Does the density of a black hole decrease after releasing a large amount of Hawking radiation?

Question

Is is true that the smaller a 'regular', stellar-mass black hole is, the denser it is inside of its event horizon? After all, if you look up (or calculate) the Schwarzschild radii of the Sun and the Earth, the Earth's resulting (theoretical) black hole is much denser than the Sun's.

I have also read that if you cross the event horizon of a supermassive black hole, you would not be immediately 'spaghettified', because of the low densities on the outskirts of those black holes.

In New Scientist magazine, on page 49 of the new Oct. 2 issue, 'Curious craters' by Jonathan O'Callaghan, it says:

"This is a result of the (primordial, Hawking-radiation-reduced) black hole being a million times denser than the moon and passing clean through without slowing down."

Unless Hawking radiation lowers the density, not just the mass, a tiny black hole with an event horizon less (far less) than a millimeter is going to be a septillion times or more denser than the moon, not just a million?

Answer 1

The smaller a 'regular', stellar-mass black hole is, the denser it is inside of its event horizon, correct?

Yes, that's correct. But the black hole density is a mean density, that is, it's just the mass of the BH divided by its Schwarzschild volume. It's not like there's a uniform ball of stuff inside the event horizon with that density.

I have also read that if you cross the event horizon of a supermassive black hole, you would not be immediately 'spaghetti-fied', because of the low densities on the outskirts of those black holes....

A BH is mostly empty space. Anything that crosses the event horizon rapidly falls towards the centre. As I mentioned in this answer:

We cannot say exactly what happens at the core of a black hole. As discussed in this answer by Florin, we need a theory that unites General Relativity and Quantum mechanics to answer such questions.

A pure GR black hole has a mathematical singularity at its core, but most astrophysicists believe that's unphysical, and that a proper quantum gravity theory will eliminate that singularity. However, it's likely that the core of a black hole is still very small, since the quantum gravity corrections probably don't become significant until the size gets smaller than an atom. 

So if a BH isn't actively accreting material, the density in its outskirts, either side of the event horizon, is zero. However, a BH with high density will spaghettify you before you cross the event horizon. You get spaghettified by the tidal force caused by the differential gravitational acceleration: if you're falling feet first towards the BH, the acceleration at your feet is greater than the acceleration at your head, so it feels like you're being pulled apart. A big BH actually has more tidal force than a small BH at the same radial distance (i.e., the distance to the centre of the BH), but a SMBH has such a large Schwarzschild radius that you have to be well within the event horizon before the tidal force is dangerous.

There are more details about spaghettification in this Physics answer by Andrew Steane. And here's a short live Python program that calculates the radial distance where a 1 m steel rod (falling vertically) would be pulled apart by the tidal force.

Eg, with a 3 $M_\odot$ BH, which has a Schwarzschild radius $r_s$ of 8.860 km, the steel rod would break at a distance of 112.176 km = 12.661353 $r_s$.

A tiny black hole with an event horizon less (far less) than a millimeter is going to be a septillion times or more denser than the moon, not just a million, correct?

Correct. BH mass is proportional to $r_s$, so the mean density is inversely proportional to the square of the mass or radius. Eg, a 1 mm BH has a million times the density of a 1 m BH.

The relevant equation for the mean density is $$\rho = \frac{3c^2}{8\pi G{r_s}^2}$$

and the Schwarzschild radius is $$r_s = \frac{2GM}{c^2}$$

You can get BH density & various other parameters of a Schwarzschild BH from the Hawking radiation calculator. FWIW, $r_s$ for the Earth is ~8.870056 mm.

Unless Hawking radiation lowers the density, not just the mass....

No, Hawking radiation doesn't do that. As the BH loses mass, it continues to obey the above equation. It might do something odd when it has lost almost all of its mass, but we need a quantum gravity theory for that.

Answer 2

Let me add an addendum to PM 2Ring's excellent answer.

Why does the New Scientist article (apparently) say that a low-mass primordial black hole is "a million times denser than the moon"? It's because it's not really referring to actual black holes, but rather to the fake black hole used in a simulation.

From reading the Yalinowich & Caplan (2021) paper that the New Scientist article is evidently based on (the non-paywalled part of the New Scientist article that I could read mentions Matthew Caplan as one of the two authors), the authors are attempting to suggest how the crater caused by a small black hole impacting the Moon might differ from a normal crater caused by an asteroid or comet. The basic idea is that an ordinary asteroid/comet will destroy itself in an explosion on the surface, spraying out lunar debris in a particular pattern. A small black hole, on the other hand, will punch through the Moon, creating a kind of continuous explosion along its passage through the Moon; the ejecta escaping from this tunnel should have a different pattern. (The idea behind the "continuous explosion" is not spelled out; it's presumably the idea that the black hole will gravitationally attract nearby lunar atoms, which will form a hot, dense accretion zone near the black hole as they try to fall into it; this will produce pressure and radiation that makes for the continuous explosion.)

As a simple test of their (simple) calculation, they used a hydrodynamical simulation code that one of the authors wrote, and they did two simulations for comparison. The first (meant to simulate an ordinary object hitting the Moon) involved a sphere of gas traveling at high velocity (the "impactor") hitting a large layer of gas with the same density (the "Moon"); they looked at how the ejecta from the impact initially traveled away from the impact site. (Since there was no gravity in the simulation, they weren't trying to simulate the resulting crater.)

For the "black hole" case, they did the exact same thing, except this time the impactor was a very compact object with a million times the density of the "Moon" (i.e., the large layer of gas). It's important to understand that this "million times denser" impactor has nothing to do with actual black holes; it's just a convenient (and very, very crude) approximation for use in their simulation.