In GR, we are working with Lorentzian metrics, which are examples of a pseudo-Riemannian metrics. That is, we are trying to find pseudo-Riemannian $g_{\mu\nu}$ that are solutions to the field equation $$R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} + \Lambda g_{\mu\nu} \propto T_{\mu\nu}.$$

Does the form of the field equations exclude positive-definite metrics, or even exclude non-Lorentzian metrics of other signatures, or is the use of Lorentzian metrics an added condition to make GR match reality?


2 Answers 2


The Einstein equations alone do not enforce any particular signature. For example, in the derivation of the Schwarzschild metric you require that the metric is asymptotically flat... but not just any flat metric: by doing so you typically require that it's the flat Minkowski metric, not the flat Euclidean one. However, you may do the same by requiring that the metric is asymptotically Euclidean! While this is mathematically possible, this wouldn't be a GR solution: if you are doing GR you have to implement a Minkowsky-like metric in the boundary conditions.

Therefore, the signature is cooked into the boundary conditions or in the initial value if you have a Cauchy problem. It is so because having a Lorentzian signature is a fundamental requirement of GR (it is fundamental for the causal structure). However, this is something that should be explicitly implemented when solving the Einstein equations. Since the Einstein equations are partial differential equations, information on the signature enters via the initial condition or boundary terms (depending on the practical problem at hand).

As a final note, a common way of implementing the Lorentzian nature in a practical way is to rely on the 3+1 split of spacetime for solving the Cauchy initial value problem but, again, this is not strictly necessary and you must have an initial condition consistent with a Lorentzian metric.

  • $\begingroup$ You are right 'The Einstein equations alone do not enforce any particular signature' but I would argue that GR demands Minkowski signature infinitesimally local. Asymptotically flat Minkowski metric results from boundary condition of the zero pressure on star surface, see equation (10). $\endgroup$
    – JanG
    Nov 8 at 19:15
  • $\begingroup$ @JanG I agree, GR must locally give special relativity. On the other hand, if one only considers Einstein equations as a set of PDFs to be solved, it is less clear where the signature comes in (Einstein equations alone are "less" than the full theoretical set-up of GR). I think that a nice alternative to answer this question would be to use the tetrad formalism: the signature is explicit since the tetrad formulation of GR is a gauge theory with a non-compact gauge group SO(3,1), the signature being in that (3,1). $\endgroup$
    – Quillo
    Nov 9 at 11:48
  • $\begingroup$ I think that's a great idea. Another question regarding EFE (PDFs). I cannot understand how Minkowski spacetime (globally, not locally) should be solution of EFE. In this manner any constant metric would be possible solution, too. $\endgroup$
    – JanG
    Nov 9 at 11:53

The Einstein field equations are derived through the variation of the Einstein-Hibert action: $$S[g] = \int \sqrt{{\rm det}(g)} R_{ab}g^{ab}$$ Although to write down this action, we utilize the metric, but we do not demand a specific form for it (however, an incorrect metric type would not result in something useful).

The Lorentzian nature of the metric can instead be attributed to the hyperbolic nature of spacetime. This guarantees that the (local) future developments in spacetime should be uniquely predicted based on the information of some initial conditions. Therefore, in my opinion, the condition that the developments in spacetime have to respect causality stands at the core of why the metric has to be Lorentzian.

  • 1
    $\begingroup$ +1 I do not understand the downvote, this answer seems to be essentially correct and in line with this one physics.stackexchange.com/a/710380/226902 $\endgroup$
    – Quillo
    Oct 3 at 6:23
  • 1
    $\begingroup$ +1 I think this answer (as well as @Quillo's answer) is concise and to the point. $\endgroup$
    – VaibhavK
    Oct 3 at 9:54

Your Answer

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

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