The Gibbons-Hawking-York Boundary Term

Every variational principle rests on a quiet bargain: fix enough data on the boundary that the leftover surface term dies, and the interior equations of motion fall out clean. For a Lagrangian built from a field and its first derivatives the bargain is cheap — hold the field fixed on the edge and you are done. General relativity refuses to pay in that currency. The Einstein-Hilbert Lagrangian density $\sqrt{-g}\,R$ hides second derivatives of the metric inside the Ricci scalar, schematically $R\sim g^{-1}\partial^2 g + (g^{-1}\partial g)^2$ , so varying the action produces not only the bulk Einstein-tensor equation but a boundary flux carrying the normal derivative of the metric variation, $n^\mu\partial_\mu\delta g_{\nu\rho}$ . Fixing the metric on the boundary — the honest Dirichlet condition — kills only tangential derivatives; the normal one survives, and the variational problem is ill-posed. You would be forced to fix the metric and its normal derivative, which over-determines the geometry and has no physical meaning. The repair, found by James York in 1972 and used by Gibbons and Hawking in 1977 to build the Euclidean path integral of gravity, is a single boundary integral: add the trace of the extrinsic curvature, $\frac{1}{\kappa}\oint_{\partial M}\sqrt{|h|}\,K$ , and its variation supplies exactly the opposite normal-derivative term, cancelling the obstruction identically. What remains is a clean Dirichlet problem whose stationary point is $G_{\mu\nu}=\kappa T_{\mu\nu}$ . This is not housekeeping: the Gibbons-Hawking-York term is what gives the gravitational action a finite numerical value, and that value is the currency of quasi-local energy, of black-hole free energy and the entropy $S=A/4G$ , and of the whole holographic dictionary. Throughout we set $c=1$ , use signature $(-,+,+,+)$ , and write $\kappa=8\pi G$ .

Key takeaways

The Einstein-Hilbert action contains second derivatives of the metric, so its variation leaves a boundary integral of $n^\mu\partial_\mu\delta g_{\nu\rho}$ that fixing $\delta g_{\mu\nu}=0$ on $\partial M$ does not annihilate — the normal derivative survives.
The Gibbons-Hawking-York term $S_{\mathrm{GHY}}=\frac{1}{\kappa}\oint_{\partial M}\sqrt{|h|}\,\epsilon\,K\,\mathrm{d}^3x$ is added so that its variation cancels the leftover exactly, making the action functionally differentiable with only the boundary metric held fixed.
Here $h_{ab}$ is the induced metric on $\partial M$ , $K=h^{ab}K_{ab}=\nabla_\mu n^\mu$ is the trace of the extrinsic curvature, and $K_{ab}$ measures how the boundary bends inside the spacetime relative to the unit normal $n^\mu$ .
The GHY coefficient is $1/\kappa$ , exactly twice the bulk coefficient $1/2\kappa$ , because the surviving Einstein-Hilbert boundary term equals $-2\epsilon\,h^{ab}\delta K_{ab}$ per unit of $1/2\kappa$ .
With a background subtraction or holographic counterterms the total on-shell action becomes finite, and in Euclidean signature it delivers the Hawking temperature $T=1/(8\pi GM)$ and the Bekenstein-Hawking entropy $S=A/4G$ of a Schwarzschild black hole.
Both the sign $\epsilon=n^\mu n_\mu=\pm1$ (timelike versus spacelike segments) and codimension-two corner terms where segments meet must be included for a fully well-posed action.
Key idea: the variation of $\sqrt{-g}\,R$ leaves a boundary term proportional to $n^\mu\partial_\mu\delta g$ , which fixing $\delta g=0$ on $\partial M$ does not annihilate; adding $\frac{1}{\kappa}\oint\sqrt{|h|}\,K$ supplies precisely the opposite boundary variation, so $\delta(S_{\mathrm{EH}}+S_{\mathrm{GHY}})$ contains only the bulk Einstein equations plus a term vanishing under Dirichlet conditions.

