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

How do scientists calculate that density? What data do they have to calculate that?

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Density of what exactly? The event horizon? But that's not a "real" thing, it's merely a limit where things happen. Density of the central singularity? In what time frame? Also, current science gives you an infinite result for that one, meaning the theory is incomplete. There are no good answers here, because the question doesn't make a whole lot of sense. – Florin Andrei Mar 15 '12 at 15:20

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...

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It helps to specify that you're talking about the mean density of the black hole. Like you say, it doesn't really make sense to talk about the "actual" density because (a) GR implies collapse to a point of infinite density and (b) we don't have a quantum theory to replace GR, although that might describe what actually happens. And the mean density can still be helpful. – Warrick Mar 15 '12 at 8:08
@Warrick: GR doesn't imply collapse to a point of infinite density, it implies an end for the infalling matter in the symmetric nonrotating collapse case. The singularity is not a spatial point of infinite density, it is a terminus for the infalling geodesics. The only meaning to a black hole density is the ratio of mass to the cube of the Schwarzschild radius. – Ron Maimon Nov 2 '12 at 17:25
What are you using for the volume here? The result you get depends on what spatial section you choose to measure. And a natural choice $\frac{4}{3}\pi r_s^3$ (for Schwarzschild radius $r_s$) isn't really the volume of anything. – Holographer May 13 '15 at 14:04

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:

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

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

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.

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We can't tell how matter behaves inside a black hole. I can think of at least several solutions, but there is no way to either confirm or deny them.

I'd say its most likely matter forms a sphere inside the event horizon equal to the radii of the black hole. Considering physics (as we know it) don't break down inside the black hole, matter can't travel faster than c and time is infinitely extended.

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A black hole is a celestial body of extreme density and high gravitational pull that not reflect or emit radiation.

The process of forming a black hole is related to the evolution of some stars. As you know, a star of similar mass to the Sun ends up becoming a white dwarf, a small star with high density.

The explosion of a nova leaves behind a new star of enormous density and small volume with a diameter not exceeding 10 km., Consisting solely of neutrons.

moreover, the density of a black hole should not be the same for all, because each has a different size depending on the original mass of the collapsed star. but that there should be no doubt that this density is very high.

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maybe this article will serve you: arxiv.org/abs/astro-ph/0612659 – jormansandoval Mar 19 '12 at 10:34

There might be no full-fledged theory of quantum gravity, but we can speculate a little on results from whatever the true theory is. Quantizing gravity usually implies quantizing spacetime- in other words, the entire universe is grainy. It is likely that you can pack no more than about one Planck mass into each Planck volume, i.e. cubic Planck length. This works out to 5.1555e96 kg/m^3. The implication of this calculation is that all black holes will have roughly the same density, and will simply increase in real volume with increasing mass.

I know I've mentioned this on another question, but I can't find it right now.

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-1: No it doesn't. What do you mean by "same density"? The "density" of the planck scale smoothing of the singularity? This answer is wrong. – Ron Maimon Nov 2 '12 at 17:27

protected by Qmechanic♦May 13 '15 at 16:40

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