---
title: Black-Hole Thermodynamics
module: Black Holes
moduleNumber: 8
lessonNumber: 3
order: 803
summary: >
  The four laws of black-hole mechanics mirror the four laws of thermodynamics
  term for term, with horizon area playing the role of entropy and surface gravity
  the role of temperature. Hawking's calculation makes the analogy literal: a black
  hole radiates at a temperature set by its surface gravity, carries a real entropy
  proportional to its horizon area, and slowly evaporates. The thermal spectrum
  raises the information paradox.
topics: [Black Holes]
draft: false
sources:
  - book: Hartle
    ref: "Gravity, Ch. 13 — Astrophysical Black Holes; §13.3 The Area Theorem and Black-Hole Thermodynamics"
  - book: Carroll
    ref: "Lecture Notes on General Relativity, §9 — Quantum Field Theory in Curved Spacetime"
---

Classically a black hole only grows: matter and radiation cross the horizon, and
the horizon area increases. Hawking's **area theorem** states this as a law — the
total horizon area of a system of black holes never decreases in any classical
process — and its resemblance to the second law of thermodynamics is exact enough
to be more than an analogy. Pushing the parallel through every law, and then
adding quantum fields, turns a black hole into a genuine thermodynamic object with
a temperature, an entropy, and a lifetime.

## The area theorem

For a Kerr–Newman hole of mass $M$, angular momentum $J$, and charge $Q$, the
horizon sits at $r_+$ and its area works out to

$$
A = 4\pi\left(r_+^2 + a^2\right)
  = 4\pi\left[\left(\frac{r_s}{2} + \sqrt{\frac{r_s^2}{4} - a^2 - r_Q^2}\right)^2 + a^2\right],
$$

with $a = J/(Mc)$ and $r_Q$ the charge length of the
[previous lesson](/relativity/black-holes/rotating-and-charged-black-holes). For
Schwarzschild this is $A = 4\pi r_s^2 = 16\pi G^2 M^2/c^4$.

> **Theorem (Area theorem).** In any classical process obeying the null energy
> condition, the total area of a black-hole event horizon is non-decreasing:
> $\d A \ge 0$. When two holes merge, the area of the final horizon exceeds the
> sum of the two initial areas.

The merger case is the sharp one. Two Schwarzschild holes of masses $M_1, M_2$
have total area $\propto M_1^2 + M_2^2$. They cannot simply combine into a hole of
mass $M_1 + M_2$ with area $\propto (M_1+M_2)^2$ and release the rest, because that
final area already exceeds the initial sum. Instead the constraint $\d A \ge 0$
caps the energy radiated as [gravitational waves](/relativity/gravitational-waves/detection-ligo-and-the-first-events):
at most the mass difference allowed by area growth can escape, a bound of order a
few percent of the total mass, matched by the observed merger signals.

$$
% caption: The area theorem in a merger. The final horizon area exceeds the sum of
% the two initial areas, so only the surplus mass can leave as radiation; area, like
% entropy, does not decrease.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % two initial holes
  \draw[black, very thick, fill=black!10] (0,1.4) circle (0.55);
  \draw[black, very thick, fill=black!10] (0,-0.2) circle (0.7);
  \node[black, anchor=east] at (-0.7,1.4) {area A1};
  \node[black, anchor=east] at (-0.85,-0.2) {area A2};
  % arrow
  \draw[black, ->] (1.4,0.6) -- (2.9,0.6);
  \node[black, anchor=south, font=\scriptsize] at (2.15,0.65) {merge};
  % final hole (bigger than sum in radius terms)
  \draw[acc, very thick, fill=acc!12] (4.4,0.6) circle (1.05);
  \node[acc] at (4.4,0.6) {area bigger};
  \node[black, anchor=west, align=left] at (5.6,0.6) {grows past\\A1 plus A2};
\end{tikzpicture}
$$

## The four laws of black-hole mechanics

The area theorem is one entry in a set of four laws that match the four laws of
thermodynamics term for term. The bridge quantity is the **surface gravity**
$\kappa$, the acceleration (measured at infinity) of a static observer hovering
just outside the horizon. For Schwarzschild,

$$
\kappa = \frac{c^4}{4GM} = \frac{c^2}{2 r_s},
$$

and for Kerr–Newman it is a function of $M, J, Q$ that is constant over the
horizon. The four laws read as follows.

- **Zeroth law.** The surface gravity $\kappa$ is constant over the horizon of a
  stationary hole — just as temperature is uniform throughout a body in thermal
  equilibrium.
- **First law.** A change in the hole's parameters obeys
  $$
  \d(Mc^2) = \frac{\kappa}{8\pi G}\,c^2\,\d A + \Omega_H\,\d J + \Phi_H\,\d Q,
  $$
  with $\Omega_H$ the horizon angular velocity and $\Phi_H$ its electrostatic
  potential. The form matches $\d U = T\,\d S + \text{work}$, pairing $\kappa$
  with temperature and $A$ with entropy.
- **Second law.** $\d A \ge 0$: horizon area never decreases, the analogue of
  $\d S \ge 0$.
- **Third law.** $\kappa$ cannot be reduced to zero in a finite sequence of
  operations; the extremal hole ($\kappa = 0$) is unreachable, as absolute zero
  temperature is unreachable.

