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?

  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/80214/2451 $\endgroup$
    – Qmechanic
    Mar 14 '17 at 11:10
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    $\begingroup$ They do not emit less radiation - it's just the ratio between mass and emission that's lower. This shouldn't be too surprising, it's everywhere in nature (e.g. bigger animals are generally more energy efficient than smaller animals (all else equal), but they still waste more energy total). $\endgroup$
    – Luaan
    Mar 14 '17 at 17:23

The Newtonian gravitational acceleration for an object of mass $M$ is given by the well known expression:

$$ a = \frac{GM}{r^2} \tag{1} $$

And the radius of the event horizon of a black hole is given by:

$$ r_s = \frac{2GM}{c^2} \tag{2} $$

Suppose we calculate the Newtonian gravitational acceleration at the event horizon. Let's not worry whether this is physically realistic for now, we'll just go ahead and do it anyway. If we rearrange equation (2) to get:

$$ GM = \frac{c^2r_s}{2} $$

then we can substitute for $GM$ in equation (1) to get:

$$ a(r_s) = \frac{c^2}{2r_s} \tag{3} $$

And what we find is that the gravitational acceleration at the event horizon is proportional to $1/r_s$ so smaller black holes have a higher surface acceleration than larger black holes. The temperature of the Hawking radiation is related to this surface acceleration. which makes intuitive sense. If the gravitational field is stronger you'd expect the Hawking radiation to be stronger. And that means:

Smaller black holes are hotter than larger black holes

To make this rigorous requires considerable work so I won't go into the details. We define a property called the surface gravity, $\kappa$, that is effectively the gravitational acceleration at the event horizon and we find for a static black hole this is given by:

$$ \kappa = \frac{1}{2r_s} \tag{4} $$

This is the proper general relativistic version of the Newtonian surface gravity I derived in equation (3). The Hawking temperature is then simply:

$$ T_H = \frac{\hbar c}{2\pi k_B} \kappa $$

So just as with our approximate calculation we find that the temperature is proportional to $1/r_s$ i.e. smaller black holes are hotter than larger black holes.


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.)


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?

Well, to begin do you know what black body radiation is?

Simply put, a black body is one that absorbs all light that it is subjected to and when such a body reaches a certain temperature $T$ will proceed to emit radiation somewhere in the electromagnetic spectrum.

Now, classically, a black hole will certainly absorb all the light that it is subjected to. However, it does not necessarily reach a temperature in which it can emit radiation? So, can we actually classify a black hole as a black body?

Well, black body radiation is a quantum effect. Thanks to Planck and his famous constant. Next, consider quantum fluctuations and the uncertainty affiliated with QM which is discussed in great detail here.

The negative energy quantum fluctuations associated with the quantum electromagnetic field are harmless and often disappear just as quick as they appear. However, if one appears before the event horizon of a black hole it may cross it. Once over the event horizon it is possible for the photon to have negative kinetic energy ( As a side note, this can lead to a forbidden region and allow for the possibility of quantum tunnelling which is discussed in Maulik K. Parikh and Frank Wilczek). The associated positive energy photon is left on the outside of the horizon which will continue moving outwards.

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?

It is important to note that nothing actually passes from in and out of the horizon it is more of the fact that the negative energy photon makes a negative contribution to the mass of the black hole. This contribution is the Hawking radiation.

Now, if you accept then that black holes radiate with the positive photon partner then the answer to your next question follows suit by simply applying some thermodynamics to black hole physics.

The formula for the luminosity of black hole radiation is inversely proportional to the square of the mass of the black hole and hence will become smaller as the mass increases. Some references for formulae are given here and here.

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    $\begingroup$ Why do we need to have certain temperature $T$ to emit radiation? Any objects with temperature greater than zero Kelvin emits electromagnetic radiation $\endgroup$
    – Paul
    Mar 14 '17 at 17:20
  • $\begingroup$ @Sherlock: formally, T = 0. But in practice, for a temperature sufficiently close to 0, the wavelength of the radiation is so large that we cannot measure it, and the heat transfer is also negligible. Also, if the temperature is below the CMB's 2.97 K, the black hole experiences the universe as "hot". $\endgroup$
    – Wtrmute
    Mar 14 '17 at 20:43
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    $\begingroup$ -1: the "negative energy particle falls into the black hole" description, while an infectious and widespread meme, doesn't actually correspond to anything in Hawking's calculation, and talk about quantum fluctuations and virtual particles "appearing and disappearing" doesn't actually correspond to anything in a QFT. $\endgroup$ Mar 15 '17 at 3:01
  • $\begingroup$ I stand corrected! Thanks for the comment. Well then what about viewing Hawking radiation as a tunnelling process? Like the reference that I have mentioned above. As far as I am aware this is the view that is taken there. $\endgroup$ Mar 15 '17 at 3:12
  • $\begingroup$ Could you add more detail as to how Hawking radiation escapes a black hole? I would love to make sure I understand what y'all are describing properly. By the way, thank you everyone for the super detailed answers! $\endgroup$ Mar 16 '17 at 7:18

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