Where this lives today

The Gibbons-Hawking-York term is a working tool across gravitational physics, not a historical footnote. In the Euclidean path-integral approach to quantum gravity and in every modern calculation of black-hole free energy — Schwarzschild, Kerr, charged, rotating, and asymptotically anti-de Sitter — the on-shell action that produces the partition function is computed with the boundary term and a subtraction or counterterm scheme; the resulting phase structure includes the Hawking-Page transition between a thermal gas and a large AdS black hole, the gravitational dual of confinement in the boundary gauge theory. In holographic renormalization the GHY term is the first layer of a systematic construction: bare action plus GHY still diverges as the conformal boundary is approached, so one adds local Balasubramanian-Kraus counterterms built from the induced metric and its intrinsic curvature, and the renormalized action's response to the boundary metric is the stress-energy tensor of the dual conformal field theory. In numerical relativity and the ADM Hamiltonian formulation, the boundary term is what makes the gravitational energy finite and correct, and varying the total action with respect to $h_{ab}$ defines the Brown-York surface stress tensor $\tau^{ab}=-(K^{ab}-Kh^{ab})/\kappa$ , whose integral returns the ADM mass at spatial infinity and the quasi-local energy of a bounded region otherwise. Even the gluing of spacetimes across a thin shell — bubbles, domain walls, matched stellar interiors — is governed by the same structure through the Israel junction conditions, in which a jump in $K_{ab}-Kh_{ab}$ equals the surface stress-energy of the shell. Wherever gravity meets a boundary, real or asymptotic, this term is doing the bookkeeping.

The physical picture

Think of the metric as the field and the boundary $\partial M$ as the place where you hand the theory its data. A good variational problem should ask you to specify the geometry of the boundary surface itself — the induced metric $h_{ab}$ , which tells you the intrinsic shape of the wall — and nothing more. Because gravity's action is second-order, the naive Einstein-Hilbert variation instead demands both the shape of the wall and the rate at which the metric changes as you step off it, the normal derivative. That is one condition too many. The extrinsic curvature $K_{ab}=\nabla_a n_b$ is exactly the geometric object that encodes "how the wall bends into the room" — how the unit normal $n^\mu$ tilts as you slide along the surface — and its trace $K$ is the single number the boundary term must carry to soak up the unwanted normal derivative. A flat sheet in flat space has $K=0$ ; a 2-sphere of radius $r$ has $K=2/r$ , larger for tighter spheres.

Figure

The variational problem on a manifold with boundary. Fixing the induced metric h_ab on ∂M is a Dirichlet condition, but the Einstein-Hilbert variation also leaves the normal derivative n^μ ∂_μ δg free. The extrinsic curvature K = ∇_μ n^μ measures how the boundary bends into the spacetime — for a 2-sphere of radius r it equals 2/r

There is a clean mechanical analogy that strips the covariance away. Take a point particle whose Lagrangian $L(q,\dot q,\ddot q)$ contains the acceleration. Varying its action produces a boundary term $\propto(\partial L/\partial\ddot q)\,\delta\dot q$ , which does not vanish when you fix only the endpoints $q$ . To make the problem well-posed with $q$ alone held fixed you add a total-time-derivative boundary term that trades $\delta\dot q$ for $\delta q$ . The GHY term is precisely the field-theoretic, generally covariant version of that trick — a boundary integral engineered so that the surviving normal-derivative variation is converted into a variation of the induced metric, which you have fixed. The prize for getting this right is a number. The value of the action, including its boundary term, is what enters $e^{iS}$ in Lorentzian signature or $e^{-I}$ after Wick rotation, and that value is decisive for tunnelling amplitudes, thermal partition functions, and the very definition of gravitational energy — which lives on boundaries because gravity has no local energy density.

Figure

The flagship payoff: the Euclidean Schwarzschild cigar. Wick-rotating t → −iτ closes time into a circle whose period β = 8πGM is fixed by smoothness at the horizon tip. Because R = 0 in vacuum the whole action is the GHY boundary term, giving I_E = 4πGM², T = 1/(8πGM) and entropy S = A/4G. For one solar mass r_s = 2.95 km, T = 6.2×10⁻⁸ K, S ≈ 1.0×10⁷⁷ k_B

The take-away from the two pictures is one sentence: the GHY term is the covariant boundary counterterm that turns "fix the metric and its normal derivative" into "fix only the boundary geometry," and the finite action it makes possible is what physics reads off as energy, temperature, and entropy.

The core equations

The starting point is the total gravitational action: the Einstein-Hilbert bulk plus the Gibbons-Hawking-York boundary term. Everything below either derives this combination or extracts a physical number from it.

S = S_{\mathrm{EH}} + S_{\mathrm{GHY}} = \frac{1}{2\kappa}\int_M \mathrm{d}^4x\,\sqrt{-g}\,R \;+\; \frac{1}{\kappa}\oint_{\partial M}\mathrm{d}^3x\,\sqrt{|h|}\,\epsilon\,K \qquad (1)

where: $\kappa$ — $8\pi G$ , Einstein's gravitational coupling in units $c=1$ ; $R$ — Ricci scalar of the bulk metric $g_{\mu\nu}$ ; $h_{ab}$ — induced (first fundamental) metric on the boundary $\partial M$ ; $K$ — trace of the extrinsic curvature, $K=h^{ab}K_{ab}$ ; $\epsilon$ — $n_\mu n^\mu=\pm1$ , equal to $+1$ for a timelike boundary and $-1$ for a spacelike one. Physical reading: the boundary term carries coefficient $1/\kappa$ , twice the bulk $1/2\kappa$ , and that factor of two is not free — it is fixed by the cancellation below. The boundary object being integrated is the trace of the extrinsic curvature. It admits two equivalent forms: a projected covariant derivative of the normal, and a plain bulk divergence of the normal field.

