February 23, 2021. A scandalously heuristic “derivation” of Bekenstein and Hawking’s expression for black hole entropy.

Introduction

Deriving thermodynamic properties of black holes is hard unless you cheat. One approach is to use dimensional analysis (see e.g. here) but even that can get a bit tedious. In this post, I’ll provide a streamlined, scandalously heuristic approach, similar in spirit to the way the Hawking temperature can be motivated by considering virtual particles near the horizon. Without resorting to dimensional analysis, we can throw around equations with sneaky abandon, starting with the Schwarzschild radius and ending with the mysterious formula for Bekenstein-Hawking entropy.

Sizing up a black hole

Suppose a star of mass $M$ runs out of nuclear fuel and collapses to form a black hole. How big is it? A running theme in this post will be different ways of estimating the energy stored in the black hole, and to start with, we invoke the most famous formula in physics:

\[E = Mc^2,\]

where $c = 3 \times 10^8 \text{ m/s}$ is the speed of light. This is how much mass-energy the black hole has, according to Einstein’s gem. We will equate this to the gravitational energy stored in the black hole. Recall Newton’s inverse square law of gravity,

\[F_\text{grav} = \frac{GMm}{r^2},\]

for masses $M$, $m$ separated by distance $r$, and Newton’s constant $G = 6.67 \times 10^{-11}$ (in SI units). By setting $m = M$, we can estimate the force the black hole applies to itself:

\[F_\text{grav} \sim \frac{GM^2}{R^2}.\]

Applying a force over a distance gives energy (work), so we estimate the gravitational energy stored in the black hole is

\[E \sim F_\text{grav} R \sim \frac{GM^2}{R}.\]

If we equate this with mass-energy, we find the famous Schwarzschild radius $R$:

\[Mc^2 \sim \frac{GM^2}{R} \quad \Longrightarrow \quad R \sim \frac{GM}{c^2}.\]

This is exactly what we find in general relativity up to a factor of two, $R = 2GM/c^2$. Just to give a sense of how small this is, the sun weights $M_\odot = 2 \times 10^{30} \text{ kg}$, so if it collapsed into a black hole, it would have a Schwarzschild radius of

\[R = \frac{2GM_\odot}{c^2} = \frac{2(6.67\times 10^{-11})(2 \times 10^{30})}{(3\times 10^8)^2} \text{ m} \approx 3 \text{ km}.\]

It would fit inside the LHC!

Fiat flux

Although a black hole traps any light that falls inside, Stephen Hawking made the remarkable discovery that just outside the boundary, black holes emit faint radiation, just like a hot lump of coal. Once again, we will rather slapdash, and try and work out only the wavelength of a typical photon coming out of the black hole. It seems reasonable to assume that, like the fundamental frequency of a violin or a flute, the black hole likes to make photons with wavelengths proportional to the Schwarzschild radius $R$.

This has various implications. Most importantly, the hotter an object is, the smaller the wavelength of the typical photons it produces. This explains why a very hot coal is orange or yellow, and becomes red (a longer wavelength of light) as it cools. The relationship is governed by something called Wien’s law:

\[\lambda \sim \frac{\hbar c}{kT},\]

where $\hbar = 1.05 \times 10^{-34}$ is Planck’s (reduced) constant, and $k = 1.38 \times 10^{-23}$ is Boltzmann’s constant (both in SI units). For a motivation from dimensional analysis, see this post. The key point is that wavelengths get shorter as temperature rises, and the rest of the constants are added to ensure things make dimensional sense. Setting $\lambda = R$ and making $T$ the subject, we obtain the Hawking temperature of a black hole:

\[\lambda \sim \frac{\hbar c}{kT} \sim \frac{GM}{c^2} \quad \Longrightarrow \quad T \sim \frac{\hbar c^3}{GMk}.\]

One of the weird things about black holes is that temperature is inversely proportional to mass $M$, as we can see directly here. As it gets smaller, it gets hotter! Before moving on, let’s calculate the temperature of our solar mass black hole. Since we already know it has a radius of around $R \sim 3$ km, the temperature from Wien’s law is

\[T \sim \frac{\hbar c}{kR} = \frac{(1.05 \times 10^{-34})(3 \times 10^8)}{(1.38 \times 10^{-23})(3 \times 10^3)} \text{ K} \sim 10^{-6} \text{ K}.\]

This is a million times colder than empty space!

