I'm experimenting with the EFE, and I ''invented'' a metric; a diagonal non-zero metric, and I discovered that the Riemann tensors are equal to zero which implies the Einstein tensor $G_{mn}$ equals zero (right?). Hence, the stress-energy-momentum tensor $T_{mn}$ also equals zero. But how can I tell what kind of space(time) I'm looking at?

Since there is "no" energy distribution in my space (I'm working in $R^2$) because $T_{mn}$ equals zero, I assume there is also no curvature. But how can i tell that my "flat" space isn't somehow curves; embedded in 3D space for example? How can I tell how my space looks like?

Also, the metric I chose is the following: $g_{11}$=$x^2$ , $g_{22}$=$9y^2$ and $g_{21}$=$g_{12}$=0. So, the only relevant Christoffel symbols are $C^x_{xx}$=$1/x$ and $C^y_{yy}$=$1/y$; which give zero-valued Riemann/Ricci tensors and hence a zero-valued Einstein Tensor. So what kind of space am I looking at, how can i tell?

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    $\begingroup$ When the stress-energy tensor is zero, you should end up with Minkowski's spacetime; i.e., Special Relativity. $\endgroup$ – Peter Diehr Mar 7 '16 at 2:27
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    $\begingroup$ @PeterDiehr Only locally. There are many spacetimes with a flat metric that are globally distinct from Minkowksi spacetime. $\endgroup$ – Timaeus Mar 7 '16 at 3:16
  • $\begingroup$ By "non-zero" metric, I assume you mean "not of the form $\eta_{\mu\nu}=diag(-1,1,1,1)$" right? If the metric were actually zero that would be a very singular situation. $\endgroup$ – Andrew Mar 7 '16 at 4:16
  • $\begingroup$ That's not a metric when either x or y is zero (and it's zero when both are zero) so are you restricting x and y to both be strictly positive for instance? And GR in two dimensional spacetime basically isn't a gravitational theory at all. And your metric doesn't have a Minkowski signature. $\endgroup$ – Timaeus Mar 7 '16 at 21:01
  • $\begingroup$ What if i add and negative component so my metric becomes (-,+) or (+,-), instead of the 4D (-,+,+,+)? and multiply the "time" component by c? Does that become a Minkowski signature? $\endgroup$ – Investor Mar 7 '16 at 21:22

Since you say that you've computed $R_{\mu\nu\rho\sigma}$ and found that it is zero, your metric is locally Minkowski (though it might have a different topology as pointed out by other commenters). As a result there exists a coordinate transformation that will take your metric into the form $\eta_{\mu\nu}=diag(-1,1,1,1)$, at least locally.

You are right that the fact that $R_{\mu\nu\rho\sigma}=0$ implies $G_{\mu\nu}=0$, so it must be that $T_{\mu\nu}=0$ if Einstein's equations are satisfied.

However, the implication doesn't go the other way: if $T_{\mu\nu}=0$ there are solutions to Einstein's equations that are not Minkowski space, even locally.

Physically, this corresponds to the possibility of having gravitational waves propagating even if no matter is present. This actually isn't surprising. In electromagnetism for example, there are non-trivial solutions to Maxwell's equations even if the charges and currents vanish, because you can have waves propagating freely through empty space.

Mathematically what I am saying is that $T_{\mu\nu}=0$ implies $R_{\mu\nu}=0$ from Einstein's equations, but this does not imply that $R_{\mu\nu\rho\sigma}=0$ [if $R_{\mu\nu\rho\sigma}=0$, then space is locally Minkowski, but the topology can still be non-trivial].

The Riemann tensor has 20 independent components (in 4 spacetime dimensions). Einstein's equations are 10 equations $G_{\mu\nu}=T_{\mu\nu}$, which fix 10 of the components of the Riemann tensor in terms of the matter stress energy tensor. There are still 10 components of the Riemann tensor that are not fixed by Einstein's equations. We can describe these 10 components with the 'Weyl tensor' (wikipedia: https://en.wikipedia.org/wiki/Weyl_tensor), which is essentially a trace-free projection of the Riemann tensor.

  • $\begingroup$ Do you (or know a website that) have/has a complete example (math included) of such a (solution? T = 0 but no Minkowski space) $\endgroup$ – Investor Mar 7 '16 at 13:37
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    $\begingroup$ @Investor If you are content with GR linearized around flat space, the answer is simple: the metric perturbation has plane wave solutions with 2 polarizations, that propagate even if T=0. If you are not happy with that, as an example of a more exact treatment there is a paper by Bondi, Pirani, and Robinson (Proc Roy Soc 251 (1267) 519-533) that discusses exact plane wave solutions to GR when no matter is present, here's a pdf: itp.kit.edu/~schreck/general_relativity_seminar/…. $\endgroup$ – Andrew Mar 7 '16 at 14:20

If the Reimann tensor is zero then a different coordinate system would have your metric look like the standard flat Minkwoski metric from special relativity.

But from this fact, you have no way to single out any particular spacetime. For instance your spacetime could be


with the metric $$\mathrm ds^2=\mathrm d a^2+\mathrm d b^2-\mathrm d x^2-\mathrm d y^2-\mathrm d z^2,$$ where time repeats itself. It could be $$\{(t,x,X,y,z):x^2+X^2=1\},$$ with the metric $$\mathrm ds^2=\mathrm d t^2-\mathrm d x^2-\mathrm d X^2-\mathrm d y^2-\mathrm d z^2,$$ where space loops back again in the $x/X$ direction. Or it could be any one of the many many many other spacetimes that locally all have a flat Minkowski metric at each event.

Just having a formula for the metric doesn't tell you what manifold you have. No one claims a metric formula tells you what manifold you have, because it can't.

  • $\begingroup$ Then how can I tell what kind of a manifold I'm looking at? $\endgroup$ – Investor Mar 7 '16 at 13:44
  • $\begingroup$ @Investor You have to actually specify the manifold. The metric is a tensor field. It's a function from the manifold to rank two tensors. When you pick some coordinates and write a formula you have a map from a coordibate patch to some rank two tensors. But the word general in general relativity covers the general case when there is no single coordinate patch that covers the whole space. So it's the physical theory where you have to specify the manifold to specify the fields. $\endgroup$ – Timaeus Mar 7 '16 at 16:01
  • $\begingroup$ How do you specify the manifold? $\endgroup$ – Investor Mar 7 '16 at 20:52
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    $\begingroup$ @Investor I specified two in my answer. You could also give a series of charts with some transition maps between parts of them. You could also take any set M and put a topology on it as a particular set T of subsets S. The point is it is up to you. Some people even do GR as an initial value problem they start with a 3d manifold and some fields and evolve them into a 4d manifold. $\endgroup$ – Timaeus Mar 7 '16 at 20:56

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