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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? Please see the below link for the podcast in question... http://www.universetoday.com/83204/podcast-supermassive-black-holes/ |
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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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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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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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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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