K_{ab} = h_a{}^{\mu}h_b{}^{\nu}\nabla_\mu n_\nu,\qquad K = h^{ab}K_{ab} = \nabla_\mu n^\mu \qquad (2)

where: $n^\mu$ — unit normal to $\partial M$ , $n_\mu n^\mu=\epsilon$ ; $h_a{}^\mu$ — projector onto $\partial M$ , $h_\mu{}^\nu=\delta_\mu{}^\nu-\epsilon\,n_\mu n^\nu$ ; $K_{ab}$ — the symmetric second fundamental form. Physical reading: $K$ is literally the divergence of the outward normal, the rate at which normals spread — positive for a convex boundary bulging into the bulk. The reason the boundary term appears at all is the Palatini identity, which says the variation of the Ricci tensor is a total covariant divergence, hence a pure boundary contribution.

\delta R_{\mu\nu} = \nabla_\lambda\,\delta\Gamma^\lambda_{\mu\nu} - \nabla_\nu\,\delta\Gamma^\lambda_{\mu\lambda},\qquad g^{\mu\nu}\delta R_{\mu\nu} = \nabla_\mu v^\mu \qquad (3)

where: $\delta\Gamma^\lambda_{\mu\nu}$ — variation of the Christoffel symbols, a genuine tensor since the difference of two connections is a tensor; $v^\mu = g^{\alpha\beta}\,\delta\Gamma^\mu_{\alpha\beta}-g^{\mu\beta}\,\delta\Gamma^\alpha_{\alpha\beta}$ — the boundary current, linear in first derivatives of $\delta g_{\mu\nu}$ . Physical reading: because $v^\mu$ contains derivatives of $\delta g$ , its flux through $\partial M$ cannot be killed by setting $\delta g=0$ alone. Varying the Einstein-Hilbert action therefore splits into the bulk Einstein-tensor piece plus a boundary flux of $v^\mu$ . This is the equation that exposes the whole problem.

\delta S_{\mathrm{EH}} = \frac{1}{2\kappa}\int_M \sqrt{-g}\,G_{\mu\nu}\,\delta g^{\mu\nu}\,\mathrm{d}^4x \;+\; \frac{1}{2\kappa}\oint_{\partial M}\sqrt{|h|}\,\epsilon\,n_\mu v^\mu\,\mathrm{d}^3x \qquad (4)

where: $G_{\mu\nu}=R_{\mu\nu}-\tfrac12 g_{\mu\nu}R$ — the Einstein tensor; $n_\mu v^\mu$ — normal projection of the boundary current. Physical reading: the bulk term gives the field equations, but the surface term does not vanish under Dirichlet data, so the variational principle is not yet well-posed. Evaluated with $\delta g_{\mu\nu}=0$ on $\partial M$ , the surviving boundary term reduces to the variation of the extrinsic curvature — and this is exactly what the GHY variation is built to cancel.

\frac{1}{2\kappa}\oint\sqrt{|h|}\,\epsilon\,n_\mu v^\mu = -\frac{1}{\kappa}\oint\sqrt{|h|}\,\epsilon\,h^{ab}\delta K_{ab},\qquad \delta S_{\mathrm{GHY}} = +\frac{1}{\kappa}\oint\sqrt{|h|}\,\epsilon\,h^{ab}\delta K_{ab} \qquad (5)

where: $\delta K_{ab}$ — variation of the extrinsic curvature at fixed induced metric $\delta h_{ab}=0$ . Physical reading: the two terms are equal and opposite, so their sum vanishes and only the bulk $G_{\mu\nu}\delta g^{\mu\nu}$ survives — the definition of a well-posed principle. With the boundary term in place the stationary condition yields the field equations, matter included.

\delta\!\left(S_{\mathrm{EH}}+S_{\mathrm{GHY}}+S_{\mathrm{matter}}\right)=0 \;\Longrightarrow\; G_{\mu\nu}=\kappa\,T_{\mu\nu} \qquad (6)

where: $T_{\mu\nu}$ — stress-energy from varying $S_{\mathrm{matter}}$ . Physical reading: the GHY term changes no bulk equation of motion, yet without it the derivation of those equations on a bounded region is illegitimate. Two numbers anchor the applications. For a 2-sphere of radius $r$ in flat space the trace is elementary, and for a Euclidean Schwarzschild black hole the on-shell action is pure boundary.

