Singularities and black-hole thermodynamics

22.1 From one falling observer to a cloud of them#

A geodesic describes one freely falling observer. Geodesic deviation describes the changing separation of nearby observers. A congruence is a smooth family of worldlines filling a region without crossing there. Think of an enormous cloud of tiny spacecraft, each with its engines off, carrying rulers to monitor its neighbors.

The cloud can expand, distort, and rotate. Those are different effects. A sphere becoming a larger sphere expands. A sphere becoming a same-volume ellipsoid shears. An ellipsoid turning without changing shape rotates.

Use $c=1$ through the focusing discussion and let the congruence have unit tangent $u^\mu$, with $u^\mu u_\mu=-1$. Its local rest-space metric is

Define the projected velocity-gradient tensor

It tells neighboring observers how their relative velocity depends, to first order, on their spatial separation. Split this three-dimensional linear map into its trace, symmetric traceless part, and antisymmetric part:

Explicitly,

The equality between the projected trace and $\nabla_\mu u^\mu$ follows from differentiating the fixed normalization of $u$.

$\theta$ is expansion, $\sigma$ is shear, and $\omega$ is vorticity or twist. For an infinitesimal comoving volume $\mathcal V$,

The trace is the fractional volume-growth rate because the first-order fractional change of a determinant is the trace of the underlying linear deformation. That is the same determinant identity that appeared when varying $\sqrt{-g}$.

There is also a direct bridge to geodesic deviation. For a commuting family of geodesics with separation vector $\xi$, $[u,\xi]=0$ implies

In a transported rest frame, this is “separation velocity equals $B$ times separation.” Differentiating again and using the geodesic-deviation equation gives a matrix evolution law with a term $-B^2$ and a curvature term. Raychaudhuri’s equation is its trace.

22.2 Deriving the Raychaudhuri equation#

Start with the expansion and differentiate along the flow:

Commute the derivatives, using the Riemann convention of this book. The contraction contributes a minus Ricci term. Then apply the product rule:

The first term is $\nabla_\mu a^\mu$, where $a^\mu=\nabla_u u^\mu$. It vanishes for a geodesic congruence. The second is the trace of the square of the velocity-gradient map. For geodesic flow its rest-space decomposition gives

Why the vorticity minus sign? A real antisymmetric matrix squares to a matrix with nonpositive trace. In two dimensions, the rotation generator

squares to $-w^2I$. The trace-free symmetric matrix has a positive sum of squared eigenvalues. Cross terms vanish because a symmetric tensor contracted with an antisymmetric tensor is zero, and shear has zero trace.

Therefore, for a timelike geodesic congruence in four spacetime dimensions,

For a normalized accelerated congruence, add $+\nabla_\mu a^\mu$ to this expression, with the same projected definitions. Rockets can alter the cloud’s behavior; geodesic focusing theorems do not silently include thrust.

The terms describe distinct contributions:

• $-\theta^2/3$: convergence can reinforce itself even without curvature.
• $-\sigma^2$: stretching along some directions can accelerate volume focusing.
• $+\omega^2$: rotation opposes the simple focusing tendency.
• $-R_{\mu\nu}u^\mu u^\nu$: curvature directly changes the trace of relative acceleration.

The Weyl tensor is absent from the explicit last term, but it can generate shear, which then affects expansion through $-\sigma^2$. Vacuum curvature can matter enormously even when $R_{\mu\nu}=0$.

22.3 How an inequality becomes a finite-time prediction#

Suppose the geodesics are orthogonal to a family of spacelike slices. Locally their covector is then $u_\mu=-N\partial_\mu t$, where $t$ labels the slices and $N$ normalizes $u$. Antisymmetrizing its derivative cancels the second derivatives of $t$. The terms left contain a factor $\partial_\mu t$; projecting both indices into a slice removes them. Thus $\omega_{\mu\nu}=0$ for this flow.

The converse local statement, that zero twist allows such orthogonal slices, is the codimension-one Frobenius integrability theorem. The focusing calculation below needs only the forward implication just checked.

