Confused about black holes and Hawking radiation

[serp777]

Why is it that the smaller the black hole is, the more quickly it supposedly evaporates? It seems like a black hole should radiate energy proportional to the surface area of the event horizon. To me it seems like the evaporation should slow down the more the black hole shrinks because the energy radiated per meter^2 of the event horizon is constant (Maybe that's wrong but i don't know why). My impression was that the the consensus was that small black holes evaporate very quickly and I'm just wondering what logic/evidence that's based on. Thanks for your time and let me know if I am just totally wrong in my understanding.

 

[Demystifier]

Why is it that the smaller the black hole is, the more quickly it supposedly evaporates? It seems like a black hole should radiate energy proportional to the surface area of the event horizon. To me it seems like the evaporation should slow down the more the black hole shrinks because the energy radiated per meter^2 of the event horizon is constant (Maybe that's wrong but i don't know why). My impression was that the the consensus was that small black holes evaporate very quickly and I'm just wondering what logic/evidence that's based on. Thanks for your time and let me know if I am just totally wrong in my understanding.

You are right that radiation power is proportional to the surface area. However, the radiation power increases also with the temperature, and it happens that smaller black hole has a larger temperature.

Let me be more quantitative. The black body with surface area $A$ and temperature $T$ radiates with the power proportional to $AT^4$. In the case of black hole we have $A\propto R^2$, $T\propto 1/R$ (where $R$ is the black hole radius), from which you can see that $AT^4\propto 1/R^2$.

 

[serp777]

You are right that radiation power is proportional to the surface area. However, the radiation power increases also with the temperature, and it happens that smaller black hole has a larger temperature.

Let me be more quantitative. The black body with surface area $A$ and temperature $T$ radiates with the power proportional to $AT^4$. In the case of black hole we have $A\propto R^2$, $T\propto 1/R$ (where $R$ is the black hole radius), from which you can see that $AT^4\propto 1/R^2$.

Thanks, but black holes don't really have a temperature right, and doesn't this black body equation only relate to a regular object that emits light? The black hole itself might have temperature but the event horizon is not a black body in the traditional sense because its a boundary where the escape velocity is equal to the speed of light rather than the surface of an object composed of atoms.

 

[Demystifier]

Thanks, but black holes don't really have a temperature right, and doesn't this black body equation only relate to a regular object that emits light?

I am not sure what do you mean by "really" having a temperature. Anyway, even if the black hole does not "really" have a temperature in some sense, here it is sufficient to know that it radiates as if it had one.

 

[accursedCursive]

You are right that radiation power is proportional to the surface area. However, the radiation power increases also with the temperature, and it happens that smaller black hole has a larger temperature.

Let me be more quantitative. The black body with surface area $A$ and temperature $T$ radiates with the power proportional to $AT^4$. In the case of black hole we have $A\propto R^2$, $T\propto 1/R$ (where $R$ is the black hole radius), from which you can see that $AT^4\propto 1/R^2$.

Seems like circular logic.
Your explanation uses T∝1/R as an axiom, but OP's question is basically asking why T∝1/R holds true.

 

[Nugatory]

Seems like circular logic.
Your explanation uses T∝1/R as an axiom, but OP's question is basically asking why T∝1/R holds true.

Hmmm... I read the original post ("It seems like a black hole should radiate energy proportional to the surface area of the event horizon") as having overlooked the relevance of the temperature altogether.

In any case, the inverse relationship between radius and temperature is derived in Hawking's papers - this one is conveniently not behind a paywall but might be a bit much for an I-level thread.

 

[Demystifier]

Seems like circular logic.
Your explanation uses T∝1/R as an axiom, but OP's question is basically asking why T∝1/R holds true.

First, as Nugatory said, T∝1/R is not an axiom but can be derived. Second, it seems quite clear to me that OP did not ask why T∝1/R holds true. But if he did, let me give a very simple heuristic derivation taken from Zee's "Einstein Gravity in a Nutshell".

In fact, the temperature of the radiation, known as the Hawking temperature $T_H$ of the
black hole, can be estimated by using dimensional analysis. You may be puzzled, since
there are two masses in the problem, the mass $M$ of the black hole and the Planck mass
$M_P$. With two masses, any function of $M/M_P$ is admissible, and so dimensional analysis
appears to be inapplicable. Indeed, we need one more piece of information. The key is that
Newton’s constant $G$ is a multiplicative measure of the strength of gravity. In Einstein’s
theory as well as in Newton’s, the gravitational field around an object of mass $M$ can only
depend on the combination of $GM$. Let us now set $c$ and $\hbar$ (but not $G$) to 1. The combination
$GM$ is a length and hence an inverse mass. On the other hand, Boltzmann and the
founding fathers of statistical mechanics had long ago revealed to us that temperature,
a highly mysterious concept at one time, is merely the average energy of the microscopic
constituents of macroscopic matter. Hence temperature has the dimensions of energy, that
is, of a mass in units with $c = 1$.

It follows immediately that $T_H\sim 1/GM$ . This “sophisticated” dimensional analysis cap-
tures an essential piece of physics: the radiation is explosive! As the black hole radiates
energy, $M$ goes down and $T_H$ goes up, and thus the black hole radiates faster. The radiative
mass loss accelerates. Certainly not something you want to see in the kitchen: an object
that gets hotter as it loses energy.

Zee expresses everything it terms of the black-hole mass $M$, but it should not be too difficult to see that it is related to the black-hole radius $R$ as $R=2GM$.