K_{\mathrm{sphere}}=\frac{2}{r},\qquad I_E = \frac{\beta M}{2}=4\pi G M^2,\qquad S_{\mathrm{BH}}=\beta M - I_E=\frac{A}{4G} \qquad (7)

where: $\beta=8\pi GM$ — inverse Hawking temperature fixed by horizon smoothness; $A=16\pi G^2M^2$ — horizon area. Physical reading: the entropy is one quarter of the horizon area in Planck units, a result that simply cannot be obtained without the boundary term.

Derivation AFrom the Einstein-Hilbert variation to the leftover boundary term

The goal of this derivation is to show, term by term, that varying the Einstein-Hilbert action on a region with boundary produces the bulk Einstein-tensor equation plus an irreducible surface flux that carries the normal derivative of the metric variation, so that the ordinary Dirichlet condition $\delta g_{\mu\nu}=0$ on $\partial M$ is not enough to make the action stationary. We vary the three factors of $\sqrt{-g}\,g^{\mu\nu}R_{\mu\nu}$ , use the Palatini identity to convert the Ricci variation into a total divergence, apply Stokes' theorem, and then isolate precisely the piece that refuses to vanish.

Step 1. Vary the three factors of the integrand: $\delta(\sqrt{-g}\,g^{\mu\nu}R_{\mu\nu})=\sqrt{-g}\,(R_{\mu\nu}\delta g^{\mu\nu}+g^{\mu\nu}\delta R_{\mu\nu})+R\,\delta\sqrt{-g}$ . Why: there are exactly three sources of variation — the inverse metric, the Ricci tensor, and the volume element — and separating them is what lets the bulk and boundary pieces be told apart.

Step 2. Use $\delta\sqrt{-g}=-\tfrac12\sqrt{-g}\,g_{\mu\nu}\delta g^{\mu\nu}$ and combine it with the $R_{\mu\nu}\delta g^{\mu\nu}$ term to assemble the Einstein tensor, $\sqrt{-g}\,(R_{\mu\nu}-\tfrac12 g_{\mu\nu}R)\delta g^{\mu\nu}=\sqrt{-g}\,G_{\mu\nu}\delta g^{\mu\nu}$ . Why: this is the bulk equation-of-motion term, the part that gives $G_{\mu\nu}=\kappa T_{\mu\nu}$ ; the entire remaining problem lives in the third factor $g^{\mu\nu}\delta R_{\mu\nu}$ .

Step 3. Apply the Palatini identity $g^{\mu\nu}\delta R_{\mu\nu}=\nabla_\mu v^\mu$ with $v^\mu=g^{\alpha\beta}\delta\Gamma^\mu_{\alpha\beta}-g^{\mu\beta}\delta\Gamma^\alpha_{\alpha\beta}$ . Why: the Ricci variation is a pure total divergence, so it contributes nothing in the interior and everything on the boundary — this is why a boundary term exists at all.

Step 4. Convert the divergence to a surface flux by Stokes' theorem, $\int_M\sqrt{-g}\,\nabla_\mu v^\mu\,\mathrm{d}^4x=\oint_{\partial M}\sqrt{|h|}\,\epsilon\,n_\mu v^\mu\,\mathrm{d}^3x$ . Why: the factor $\epsilon=n_\mu n^\mu$ appears here from the orientation of the outward normal, and it is what makes the sign flip between timelike and spacelike boundary segments.

Step 5. Substitute $\delta\Gamma^\lambda_{\mu\nu}=\tfrac12 g^{\lambda\rho}(\nabla_\mu\delta g_{\nu\rho}+\nabla_\nu\delta g_{\mu\rho}-\nabla_\rho\delta g_{\mu\nu})$ into $n_\mu v^\mu$ ; the projection contains the piece $h^{ab}\,n^\mu\partial_\mu\delta g_{ab}$ . Why: this exhibits the normal derivative of the metric variation explicitly, the object that a Dirichlet condition leaves untouched.

Step 6. Impose $\delta g_{\mu\nu}=0$ everywhere on $\partial M$ . Tangential derivatives die because $\delta g_{ab}$ is identically zero across the surface, but the normal derivative $n^\mu\partial_\mu\delta g_{ab}$ is unconstrained and survives, reorganizing into $n_\mu v^\mu\big|_{\delta g=0}=-2\epsilon\,h^{ab}\delta K_{ab}$ . Why: this residual is the obstruction; it is nonzero for generic variations, so $\delta S_{\mathrm{EH}}\neq0$ and the principle is ill-posed until a counterterm cancels it. *Verified: reproduces Wald Appendix E.1 and Poisson §4.1–4.3; the bulk term yields the vacuum Einstein equations $G_{\mu\nu}=0$ as required, and the residual $-2\epsilon\,h^{ab}\delta K_{ab}$ matches the standard normalization.*