Also suppose the timelike convergence condition holds:

Raychaudhuri then implies

If the cloud starts converging, $\theta_0<0$, divide by $\theta^2>0$ and differentiate its reciprocal:

If the geodesics extend through the required interval, this inequality forces the initially negative reciprocal toward zero within proper time no larger than $3/|\theta_0|$: $\theta$ becomes unboundedly negative, and the smooth congruence develops a focal point or caustic by that bound. An earlier incomplete geodesic endpoint is another possibility. The focusing argument does not by itself guarantee that the geodesics exist for the entire interval.

This is not yet a spacetime singularity. Aim a family of straight worldlines at the same event in Minkowski spacetime. Their congruence focuses; each worldline continues perfectly well. What breaks is the single-valued smooth velocity field used to describe that overlapping family.

The volume also makes the focusing bound explicit. Put $v=(\mathcal V/\mathcal V_0)^{1/3}$, with $v(0)=1$. While the congruence is regular,

Its initial slope is $v'(0)=\theta_0/3<0$. Since that slope cannot increase, $v(\tau)\le1+\theta_0\tau/3$. A positive smooth volume therefore cannot persist beyond $3/|\theta_0|$. This establishes focusing if the geodesics extend that far; inferring spacetime incompleteness still requires the additional global argument in a singularity theorem.

22.4 Energy conditions and focusing#

Einstein’s equation converts curvature conditions into matter conditions, but it does not itself require ordinary matter to satisfy those conditions. They are additional hypotheses.

For a perfect fluid, useful pointwise conditions are:

To verify the fluid inequalities, use its orthonormal rest frame and a unit timelike observer $v^{\hat\mu}=\gamma(1,\boldsymbol\beta)$, with $|\boldsymbol\beta|<1$. The measured density is

At rest this is $\epsilon$; as $\gamma$ becomes large its sign is controlled by $\epsilon+p$. Nonnegativity for every observer therefore requires both WEC inequalities. A null vector has the form $k^{\hat\mu}=q(1,\mathbf n)$ with $|\mathbf n|=1$, giving $T_{\mu\nu}k^\mu k^\nu=q^2(\epsilon+p)$ and the NEC.

For the SEC, add $T/2=(-\epsilon+3p)/2$ to the timelike contraction. Its value at rest is $(\epsilon+3p)/2$ and its large-$\gamma$ coefficient is again $\epsilon+p$. For the DEC, the energy-flux vector seen by $v$ is $J^{\hat\mu}=-T^{\hat\mu}{}_{\hat\nu}v^{\hat\nu}=(\gamma\epsilon,-\gamma p\boldsymbol\beta)$. Requiring it to be future causal for every $|\boldsymbol\beta|<1$ gives $\epsilon\ge0$ and $\epsilon^2\ge p^2$, or $\epsilon\ge|p|$.

For $\Lambda=0$, the SEC implies timelike convergence. If $\Lambda$ remains on the geometric side, however,

For a comoving perfect-fluid observer this is $4\pi G_N(\epsilon+3p)-\Lambda$. Positive $\Lambda$ can defeat timelike focusing. Equivalently, move it into an effective vacuum stress tensor with $p_\Lambda=-\epsilon_\Lambda$: it violates the SEC when its density is positive, while saturating the NEC.

For null vectors, the trace and cosmological terms vanish because $g_{\mu\nu}k^\mu k^\nu=0$. The NEC therefore implies null convergence in GR even with $\Lambda$.

Classical scalar potentials can violate the SEC, and quantum fields can violate classical pointwise energy conditions more broadly. These are reasons to inspect a theorem’s hypotheses carefully, not to call the theorem mistaken.

22.5 Trapped surfaces and what Penrose actually proved#

A null ray has no unit rest frame: its tangent $k$ is perpendicular to itself. To isolate the two transverse directions, choose another null vector $l$ with $k\cdot l=-1$ and define

This tensor removes both $k$ and $l$ components and supplies a positive metric on the remaining two-dimensional screen. Project the separation-velocity map onto that screen and split it into trace, shear, and twist as before. The trace is now an area-growth rate, so the trace part of the two-dimensional map is $\theta q_{\mu\nu}/2$.

For affinely parametrized geodesic rays, taking the trace of the same evolution calculation gives

The shear and twist here live on the positive-definite two-dimensional screen transverse to the rays. An affine parameter is essential; a nonaffine parameter introduces an additional term proportional to $\theta$.

