekar equations, which allow a change of signature, I guess this raises the question about how you describe the physics of the matter fields when the signature changes. You clearly can't do that using the Standard Model. But I think the main interest of these alternative formulations is that they might be paths toward quantizing gravity. Presumably a theory of quantum gravity would relate the observed properties of matter to something more fundamental.

It's also unclear to me what it would mean to make measurements or predictions in a Euclidean spacetime. You might be able to solve wave equations using known boundary conditions, but the boundary conditions wouldn't be the kind we're used to, since there would be no such thing as a Cauchy surface. And in general our notion of measurement is closely tied to specific features of the Minkowski signature, e.g., we have theorems that say that you can only have bradyonic observers, not luminal or (in 3+1 dimensions) tachyonic ones (V. Gorini, "Linear Kinematical Groups," Commun Math Phys 21 (1971) 150 http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.cmp/1103857292 )

11. Jan 6, 2015

### CSnowden

The question came up in thinking of how GR reduces to SR (in the simplest case of zero matter). As a thought experiment (not a viable theory) , if one could create an accurate model of gravity as spacetime curvature using only a positive definite metric it would clearly show that In the SR metric, the pseudo-riemann characteristic seems to introduce some important properties such as causal structure (null cones), etc. - was just wondering if this was pushed down as a mathematical requirement from GR (to reduce to SR), or if it is truly a separate empirical assumption. It seems that asserting a mixed-signature metric is equivalent to asserting the constancy of light in all reference frames (can't seem to have one without the other?), so was thinking that it was a separate postuate.

12. Jan 6, 2015

### CSnowden

Perhaps a step back to clarify the context of the question might be beneficial. The question came up in thinking of how GR reduces to SR
(in the simplest case of zero matter). As a thought experiment (not a viable theory) , if one could create an accurate model of gravity as spacetime
curvature using only a positive definite metric it would clearly show that gravity is a very distinct and separable characteristic of the ST manifold as compared to pseudo-riemann characteristics. On the surface it seems that the description of curvature required for gravity and description of stress-energy to generate curvature (and define ST) could be adequately described without a mixed signature metric; wasn't sure on that point though and was looking for clarification. If there are issues in formulating an equivalent, meaningful stress energy tensor with a (non-flat) riemann metric could you please expand on that?

On the other hand the properties emerging from a mixed signature metric seem very different from curvature and gravity, such as causal structure (null cones), disjoint causal areas, large temporal and spacial coordinates cancelling to small or zero total metric separation, etc.. If the need for a mixed signature metric really is a separate empirical fact, not driven as a mathematical requirement to describe gravity as curvature, then one could amusingly imagine a different history in which someone discovered gravity as spacetime curvature (say Clifford as he was onto the core concept, formulated positive definite) and then someone coming along later and discovering (separately) that the metric happens to be pseudo-Riemann (say from the observation that the speed of light is equivalent in all frames) - obviously the reverse of how things actualy happened. These are very basic and simple questions but I appreciate the feedback from those better positioned mathematically to know if these aspects are really as separable as is being proposed.

13. Jan 6, 2015

### bcrowell

Staff Emeritus
In relativity, we have a four-vector for mass-energy-momentum. The stress-energy tensor is the density of mass-energy-momentum. Locally, in a given frame, the components of the mass-energy-momentum vector are (E,p), where E is mass-energy and p is momentum. The mass of an object is defined by the equation $m^2=E^2-p^2$, which is the squared norm of the mass-energy-momentum vector. The minus sign is there because we're in a semi-Riemannian signature. If we had a Euclidean signature, this definition would not be useful. So even something as basic as our normal concept of mass doesn't translate to a Euclidean signature.

Another fundamental property of the stress-energy tensor is that it has a zero divergence, the interpretation being that mass-energy-momentum is locally conserved. The same equation can be applied in a Euclidean space, but it doesn't have the interpretation of being a conservation law.

14. Jan 6, 2015

### robphy

It might be good to note that the spacetimes in the Galilean/Newtonian limit have a nonEuclidean metric (which happens to be degenerate).

For example, the position vs time graph in phy101 has a nonEuclidean geometry (which isn't widely appreciated).
If velocities can be arbitrarily large but not infinite (which defines timelike in this case), there is causal structure that also partially-ordered.