Go deeper · Derivation B · 1 minThe Gibbons-Hawking-York term cancels the obstruction

Derivation BThe Gibbons-Hawking-York term cancels the obstruction

The goal here is to prove that adding $S_{\mathrm{GHY}}=\frac{1}{\kappa}\oint\sqrt{|h|}\,\epsilon\,K$ to the action produces a variation that is exactly equal and opposite to the leftover Einstein-Hilbert boundary term of the previous derivation, so that their sum vanishes for every metric variation obeying $\delta g_{\mu\nu}=0$ on $\partial M$ , and the full action is at last functionally differentiable with only the induced boundary metric held fixed. We compute $\delta S_{\mathrm{GHY}}$ holding $\delta h_{ab}=0$ , note that the induced volume element is then also unvaried, and add the two surface terms.

Step 1. Record the surviving Einstein-Hilbert boundary term from equation (5): $\frac{1}{2\kappa}\oint\sqrt{|h|}\,\epsilon\,n_\mu v^\mu=-\frac{1}{\kappa}\oint\sqrt{|h|}\,\epsilon\,h^{ab}\delta K_{ab}$ . Why: the factor of $2$ from $-2\epsilon\,h^{ab}\delta K_{ab}$ combines with the bulk coefficient $1/2\kappa$ to give a clean $-1/\kappa$ , which pins the GHY coefficient.

Step 2. Vary the GHY integrand at fixed induced metric: $\delta(\sqrt{|h|}\,\epsilon K)=\epsilon\,K\,\delta\sqrt{|h|}+\epsilon\sqrt{|h|}\,\delta K$ , and observe $\delta\sqrt{|h|}=\tfrac12\sqrt{|h|}\,h^{ab}\delta h_{ab}=0$ . Why: the Dirichlet condition $\delta g_{\mu\nu}=0$ on $\partial M$ forces $\delta h_{ab}=0$ , so the volume-element term drops and only $\delta K$ contributes.

Step 3. Use $\delta K=\delta(h^{ab}K_{ab})=h^{ab}\delta K_{ab}$ since $\delta h^{ab}=0$ , giving $\delta S_{\mathrm{GHY}}=\frac{1}{\kappa}\oint\sqrt{|h|}\,\epsilon\,h^{ab}\delta K_{ab}$ . Why: the GHY variation reduces to exactly the same tensor structure $h^{ab}\delta K_{ab}$ that plagued the Einstein-Hilbert term, but with the opposite sign.

Step 4. Add the two surface terms: $-\frac{1}{\kappa}\oint\sqrt{|h|}\,\epsilon\,h^{ab}\delta K_{ab}+\frac{1}{\kappa}\oint\sqrt{|h|}\,\epsilon\,h^{ab}\delta K_{ab}=0$ . Why: the cancellation is exact and identity-level — it holds for any boundary, any metric, any admissible variation, which is what "well-posed" means.

Step 5. Conclude that $\delta(S_{\mathrm{EH}}+S_{\mathrm{GHY}})=\frac{1}{2\kappa}\int_M\sqrt{-g}\,G_{\mu\nu}\delta g^{\mu\nu}\,\mathrm{d}^4x$ , so demanding stationarity for arbitrary boundary-vanishing $\delta g^{\mu\nu}$ enforces $G_{\mu\nu}=0$ in vacuum, or $G_{\mu\nu}=\kappa T_{\mu\nu}$ with matter. Why: the only survivor is the bulk term, so the correct field equations emerge from a genuinely differentiable action. *Verified: the coefficient $1/\kappa$ (versus $1/2\kappa$ in the bulk) is the standard York-Gibbons-Hawking normalization, matching MTW §21.5 and Carroll's appendix; the cancellation is exact for both signs of $\epsilon$ .*

Go deeper · Derivation C · 2 minEuclidean action of the Schwarzschild black hole and the entropy A over 4G

Derivation CEuclidean action of the Schwarzschild black hole and the entropy A over 4G

The goal of this derivation is to evaluate the on-shell gravitational action of the Schwarzschild solution in Euclidean signature, using the GHY boundary term and a flat-space background subtraction to render it finite, and to extract from it the Hawking temperature and the Bekenstein-Hawking entropy $S=A/4G$ — the single most important application of the boundary term and the reason it is not a technicality. We Wick-rotate to Euclidean time, fix the period by demanding smoothness at the horizon, note that the bulk action vanishes because the solution is Ricci-flat, and evaluate the subtracted GHY integral.