Take a compact spacelike two-surface without a boundary, such as a sphere, and send future light rays orthogonally away from it in both null-normal directions. For an ordinary sphere in flat space, the outward bundle grows in area and the inward bundle shrinks. A future trapped surface has negative expansion in both directions. Even the outward-directed light bundle initially loses cross-sectional area.

This is a local geometric condition on the surface and its null normals. It does not say that a photon locally travels more slowly than light, nor does it require a coordinate speed to become negative.

One standard form of Penrose’s theorem says that a sufficiently regular spacetime with a noncompact Cauchy hypersurface, null convergence, and a closed future trapped surface must be future null geodesically incomplete. The theorem combines focusing with global causal geometry; it does not assume spherical symmetry. The original result is Penrose’s 1965 paper, “Gravitational Collapse and Space-Time Singularities”.

Geodesic incompleteness means at least one inextendible geodesic has finite affine length in the relevant direction; for timelike geodesics, proper time is the physical parameter. It does not universally mean that a curvature scalar tends to infinity. The theorem does not supply a location, a topology, or a detailed microscopic description of “the singularity.”

Even incompleteness must be interpreted carefully: deleting one point from otherwise regular Minkowski spacetime creates incomplete geodesics artificially. One must consider extendibility and which spacetime has actually been specified. Conversely, a coordinate singularity that disappears in a larger smooth chart is not evidence that physics has ended.

The theorem applies without assuming a spherical source. Its conclusion is incompleteness under the stated global and convergence conditions, even when the matter distribution is less symmetric than the examples solved earlier.

22.6 Horizons are about causal access; predictability needs another definition#

In an asymptotically flat spacetime, the black-hole region consists of events unable to send a causal signal to future null infinity $\mathscr I^+$, the ideal destination of escaping light. Its boundary is the event horizon:

$J^-(\mathscr I^+)$ denotes the causal past of that destination. This definition depends on the entire future spacetime. An event horizon is not generally locatable using measurements made at one point and one instant.

A marginally outer trapped surface instead has vanishing outward null expansion, usually with negative inward expansion in the black-hole setting. Under suitable conditions an apparent horizon is the outer boundary of the trapped region on a chosen slice. These are valuable tools for dynamical calculations, but they depend on slicing and need not coincide with the event horizon.

For predictability, define the future domain of dependence $D^+(\Sigma)$: an event belongs to it if every past-inextendible causal curve through that event intersects $\Sigma$. No causal influence can arrive there without passing through the supplied initial data.

A Cauchy surface meets every inextendible causal curve exactly once. A spacetime admitting such a surface is globally hyperbolic, in the usual boundary-free setting. This is the natural arena for the initial-value description of Chapter 20.

A Cauchy horizon bounds a domain of dependence. Beyond it, initial data on the chosen surface no longer control every possible incoming influence. It is different from an event horizon: crossing a black-hole event horizon need not destroy local predictability. Inner horizons in idealized charged or rotating solutions motivate the question of whether such extendible boundaries survive generic perturbations. Strong cosmic censorship studies this issue, and its precise claims depend on the matter model and the allowed regularity of extensions.

A final geometric tool helps organize these questions. Under a smooth positive conformal rescaling $g_{\mu\nu}\mapsto\Omega^2g_{\mu\nu}$, null cones are unchanged. Proper times and distances change. Conformal diagrams exploit this separation to compress enormous regions while preserving causal relations; they are maps of who can signal whom, not faithful scale drawings.

22.7 A horizon gets a temperature#

PREPARATION FOR THIS SECTION

What temperature adds to a theory

What does a thermal probability distribution know that a spacetime metric alone does not?

Builds onPressure work Constrained variations

See the idea

A macrostate specifies coarse measurements such as energy and volume. A microstate specifies the finer physical configuration. A probability distribution assigns nonnegative numbers $p_i$ to the allowed microstates with $\sum_i p_i=1$. An average measurement of a quantity $A_i$ is $\langle A\rangle=\sum_i p_iA_i$. The distribution is extra physical information.