Read purely classically, the pairing $(T, S) \leftrightarrow (\kappa, A)$ is a
formal coincidence with an embarrassing flaw: a classical black hole has
temperature zero, since it absorbs everything and emits nothing, so no finite $T$
should appear. The analogy demands that the hole radiate.

| Thermodynamics | Black-hole mechanics |
| --- | --- |
| Temperature $T$ uniform in equilibrium | Surface gravity $\kappa$ uniform on horizon |
| $\d U = T\,\d S + \text{work}$ | $\d(Mc^2) = \tfrac{\kappa c^2}{8\pi G}\,\d A + \Omega_H\,\d J + \Phi_H\,\d Q$ |
| Entropy $\d S \ge 0$ | Area $\d A \ge 0$ |
| $T = 0$ unreachable | $\kappa = 0$ (extremal) unreachable |

$$
% caption: The four laws paired term for term. Each thermodynamic law on the left
% matches a black-hole law on the right under the single dictionary that maps
% temperature to surface gravity and entropy to horizon area.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % left column: thermodynamics
  \foreach \y/\t in {3.0/{T uniform}, 2.0/{dU equals T dS}, 1.0/{dS at least 0}, 0.0/{T equals 0 hard}} {
    \node[black, draw=black, minimum width=2.4cm, minimum height=0.7cm, anchor=east] at (0,\y) {\t};
  }
  \node[black, anchor=east] at (0,3.9) {thermodynamics};
  % right column: black-hole mechanics
  \foreach \y/\t in {3.0/{kappa uniform}, 2.0/{dMc2 equals term}, 1.0/{dA at least 0}, 0.0/{kappa 0 hard}} {
    \node[draw=black, very thick, minimum width=2.6cm, minimum height=0.7cm, anchor=west] at (3.4,\y) {\t};
  }
  \node[black, anchor=west] at (3.4,3.9) {horizon mechanics};
  % pairing arrows
  \foreach \y in {3.0,2.0,1.0,0.0} {
    \draw[black, <->] (0.15,\y) -- (3.25,\y);
  }
\end{tikzpicture}
$$

## Hawking radiation

Quantizing a field on the fixed Schwarzschild background removes the flaw.
Hawking found that the collapse geometry mixes the positive- and negative-frequency
modes of a quantum field, so that a state with no incoming particles evolves into
an outgoing state that is thermally populated. The distant observer detects a
blackbody flux at the **Hawking temperature**

$$
T_H = \frac{\hbar\kappa}{2\pi k_B c}
    = \frac{\hbar c^3}{8\pi G M k_B}.
$$

The identification is now literal: $T_H$ really is a temperature, with the same
$\kappa$ that appeared in the first law, and the proportionality constant fixes
the entropy. Matching $\d(Mc^2) = T_H\,\d S$ to the first law gives the
**Bekenstein–Hawking entropy**

$$
S_{\text{BH}} = \frac{k_B c^3}{4 G \hbar}\,A
             = \frac{k_B}{4}\,\frac{A}{\ell_P^2},
\qquad
\ell_P^2 = \frac{G\hbar}{c^3}.
$$

The entropy is one quarter of the horizon area in Planck units — an enormous
number. For a solar-mass hole, $A \approx 1.1\times10^8\,\text{m}^2$ and
$\ell_P^2 \approx 2.6\times10^{-70}\,\text{m}^2$, so
$S_{\text{BH}}/k_B \approx 1\times10^{77}$, dwarfing the thermodynamic entropy of
the star that formed it.

A heuristic picture makes the emission plausible without the full mode
calculation. Vacuum fluctuations near the horizon produce virtual particle pairs;
occasionally one member has negative energy (possible just outside the horizon, as
in the [Penrose process](/relativity/black-holes/rotating-and-charged-black-holes))
and falls in while its positive-energy partner escapes to infinity. The escaping
partners form the thermal flux, and the negative-energy infall lowers the hole's
mass. The picture is only a mnemonic — the real effect is a global property of the
quantum field on the collapse background — but it gives the right sign and the
right qualitative behavior.

$$
% caption: Hawking emission mnemonic. A vacuum fluctuation near the horizon splits;
% the negative-energy partner falls in and lowers the mass while the positive-energy
% partner escapes as thermal radiation.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \fill[black] (0,0) circle (1.3);
  \draw[acc, very thick] (0,0) circle (1.3);
  \node[black, font=\scriptsize] at (0,-0.55) {hole};
  \node[acc, anchor=south, font=\scriptsize] at (0,1.32) {horizon};
  % pair creation dots just outside
  \fill[black!70] (1.3,0.55) circle (1.3pt);
  \fill[black!70] (1.55,0.75) circle (1.3pt);
  % escaping partner
  \draw[black, very thick, ->] (1.55,0.75) -- (3.2,1.6);
  \node[black, anchor=west, font=\scriptsize] at (2.4,1.55) {escapes};
  % infalling partner
  \draw[black, very thick, ->] (1.3,0.55) -- (0.35,0.15);
  \node[black, anchor=north, font=\scriptsize] at (0.9,0.2) {falls in};
  % several outgoing thermal quanta
  \foreach \ang in {20,55,90} {
    \draw[black, ->] ({1.3*cos(\ang)},{1.3*sin(\ang)}) -- ({2.2*cos(\ang)},{2.2*sin(\ang)});
  }
\end{tikzpicture}
$$

