On-Shell Action and Boundary Terms

The Euclidean bulk action alone neither defines a Dirichlet gravitational variational problem nor produces a finite thermodynamic potential. One must add the Gibbons–Hawking–York term, regulate the outer boundary, choose either matched-background subtraction or local counterterms, and include any ensemble-changing Legendre term. Only the complete on-shell action may be identified with $-\log Z$. Its value depends on which boundary geometry and chemical potentials are held fixed.

Required background. Gravitational Actions and Boundary Terms supplies the Dirichlet action and outward-normal convention; Variational and Symplectic Geometry supplies boundary charges and Legendre transforms.

Helpful background. Quantum Fields in Curved Spacetime supplies the thermal functional integral; Einstein Equations supplies the on-shell bulk relation; and Black-Hole Geometry and Mechanics supplies horizon charges.

The regulated Euclidean Dirichlet action

Continue $t=-i\tau$ to a positive-definite real section. With outward unit normal at the outer boundary and the corresponding definition $K=h^{ab}\nabla_a n_b$, take

$$\begin{aligned} I_E[g] ={}&-\frac{1}{16\pi G} \int_{\mathcal M}\!d^4x\,\sqrt g\,(R-2\Lambda)\\ &-\frac{1}{8\pi G} \int_{\partial\mathcal M}\!d^3x\,\sqrt h\,K +I_{\mathrm{matter}}+I_{\mathrm{ren}} . \end{aligned}$$

Here $I_{\mathrm{ren}}$ denotes either a reference subtraction or a set of local boundary counterterms. The Gibbons–Hawking–York term cancels normal derivatives of $\delta g_{\mu\nu}$, so fixing the induced metric $h_{ab}$ makes the variational problem differentiable. Gibbons and Hawking use this completed action to connect stationary saddles with the gravitational partition function Gibbons and Hawking 1977, pp. 2752–2756.

The sign of the displayed boundary term is tied to the stated Euclidean and outward-normal conventions. Reversing the normal reverses $K$ and must be accompanied by the corresponding boundary orientation; copying only one sign from another convention gives a wrong free energy.

Finite action requires a declared subtraction

For an asymptotically flat saddle, put the outer boundary at $r=R$ and compare with a reference geometry whose induced metric agrees there. In particular, the proper circumference $\beta_R$ of the Euclidean time circle—not merely its coordinate period—must match. Then replace

$$\int_{\partial\mathcal M}\sqrt h\,K \quad\longrightarrow\quad \int_{\partial\mathcal M}\sqrt h\,(K-K_0)$$

before sending $R$ to infinity. For asymptotically AdS geometries, local intrinsic counterterms provide a different renormalization prescription. Reference subtraction and AdS counterterms may yield equivalent differences when applied consistently, but their zero points cannot be interchanged mid-comparison.

A finite cavity is yet another problem. The boundary metric fixes the wall area and proper inverse temperature $\beta_B$; the relevant energy is the quasilocal charge conjugate to wall proper time. York derives this canonical cavity construction and its two Schwarzschild branches in York 1986, §§II–V. Replacing $\beta_B$ by the period normalized at infinity changes the boundary data and therefore changes the ensemble.

Ensemble terms determine the potential

At leading saddle order,

$$Z(\beta,\Omega,\Phi)\simeq e^{-I_{E,\mathrm{on-shell}}}, \qquad I_{E,\mathrm{on-shell}} =\beta(E-\Omega J-\Phi Q)-S.$$

Thus $I_E/\beta$ is the Gibbs-type potential for fixed $(\beta,\Omega,\Phi)$. Fixing $J$ or $Q$ instead requires a Legendre transform. For example, in Maxwell theory the action appropriate to fixed boundary potential is not the action appropriate to fixed electric flux. Adding the electromagnetic boundary Legendre term changes the on-shell potential by $+\beta\Phi Q$ in the convention above, converting $E-TS-\Phi Q$ to $E-TS$. The endpoint-difference sign in $\Phi=\chi^\mu A_\mu|_{\infty}-\chi^\mu A_\mu|_{\mathrm H}$ must use the same gauge and horizon generator as the first law.

