# Why is a black hole's Hawking radiation temperature inversely proportional to its mass/size?

As a black hole shrinks in volume and mass, shouldn't its temperature get lower? Shouldn't it evaporate more slowly?

Naively, (very naively), I think that with a smaller surface area (as per its event horizon), it would have fewer and fewer random quantum foam particles popping up just outside its surface....

## 2 Answers

In a sense, you're right. Vacuum energy over a surface area is proportional to the surface area so larger area means more of the particle-antiparticle occurrences and that's the source of Hawking radiation, that is, if you don't mind the virtual particle explanation, which is the only explanation I sort of understand. I look at the equations and my brain goes blank.

But surface area isn't the only factor. The more important factor is how quickly the gravitational field drops off, which, for stellar mass black holes or larger, the drop off over particle-anti particle distance is tiny, given a radius of at least a few miles and a very small particle-anti particle distance or high probability of a small distance. That small drop-off in gravitational field leaves little chance for the outermost of the two virtual particles to escape.

That's why the only black holes that are thought to be hot are theoretical micro black holes, which are tiny. The smaller the theoretical black hole, the faster it's gravitation falls with distance, so, even small distances like those between a particle and an anti-particle, make it much easier for one of the particles to escape the gravity, effectively stealing energy from the black hole and radiating outwards.

Þe olde goode dimensional analysis can help.

The Hawking radiation is essentially a quantum field effect of the same nature as the Unruh effect. Quantum fields “live” in a distorted space-time of the hole and do not need to know anything about gravitation. Like for the Unruh effect (having $$c/a$$ as its characteristic time), there is a natural time scale for this distorted space-time – the proper time since crossing the event horizon until reaching singularity for an in-falling observer having zero initial speed (at the infinite distance from the hole). Alternatively (up to reasonably small factors), one may think of the period of light’s circular orbit around the hole. By multiplication by $$c$$, it can be thought of as characteristic size – the hole’s size (Schwarzschild radius, size of the event horizon, or so). Frequencies of the Hawking radiation are of the same order as inverse characteristic time and respective wavelengths are of the same order as characteristic size. Whereas energy of a quantum is proportional to the frequency and inversely proportional to the wavelength. It goes from quantum physics.

Schwarzschild radius is proportional to the hole’s “mass”. It goes from general relativity.

• This answer really just shifts the question to "why is Hawking radiation of the same wavelength as the hole's size?" It's not at all obvious that this must be true just from dimensional analysis. For example, why can't the energy of the quanta just be proportional to the mass of the black hole? That would lead to exactly the opposite conclusion. – Michael Seifert Dec 22 '20 at 23:11
• @Michael Seifert: edited. As for “why can't the energy of the quanta just be proportional to the mass of the black hole”, Ī̲ see two possible implementations of such conjecture. Firstly, you can assume that the energy has the same order as the hole’s “mass”. It predicts near-instant shrinking of black holes which egregiously contradicts to observations. Secondly, a small dimensionless factor can be postulated, but such factor should have some grounds and and it is your onus to invent substantiation for it, not mine. – Incnis Mrsi Dec 23 '20 at 9:35