<p>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.</p>

<p>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<span class="math-container">$^{*}$</span>.</p>

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

<p>So, if you have a black hole of mass <span class="math-container">$1.4 \times 10^{8} M_{\odot}$</span>, then it has the "density" of water if you calculate it like this.</p>

<p>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 <span class="math-container">$10^{8} M_{\odot}$</span>, so fits the bill perfectly.</p>

<ul>
<li>Note that it is a vast simplification (and incorrect) to use <span class="math-container">$4\pi r_s^3/3$</span> as the volume inside the event horizon. There is in fact no uniquely defined volume; it depends on choice of coordinate system (see <a href="https://arxiv.org/abs/0801.1734" rel="noreferrer">DiNunno &amp; Matzner 2008</a>).</li>
</ul>

CC BY-SA 4.0
2020-01-31T17:51:23.320Z
Thus demonstrating why we shouldn't depend on <x> (mean of x).  So,  what's the median density of this black hole?  :-)
Carl Witthoft
CC BY-SA 4.0
2020-02-01T02:56:46.043Z
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).
PM 2Ring
CC BY-SA 4.0
2020-02-01T07:42:28.510Z
@PM2Ring indeed I did. Hence the careful wording I used.
ProfRob
CC BY-SA 4.0
2020-01-31T16:11:43.543Z
2020-02-01T07:58:06.853Z
2020-02-01T07:58:06.853Z
ProfRob
<p>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. </p>

<p>Now indeed, the schwarzschild radius of such a black hole would be
<span class="math-container">$R = c\sqrt{\frac{3}{8 \pi G \rho}}$</span>
with <span class="math-container">$\rho$</span> the density of water. </p>

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

CC BY-SA 4.0
2020-01-31T16:00:53.157Z
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.)
antlersoft
<black-hole><astrophysics>
Can a black hole with the density of water exist?
usernumber
# Can a black hole with the density of water exist?

(Asked by: username_0 on 2020-01-31)

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?

**Tags:** black-hole, astrophysics
> **Comments:**
> **username_1**: 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.)  

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## 1 answers

### Answer by username_2 on 2020-01-31. Score: 10 | Accepted answer

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](https://arxiv.org/abs/0801.1734)).

> **Comments:**
> **username_3**: Thus demonstrating why we shouldn't depend on <x> (mean of x).  So,  what's the median density of this black hole?  :-)  
> **username_4**: 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).  
> **username_2**: @PM2Ring indeed I did. Hence the careful wording I used.  

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