Step 1. Wick-rotate $t\to-i\tau$ in the Schwarzschild metric; near the horizon $r=2GM$ the $(\tau,r)$ plane looks like a plane in polar coordinates, and smoothness with no conical singularity fixes the period $\beta$ of $\tau$ . Why: a cone at the tip would be a curvature delta-function, so regularity is a physical condition, and it is what quantizes the temperature.

Step 2. Relate the period to the surface gravity $\kappa_{\mathrm{sg}}=1/(4GM)$ : $\beta=2\pi/\kappa_{\mathrm{sg}}=8\pi GM$ , so the temperature is $T=1/\beta=1/(8\pi GM)$ . Why: this is the Hawking temperature, derived here purely from Euclidean geometry rather than from quantum field theory in curved space, and it agrees with Hawking's 1974 result.

Step 3. Note that Schwarzschild is a vacuum solution, $R=0$ , so $S_{\mathrm{EH}}=0$ and the entire on-shell action is the GHY boundary term evaluated at a large radius $r$ , minus the same term for flat space at the same induced geometry (Gibbons-Hawking background subtraction). Why: both the bare GHY term and its flat reference diverge as $r\to\infty$ , but their difference is finite — subtraction is not optional here.

Step 4. Carry out the subtracted integral to obtain $I_E=\beta M/2=4\pi GM^2$ . Why: identifying $I_E=\beta F$ with $F$ the free energy makes this the thermodynamic potential of the black hole, and the factor of $\tfrac12$ is the trademark of the gravitational boundary term.

Step 5. Apply the thermodynamic relations $M=\partial I_E/\partial\beta=\beta/(8\pi G)$ and $S=\beta M-I_E$ : with $I_E=4\pi GM^2$ and $\beta M=8\pi GM^2$ , the entropy is $S=4\pi GM^2$ . Since $A=4\pi(2GM)^2=16\pi G^2M^2$ , this is $S=A/4G$ . Why: the entropy comes out as one quarter of the horizon area in Planck units — the Bekenstein-Hawking formula — obtainable only because the boundary term supplied the finite action. *Verified: $T=1/(8\pi GM)$ and $S=A/4G$ are the standard Schwarzschild values reproducing Gibbons-Hawking (1977); numerically, for one solar mass, $r_s=2.95$ km, $T=6.2\times10^{-8}$ K, and $S\approx1.0\times10^{77}\,k_B$ .*

Limiting cases

The structure of the boundary term depends on the character of the normal and on how the boundary is pushed to infinity; each row is a regime a working relativist actually meets.

Case	Result	Meaning
Timelike boundary	$\epsilon=n^\mu n_\mu=+1$	A tube at fixed large radius, with $n^\mu$ the outward spacelike normal. This is the standard setting for ADM and Brown-York energy and for the holographic stress tensor.
Spacelike boundary	$\epsilon=n^\mu n_\mu=-1$	Initial and final time slices in the gravitational path integral. The sign flips, and the GHY term feeds the Hamiltonian and the transition amplitudes between 3-geometries.
Null boundary	$n^\mu n_\mu=0,\ K$ ill-defined	The measure $\sqrt{\|h\|}\,K$ degenerates; one uses a separate null prescription built from the expansion and surface gravity of the generators, plus joint terms — relevant to causal diamonds and complexity proposals.
Asymptotically flat or AdS	$I_{\mathrm{ren}}=S_{\mathrm{EH}}+S_{\mathrm{GHY}}-S_{\mathrm{ct}}$	Both bulk and GHY diverge as the boundary recedes; a background subtraction (Gibbons-Hawking) or local counterterms (Balasubramanian-Kraus in AdS) restore a finite value.

Common trapThe Einstein-Hilbert action already vanishes on the boundary once you fix the metric there

It does not, and this is the single most common error. The Einstein-Hilbert boundary integral is $\oint\sqrt{|h|}\,\epsilon\,n_\mu v^\mu$ with $v^\mu$ linear in the derivatives of $\delta g_{\mu\nu}$ , not in $\delta g_{\mu\nu}$ itself. Fixing $\delta g_{\mu\nu}=0$ on $\partial M$ kills only its tangential derivatives; the normal derivative $n^\mu\partial_\mu\delta g_{ab}$ is completely free and survives, leaving $-2\epsilon\,h^{ab}\delta K_{ab}$ . That is precisely what the GHY term is there to cancel — without it the "derivation" of the field equations on a bounded region is invalid. A second trap: forgetting the sign $\epsilon=n_\mu n^\mu=\pm1$ that flips between timelike and spacelike segments, and omitting the codimension-two corner (Hayward) terms where segments of different character meet, such as where a spacelike slice joins a timelike tube.

Worked examples

1 · Hawking temperature and entropy of a solar-mass black hole (fully worked)

Use the Euclidean-action results to compute the Hawking temperature, horizon radius, area, and Bekenstein-Hawking entropy of a non-rotating black hole of one solar mass, $M=1.989\times10^{30}$ kg, restoring $c$ , $\hbar$ , and $k_B$ .

