The Schwarzschild solution in GR only has a singularity at the origin $r=0$: otherwise there is no matter content. The right-hand side of Einstein's equation is hence almost everywhere zero, but I would expect a Dirac-delta like matter distribution at the origin. The vacuum Einstein's equations are

$$R_{\mu\nu} - \frac{g_{\mu\nu}}{2}R = 0$$

But with the Schwarzschild solution I would expect something like

$$R_{\mu\nu} - \frac{g_{\mu\nu}}{2}R = const \; g_{\mu\nu} \; \delta^3(r)$$

which would correspond to the singularity at the origin. What is, precisely, the right-hand side?

Put it differently, what kind of solution would I get if I started out solving

$$R_{\mu\nu} - \frac{g_{\mu\nu}}{2}R = const \; g_{\mu\nu} \; \delta^3(r)$$

from the beginning? Or potentially other type of singularities on the right, like $\delta^\prime(r)$, etc.

Usually Dirac-delta type singularities in 3D can be detected by doing a surface integral around them either very close or at infinity and observing that the result is non-zero. Hence in these cases an exact zero is not possible, only something which is zero almost everywhere.

The analogy I have in mind is the situation of a point charge in electrostatics. The potential is almost everywhere harmonic, but there is a Dirac-delta like singularity at the origin where the point charge is sitting

$$\Delta \phi(r) = const \; \delta^3(r)$$

  • $\begingroup$ I am not sure the accepted answer is correct. For example, please see appendix A of Øyvind Grøn's "Celebrating the centenary of the Schwarzschild solutions" in AJP which briefly discusses the requisite delta-function. $\endgroup$
    – user196574
    Mar 20, 2022 at 6:32

4 Answers 4


Your question is understandable: if the metric is curved by a mass, shouldn't there be a mass in the center, like there is a charge in the center of an electrostatics problem?
But the answer is no, because there is no mass. Let me explain.

In deriving the Schwarzschild solution, we impose that, in the Einstein Equations (EE), $T^{\mu\nu}=0$: it's a vacuum, spherically symmetric, solution (it's the only one actually, according to Birkhoff's theorem). We then impose the symmetries to the metric tensor, and plug it into our vacuum EE to find that there is only one free parameter, $m_0$, the Misner-Sharp """mass""", that is completely invariant
$\frac{\partial}{\partial t}m_0=\frac{\partial}{\partial r}m_0=0$
and that has a form of
$1-\frac{2m_0}{R}=-g^{\mu\nu}\partial_\mu R \partial_\nu R$
where $R$ is the coefficient of the angular part of the metric. The Schwarzschild metric then takes the form
$ds^2=\left(1-\frac{2m_0}{r}\right)dt^2-\left(1-\frac{2m_0}{r}\right)^{-1}dr^2-r^2\left(d\theta^2+sin^2\theta d\phi^2\right)$.
Now one would like to find a physical meaning for $m_0$, and by comparison with the Newtonian limit, $g_{00}\simeq1+\frac{2}{c^2}\frac{-GM}{r}$ where $M$ is the mass of the Newtonian source. In God-given-units $(c=G=1)$ one can see that $M$ is the Newtonian interpretation of $m_0$. This is the only reason why we say that $m_0$ is a mass: if we don't take Newton into account, it is only a parameter in the metric.

As I said in the comments, one could decide to study a different, modified Schwarzschild metric, where one imposes $T^{\mu\nu}\sim m\delta (\vec{r})$: there is an interesting paper by Fiziev where this alternative is considered. Please note: this isn't the mathematical ideal framework that we call Schwarzschild metric, so the answer to your question would still be no, there is no Dirac-delta in the stress-energy tensor of the Schwarzschild solution.

  • 4
    $\begingroup$ This is all fine, I completely agree. But this doesn't answer the question whether a Dirac-delta type singularity is there on the right hand side or not. Example: charged particle again, you solve $\Delta\phi(r)=0$ after imposing rotational symmetry. You get $\phi(r) \sim 1/r$ in a unique way. Then you go back to $\Delta\phi(r)$ and notice that it is not completely zero, only almost everywhere. $\endgroup$ Nov 11, 2020 at 11:00
  • $\begingroup$ @DanielFetchinson There isn't a Dirac-delta because there isn't one. The solution is in the vacuum a priori. You could say, a posteriori, that since the metric is singular at $r=0$ you can see it as a Dirac-delta, but it's a different procedure than the electrostatics case. Electrostatics: "there is a charge causing this field, but where? I can get arbitrarily close to the source. I solve the equations and find a delta" GR: "I want to find a vacuum spherical solution. But what is causing it? There is a parameter that I can see as mass (or gravitational charge, if you wish)" $\endgroup$ Nov 11, 2020 at 11:43
  • $\begingroup$ Also (I hadn't thought of that) you could impose the delta distribution from the beginning. You would probably still find something very similar to Schwarzschild, but I'm not sure that would be classified as a "Schwarzschild solution" $\endgroup$ Nov 11, 2020 at 11:45
  • 2
    $\begingroup$ I guess my question is a purely mathematical one, regardless of how one interprets the parameters. The way you posed it is useful I think: "You could say, a posteriori, that since the metric is singular at r=0 you can see it as a Dirac-delta" That's exactly my question. Is there a Dirac-delta in the curvature or not? $\endgroup$ Nov 11, 2020 at 12:05
  • 2
    $\begingroup$ There is such a thing as the ADM mass, and it is $M$. Also, this answer should explain why the same reasoning does or does not apply to the electric field of a point charge. $\endgroup$
    – Javier
    Nov 11, 2020 at 14:12

The Schwarzschild solution is a vacuum solution. There is no mass in it anywhere. If you are using it to model e.g. a planet then it only applies in the vacuum region outside of the planet. For this reason, I prefer to think of the free parameter in the Schwarzschild spacetime as a length (the Schwarzschild radius) rather than a mass.