Atoms of spacetime

We’ve measured the energy in a black hole using Einstein’s mass-energy formula, and Newton’s formula for gravitation. We’ve said that a black hole is hot, so let’s calculate the energy a third way, in terms of heat. People often say that the temperature of the substance measures the amount of “molecular motion”, or kinetic energy per molecule. This sounds fuzzy, but can be written as a perfectly rigorous equation called the equipartition theorem:

\[\epsilon_\text{avg} \sim kT \quad \Longrightarrow \quad E \sim N kT,\]

where $\epsilon_\text{avg}$ is the average energy per particle, and $N$ is the total number of particles, which we multiply by to get the total energy $E$. Basically, hot systems are democratic, and try to spread their energy evenly between their constituents. Since we have already estimated the energy in a black hole, we can use equipartition to estimate the total number of particles. Using the mass-energy for instance, we have

\[E = Mc^2 \sim Nk T \sim \frac{N\hbar c}{R} \quad \Longrightarrow \quad N \sim \frac{RMc}{\hbar} \sim \frac{R^2c^3}{G\hbar}.\]

This is a very strange result. The first thing to notice is that the number of particles is proportional to $R^2$, and hence the surface area $4\pi R^2$ of the black hole. Usually, the number of particles is proportional to the volume of a material, not the surface area, so something weird appears to be happening. The second interesting thing is that we are dividing the surface area by a particular combination of constants

\[A_P = \frac{G \hbar}{c^3}\]

called the Planck area. It is the square of the Planck length $\ell_P = \sqrt{G\hbar/c^3}$. For various reasons, this can be regarded as the smallest length that makes sense when quantum gravity effects are taken into account. For a sun-sized black hole, the number of particles is

\[N \sim \frac{R^2c^3}{G\hbar} = \frac{(3 \times 10^3)^2(3 \times 10^8)^3}{(6.67 \times 10^{-11}) (1.05 \times 10^{-34})} \sim 10^{76}.\]

This is a stupendously large number! A typical galaxy is estimated to have around $10^{67}$ atoms, and a solar mass black hole seems to have a billion times more. How can a black hole have so many more atoms than the object it collapses from? The answer is that the nature of the atom is different! A black hole is not made from the atoms you find in chemistry textbooks. Instead, these are “Planck atoms”, somehow the indivisible units of spacetime itself.

Bekenstein-Hawking entropy

Of course, this raises the question: what the heck is a spacetime atom?! We know from chemistry textbooks how regular atoms combine and create matter. Trying to understand how Planck atoms combine and create spacetime is much harder! But there are a few demystifying statements we can make. Suppose each Planck atom can be in $f$ states. If there are $N$ Planck atoms in our black hole, then the total number of configurations (assuming all are allowed) $\Omega$ is

\[\Omega = f \times f \times \cdots \times f = f^N,\]

since each atom gets $f$ options. The entropy is just a convenient measure of the number of ways to be, conveniently defined by Ludwig Boltzmann as

\[S = k \log \Omega =Nk \log f,\]

where $\log$ is the logarithm to the base $e$. This is often talked about as a measure of disorder, since there are fewer ways to be highly ordered than disordered. As an example, if we now demand that all the Planck atoms have the same state, there are only $f$ different ways to be! We can connect this to thermodynamics as follows. Suppose we change the number of particles by $\Delta N$. Then by definition, the entropy changes as

\[\Delta S = k \log f \Delta N.\]

From the equipartition theorem, the total system energy changes as

\[\Delta E \sim kT\Delta N.\]

Keeping the temperature $T$ fixed, and ignoring numbers involving $f$, this yields

\[\Delta S \sim \frac{\Delta E}{T}.\]

With a bit more care, this can be turned into a definition of entropy, but the point is that there is a microscopic view of entropy (ways to be) and a macroscopic view (change in energy divided by temperature). Either way, the entropy of a black hole is

\[S \sim Nk \sim \frac{k A}{A_P},\]

up to numerical constants, where $A$ is the black hole horizon area and $A_P$ is the Planck area. This result was worked out by Jacob Bekenstein and Stephen Hawking in the 70s, so it’s called the Bekenstein-Hawking entropy. This is the starting point for exploring the holographic principle and other wonders of quantum gravity, but I must leave that for another time!