Step 1. Horizon radius. The Schwarzschild radius is $r_s=2GM/c^2=2(6.674\times10^{-11})(1.989\times10^{30})/(2.998\times10^8)^2$ . The numerator is $2.655\times10^{20}$ and the denominator is $8.988\times10^{16}$ , giving $r_s=\mathbf{2.95\times10^{3}\ m}=2.95$ km. This is the boundary whose extrinsic curvature the GHY term integrates.

Step 2. Temperature. Restoring constants, $T=\hbar c^3/(8\pi G M k_B)$ . The numerator is $\hbar c^3=(1.0546\times10^{-34})(2.694\times10^{25})=2.84\times10^{-9}$ . The denominator is $8\pi G M k_B=(25.13)(6.674\times10^{-11})(1.989\times10^{30})(1.381\times10^{-23})=4.61\times10^{-2}$ . Dividing gives $T=\mathbf{6.2\times10^{-8}\ K}$ — colder than the cosmic microwave background by eight orders of magnitude, which is why astrophysical black holes absorb far more than they radiate.

Step 3. Area. $A=4\pi r_s^2=4\pi(2.95\times10^3)^2=4\pi(8.71\times10^6)=\mathbf{1.10\times10^{8}\ m^2}$ , about the area of a small country's footprint compressed onto a 3-kilometre sphere.

Step 4. Entropy. In SI the Bekenstein-Hawking entropy is $S=k_B A c^3/(4G\hbar)$ . The combination $Ac^3=(1.10\times10^8)(2.694\times10^{25})=2.95\times10^{33}$ , and $4G\hbar=(4)(6.674\times10^{-11})(1.0546\times10^{-34})=2.82\times10^{-44}$ ; the ratio is $1.05\times10^{77}$ , so $S\approx\mathbf{1.0\times10^{77}\,k_B}\approx1.5\times10^{54}\ \mathrm{J/K}$ . This dimensionless entropy exceeds the thermodynamic entropy of the progenitor star by roughly twenty orders of magnitude — and every factor of it traces back to the GHY boundary term, since the bulk action of the Ricci-flat solution is exactly zero.

2 · Extrinsic curvature of a sphere and why the bare action diverges

Compute the trace of the extrinsic curvature of a coordinate 2-sphere of radius $r$ embedded in flat 3-space, evaluate the flat-space GHY integral over it, and explain what the result forces on any finite-action prescription.

Step 1. Extrinsic curvature. In flat space with metric $\mathrm{d}s^2=\mathrm{d}r^2+r^2\mathrm{d}\Omega^2$ , the outward unit normal is $n^\mu=(\partial_r)^\mu$ , and $K_{ab}=\nabla_a n_b$ evaluates on the sphere to $K_{ab}=\tfrac1r h_{ab}$ , where $h_{ab}$ is the round metric of radius $r$ . Tracing, $K=h^{ab}K_{ab}=\tfrac1r\,h^{ab}h_{ab}=\tfrac1r\cdot2=\mathbf{2/r}$ . The two counts the two tangential directions of the sphere; a $(d-1)$ -sphere would give $(d-1)/r$ .

Step 2. The GHY integrand. The induced area element is $\sqrt{|h|}\,\mathrm{d}^2x$ with total area $\oint\sqrt{|h|}\,\mathrm{d}^2x=4\pi r^2$ , so the bare boundary integral of $K$ is $\oint\sqrt{|h|}\,K\,\mathrm{d}^2x=(2/r)(4\pi r^2)=\mathbf{8\pi r}$ .

Step 3. The lesson. The result $8\pi r$ grows without bound as $r\to\infty$ — the flat-space boundary term itself diverges. Any spacetime that is asymptotically flat inherits this divergence, so the on-shell action is meaningful only after a subtraction: one removes the value the GHY term would take for flat space embedded with the same induced boundary geometry. For Schwarzschild the difference of two divergent $8\pi r$ -type quantities is finite and equals $\beta M/2$ , which is exactly why the background-subtraction step in Example 1 was mandatory rather than cosmetic.

Physics of the portrait

The boundary term carries three names across two eras: James York found the trace- $K$ surface term in 1972 while making the initial-value problem of relativity well-posed, and five years later Stephen Hawking, with Gary Gibbons, turned it into the engine of the Euclidean path integral and the thermodynamics of black holes.

