The usual way to derive metric of a point mass in general relativity is (to my knowledge) based on assuming specific form of the metric that reflects spherical symmetry and independence on "time" (Schwarzschild metric).

I don't quite like this approach because the argument seems valid and natural, however after the solution is obtained, it turns out that it's incomplete. Complete metric is said to be given by Kruskal coordinates, usually they are introduced as a transformation of Schwarzschild coordinates.

I've found on the Internets a derivation of Kruskal metric directly from field equations. It's assuming spherical symmetry, but "radius" itself is not used as an independent variable and this accounts for it's implicit form and possible "time"-dependence. This approach in my opinion is much better, however it still leaves a feeling that something might not have been taken into account.

Is there a way to solve GR equations starting just with the fact that we have a point mass and that metric should be flat at infinity, without any prior assumptions about its form? Frankly, I'm not even sure how to write down stress-energy tensor without assuming something coordinates.

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    $\begingroup$ Schwarzschild metric does not describe “a point mass”. And stress-energy tensor of Schwarzschild solution is zero. As for a feeling that something might not have been taken into account would Birkhoff's theorem help with the feeling? $\endgroup$
    – A.V.S.
    Oct 8, 2019 at 15:16
  • $\begingroup$ @A.V.S. The Birkhoff's theorem makes Schwarzschild unique in GR, but GR does not seem to be unique or needed for Schwarzschild. For example Friedrich Kottler 1918 or mathpages.com/rr/s5-05/5-05.htm - Are any field equations other than GR explicitly known that also produce the Schwarzschild solution? Thanks! $\endgroup$
    – safesphere
    Oct 10, 2019 at 14:27
  • $\begingroup$ @safesphere: Sure, for example Lovelock gravity. Schwarzschild metric is Ricci flat, so any theory that admits Ricci-flat solutions would do. Also, one could do something along the lines of hep-th/0003271, write a general Lagrangian with lots of coefficients and select those values that admit a specific set of solutions. $\endgroup$
    – A.V.S.
    Oct 10, 2019 at 15:47
  • $\begingroup$ @A.V.S. thanks, I heard about Birkhoff's theorem. Not sure why you say that stress-energy tensor is zero, is it zero at $r=0$? Anyway, what confuses me (maybe I'm wrong) is that all these derivations start by assuming some explicit form of the metric, which is based on an analogy with flat-space about what spherical symmetry is. What if there are other possible forms which are spherically-symmetric at infinity (thus, resembling newtonian gravity) but have more complicated structure close to singularity? $\endgroup$
    – xaxa
    Oct 11, 2019 at 11:36
  • $\begingroup$ @A.V.S. Kerr metric is probably an example of what I'm talking about. Also, we can add non-zero charge. And then we seem to stop because of "no-hair" theorem. You see, what confuses me is that all these solutions work backwards: they start from a specific form of the metric, solve GR equations with that metric and then interpret the result. Can we miss any other solutions? Shouldn't it be the other way round - we start from providing boundary conditions and expression for stress-energy tensor, then we obtain the metric. I hope I'm making sense to you. $\endgroup$
    – xaxa
    Oct 11, 2019 at 11:44

2 Answers 2


It is possible to describe a point mass in general relativity. It is just extremely unpleasant and rarely done to do so.

The theory generally used for point masses in general relativity is the theory of Colombeau algebras. This is a theory used for products of distributions in which the functions are mapped to generalized functions. Those algebras get around Schwarz's impossibility theorem by not being in general mappable to the usual product in $C(\mathbb{R}^n)$. It can indeed be shown that, given the Schwarzschild metric, this corresponds to the stress-energy tensor of a point particle.

As far as I know, showing the reverse isn't typically done. There are many reasons for this I can guess at : Just assuming a point particle may not in general give you the Schwarzschild metric, as other metrics could also generate such a stress-energy tensor (this depends on the initial conditions of the metric). It is possible that simply assuming asymptotical flatness (along with some other standard assumptions) may be sufficient to show this. The main obstacle I'm seeing for this being random gravitational waves

Also, the whole Colombeau algebra treatment is fairly rare (it is usually confined to very specific cases like topological defects or shockwaves), and doing it in such a general setting may be nightmarish. Without any assumption of symmetry or initial conditions, you are doing general relativity under the worst possible conditions.


It is impossible to describe a point-like particle in general relativity (and similarly non-linear theories). The mathematical structure of the theory simply does not allow for such constructs. (See the work by Geroch and others).

One can solve the Einstein field equations perturbatively for point masses. This is what is done in post-Newtonian and post-Minkowskian theory. One can also find the perturbation to a curved background due to a point source. However, this approach breaks down at higher orders in perturbation theory, and one can no longer the source as a point particle (there are various technical ways of working around this issue).

Note that this is not inconsistent with the existence of the Schwarzschild or Kerr solutions. The singularity in (the maximal extension of) the Schwarzschild solution is not pointlike. In fact it is not even timelike. Kerr has a timelike singularity, but it is not a point but a ring. (Also it is not clear if this singularity is stable under perturbations).

Finally, as pointed out in the comments, it may be helpful to refer to Birkhoff's theorem, which tells use that any spherical symmetric solution to the vacuum Einstein field equations is static, and therefore isometric to the Schwarzschild solution.


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