Last edited: Jan 6, 2015
15. Jan 6, 2015

### pervect

Staff Emeritus
My \$.02. First ask if a pseudo-Riemann metric is intrinsic to special relativity. I'd answer yes, but one could argue about what the question really meant in the first place. Clearly the Lorentz interval has a pseudo-Riemann signature and also the Lorentz interval represents an observer independent quantity. I suppose one might attempt to argue that these two facts are not "intrinsic" to SR and handwave them away as "something that just happens, not really anything vital". Instead of a closely reasoned argument, I'll summarize my emotional reaction. Blechhh.

Next ask if it is intrinsic to GR that it should reduce to SR in the limit of no gravity. I'd answer yes again (the same caveats would apply).

16. Jan 7, 2015

### bcrowell

Staff Emeritus
I don't quite agree with this. A degenerate metric is no metric at all. What we really have in galilean spacetime is two *separate* metrics, one for time and one for space. They aren't unified into a single metric. They're two separate systems of measurement, and you can't compare one with the other in any absolute sense.

Do you mean that it's an affine geometry? That's true, but affine geometry isn't metric.

17. Jan 7, 2015

### CSnowden

Many thanks, this feedback has been very helpful! With respect to the stress energy tensor, it does make sense that the mixed signature implications have an impact on this tensor (and thus generated geometry/gravity) at high velocity/energies and hence things are not 'truly separable' - in retrospect I guess that that isn't unexpected as we do exist in a semi-Riemann world and no theory will be completely correct that does reflect that reality.

Do you think it is the case that in the non-relativistic domain an accurate theory of gravity as ST curvature could have been constructed, with positive-definite metric, accurately predicting experimentally verifiable differences - ie. with a positive definite stress-energy tensor but still producing veritably different predictions like the precession of Mercury? If so, one could still make the case the essential characteristics of the mixed-signature metric (emerging at high energies/velocities) are somewhat distinct from the essential characteristics of GR/gravity?

18. Jan 7, 2015

### robphy

As suggested by http://en.wikipedia.org/wiki/Metric_signature, a degenerate metric is a metric in some circles [pun intended?]... otherwise why bother with the "nondegenerate" adjective? While it's true that a second degenerate metric is needed in the Galilean spacetime, adding additional structure doesn't disqualify it from being "metrical" (in the sense of that wikipedia article). If the implied term of "metric tensor" is too strong, a looser "metrical structure" might be more acceptable.

(It reminds me of discussions where "metric" was only supposed to refer to positive-definite metrics... but even after "pseudo-riemannian" was added to include (say) a Lorentz-signature metric, there was still resistance to accepting it as a "metric".)
However, in the end, call it whatever you want, but the point is that there is an effort to try to understand the structure of theories, possibly introducing variants of current structures or viewpoints: http://www.google.com/search?q="degenerate+metric"
(In some theories that explore signature-change, one might have to pass through a degenerate case.)

Yes, there is no limiting case in an _absolute_ sense.
It's fair to say that handling degeneracy in this and other situations
requires a specification of the limit one takes.

Yes, affine isn't metric... unless addtional structure is added, which is done.
In the Cayley-Klein classification of geometries, Euclidean, Minkowski, and Galilean are affine geometries... in the sense that they have a notion of parallelism.
But additional structure present give rise to the metrical notions.

Last edited: Jan 7, 2015
19. Jan 7, 2015

### PAllen

I don't see how to give the bolded part any meaning at all. ST intrinsically means pseudo-riemannian metric (IMO), so ST with positive definite metric is a self contradiction.

[edit: I see you mean ST as not equivalent to SR. However, then, forget gravity. I don't see how you can have a positive definite ST metric that describes any physics remotely resembling what we observe, in any domain. Note carefully the discussion that Newtonian physics does not have a positive definite ST metric. Instead, at best, it has two metrics, and is, in some sense, a limiting case of pseudo-riemannian metric.]

Last edited: Jan 8, 2015
20. Jan 8, 2015

### bcrowell

Staff Emeritus
It seems to me that we've already answered this question, there is a consensus, and the answer is no. If you are really strongly motivated to pursue this, my advice would be to take a few weeks and teach yourself about some of the alternative formulations of GR that I described in #6.