The horizon is not an inner boundary on a smooth nonextremal Euclidean manifold: it is the polar origin. Its entropy nevertheless appears in the on-shell identity because regularity relates the period to the solution parameters. Introducing a small excised disk is a useful derivational device, but its limiting boundary contribution must be restored with the correct orientation.

Schwarzschild check and regime

For four-dimensional Euclidean Schwarzschild, write $f(r)=1-2GM/r$ and denote the outer cutoff radius by $r_B$ to distinguish it from the curvature scalar. The bulk scalar curvature vanishes. At the boundary,

$$\begin{aligned} K&=\frac{2\sqrt{f(r_B)}}{r_B} +\frac{f'(r_B)}{2\sqrt{f(r_B)}},\qquad K_0=\frac{2}{r_B},\\ \sqrt h&=r_B^2\sqrt{f(r_B)}\sin\theta . \end{aligned}$$

The reference flat metric has period $\beta_0=\beta\sqrt{f(r_B)}$ so that both induced boundary metrics agree. Performing the angular integral gives

$$\begin{aligned} I_E(r_B) &=-\frac{\beta}{2G} \left[2r_B f(r_B)+\frac{r_B^2 f'(r_B)}2 -2r_B\sqrt{f(r_B)}\right],\\ 2r_B f(r_B)+\frac{r_B^2 f'(r_B)}2 -2r_B\sqrt{f(r_B)} &=-GM+O\!\left(\frac{(GM)^2}{r_B}\right). \end{aligned}$$

Thus the finite boundary term, even with zero bulk curvature, yields

$$I_E=\frac{\beta M}{2} =4\pi GM^2 =\beta M-S_{\mathrm{BH}}.$$

With $\beta=8\pi GM$, this yields $S_{\mathrm{BH}}=4\pi GM^2=A/(4G)$. The equality independently checks the subtraction, boundary orientation, and thermal period. Dropping the reference contribution makes the action divergent; fixing an arbitrary period while retaining a smooth Schwarzschild mass produces a conical defect rather than the same saddle.

This calculation is tree level in the gravitational effective action. One-loop determinants, matter effective actions, and renormalized higher-curvature couplings add further bulk and boundary terms. The geometry must be smooth relative to the cutoff and the selected saddle must satisfy the declared boundary data. The Euclidean saddle supplies a term in a formal equilibrium partition function; its existence alone does not ensure a convergent canonical ensemble. It does not by itself describe real-time evaporation or prove dynamical stability.

Exercise

Suppose the properly subtracted Schwarzschild action is $I_E(\beta)=\beta^2/(16\pi G)$. Using canonical thermodynamic derivatives, recover the energy, entropy, and heat capacity. What does the sign of the heat capacity say—and not say—about the action calculation?

Solution

The canonical energy is

$$E=\frac{\partial I_E}{\partial\beta} =\frac{\beta}{8\pi G}=M.$$

The entropy is

$$S=\beta E-I_E =\frac{\beta^2}{16\pi G} =\frac{A}{4G}.$$

Since $T=\beta^{-1}$,

$$C=\frac{dE}{dT} =\frac{dE/d\beta}{dT/d\beta} =-\frac{\beta^2}{8\pi G}\lt0.$$

The negative heat capacity shows that asymptotically flat Schwarzschild is not locally stable in an ordinary canonical bath. It does not invalidate the regulated on-shell action or its entropy; stability is a further question about the second variation and the ensemble.

References

• Gibbons, G. W., and S. W. Hawking (1977), “Action Integrals and Partition Functions in Quantum Gravity,” Physical Review D 15, 2752–2756. DOI.
• York, J. W., Jr. (1986), “Black-Hole Thermodynamics and the Euclidean Einstein Action,” Physical Review D 33, 2092–2099. DOI.