# What size is the smallest black hole stable at Standard temperature & pressure?

I have seen several questions regarding the size of the absolute smallest black hole, the smallest stable black hole and similar. These made me wonder; what is the smallest stable black hole if it is immersed in a standard atmosphere?

I assume there is some maximum rate at which the black hole can consume matter, and that this is related to its size. Therefore, since its Hawking radiation is inversely related to its size, logically there should be an unstable equilibrium point when these two equal each other. I could, of course, be completely wrong - however I'm afraid my knowledge of how to calculate anything regarding black holes is limited to working out the Schwarzschild radius.

The rate of decrease of mass $$m$$ of a black hole due to Hawking radiation goes as

$$\frac{dm}{dt} = -\frac{\hbar c^4}{15360 \pi G^2} \frac{1}{m^2} \,,$$

where the symbols have their usual meaning. If you assume that the mass growth is totally due to (isotropic) accretion of baryonic matter or cold-dark matter, then the mass growth takes the form (see Bondi-Hoyle accretion)

$$\frac{dm}{dt} = \frac{4 \pi G^2 m^2 \rho_{\infty}} {\left(c_{\infty}^2+v_{\infty}^2\right)^{3 / 2}} \,,$$

where $$\rho_\infty$$ is the density of the medium a large distance from the black hole, $$v_\infty$$ is the speed at which the black hole movies w.r.t. the static medium and $$c_\infty$$ is the speed of the sound in the medium.

A black hole would be stable against Hawking evaporation if the second equation dominates the first one. To find the answer you were looking, you should equate the above two equations. This will give

$$m = \left[ \frac{\hbar c^4}{15360} \frac{\left(c_{\infty}^2+v_{\infty}^2\right)^{3 / 2}} {4 \pi^2 G^4 \rho_{\infty}} \right]^{1/4}$$

You can assume $$v_{\infty} = 0$$ but would need to know the other values of the ambient medium.

For the medium you asked about, air at STP, the values are: $$c_{\infty} = 331 m/s$$, and $$\rho_{\infty} = 1.29 kg/m^{3}$$.

Using these numbers results in a black hole with the following mass and Schwarzschild radius: $$m = 3.8×10^{10} kg$$ $$r_{s} = 5.6×10^{-17} m$$

However, as noted in the comments, this Schwarzschild radius is too small for the gas to be treated as a perfect fluid. A more accurate value of the mass limit in this medium would require an equation other than the standard Bondi-Hoyle equation, which accounts for the quantum-mechanical nature of gas particles at that scale.

• This is a great answer with lots of useful ideas but I came up with some issues. For $v_\infty=0$ this evaluates to $4\times10^{10}$ kg (using the density/speed of sound in air). This gives a radius of $6\times 10^{-17}$ m. Being much smaller than the mean free path in air, this makes the Bondi-Hoyle accretion formula invalid. It's so small that one would need to treat collisions between the black hole and an incoming air molecule quantum mechanically (I don't think it requires quantum gravity because the black hole could be treated classically). It just requires QM in a curved spacetime. Commented Sep 8, 2023 at 8:23
• @AXensen I think since the mass of the black hole is still $\approx 10^{10}$ kg so the black hole would have a tiny de Broglie wavelength to be classified as a quantum mechanical system. Also, the infalling particle won't be able to tell the size of the black hole when it is far away (Birkoff's theorem). But once it gets really close, then it will experience strong tidal forces.
– S.G
Commented Sep 8, 2023 at 11:34
• I wasn't intending to treat the black hole quantum mechanically, I was intending to treat an air molecule that might be swallowed by the black hole quantum mechanically. You dont need quantum gravity - the black hole is classical because its mass is well above the Planck scale. The only thing I'd want to do is calculate the cross section for an incoming air molecule to be brought into the black hole. For that calculation, of course the full Schwarzschild solution will be needed. Commented Sep 8, 2023 at 11:50
• @AXensen I see, thank you!
– S.G
Commented Sep 8, 2023 at 11:52
• Thank you for the very useful answer. So at STP and assuming no quantum effects on the air particles; the evaporation limit will be a black hole with mass 3.8×10¹⁰kg. The radius of its event horizon will be 5.6×10⁻¹⁷m. @S_G would you mind if I edited these values onto the end of your answer? Commented Sep 13, 2023 at 22:16