Suppose we use the signature of a Riemannian manifold

$$ \eta^{\mu\nu}=\operatorname{diag}(+,+,+,+) $$

as the starting point to describe a 4d Euclidean version of general relativity. Alternatively one may use a Wick rotation on time $t\to i \tau$ on normal general relativity to get this metric.

But instead of remaining in Euclidean GR, one then further flips the metric back to Lorentz $\eta^{\mu\nu}=\operatorname{diag}(+,-,-,-)$ but this time using the Stress-Energy tensor, instead of a wick rotation.

One will end up with the correct Lorentz signature $\eta^{\mu\nu}=\operatorname{diag}(+,-,-,-)$ but also with a non-zero stress-energy tensor.

My question is; what is the simplest stress-energy tensor which curves a $(+,+,+,+)$ flat metric into one with curvature $(+,-,-,-)$?

  • 2
    $\begingroup$ There is no mechanism for a signature change in GR. I'm not sure if this has been studied in mathematics but there is no mechanism for it in physics. $\endgroup$ Commented Jul 5, 2020 at 13:43
  • $\begingroup$ What do you mean by "flips the metric back to Lorentz using the Stress-Energy tensor"? $\endgroup$
    – J. Murray
    Commented Jul 5, 2020 at 13:55
  • $\begingroup$ @J.Murray Well the stress-energy tensor is correlated to the metric of space-time, right? For example, if you place a lot of matter/energy in a small volume you get a black hole (Schwarzschild metric), etc. My question is; does there exist a stress-energy tensor such that a (+,+,+,+) metric is "curved" to a (+,-,-,-) metric? In that case, the Pseudo-Riemannian feature of the otherwise Riemannian metric is induced by the distribution of matter and energy in the system, rather than by postulating the metric to be Pseudo-Riemannian. I hope this is clearer... $\endgroup$
    – Anon21
    Commented Jul 5, 2020 at 14:10
  • $\begingroup$ @J.Murray Instead of placing a lot of matter/energy in a small volume to get the Schwarzschild metric, you would distribute matter/energy appropriately in all of the space-time such that you get a parity inversion from (+,+,+,+) to (+,-,-,-). $\endgroup$
    – Anon21
    Commented Jul 5, 2020 at 14:14
  • $\begingroup$ @JohnRennie Well, in my question I ask for a parity change with does not change the full signature, but only flips the sign of the space elements. Does your statement also applies to parity inversion or just signature changes? A signature change would be (-,+,+,+) to (+,-,-,-) and an example of a parity inversion would be (+,+,+,+) to (+,-,-,-). $\endgroup$
    – Anon21
    Commented Jul 5, 2020 at 14:21

1 Answer 1


In the standard formulation of GR, this is not possible. This is because the metric tensor field is assumed to be everywhere non-degenerate and continuous, and in order to pass from a region of signature (++++) to a region of signature (+---), you need to pass through some regions at which the metric is either degenerate or discontinuous.

As a toy model, one could imagine a metric tensor field of the form $$ds^2 = dt^2+x (dx^2+dy^2+dz^2)$$

which would have signature (++++) for $x\gt 0$ and $(+---)$ for $x\lt 0$. However, on the hyperplane $x=0$, the metric would have signature $(+000)$ and would therefore be degenerate.

These assumptions are baked into Einstein's equations. Recall that

$$\Gamma^i_{\ \ j k}= \frac{1}{2} g^{i\alpha}(\partial_j g_{\alpha k} + \partial _k g_{\alpha j} - \partial_\alpha g_{jk})$$

and $$R = g^{\alpha \beta}R_{\alpha \beta}$$

If the metric is not at least continuous, then the $\Gamma$'s are not well-defined (or involve things like delta functions) at the points of metric discontinuity. Furthermore, the components of the inverse metric are the matrix inverse of the components of the metric; if the metric is degenerate, then it is non-invertible, so the "inverse metric" doesn't exist. The entire structure of raising and lowering indices (or at least raising them) depends on the non-degeneracy of the metric, so we must abandon it if we wish to allow metrics with spatially varying signatures.

That being said, apparently the idea of spacetimes which contain multiple regions of different metric signature has been kicked around for a while. A quick search of the arXiv yields this paper, and the subsequent 23 papers which have cited it which you might find interesting. I am nowhere close to an expert on even standard GR, much less fairly dramatic extensions of it, so this is where my ability to speak coherently on the topic ends.

  • $\begingroup$ Hello, I was more thinking of a case where, using your "toy" metric, $x=-1$. Thus, $x$ would not change. The metric would be $(+,-,-,-)$ everywhere and thus "unique". Is there a stress-energy tensor that can fix $x$ to $-1$ for metric initially defined as $(+,+,+,+)$? If I then define this stress-energy tensor as the "true vacua of space" then I do not need to worry about the possibility of my metric switching to a different signature. Does a stress-energy tensor that fixes $x$ to $-1$ change anything to your conclusion? In this case there would be no discontinuity nor degeneracy. $\endgroup$
    – Anon21
    Commented Jul 5, 2020 at 16:01
  • $\begingroup$ @AlexandreH.Tremblay I don't understand what you mean when you say a metric initially has signature (++++) but also has signature (+---) everywhere. Everywhere means everywhere in spacetime - at all points, and at all times. $\endgroup$
    – J. Murray
    Commented Jul 5, 2020 at 16:05
  • $\begingroup$ I apologize if I am using the wrong terms. My understanding of General Relativity is that it recovers Special Relativity in the absence of matter/energy; that is when the stress-energy tensor is zero everywhere. Now imagine instead that we define an analog to General Relativity, called Euclidean gravity, that instead recovers a metric with signature (+,+,+,+) when the stress-energy tensor is zero. Now I ask, in Euclidean gravity, what is the stress-energy tensor whose corresponding metric is (+,-,-,-) everywhere? In this case, there is no change of metric signature in spacetime. $\endgroup$
    – Anon21
    Commented Jul 5, 2020 at 16:19
  • $\begingroup$ @AlexandreH.Tremblay Let's continue this in chat. $\endgroup$
    – J. Murray
    Commented Jul 5, 2020 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.