# Density of a black hole

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[Mentors' note: split off from this thread]
Do all black holes have the same density?

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George Jones
Staff Emeritus
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I am not sure what you mean by "density". The following may or may not be relevant. It certainly is not rigourous..

As mass increases, the density of matter necessary to form a black hole decreases. If fact, if enough stars are used, they don't even have to touch for a black hole to form. In other words the stars have to be close together, but there still can be space between them. Below, I calculate a quantity that I'll call density, but, in reality, the quantity is only suggestive of density.

Setting this "density" to the average density of the sun, about 1400 kg/m^3, gives a black hole mass of about 100 million solar masses. So, if more than 100 million or so (within an order of magnitude) sunlike stars congregate in the centre of a galaxy, they don't have to touch (initially) to form a black hole.

The following calculation is only suggestive, and it is in no way rigorous. Because of the curvature and nature of spacetime, it probably doesn't make sense to calculate the spatial volume inside the event horizon of a black hole.

Density is mass over volume, i.e.,

$$\rho = \frac{M}{V},$$

and the volume of a spherical object of radius $R$ is given by $4\pi R^3/3$, so the density of a uniform sphere is

$$\rho = \frac{3M}{4\pi R^3}.$$

A spherical black hole has event horizon (boundary) located at

$$R = \frac{2GM}{c^2},$$

where $G$ is Newton's gravitational constant and $c$ is the speed of light.

Subsituting this equation into the density equation for a spherical black hole gives

$$\rho = \frac{3c^6}{32\pi G^3} \frac{1}{M^2}.$$

The first bit is just a constant number, while the second bit shows that the "density" of a spherical black rapidly decreases as mass increases.

Inverting this equation gives

$$M = \frac{c^3}{4}\sqrt{\frac{3}{2\pi G^3}}\sqrt{\frac{1}{\rho}},$$

and using the Sun's density for $\rho$ gives the result I mentioned at the top.

Density is a derived quantity, not a fundamental one, equal to mass divided by volume. The volume of a singularity is not defined, so neither is the density. It just isn't a useful concept for black holes.

Nugatory
Mentor
Short answer: No, as you can see just by looking at the relationship between the mass and the Schwarzschild radius. Doubling the mass doubles the "radius" so increases the "volume" eight-fold, commensurately reducing the "density".

Longer answer: Those scare-quotes are in the short answer for a reason. The volume of a black hole, and hence its density, aren't especially well defined. The first and most obvious problem is that the mass is not distributed evenly inside the black hole; as far as GR will tell us (and there's no reason to believe this particular prediction) it is concentrated in a point of zero volume at the singularity in the "center" so there's no physical significance to any density you might calculate. Second, the Schwarzschild radius is not the length of a line segment between the central singularity and the event horizon (in fact, there is no such thing) so you can't just plug it into the formula for the volume of a sphere.

Usually when someone speaks of the "density" of a black hole, they mean the ratio of the mass to the volume of a hypothetical sphere whose surface area is equal to the surface area of the event horizon. And if that's what you mean by the density, you can go with the short answer above: the bigger the black hole, the lower the density.

PAllen