The Schwarzschild metric is $$ds^2 = - \left( 1 - \frac{2GM}{r} \right) dt^2 + \left(1-\frac{2GM}{r}\right)^{-1} dr^2 + r^2 d\Omega^2.$$ In all GR books, it is stated that $M$ is the mass of the black hole. The proof is that, in a weak gravitational field, $$g_{tt} = -(1+2\Phi)$$ where $\Phi$ is the gravitational potential, and in spherical coordinates, the potential of a point mass of mass $M_0$ is $$\Phi = -\frac{GM_0}{r}.$$ Comparing these expressions, we find $M = M_0$.

I don't buy this argument. In the last equation, $r$ is the radial spherical coordinate. In the Schwarzschild metric, $r$ is simply the name of one of the coordinates. I could have applied some coordinate transformation (like substituting $r' = 2r$) and gotten a different answer.

One can argue that the Schwarzschild $r$ is naturally "the" radial coordinate, because it makes the areas of spheres behave like in regular spherical coordinates (i.e. $ds^2$ contains $r^2 d\Omega^2$). But I could have chosen another coordinate, $\bar{r}$, that made radial distances work regularly (i.e. $ds^2$ would contain $\bar{r}^2$). Both of these feel like a radial coordinate to me.

What allows us to identify the Schwarzschild $r$ with the spherical radial coordinate $r$? Is there another way to conclude the Schwarzschild solution has a mass $M$?


Conserved quantities in GR

In GR, energy (or mass) is typically an ill-defined concept. In flat spacetime, we define energy as the conserved quantity corresponding to time translational symmetry. Extending this to GR is quite tricky mainly because, what one is calling time is already observer dependent (this is of course also true in flat spacetime, but at least there we have a canonical definition of time given by inertial observers). A second problem in GR is that time translation may not be a symmetry of the space-time, making it impossible to define energy. In particular, recall that the metric in GR is a fluctuating field, which makes it doubly hard to define timelike Killing vectors when the background itself is fluctuating.

Anyway, I hope what you can get from this is that defining energy and in fact any conserved quantity that depends on isometries of the space-time is not really something one can talk about in general relativity. So what do we do? How do we define such quantities?

How to define energy in GR?

One possible solution is to go very very far away from all forms of matter in a region where only radiation may exist. In this region - known as asymptotic infinity - spacetime is approximately flat, and one may hope to define energy here. In this region, we have a well defined notion of inertial observers w.r.t. whom we may define time and hence energy. The energy/mass so defined is called the ADM (Arnowitt, Deser, Misner) energy of the space-time. It describes the mass of the system as measured by an inertial observer sitting at infinity.

ADM mass of the Schwarzschild Black Hole

The precise formulae for the ADM mass can be read off for instance in Carroll. Using that formula, we can compute the ADM mass of the Schwarzschild black hole and we find that it is $M$. This is how we know that the quantity $M$ represents the mass of the Schwarzschild Black Hole. In other words, the statement is, place an inertial observer very far away from the black hole and ask him/her to measure the energy of the system which he/she will do w.r.t. the time that he/she is experiencing. The result they will find is that the energy of the system $=M$.

A caveat here is that they must make sure that they are themselves at rest w.r.t. the black hole. There is a wide class of inertial observers at infinity, some (actually, most) of which are moving relative to the black hole. We would like to define mass as the energy of the system at rest. Thus, we must choose our inertial observer so that the momentum that he/she measures is zero. In this frame, the energy that he/she measures will be the mass. When this is done for Schwarzschild, the answer we get is $M$.

A side note

The ADM mass is what we would typically like to call mass of a system, except that it lacks in one respect. An inertial observer at infinity, is not able to measure energy in gravitational or electromagnetic radiation that is emitted. For instance, if the Schwarzschild black hole were to start radiating energy via gravitational waves and eventually disappear, the ADM mass measured by the observer at infinity would still be $M$.

When gravitational radiation is important (for instance, when studying scattering of gravitational waves) for the problem, a more convenient definition of the mass is the Bondi mass $m_B$ which is defined as the mass measured by a Bondi observer at infinity. A Bondi observer is one that moves at the speed of light along null infinity. The Bondi mass is a function of (null) time $m_B(u)$ so that it captures not only the current mass, but also the change in the mass of the system due to radiation.

  1. Let us for simplicity work in units where the speed of light $c=1$ is equal to one, and assume that there is no cosmological constant $\Lambda=0$. A spherically symmetric vacuum solution to the EFE of the form $$\tag{1} ds^2~=~g_{tt}(r)dt^2 + g_{rr}(r)dr^2 +r^2 d\Omega^2,$$ and such that it asymtotically becomes Minkowski space $$\tag{2} -g_{tt}(r\!=\!\infty)~=~ 1~=~g_{rr}(r\!=\!\infty), $$ is then uniquely given by $$\tag{3} -g_{tt}(r)~=~ 1-\frac{R_S}{r} ~=~\frac{1}{g_{rr}(r)}, $$ where $R_S$ is a length parameter, cf. Birkhoff's theorem and this Phys.SE post.