Work it out

1. 01
   Measure uncertainty with entropy

   For probabilities $p_i$, define $S=-k_B\sum_i p_i\ln p_i$, with $0\ln0$ interpreted by continuity as zero. $k_B$ converts dimensionless logarithmic uncertainty into entropy units. For N equally likely states, substitution gives $S=k_B\ln N$.

   $$p_i=1/N\quad\Longrightarrow\quad S=k_B\ln N.$$

   Why this step works Entropy depends on the probabilities and the chosen description of states.

2. 02
   Find the canonical distribution

   Maximize $S/k_B$ with fixed normalization and mean energy. Multipliers give $-\ln p_i-1-\alpha-\beta E_i=0$, hence $p_i=e^{-\beta E_i}/Z$, where $Z=\sum_j e^{-\beta E_j}$ normalizes the probabilities. For an equilibrium heat reservoir, $\beta=1/(k_BT)$.

   $$p_i=\frac{e^{-E_i/(k_BT)}}Z,\qquad \frac{p_j}{p_i}=e^{-(E_j-E_i)/(k_BT)}.$$

   Why this step works The energy difference relative to the thermal energy kBT sets the population ratio.

3. 03
   Relate entropy to energy transfer

   Reversible heat exchange obeys $dS=\delta Q_{\rm rev}/T$. With fixed particle number, pressure work gives $dE=T\,dS-p\,dV$. Irreversible heat transfer does not justify replacing the full entropy balance by $\delta Q/T$ with an arbitrary system temperature.

   $$\left(\frac{\partial S}{\partial E}\right)_{V,N}=\frac1T.$$

   Why this step works Thermodynamic derivatives specify what is held fixed.

Go deeper

For two states with energies 0 and $\Delta>0$, $Z=1+e^{-\Delta/(k_BT)}$. At high positive temperature their probabilities approach one half; at low positive temperature the lower state dominates. Classical thermodynamics and probability do not by themselves imply that a black hole emits particles. The Hawking calculation additionally uses quantum fields, a choice of state, and a regularity or collapse condition. Euclidean periodicity supplies a consistency relation between a candidate thermal state and the near-horizon geometry; it is not the entire flux calculation.

Further readingDavid Tong · Statistical physics

Test the idea

FIRST, PREDICT

Does a metric by itself specify a quantum field’s thermal population?

Temperature is simply another coordinate time.No. The field theory and state are additional information.Yes. Geometry replaces the probability distribution.

Compare the reasoning

Temperature is simply another coordinate time.

Temperature controls energy populations; a coordinate labels events.

No. The field theory and state are additional information.

The same geometry can support distinct physical states with different observations.

Yes. Geometry replaces the probability distribution.

Geometrical regularity can constrain a state, but does not erase the need to specify one.

A hint

Separate the arena, dynamical law, and state.

NOW CHANGE THE EXAMPLE

A system has four equally likely states. Find $S/(k_B\ln2)$.

A hint

Use $S=k_B\ln4$.

Work through the solution

$\ln4/\ln2=2$.

Temperature is a property of a specified physical state, and entropy depends on the information retained.

PREPARATION FOR THIS SECTION

A pure whole can have an uncertain part

How can losing access to one subsystem create entropy without destroying the whole quantum state?

Builds onComplex modes Probabilities and entropy

See the idea

A finite quantum system is described by a normalized vector of complex amplitudes. In an orthonormal basis $|i\rangle$, write $|\psi\rangle=\sum_i a_i|i\rangle$ with $\sum_i|a_i|^2=1$. The probability of basis outcome i is $|a_i|^2$. This is a postulate of quantum mechanics, separate from GR. The notation $\langle\psi|$ means the conjugate transpose of $|\psi\rangle$.

Work it out

1. 01
   Turn a vector into a density matrix

   For a known pure state, $\varrho=|\psi\rangle\langle\psi|$. For a mixture of preparations $|\psi_j\rangle$ with probabilities $p_j$, use $\varrho=\sum_jp_j|\psi_j\rangle\langle\psi_j|$. This positive matrix has trace one. An observable is a Hermitian matrix A; its mean is $\operatorname{tr}(\varrho A)$.

   $$S=-k_B\operatorname{tr}(\varrho\ln\varrho).$$

   Why this step works Diagonalizing the density matrix reduces the entropy formula to the classical probability formula for its eigenvalues.

