# How can super massive black holes have a lower density than water?

I'm new here, so go easy!

I heard on a podcast recently that the supermassive black holes at the centre of some galaxies could have densities less than water, so in theory, they could float on the substance they were gobbling up... can someone explain how something with such mass could float?

http://www.universetoday.com/83204/podcast-supermassive-black-holes/

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Another way of sort of saying the same thing is that if you think of gravity as a flux, as would be suitable for a 1/r^2 gravity theory like Newton's, then there's no limit to the flux you can obtain no matter the lack of density of the mass you make it from; just make it big enough. – Carl Brannen Feb 26 '11 at 0:29
Floatation makes no sense in this context.. A supermassive black hole has more grav arrtn than the earth.	So the blach hole will stay put at half the height of the fluid. – Manishearth♦ Feb 13 '12 at 15:45

## 4 Answers

Well, it can't (float), since a Black Hole is not a solid object that has any kind of surface.

When someone says that a super massive black hole has less density than water, one probably means that since the density goes like $\frac{M}{R^3}$ where M is the mass and R is the typical size of the object, then for a black hole the typical size is the Schwarzschild radius which is $2M$, which gives for the density the result

$$\rho\propto M^{-2}$$

You can see from that, that for very massive black holes you can get very small densities (all these are in units where the mass is also expressed in meters). But that doesn’t mean anything, since the Black Hole doesn’t have a surface at the Schwarzschild radius. It is just curved empty space.

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You should be clear that "it can't" refers to floating, not to having a low density. – dmckee♦ Feb 25 '11 at 17:23

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As I’ve pointed out to your answer, since there is no surface for the BH and the matter is freefalling in the BH, there is no pressure. Thus there is no floating in any conventional way. The only thing that matters in that case is the total momentum transfer from the fluid, which will be pointing on the same way as the exterior field. It is a wrong, bad analogy. Sorry. – Vagelford Sep 3 '11 at 22:35

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Free falling does not mean no pressure! It means that there is no pushing on the black hole itself. The pressure accelerates the infalling water, which enters with a greater kinetic energy (and up-momentum) at the bottom than at the top. The difference is exactly the bouyancy force in the fluid, as can be seen by analying the flow of up-momentum in a big sphere surrounding the black hole. – Ron Maimon Sep 3 '11 at 22:51

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@vagelford: the whole construction is not oversimplified, it is not misleading, and it can be done in principle with a viscous enough liquid pool and a small enough black hole. The pool has to be in gravity or accelarated, to give it a pressure gradient. A black hole falling through such a pool (in gravity or accelerated) will absorb fluid in such a way that it feels a bouyancy force. The bouyancy force is the same as any other floating/sinking object, because it is determined by conservation laws. – Ron Maimon Sep 5 '11 at 14:28

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@Vagelford: I did not make a mistake. – Ron Maimon Sep 6 '11 at 1:16
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The black hole would float in water, if you could make a large enough pool to submerge it, and with enough replenishment to replace the water that the black hole will sucks up. The black hole will remove water from its surroundings, but the water below will come into the horizon at higher pressure than the water above, so the velocity inward will not be uniform.

If the black hole is denser than water, it will sink for a while, because the pressure difference is not enough to compensate for the pull of gravity. If the black hole has less density than water, it will float. It's like a balloon that sucks in water and expands, always maintaining a volume which is big enough to keep itself lighter than water.

The problem is that when the black hole density is as that of water, a volume of water equal to the black hole's volume will not be stable to gravitational collapse, so it will be impossible to set up the pool.

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Well, you are forgetting the effect of the self-gravity of the black hole which in the case of a balloon submerged in a fluid is negligible. In order to have the setting you are proposing the water and the BH would have to be embedded in an exterior, lets say uniform, gravitational field which would also define the difference between up and down. The first thing to point out is that in the free falling matter in the BH the important thing is the fluids kinetic energy and not the pressure. So, “under” the BH the gravitational pull towards the BH would be smaller relative to “over” the BH. – Vagelford Sep 3 '11 at 22:26
Thus the flow coming from the “up side” would have grater kinetic energy and thus it would excerpt a grater dynamic pressure (transfer more momentum to the BH). Whatever the case, it is a bad and misleading analogy the whole “floating” thing. – Vagelford Sep 3 '11 at 22:27
Ignoring the issue that the water would collapse under gravity, which invalidates the whole thing, the greater pressure below ensures that the water below will enter with a greater kinetic energy, and this is the reason for "floating". The water at the top will enter at a lower velocity, so there will be a total momentum transfer which is exactly as for any other submerged object. The problem with the density of water being unsustainable might be fixable by going to a more viscous fluid which will ooze into the black hole slowly, so you could have slow sucking up of material. – Ron Maimon Sep 3 '11 at 22:49
There is no hydrostatic pressure difference in freefall. – Vagelford Sep 4 '11 at 9:33

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@Vagelford: it's not funny. The buoyancy force on a region is determined by conservation laws on a big sphere far away from the horizon, where the water is essentially stationary. On such a sphere, the total force is equal to the mass of displaced water, and the total mass is less than this because of the black hole's lower density. There is no way around it, because it is determined by conservation laws. – Ron Maimon Sep 7 '11 at 16:37
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The Schwarzschild radius scales with mass as $r~=~2GM/c^2$. What might be defined as a Schwarzschild volume would then be $V~=~4\pi r^3/3$ $=~(32/3)\pi(GM/c^2)^3$. So the density of matter defined by the horizon is $\rho~=~(3/32)(c^2/G)^3M^{-2}$. So density scales as the inverse square of the mass. A 10 billion solar mass black hole has a radius about $10^{10}$km, or a volume $V~\sim~10^{30}km^3$ $=~10^{39}m^3$. A solar mass is $10^{30}$kg and the density defined by the horizon is then $\rho~=~10^{-9}kg/m^3$. That is actually quite small.

Of course if you fall into a black hole of any mass you encounter a region with enormous Weyl curvature and tidal forces. The source of this is in your future, and eventually you reach it --- it is inescapable. This region where curvature diverges is a spatial surface of infinite extent.

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You should add some closing [/math] tags 'Sothedensityofmatterdefine...' yeah after reading that two times I could decipher it :) – datenwolf Feb 25 '11 at 17:31

I think it is actually misleading to make the claim that is puzzling you. "Density" suggests that the mass is distributed more or less uniformly within the black hole, and this is non-sense. The black hole is mostly empty, and all the mass is concentrated within a tiny region (clasically a point) in the center of the black hole.

If you ignore this and pretend a black hole of mass $M$ and volume $V\propto r^3$ had a uniform density $\rho$ then you can calculate it, simply using $\rho=M/V$. Since for Schwarzschild black holes the radius of the black hole is proportional to its mass you obtain finally $\rho\propto 1/M^2$, so the heavier a black hole the smaller its density. But again, this provides a highly misleading picture of the mass distribution within the black hole. All its mass is in the center, so classically the density is infinite.

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