In the case of a black hole, the coordinate r=0 with the singularity must be removed from the manifold. The infinite curvature at the singularity ruins the manifold structure. So you cannot put a delta function at r=0 since r=0 is not a part of the manifold.

For the case of a spherically symmetric planet or star, the Schwarzschild spacetime only describes the vacuum region outside the mass, and inside the mass you must use an appropriate interior Schwarzschild solution. The vacuum solution is unique, but the interior solution will depend on the details of the stress-energy distribution


Singularities in GR are little more complicated than just putting some distribution somewhere in the equations. In Maxwell theory, the singularity has no influence on underlying structure. In GR, singularity means problems with underlying structure itself. Singularity in GR literally means some border, where geodesics end in finite amount of affine parameter and yet it is mathematically impossible to extend the spacetime to make the geodesics continue. It is border where the spacetime itself ends.

The distribution like dirac delta function is defined by its behavior under integration. In Maxwell theory you can write and compute charge $q=\int\rho(\vec{x})dV,$ even if the charge distribution $\rho$ is delta function, because the space over which you are integrating is still well-behaved. This makes dirac delta function useful concept. Not so in general relativity. In GR there is problem in volume integral itself, not just the integrand. The integral is defined only on spacetime, but the singular point you would like to describe by some generalized function does not belong to spacetime. The integral is simply undefined and you would really need to generalize the notion of integral (or manifold) itself. I do not have any knowledge that such generalization exists.


Just to describe the problem more closely, imagine you would wish to get rid of the problem of singular point not being in spacetime by gluing the singular border together. Simply put, you would need to decide how will the geodesic that ends in singularity emerges at the other side. Now take for example timelike geodesic. In Schwarschild spacetime, this would mean that ingoing particle will start to move away from singularity. This means, it would either become spacelike, or it would start to travel back in time. So you see, the whole idea of continuation through singularity is quite problematic. The best you could do, is that the geodesic will remain at singularity, but the time for it will go on. But because the point is singular, this makes no sense. What is the notion of time in singularity itself? There is no such thing, you would need to invent it. And then you would be stuck with trying to build your concepts inside the singular point. Schwarzschild spacetime is simply irremediable without extensive (and unnatural) modifications.

But there are "distribution-like" singularities in GR like for example conical singularity. The spacetime for conical singularity is everywhere flat except the fact that there is some central (spacelike) line around which angle is not $2\pi$, but some different number. How will the infalling geodesic continue after reaching the central line? Sensible choice is to just make a straight line in your cartesian coordinate chart. There is no problem doing this. And if you do, you can use Einstein field equations and you will get that there is some distribution in stress-energy in central line. It will make sense. But you cannot do this for Schwarzschild, there the singularity is much more serious.


The Schwarzschild solution in GR only has a singularity at the origin 𝑟=0:

Inside the horizon, the Schwarzschild r coordinate is timelike, not spacelike. Therefore it's not correct to imagine that as $r\rightarrow0$, one is getting close to a point in space that is the center of symmetry. As $r\rightarrow0$, one is getting close to the end of time, and this is the way it's depicted on a Penrose diagram.

As other answers have pointed out, there is no underlying spacetime that extends all the way to the singularity. Topologically, the singularity is a hole cut out of spacetime.

Usually Dirac-delta type singularities in 3D can be detected by doing a surface integral around them either very close or at infinity and observing that the result is non-zero.

Gauss's theorem fails in curved spacetime for nonscalar quantities, because adding fluxes requires comparing vectors at different places, but that can only be done by parallel transport, which is path-dependent. The mass-energy density is not a scalar, it's a component of a rank-2 tensor, so Gauss's theorem doesn't give you a way to find the mass-energy inside a surface. We can observe the distant gravitational field of a black hole and use it to find a mass. Conceptually, you can think of this mass as the mass-energy of the gravitational field in and around the black hole. But the equivalence principle guarantees that we can never localize the mass-energy of the gravitational field, since the gravitational field at a given point can always be zero if you are an inertial (free-falling) observer.

A Schwarzschild black hole is not an astrophysical black hole. In spacetimes that describe a black hole that formed by gravitational collapse, we can try to trace what happens to the infalling matter. There is a lot of variety in the things that can happen, and it's not really well established, for example, whether the singularity starts out being a naked singularity and then later on becomes a spacelike singularity more like a Schwarzschild singularity. We also don't really know for sure whether there is initially a strong curvature singularity, in which the volume of a cloud of infalling test particles goes to zero. A Schwarzschild black hole doesn't have a strong curvature singularity, it has a spaghettifying singularity.

Although the details of the process of realistic astrophysical gravitational collapse are an open question, it's probably roughly correct conceptually to say the following. As a black hole collapses, the mass-energy of the infalling matter gets converted to a mass-energy of the black hole's gravitational field. Because of the technicalities of how the Einstein field equations are expressed, this gravitational field energy is inherently delocalized, and does not appear as a term in the stress-energy.


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