  2. Asymptotically for weak gravitational fields $|\Phi|\ll 1$, it is well-known that we may identify the $tt$-component of the metric $$\tag{4} -g_{tt}~\approx~e^{2\Phi}~\approx~1+2\Phi, \qquad r\to \infty, $$ with the Newtonian potential $$\tag{5}\Phi~=~-\frac{GM}{r},$$ where $M$ is a mass parameter, cf. e.g. Refs. 1 & 2.

  3. Comparing eqs. (3), (4), and (5), we deduce a relation $$\tag{6} R_S~=~ 2GM $$ between the length parameter $R_S$ and the mass parameter $M$.

  4. OP is essentially pondering whether it is possible to upset the previous conclusion (6) by considering a reparametrization $\bar{r}=f(r)$ of the radial coordinate $r$. The answer is No. Birkhoff's theorem and the requirement of asymtotical Minkowski space put too strong conditions on a radial reparametrization.


  1. Sean Carroll, Spacetime and Geometry: An Introduction to General Relativity, 2003.

  2. Sean Carroll, Lecture Notes on General Relativity, Chapter 4. The pdf file is available here.

  • $\begingroup$ But is it possible to actually get the higher-order approximation? I ask because I'm dealing with a situation (the vacuum Levi-Civita metric, or a static cylindrically-symmetric spacetime) where $-g_{tt} = r^a$ for some $a$ and I'm trying to match this with the Newtonian potential $\Phi = k \ln r$ that one gets for an infinite line mass. $\endgroup$
    – user76284
    Aug 3 '19 at 7:25

Is there another way to conclude the Schwarzschild solution has a mass M

It's not so much a conclusion as a definition. From Schutz in "A first course in general relativity", section 8.4 "Newtonian gravitational fields", pages 207 - 208:

Any small body, for example a planet, that falls freely in the relativistic source's gravitational field but stays far away from it will follow the geodesics of the metric, Eq. (8.49), with $\phi$ given by Eq. (8.59). In Ch. 7 we saw the these geodesics obey Kepler's laws for the gravitational field of a body of mass $M$. We therefore define this constant $M$ to be the total mass of the relativistic source.

Notice that this definition is not an integral over the source: we do not add up the masses of its constituent particles. Instead, we simply measure its mass - 'weigh it'- by the orbits it produces in test bodies far away. This definition enables us to write Eq. (8.50) in its form far from any stationary source:

$$\mathrm{d}s^2 = -[1 - 2M/r + O(r^{-2})]\mathrm{d}t^2 + [1 + 2M/r + O(r^{-2})](\mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2)$$

Then, in section 10.4, "The exterior geometry", pages 257 - 258, we have

We therefore see that the the exterior metric has the following form, called the Schwarzschild metric:

$$\mathrm{d}s^2 = -\left(1 - \frac{2M}{r}\right)\mathrm{d}t^2 + \frac{\mathrm{d}r^2}{1 - \frac{2M}{r}} + r^2\mathrm{d}\Omega^2$$

For large $r$, this becomes

$$\mathrm{d}s^2 = -\left(1 - \frac{2M}{r}\right)\mathrm{d}t^2 + \left(1 + \frac{2M}{r}\right)\mathrm{d}r^2 + r^2\mathrm{d}\Omega^2$$

One can find coordinates $(x,y,z)$ such that his becomes

$$\mathrm{d}s^2 = -\left(1 - \frac{2M}{R}\right)\mathrm{d}t^2 + \left(1 + \frac{2M}{R}\right)(\mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2)$$

where $R \equiv (\mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2)^{1/2}$. We see that this is the far-field metric of a star of total mass $M$ (see Eq.(8.60)). This justifies the definition, Eq. (10.28), and the choice of the symbol $M$.

  • $\begingroup$ Your summary of Schutz's discussion seems potentially misleading to me. He's saying that we define it by its distant field, not that it's a totally arbitrary definition. $\endgroup$
    – user4552
    Oct 28 '18 at 21:22

One way to establish that $r$ in Schwarzschild coordinates is equivalent to the spherical radial coordinate is its asymptotic behaviour, namely for $r\to\infty$ the metric tends to Minkowski, and the weak field approximation yields Newton's gravitational potential.

One way to establish that the parameter $M$ is indeed the mass of the spacetime is to calculate the Komar mass: $$M_K \equiv -\frac{1}{8\pi}\int_S \epsilon_{abcd} \nabla^c\xi^d = - \frac{1}{8\pi} \int_S \star d\boldsymbol{\xi},$$ where $\xi^a$ is the timelike Killing vector ($\boldsymbol{\xi}$ denotes the respective form), $\epsilon_{abcd}$ the volume element, $\star$ is the Hodge dual, and $S$ an arbitrary spherical shell that includes the black hole. Let $\alpha \equiv \epsilon^{ab} \epsilon_{abcd}\nabla^c \xi^d = - 2M/r^2$, then the integral becomes, $$M_K = -\frac{1}{8\pi} \int_S \alpha \epsilon_{ab} = -\frac{1}{8\pi} \int_0^{2\pi} \int_0^\pi \alpha r^2 \sin\theta d \theta d \phi = M.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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