# Black Hole Temperature and Entropy

1. Dec 22, 2007

### George Jones

Staff Emeritus
A given body of knowledge can often be presented in a number of orders. What follows is one order for black hole temperature and entropy.

The relative intensities of the different parts of the spectrum of a hot body depends on its temperature. For example, the colour of a glowing metal (think stove element) depends on its temperature. The spectrum of a radiating black hole is (almost) thermal, so relative intensities give the temperture of a black hole.

Small black holes are hotter than large black holes because tidal forces are larger near the event horizon of a small black hole than they are near the horizon of a large black hole. The temperature $T$, as measured by an observer far from the black hole, of a black hole is inversely proportional to its mass $M$, with
$$T = \frac{\hbar c^3}{8 \pi k G M},$$
where $k$ is Boltzmann's constant.

The Stefan-Boltzmann law for radiation by black bodies is
$$P = \sigma A T^4,$$
where $P$ is the power (energy per unit time) radiated, $\sigma$ is Stefan's constant, and $A$ is the radiating surface area.

Putting these together gives that a black hole radiates power according to
$$P = \frac{dE}{dt} = \sigma A \left( \frac{\hbar c^3}{8 \pi k G M} \right)^4.$$
Both of Wald's technical books say that the Stefan-Boltzmann formula holds only approximately. A more accurate calculation sums over the modes radiated. As the temperature of black hole increases, species of particles with higher and higher masses come into play, as do the effects of the particles' energies and masses on curvature. I think Page, starting in the mid-seventies, published realistic calculations.

Assume that Stefan-Boltzmann holds.

The r-coordinate in Schwarzschild spacetime is defined such that for $r > 2GM/c^2$, the proper surface area of a sphere of constant $r$ is given by $A = 4 \pi r^2.$

Assume that for a black hole $A = 4 \pi r_S^2,$ where $r_S = 2GM/c^2$ is the Schwarzschild radius. (Wald, however, says that, due to light bending, the effective radius of a radiating black hole is $3^{3/2} GM/c^2.$)

Using this in the above radiation radiation formula for black holes gives
$$P = \frac{dE}{dt} = \sigma 4 \pi \left( \frac{2GM}{c^2} \right)^2 \left( \frac{\hbar c^3}{8 \pi k G M} \right)^4.$$
Now, the rate at which the black hole emits energy equals the rate at which the black hole loses mass times $c^2$. Thus, after simplifying,
$$\frac{dM}{dt} c^2 = -\frac{\sigma c^{8}\hbar ^{4}}{256\pi ^{3}k^{4}G^{2}}\frac{1}{M^{2}}$$
Suppose that the black hole has initial mass $M$, and that after a time $\tau$ the black hole has evaporated completely. After rearranging the above, this is expressed in terms of integrals as
$$\begin{equation*} \begin{split} \int_M^0 m^2 dm &= - \alpha \int_0^\tau dt,\\ \end{split} \end{equation*}$$
which gives
$$\tau = \frac{1}{3\alpha} M^3,$$
where
$$\alpha = \frac{\sigma c^{6}\hbar ^{4}}{256\pi ^{3}k^{4}G^{2}}.$$
The lifetime of a black hole is proportional to the cube of its mass, so realistically sized black holes (even ignoring infalling radiation and matter) take a long, long time to evaporate.

If I have done the calculations correctly, this gives
$$\tau = 10^{76} \left( \frac{M}{M_\mathrm{Sun}} \right)^3 \mathrm{ seconds} = 10^{68} \left( \frac{M}{M_\mathrm{Sun}} \right)^3 \mathrm{ years}$$
Carroll says the real calculation, which he doesn't show, gives
$$\tau = 10^{71} \left( \frac{M}{M_\mathrm{Sun}} \right)^3 \mathrm{ seconds}$$
The nice thing about this calculation is that it can be included as an application of the Stefan- Boltzmann formula in calculus-based first and second-year physics courses.

Students are already familiar with $E=mc^2$, and easily buy that the "radius" of a black hole is proportional to its mass. The inverse relationship between temperature and mass can be explained by tidal effects.

The whole presentation takes, at most, one lecture. However, finding space for an extra lecture in this already packed course can be difficult.

The thermal spectrum and non-zero temperature of a radiating black hole suggest a connection with thermodynamics. Consider the "surface area" (limit of proper surface areas as $r \rightarrow r_S$) of a spherical black hole:
$$\begin{equation*} \begin{split} A &= 4 \pi r_S^2 \\ &= 4 \pi \left( \frac{2GM}{c^2} \right)^2 \\ &= \frac{16 \pi G^2}{c^4} M^2 \\ dA &= \frac{32 \pi G^2}{c^4} M dM. \\ \end{split} \end{equation*}$$
Using this with the energy content $E=Mc^2$ of a black hole gives
$$\begin{equation*} \begin{split} dE &= c^2 dM\\ &= \frac{c^6}{32 \pi G^2 M} dA\\ &= \frac{\hbar c^3}{8 \pi k G M} \frac{c^3 k}{4 \hbar G} dA\\ &= T \frac{c^3 k}{4 \hbar G} dA. \end{split} \end{equation*}$$
This together with the first law of thermodynamics with no work terms, $dE=TdS$, suggest the identification of the entropy of a black hole with its surface area:
$$S=\frac{c^3 k}{4 \hbar G} A.$$
Historically, results happened in a different order.

Last edited by a moderator: Apr 23, 2017
2. Dec 22, 2007

### marcus

That's a nice exposition and it would, as you say, make an interesting lecture to illustrate thermodynamics, in a first or second year general physics course. key formulas not to complicated, and it is an impressive illustration of how far reaching things like Stefan-Boltzmann radiation law can be. Special thanks for adding it to the PF Astro stewpot!
Something we can link to. (as I think you already may have done.)

3. Jan 2, 2008

### Garth

Hi George,

I second those thanks from marcus, however is there an explanation for the above discrepancy of 105?

Or is it simply that Carroll has got it wrong?

Garth

4. Jan 5, 2008

### George Jones

Staff Emeritus
At first I thought Carroll was giving the result of a more careful calculation, but after looking more carefully at Carroll and at Wald's General Relativity (which gives the same number), I realized that this isn't so.

Wald gives essentially the same argument that I did, but he neglects all constants by writing proportionality statements instead of equalities, with final result

$$\tau \sim M^3.$$

He gives his argument in units such that $c = G =\hbar =1$. Conventional units are restored by writing

$$\tau_\mathrm{Sun} \sim \left( \frac{M_\mathrm{Sun}}{m_P} \right)^3 t_P \medspace \mathrm{seconds},$$

where $m_P$ and $t_P$ are the Planck mass and time.

Taking $M_\mathrm{Sun}= 1.989 \times 10^{30} \medspace \mathrm{kg}$, $m_P = 2.177 \times 10^{-8} \medspace \mathrm{kg}$, and $t_P = 5.391 \times 10^{-44} \medspace \mathrm{s}$ gives their result.

Last edited: Oct 28, 2008
5. Jan 9, 2008

### Kurdt

Staff Emeritus
Thats a nice approach George. In a normal ThermoD class these things would be approached by analogy.

6. Jan 31, 2008

### pseudo

can you give me an expression for the emission of hawking radiation from a blackhole

Last edited by a moderator: Apr 23, 2017