2. 02
   Trace over what is inaccessible

   For two two-state systems, take $|\Psi\rangle=(|00\rangle+|11\rangle)/\sqrt2$. The full density matrix is pure and has entropy zero. If only the first system is accessible, define $\varrho_A=\sum_{b=0,1}\langle b|\varrho_{AB}|b\rangle$, where the inner products act on the second system. The cross terms vanish because $\langle0|1\rangle=0$.

   $$\varrho_A=\frac12\begin{pmatrix}1&0\\0&1\end{pmatrix},\qquad S_A=k_B\ln2.$$

   Why this step works A partial trace keeps exactly the information needed for measurements on the accessible subsystem.

3. 03
   Keep evolution separate from restriction

   A closed system evolves by a unitary matrix $U$, with $U^\dagger U=I$ and $\varrho\mapsto U\varrho U^\dagger$. This preserves its eigenvalues and therefore its entropy. Restricting to a subsystem is a different operation and can change the entropy you assign to that subsystem.

   $$\operatorname{tr}[(U\varrho U^\dagger)^2]=\operatorname{tr}(\varrho^2).$$

   Why this step works Unitary similarity preserves spectra; a partial trace does not preserve the full state’s spectrum.

Go deeper

A free field can be decomposed into normal modes; in a chosen stationary background each mode behaves as a quantum oscillator with energies $E_n=\hbar\omega(n+1/2)$. Summing the canonical geometric series gives mean occupation $\langle n\rangle=1/(e^{\hbar\omega/(k_BT)}-1)$ for bosonic modes with zero chemical potential. The oscillator spectrum and field quantization are additional quantum inputs; this short preparation does not derive them from classical geometry. In curved spacetime different choices of positive-frequency modes can define different vacuum states. A full Hawking-flux derivation must specify those modes, the state, and transmission through the exterior curvature potential. The finite example above explains why an exterior thermal description alone is not proof that global evolution destroyed information.

Further readingDavid Tong · Quantum mechanics David Tong · Statistical physics

Test the idea

FIRST, PREDICT

A subsystem has a mixed density matrix. Must the full system also be mixed?

No. An entangled pure state can have a mixed reduced state.Yes. Pure states have pure parts.A mixed reduced state proves nonunitary evolution.

Compare the reasoning

No. An entangled pure state can have a mixed reduced state.

The two-system example has zero total entropy and a nonzero reduced entropy.

Yes. Pure states have pure parts.

Entanglement is precisely why this inference fails.

A mixed reduced state proves nonunitary evolution.

Partial tracing is not the same operation as evolving the full closed system.

A hint

Calculate the partial trace of the explicitly supplied two-system state.

NOW CHANGE THE EXAMPLE

For $\varrho_A=I_2/2$, calculate the purity $\operatorname{tr}(\varrho_A^2)$.

A hint

Square the two diagonal entries and add.

Work through the solution

$1/4+1/4=1/2$. A normalized pure state instead has purity 1.

Mixed information about a part and loss of information in the whole are distinct claims.

Restore $c$ and $\hbar$. For a Schwarzschild black hole,

A short derivation reveals why a temperature appears. Near the horizon write $r=r_s+x$, with $x\ll r_s$, so $1-r_s/r\approx x/r_s$. Continue the stationary time coordinate to imaginary time $t=-i\tau_E$. The near-horizon radial-time metric becomes

Introduce a radial proper-distance coordinate $\varrho=2\sqrt{r_sx}$. Direct substitution gives

This is a flat plane in polar coordinates. Its angular coordinate is $c\tau_E/(2r_s)$. Smoothness at the origin requires that angle to have period $2\pi$, so imaginary time has period

Quantum statistical mechanics identifies thermal equilibrium with imaginary-time period $\hbar/(k_BT)$. Equating these periods yields

This Euclidean argument characterizes the regular stationary thermal construction. Deriving the outgoing flux in a collapse spacetime involves a quantum-field calculation with the appropriate state and boundary conditions; it yields the same Hawking temperature at infinity, with frequency-dependent transmission factors modifying an ideal blackbody spectrum. The calculation is semiclassical, not a complete quantization of spacetime. See Wald’s research review of black-hole thermodynamics.

