20
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Florin Andrei Mar 15 '12 at 15:20
14
$\begingroup$

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...

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Warrick Mar 15 '12 at 8:08
  • 5
    $\begingroup$ @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. $\endgroup$ – Ron Maimon Nov 2 '12 at 17:25
  • $\begingroup$ 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. $\endgroup$ – Holographer May 13 '15 at 14:04
  • $\begingroup$ This is not a very good answer. It neglects the fact that the volume inside the event horizon is not well defined, and it also assumes, incorrectly, that the singularity has zero volume. $\endgroup$ – Ben Crowell Jun 20 at 18:46
16
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ So basically if you were to have several chunks of some heavier-than-iron material (so that it won't start to fuse), which (in total) weigh 4.3 bln solar masses, and merge them all, the total object would become less dense? Or in fact if you merge two 2bln solar mass black holes, the total volume of the resulting black hole would more than double the original size? $\endgroup$ – Joeytje50 Jun 5 '16 at 19:57
  • $\begingroup$ @Joeytje50: yes, but note my warnings about the interpretation of the density. The event horizon is not a physical object - if you were falling through it you wouldn't even notice it was there. So defining the density by the volume inside the horizon doesn't have any special physical significance. $\endgroup$ – John Rennie Jun 6 '16 at 4:05
  • $\begingroup$ This is the correct answer since, obviously, you actually answered the question. $\endgroup$ – CoolHandLouis May 4 '17 at 3:49
  • $\begingroup$ First you say that the volume is coordinate-dependent, which is correct, but then you try to calculate it using the Euclidean formula for the volume of a sphere. That's just wrong. The spatial geometry is not Euclidean. And there is no such thing as "the Schwarzschild observer" in the sense you seem to imply. $\endgroup$ – Ben Crowell Jun 20 at 19:15
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The paper you quote defines the inner volume as "large" for a real BH that hasn't existed forever. Using the same argument, as recently mentioned, the inner volume of an eternal Schwarzschild BH is infinite. The outer BH volume is $\int_{r_{min}}^{r_s}\sqrt{g}d^3x$ where $g=\frac{r^4\sin^2{\theta}}{1-r_s/r}$ is the determinant of the matrix of the Schwarzschild metric components for a constant time slice of $dt=0$. Because the horizon is lightlike, there exists no such Schwarzschild slice, in which $r<r_s$. Thus $r_{min}=r_s$ and the volume is zero ($g$ is singular, but integrable at $r=r_s$). $\endgroup$ – safesphere Jun 21 at 7:47
  • $\begingroup$ The outer volume in special coordinates (like Kruskal-Szekeres and others) is not zero, but it has no physical meaning, because such coordinates don't represent time and space of any real observer. For any real external observer the volume of any black hole is zero, because the horizon is lightlike $\endgroup$ – safesphere Jun 21 at 7:48
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ This is very short and vague and doesn't answer the question. $\endgroup$ – Ben Crowell Jun 20 at 18:44
-1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ -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. $\endgroup$ – Ron Maimon Nov 2 '12 at 17:27

protected by Qmechanic May 13 '15 at 16:40

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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