A Dyson sphere around a black hole

The temperature T associated with a black hole's Hawking radiation goes as 1/M, where M is the black hole's mass, i.e. it decreases with M. Meanwhile, its surface area A goes as M². Plugging this into the Stefan-Boltzmann law yields a total power output P of

P ~ A · T⁴ ~ 1/M²

So while a larger total amount of energy can be extracted from a supermassive black hole (because it's more massive – duh), it also takes much (much!) longer to extract a given amount of energy: A supermassive black hole's mass is in the order of 10⁶ to 10⁹ solar masses, i.e. 10⁶ to 10⁹ times the mass of a stellar-mass black hole, meaning that its power is just 10^-12 to 10^-18 × the power of a stellar mass black hole.

Cool! Thanks for deriving the equation. A lot about super massive black holes that become unintuitive at that scale. For example, I understand that the black hole at the center of the Milky Way has a lower density than iron.

> I understand that the black hole at the center of the Milky Way has a lower density than iron.

That doesn't sound like it could be true. Black holes are densest possible objects in the universe. For instance, take a piece of iron of mass M = 1kg. Compress it down till it's diameter is smaller than twice the Schwarzschild radius associated with this mass (which is 2GM/c² = 1.4· 10^(-27) meters, where G is the gravitational constant and c is the speed of light[0]). You now have a black hole – one which has the same mass as your original piece of iron but which occupies a much smaller space than that piece of iron did.

For a supermassive black hole like Sagittarius A* (4.3·10⁶ solar masses, radius 11.25·10⁹ m) one obtains a density of volume / mass ~ 4.5·10⁶ kg/m³ – clearly this is much higher than the density of any iron that you can find on the surface of Earth.

Side note: Terms like "density" (= mass divided by volume) and "volume" don't really make sense in the context of black holes – the region "behind" the event horizon (the "inside" of the black hole) is not of "space" type (in physics lingo: it's not a 3-dimensional spacelike hypersurface). Instead, it's a 4-dimensional region of spacetime, meaning that it includes a time direction. It makes little sense to talk about the volume of such a thing (let alone of the density of mass "spread out" in such a spacetime).

But even more crucially, the region of spacetime behind the event horizon contains spacelike hypersurfaces of infinite volume. So it makes even less sense to assign a volume to it.

What people usually mean when they refer to the volume (and, by extension, the density) is the following: It's (roughly speaking) the volume you have to compress a mass down to for it to become a black hole. More precisely, for an observer at infinity, at any given instant of time a (static, spherically symmetric) black hole's event horizon appears to be a sphere of surface area 4π·(2GM/c²)², i.e. radius r = 2GM/c² [1]. Now take that radius and plug it into the usual formula for the volume of a ball of radius r. You have now defined the "volume" for a (static, spherically symmetric) black hole.

[0] https://www.wolframalpha.com/input/?i=schwarzschild+radius+1...

[1] This is really the definition of the radius. This is because there is no center of the black hole on the "inside", so in particular it doesn't make sense to talk about the radius (= distance of the black hole's spherical surface to its "center").