## Evaporation

A radiating hole loses mass, and because $T_H \propto 1/M$ the loss accelerates.
Modeling the horizon as a blackbody of area $A \propto M^2$ and temperature
$T_H \propto 1/M$, the Stefan–Boltzmann law gives a luminosity
$L \propto A\,T_H^4 \propto M^2 \cdot M^{-4} = M^{-2}$, so

$$
\frac{\d M}{\d t} \propto -\frac{1}{M^2},
\qquad
M(t)^3 = M_0^3 - K t,
$$

for a constant $K$. The hole shrinks slowly at first and then runs away, emptying
in a finite time

$$
t_{\text{evap}} = \frac{5120\,\pi\,G^2 M_0^3}{\hbar\,c^4}.
$$

The cubic dependence spans an absurd range. The numbers below make the regime
clear: astrophysical holes are colder than the cosmic microwave background and
grow rather than evaporate, while only a hole far lighter than a mountain would
have evaporated within the age of the universe.

| Hole mass | $T_H$ | $t_{\text{evap}}$ |
| --- | --- | --- |
| $1\,M_\odot$ | $6.2\times10^{-8}\,\text{K}$ | $2.1\times10^{67}\,\text{yr}$ |
| $10^{12}\,\text{kg}$ (primordial) | $10^{11}\,\text{K}$ | $\sim$ age of universe |
| $10^{9}\,\text{kg}$ | $10^{14}\,\text{K}$ | $\sim$ a few years |

$$
% caption: Hawking temperature against mass. Because the temperature scales as the
% inverse mass, light holes are hot and evaporate fast while heavy holes are cold and
% effectively stable on cosmic timescales.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (5.8,0) node[right] {mass};
  \draw[->, black] (0,0) -- (0,3.6) node[above] {temperature};
  % 1/M curve via bezier segments
  \draw[acc, very thick] (0.5,3.3) .. controls (0.9,1.6) and (1.5,0.85) .. (2.4,0.55)
     .. controls (3.4,0.28) and (4.6,0.16) .. (5.6,0.12);
  \node[acc, anchor=west] at (2.4,1.0) {T scales as one over M};
  \node[black, anchor=west, font=\scriptsize] at (0.6,3.0) {light and hot};
  \node[black, anchor=south, font=\scriptsize] at (4.6,0.2) {heavy and cold};
\end{tikzpicture}
$$

No astrophysical hole has been observed to evaporate. A stellar-mass hole's
Hawking temperature, $\sim10^{-8}\,\text{K}$, lies far below the $2.7\,\text{K}$
microwave background, so it absorbs far more than it emits and its mass grows.
Evaporation would dominate only after the universe cools below $T_H$, in the
extraordinarily distant future, or for hypothetical primordial holes light enough
to have finished evaporating already.

## The information paradox

The thermal character of the radiation creates a conflict with quantum mechanics.
A blackbody spectrum is fixed entirely by one number, the temperature, hence by
the hole's mass; it carries no other detail. If a hole forms from matter in a
definite quantum state and then evaporates completely into thermal radiation, the
final state is thermal — the same for every initial configuration of the same
mass, spin, and charge.

> **The information paradox.** Unitary quantum evolution maps distinct initial
> states to distinct final states and preserves information. Complete evaporation
> into a thermal spectrum appears to map many distinct initial states to the same
> final state, destroying information. The two statements cannot both hold as
> stated; reconciling them is an open problem at the boundary of general
> relativity and quantum theory.

The no-hair theorem sharpens the tension: the collapse already hid every detail
of the progenitor behind three numbers, and the Bekenstein–Hawking entropy
$S_{\text{BH}} \sim 10^{77} k_B$ counts an enormous number of internal
microstates that the external observer cannot resolve. Whether the information
returns encoded in subtle correlations of the late radiation, remains in a
remnant, or requires a revision of the semiclassical picture is not settled by the
tools of this course. The result stands as the clearest signpost that a full
theory of gravity must be quantum, and it closes the general-relativistic
treatment of black holes: the classical geometry of horizons ends at a question
that only quantum gravity can answer.

[^hartle-thermo]: **Hartle**, _Gravity_, Ch. 13 — astrophysical black holes, the area theorem, the four laws of black-hole mechanics, the Hawking temperature and Bekenstein–Hawking entropy, and evaporation.

[^carroll-hawking]: **Carroll**, _Lecture Notes on General Relativity_, §9 — quantum field theory in curved spacetime, the derivation of the Hawking temperature $T_H = \hbar\kappa/2\pi k_B c$, the entropy $S = k_B A c^3/4G\hbar$, and the information paradox.