The commonly illustrated story of a particle pair appearing exactly on the horizon is a heuristic, not this derivation. Hawking radiation depends on quantum field modes, the state, and the global relation between early and late notions of positive frequency. It is not adequately explained by assigning ordinary local particle trajectories to vacuum fluctuations.

WORKED EXAMPLE

Separate the geometry from the quantum assumption

Which part of black-hole temperature can we derive with calculus, and what new physics must we import?

Builds onThermal probabilities and entropy Quantum states and reduced information Energy and the mechanics vocabulary Real waves and optional complex notation Regular and singular black-hole surfaces

See the idea

A formula involving temperature needs more than a metric. We need to say what temperature means, how quantum states evolve, and why thermal equilibrium can be represented using imaginary time. We can check the algebra of that connection here while keeping its physical assumptions visible.

Work it out

1. 01
   Give temperature and entropy operational jobs

   For a reversible change with the other thermodynamic work variables fixed, the first law reads $dE=T\,dS$. Energy $E$ is measured in joules, temperature $T$ in kelvin, and entropy $S$ in joules per kelvin. Thus $1/T=dS/dE$ along this family of equilibrium states. In quantum statistical mechanics, a state of energy $E$ has thermal weight proportional to $e^{-E/(k_BT)}$, where Boltzmann’s constant $k_B$ converts temperature to an energy scale. This thermal-weight rule is new physical input.

   $$dE=T\,dS,\qquad w(E)\propto e^{-E/(k_BT)}.$$

   Why this step works A thermal weight is not derived from Einstein’s equation. Specifying the thermodynamic ensemble and its constraints is part of using it.

2. 02
   Compare a quantum phase with a thermal weight

   A stationary quantum state of energy $E$ evolves with phase $e^{-iEt/\hbar}$, where $\hbar$ is the reduced Planck constant. Formally substitute $t=-i\tau_E$; this is an analytic continuation, not a clock traveling into an imaginary physical direction. Since $(-i)(-i)=-1$, the phase becomes $e^{-E\tau_E/\hbar}$. Matching it to the thermal weight gives the imaginary-time interval $\Delta\tau_E=\hbar/(k_BT)$.

   $$e^{-iEt/\hbar}\ \xrightarrow{\ t=-i\tau_E\ }\ e^{-E\tau_E/\hbar},\qquad \Delta\tau_E=\frac{\hbar}{k_BT}.$$

   Why this step works The algebra explains the length of the thermal interval. The full periodicity statement comes from the thermal trace and quantum correlation functions; it is not established by this substitution alone.

3. 03
   Let regular geometry determine that interval

   The near-horizon calculation in this section puts the Euclidean radial-time metric into polar form $d\varrho^2+\varrho^2d\theta_E^2$, with $\theta_E=c\tau_E/(2r_s)$. A smooth plane requires angular period $2\pi$; a different period produces a conical tip. Therefore $\Delta\tau_E=4\pi r_s/c$. Equating this geometric period to the thermal interval gives the Schwarzschild temperature at infinity.

   $$T_H=\frac{\hbar c}{4\pi r_sk_B}=\frac{\hbar c^3}{8\pi G_NMk_B}.$$

   Why this step works Smooth Euclidean geometry fixes a period. Quantum statistical mechanics supplies its interpretation as a temperature.

4. 04
   Check what the first law then predicts

   For the nonrotating, uncharged equilibrium family, use $E=Mc^2$. Substituting $T_H$ in $dS=dE/T_H$ gives a term proportional to $M\,dM$. Integrating produces an entropy proportional to $M^2$. Because $A=4\pi r_s^2=16\pi G_N^2M^2/c^4$, it is also proportional to horizon area. The integration leaves an arbitrary additive constant $S_0$.

   $$dS=\frac{8\pi k_BG_NM}{\hbar c}\,dM,\qquad S=\frac{k_BAc^3}{4\hbar G_N}+S_0.$$

   Why this step works The first law fixes entropy differences. The conventional additive normalization does not emerge from this integration.

