What is exactly the density of a black hole and how can it be calculated?

The obvious interpretation of black hole density is the mass of the black hole divided by the volume inside the event horizon. We need to be a bit cautious about taking this too literally because the volume inside the horizon is not coordinate independant so different observers will measure different densities. However we can easily calculate the density measured by the Schwarzschild observer.

The volume inside the event horizon is:

$$ V = \tfrac{4}{3}\pi r_s^3 $$

where $r_s$ is the Schwarzschild radius, so the density is just:

$$ \rho = \frac{M}{V} = \frac{M}{\tfrac{4}{3}\pi r_s^3} $$

The Schwarzschild radius is:

$$ r_s = \frac{2GM}{c^2} $$

Putting this value into the equation for the density and rearranging we get:

$$ \rho = \frac{3c^6}{32 \pi G^3 M^2} $$

So the density is dependent only upon the mass of the black hole, which makes sense because we know that black holes are entirely characterised by their mass, spin and charge.

There are an awful lot of constants in that equation, and it might be a bit easier to grasp if we write it in the form:

$$ \rho \approx 1.85 \times 10^{19} \frac{1}{m^2} $$

where now $m$ is the mass of the black hole in solar masses i.e. units where $1$ means the same mass as the Sun. With this equation we can see immediately that a black hole with the same mass as the Sun would have the (enormously high) density of $1.85 \times 10^{19}$ kg/m$^3$. Alternatively, a super supermassive black hole with the mass of 4.3 billion Suns would have a density equal to one i.e. the same density as water.


Black holes are really hard to get a density. Basically, they are so dense that there is no known mechanism for providing sufficient outward force to counterbalance the inward pull of gravity, so they will collapse into an infinitesimally small size. Of course, that doesn't seem likely, it seems likely there is something that will keep the volume from being 0, but it is extremely dense.

An alternative method of measuring the volume of a black hole is to take the radius beyond which light can't escape, also commonly known as the Event horizon. Wikipedia has a great article on potential black hole sizes and masses, using the event horizon. Here's a few example values:

Stellar black hole: mass = 2$\times$10$^{31}$ kg, volume = 3.4$\times$10$^{12}$ m$^3$. The density would then be mass/volume, or 6$\times$10$^{18}$ kg/m$^3$.

Galactic sized: Mass is 2$\times$10$^{39}$ kg, volume= 10$^{37}$ m$^3$, density= 200 kg/m$^3$.

It seems that the larger they are, the less dense they would be, but only if you consider the event horizon as the limit. Of course, we don't know what is beyond an event horizon, so...


The density of a black hole is not a well-defined thing. Depending on what you mean by density, and what kind of black hole you're talking about, the "density" can be zero, infinity, or anything in between.

A Schwarzschild black hole is a vacuum solution to the Einstein field equations, meaning that this type of black hole spacetime consists of nothing but empty space, everywhere. Therefore in this sense a black hole's density can be zero.

Real astrophysical black holes have to form by gravitational collapse, and the ones we observe also seem to be accreting additional matter at some rate. The density of the infalling matter is quite low, probably comparable to a pretty good laboratory vacuum on earth. As this matter approaches the singularity, you might think that it would get compressed, but actually that's not the case. The Einstein field equations say that when you start with a cloud of particles with some size, and let it free-fall through a vacuum, it always maintains constant volume. This is basically a statement that gravitational fields in a vacuum are tidal forces. Infalling objects don't get squished, they get spaghettified.

There can be singularities in general relativity, called strong curvature singularities, that do compress infalling matter infinitely, and it's possible that a black hole's singularity is a strong curvature singularity during its initial formation -- but we don't really know.

In this type of discussion, you will often hear people say that the singularity has zero volume, so the density of the singularity must be infinite. Not so. The volume of the singularity is not well defined, basically because the machinery for measuring the sizes of things breaks down at the singularity -- that's pretty much the definition of a singularity. In fact, we can't even define how many spatial dimensions a black-hole singularity has, so we shouldn't even think of it as a point.

The event horizon of a black hole does have a definable size. When we try to define things like this in general relativity, it gets tricky because GR doesn't have a preferred set of coordinates. However, the area of a black hole is well defined and coordinate independent. So we can certainly take the mass of a black hole and divide by its area, and get a meaningful result, but this doesn't have units of density. For a sphere in Euclidean space, it's a matter of trivial algebra to find its volume in terms of its area. But the space inside a black hole is not at all Euclidean. It isn't even static, so the volume depends on the choice of spacelike surface. A paper by Christodoulou and Rovelli, "How big is a black hole?," argues that the volume is in some sense time-dependent and can be many, many orders of magnitude larger than the Euclidean value --- in their analysis, it diverges to infinity for $t\rightarrow\infty$. So we could try to divide the black hole's mass by the volume and get some sort of average density, but it wouldn't be a well-defined, finite number.