temperature - Why do larger black holes emit less Hawking Radiation than smaller black holes?

Pedestrian question from a non-physicist:

I read on Wikipedia that larger black holes emit less net Hawking radiation than smaller black holes.

This seems counterintuitive to me. If black holes are essentially sucking in mass and converting it to Hawking radiation, why wouldn't a more massive black hole emit more Hawking radiation?

Moreover, how is it that Hawking radiation can escape black holes in the first place when visible light (which I know to be another form of radiation) cannot?

Answer

Here is an answer more based on thermodynamics. It all boils down to this: The entropy of a black hole is proportional to its surface area, but its mass = energy to the radius.

You can think of it in the following way: all states inside the black hole are the same to observers outside. Information about the interior is encoded on the surface because you can solve boundary value problems. So the larger the surface, the larger the entropy. This is far from a proof, but you can make this rigorous.

Now, what are heat and temperature? Heat is the spontaneous transfer of energy, without work being done. The second law of thermodynamics says that entropy can never decrease. Since energy is conserved, for two systems in equilibrium, $$\frac{\Delta S_1}{\Delta U} = \frac{\Delta S_2}{\Delta U}$$ that is, the the change in the entropy $S_1$ of system 1 due to changing its energy by $\Delta U$ is equal to the change in entropy $S_2$ of system 2 due to changing its energy by the same amount. Hence we define $$\beta = \frac{\partial S}{\partial U}$$ and you can realize that $\beta = 1/T$ where $T$ is the temperature. (Heat flows from low $\beta$ to high $\beta$, but from high $T$ to low $T$.)

From general relativity, we have that the mass $M$ and radius $r$ of a black hole are proportional, $r = 2GM/c^2$. If we take the energy to be $U = E = Mc^2$, then $U \propto r$, so $S \propto U^{2}$ which gives $\beta \propto U $ or $T \propto 1/r \propto 1/M$. Thus, a smaller black hole is hotter, and consequently radiates more.

If we invert the relation between entropy and energy, we get $U \propto \sqrt S$. Plot $\sqrt S$. It is very steep for small $S$, and quite flat for large $S$. That means for small black holes, the Hawking radiation only needs to have a little entropy, even if the intensity is large, for the process to be allowed by thermodynamics.

You can learn more from these talks by Hawking himself. Hawking uses some technical concepts, but I hope they are fairly accessible. (They are, at least, much more so than the original papers.)