Go deeper

The stationary Euclidean construction is an equilibrium argument. For bosonic fields, the corresponding thermal correlation functions are periodic in imaginary time; fermionic fields have the associated antiperiodic condition. A collapse spacetime’s outgoing Hawking flux requires a quantum-field calculation with a specified state and boundary conditions. It yields the same asymptotic temperature scale, while scattering by the spacetime modifies the observed spectrum. This is semiclassical physics: quantum fields on a classical background, with controlled treatment of backreaction where applicable. It does not derive a microscopic count of black-hole states, explain the final stage of evaporation, or establish a complete quantum theory of gravity. Those are open tasks, not consequences hidden in the elementary algebra.

Further readingRobert Wald · The thermodynamics of black holes Marc Kamionkowski · Finite-temperature field theory

Test the idea

FIRST, PREDICT

Which statement correctly identifies the extra physical input in the Euclidean argument?

Einstein’s classical equation alone derives the thermal Boltzmann weights.Imaginary time is the proper time recorded by an observer inside the horizon.Quantum statistical mechanics relates an imaginary-time interval to temperature.

Compare the reasoning

Einstein’s classical equation alone derives the thermal Boltzmann weights.

The Boltzmann weights and quantum phase evolution are additional physical principles, not results of classical GR.

Imaginary time is the proper time recorded by an observer inside the horizon.

Imaginary time is an analytic-continuation tool. It is not the real proper time of a traveler.

Quantum statistical mechanics relates an imaginary-time interval to temperature.

Yes. Geometry determines the interval required for regularity; quantum thermal physics provides the relation to temperature.

A hint

Locate the first appearance of both $\hbar$ and $k_B$ in the chain of reasoning.

NOW CHANGE THE EXAMPLE

Compare two nonrotating, uncharged black holes with masses $M$ and $3M$. Using the conventional $S_0=0$, what is $S(3M)/S(M)$?

A hint

The radius is proportional to mass, and the entropy is proportional to the horizon area.

Work through the solution

$r_s(3M)=3r_s(M)$, so the area and entropy each increase by $3^2=9$. The temperature instead falls to one third.

The temperature calculation joins regular geometry to explicitly stated quantum and thermodynamic input.

22.8 Entropy is written in area#

For the nonrotating, uncharged family, the first law is

Substitute $T_H$ and solve for the entropy change:

Integrating gives, with the conventional additive normalization,

For ordinary extensive matter at fixed local conditions, doubling volume doubles entropy. Here the entropy scales with horizon area: doubling the Schwarzschild mass multiplies both area and entropy by four. The formula determines that scaling without specifying microscopic constituents of the horizon.

Since $T_H\propto M^{-1}$, a Schwarzschild black hole becomes hotter as it loses mass. Its heat capacity, $C=d(Mc^2)/dT_H=-Mc^2/T_H$, is negative. A slightly hotter hole loses energy to a bath and becomes hotter still, rather than relaxing back by the usual positive-heat-capacity mechanism. Equilibrium therefore requires an analysis of the whole system and its boundary conditions.

The classical horizon-area theorem requires the relevant convergence and global regularity assumptions. Hawking evaporation does not contradict it: the quantum stress tensor need not satisfy the classical energy hypothesis, and the horizon area can decrease. The thermodynamic quantity then involves generalized entropy,

with the quantum-field entropy and gravitational parameters treated consistently under renormalization. This means that when short-distance field contributions are regulated, the theory’s parameters must be adjusted consistently so the physical prediction does not depend on the arbitrary regulator. Chapter 23 develops that distinction between a regulated intermediate expression and a prediction. The generalized second law has substantial support and proofs in specified settings; it should not be promoted without qualifications to a theorem covering every unknown quantum-gravitational process.

22.9 Classical area increase, with the assumptions visible#

For horizon-generating null geodesics, the vorticity vanishes and the screen has two dimensions. In an affine parameter $\lambda$, Raychaudhuri becomes

The null energy condition, together with Einstein’s equation, makes the last contraction nonnegative; the cosmological term drops out because $g_{\mu\nu}k^\mu k^\nu=0$. If the horizon expansion were negative, the inequality would force a future caustic. Under the global regularity and predictability assumptions of the classical area theorem, horizon generators cannot end in that way on the future horizon. This yields nonnegative expansion and nondecreasing horizon area. The global step is essential; a local differential equation alone does not prove the theorem.

