4
$\begingroup$

During an outreach event, the speaker mentioned the fact that black holes weren't necessarily dense objects and that black holes of any density could exist, even black holes with the density of water.

Now indeed, the schwarzschild radius of such a black hole would be $R = c\sqrt{\frac{3}{8 \pi G \rho}}$ with $\rho$ the density of water.

But could such a non-dense black hole exist in practice? If not, what would prevent it from forming or from being stable?

$\endgroup$
  • $\begingroup$ If you just do the math you can find the mass of a Schwarzchild black hole with the density of water. I'm too lazy, but I believe it is in the range of known super-massive black holes (so it not only can exist, but some have been observed.) $\endgroup$ – antlersoft Jan 31 at 16:00
9
$\begingroup$

Applying a numerical density to a black hole isn't possible. The material inside the event horizon will fall to a "singularity" (or some other ultrahigh density state that we currently have no adequate theory to describe) on a relatively short timescale.

What you can do, is exactly what you have done, which is divide the gravitational mass of a black hole by the volume defined by a simple Euclidean estimate using the Schwarzschild radius$^{*}$.

If instead of substituting out the mass as you have done, you instead substitute out the Schwarzschild radius, then $$ M = \left(\frac{3}{4\pi \rho}\right)^{1/2} \left(\frac{c^2}{2G}\right)^{3/2} = 1.4\times10^{8} \left(\frac{\rho}{\rho_w}\right)^{-1/2} \ M_{\odot},$$ where $\rho_w = 1000$ kg/m$^3$.

So, if you have a black hole of mass $1.4 \times 10^{8} M_{\odot}$, then it has the "density" of water if you calculate it like this.

There are lots of black holes with this mass or even higher that are situated in the centres of galaxies. The black hole at the centre of the Andromeda galaxy has a mass of about $10^{8} M_{\odot}$, so fits the bill perfectly.

  • Note that it is a vast simplification (and incorrect) to use $4\pi r_s^3/3$ as the volume inside the event horizon. There is in fact no uniquely defined volume; it depends on choice of coordinate system (see DiNunno & Matzner 2008).
| improve this answer | |
$\endgroup$
  • $\begingroup$ Thus demonstrating why we shouldn't depend on <x> (mean of x). So, what's the median density of this black hole? :-) $\endgroup$ – Carl Witthoft Jan 31 at 17:51
  • 2
    $\begingroup$ I'd also like to mention that the Schwarzschild radius $r_s$ is defined in terms of the circumference of the EH (event horizon), but it's not correct to think of $r_s$ as measuring the radius of a Euclidean sphere (i.e., the distance from the EH to the "singularity"), since space in that region isn't flat, and it's also dubious to use it with the standard Euclidean formula for the volume of a sphere to calculate the volume inside the EH. (I'm sure you know this Rob, I'm just mentioning it for the benefit of other readers). $\endgroup$ – PM 2Ring Feb 1 at 2:56
  • 1
    $\begingroup$ @PM2Ring indeed I did. Hence the careful wording I used. $\endgroup$ – Rob Jeffries Feb 1 at 7:42

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.