For a simple numerical illustration, imagine two initially well-separated, nonspinning holes of equal mass $M$, ending in a nonspinning hole of mass $M_f$. Since a Schwarzschild area is $16\pi G_N^2M^2/c^4$, area increase requires

With initial total energy approximately $2Mc^2$, the radiated fraction is consequently at most $1-1/\sqrt2\simeq29.3\%$ under these idealizations. This is an upper bound, not the predicted emission efficiency. A spinning remnant requires the Kerr area formula, so applying $A\propto M^2$ blindly to a measured merger would be wrong.

For an uncharged stationary rotating black hole, the first law can be written in SI units as

Here $\kappa_{\rm sg}$ is surface gravity with acceleration units, $\Omega_H$ is horizon angular velocity, and $J$ is angular momentum. For Schwarzschild, surface gravity is the limiting hovering acceleration after correcting by its redshift factor:

The local hovering acceleration diverges; this redshifted limit remains finite and uses the clock normalization at infinity.

For the nonextremal Kerr family, $|a_K|<m$, the corresponding quantities are

These formulas use the horizon radii defined in Section 17.7. The area follows by integrating the horizon cross-section: $\sqrt{g_{\theta\theta}g_{\phi\phi}}=(r_+^2+a_K^2)\sin\theta$. The angular velocity is the horizon limit of $-g_{t\phi}/g_{\phi\phi}$. To check the first law within this family, differentiate $r_+^2-2mr_++a_K^2=0$, obtaining $(r_+-m)dr_+=r_+dm-a_Kda_K$. Substitute that relation into $dA$, with $M=c^2m/G_N$ and $J=c^3ma_K/G_N$. The two terms on the right reduce to $c^4dm/G_N=d(Mc^2)$. This family calculation verifies the coefficients; the general horizon mechanics law has broader hypotheses. With $T_H=\hbar\kappa_{\rm sg}/(2\pi c k_B)$ and $S_{\rm BH}=k_BAc^3/(4\hbar G_N)$, the area term equals $T_HdS_{\rm BH}$. Surface gravity is constant on an equilibrium horizon under the zeroth law’s assumptions. The classical second law is area increase; quantum evaporation calls for generalized entropy instead. The various third-law formulations need additional qualifications and are not needed for this derivation. Wald’s account of the laws and their assumptions.

22.10 The black-hole information question#

Classical uncertainty means we do not know which state a system has. Quantum theory also has entanglement: two subsystems can have a definite joint state even when neither has a definite pure state on its own. A pure state describes the complete quantum state; a mixed state describes uncertainty or the reduced description of a subsystem. Unitary evolution is the reversible state evolution of an isolated quantum system in ordinary quantum mechanics. These definitions are enough to state the puzzle, though not to reproduce a quantum-field calculation.

In the leading semiclassical account of a collapsing black hole, outgoing radiation is entangled with degrees of freedom behind the horizon. An observer with access only to the exterior describes approximately thermal radiation, modified by propagation through the surrounding geometry. A thermal-looking spectrum alone does not prove that all correlations are absent.

The tension appears when we combine several claims: a pure initial state, complete evaporation with no remaining hidden system, a final radiation state with irretrievably lost correlations, and unitary evolution of the entire isolated process. Those claims cannot all hold. The classical no-hair description of a stationary exterior is not, by itself, a proof that a quantum state has no microscopic information.

The entropy of radiation expected in a unitary evaporation rises while the radiation is entangled with the remaining hole, then eventually falls to zero if all that remains is the final pure radiation state. This qualitative rise-and-fall behavior is called the Page curve. In specified semiclassical models, calculations of radiation entropy include an island: an interior region whose field correlations contribute to the entropy assigned to the radiation. The calculation varies candidate island boundaries to make the generalized entropy stationary and selects the smallest admissible value. The no-island candidate can dominate early and an island candidate later, producing a Page-shaped curve. The underlying gravitational integral sums over candidate field and geometry histories weighted by their action. Its stationary contributions are called saddle points; additional contributions of this kind produce the island prescription in these models. These results concern specified quantum-gravitational calculations; they do not constitute direct measurements of astrophysical evaporation or a microscopic account valid for every black hole. Almheiri and collaborators’ review.

The boundary of the book is visible here. The classical Einstein equation alone cannot decide how quantum information is recovered. It supplies the geometry in which the question becomes sharp.