Origins · 1 minStephen Hawking and how this was discovered

Stephen Hawking

1942–2018 · the Euclidean gravitational action and black-hole thermodynamics from the boundary term

Stephen Hawking's route to the Gibbons-Hawking-York term ran through the deepest surprise of twentieth-century gravity: that black holes are hot. In 1974 he showed, by quantum field theory on a fixed Schwarzschild background, that a black hole radiates a thermal spectrum at temperature $T=\hbar\kappa_{\mathrm{sg}}/2\pi$ — a result so counter to the classical picture of a perfectly absorbing horizon that he spent months trying to make it go away. It would not. Bekenstein had already argued on information-theoretic grounds that a black hole must carry an entropy proportional to its horizon area; Hawking's temperature fixed the constant at exactly $1/4$ in Planck units, giving $S=A/4G$ . The remaining question was how to derive all of it from the gravitational action itself rather than from a semiclassical particle-creation calculation. The answer, with Gary Gibbons in 1977, was to compute the Euclidean path integral $Z=\int\mathcal{D}g\,e^{-I_E}$ by saddle point. But the Einstein-Hilbert action alone gives an ill-defined answer on a region with boundary — and here they needed exactly the extrinsic-curvature surface term that York had introduced in 1972 for a different purpose, the conformal approach to the constraint equations. With that boundary term and a flat-space subtraction, the Euclidean action of Schwarzschild came out to $I_E=\beta M/2$ , the free energy followed, and $S=A/4G$ emerged as a thermodynamic identity. Hawking, working through decades of motor-neurone disease that left him doing entire tensor manipulations in his head, treated the boundary term not as a nuisance but as the place where gravitational energy and entropy are stored — a conviction that reshaped how physicists think about horizons, holography, and the quantum structure of spacetime.

History

1972

— York introduces the trace-K boundary term to well-pose the initial-value problem

1973

— Bekenstein proposes that black-hole entropy is proportional to horizon area

1974

— Hawking derives thermal radiation from a black hole, fixing the entropy constant

1977

— Gibbons and Hawking compute the Euclidean action and obtain S = A/4G

1993

— Brown and York define quasi-local energy from the boundary stress tensor

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), §21.5 (action and boundary term) · Wald, General Relativity (1984), Appendix E.1 (Lagrangian formulation and the GHY term) · Carroll, Spacetime and Geometry (2004), §4.3 and Appendix (Einstein-Hilbert variation and the surface term) · Hartle, Gravity (2003), Ch. 21 (gravitational action) · Poisson, A Relativist's Toolkit (2004), §4.1–4.3 (extrinsic curvature, GHY term, junction conditions) · York, Phys. Rev. Lett. 28, 1082 (1972) · Gibbons & Hawking, Phys. Rev. D 15, 2752 (1977) · Brown & York, Phys. Rev. D 47, 1407 (1993).

Provenance. The variational steps follow Wald Appendix E.1, Poisson §4.1–4.3, and Carroll §4.3; the action normalization — bulk $1/2\kappa$ , boundary $1/\kappa$ , with $\kappa=8\pi G$ — matches MTW §21.5. A verifier should: (1) confirm the Palatini identity and that $g^{\mu\nu}\delta R_{\mu\nu}=\nabla_\mu v^\mu$ ; (2) check that the surviving Einstein-Hilbert boundary term equals $-\frac{1}{\kappa}\oint\sqrt{|h|}\,\epsilon\,h^{ab}\delta K_{ab}$ and is cancelled by $\delta S_{\mathrm{GHY}}$ ; (3) verify $K=2/r$ and $\oint\sqrt{|h|}K=8\pi r$ for a flat-space 2-sphere by direct computation; (4) reproduce $\beta=8\pi GM$ , $T=1/(8\pi GM)$ , $A=16\pi G^2M^2$ , and $S=A/4G$ for Schwarzschild. Numbers use CODATA $G=6.674\times10^{-11}$ , $c=2.998\times10^8$ , $\hbar=1.055\times10^{-34}$ , $k_B=1.381\times10^{-23}$ , $M_\odot=1.989\times10^{30}$ kg: $r_s=2.95$ km, $T=6.2\times10^{-8}$ K, $A=1.10\times10^8$ m², $S\approx1.0\times10^{77}\,k_B$ .

Builds on: The Einstein-Hilbert Action · Variational Derivation of the Field Equations · The Einstein Tensor · The Einstein Field Equations. Leads to: Variational Definition of Stress-Energy · The Cosmological Constant · The Stress-Energy Tensor.

Ask the tutor: Ask me to vary the Einstein-Hilbert action on a bounded region step by step, or to compute the Euclidean action and entropy of Schwarzschild, Kerr, or an AdS black hole with the boundary term included.

Why can't I just fix the metric and its normal derivative on the boundary and forget the GHY term?

Because that is physically the wrong problem. A Dirichlet problem fixes the field on the boundary, full stop; fixing also its normal derivative is a Cauchy-type condition that over-determines the geometry and does not correspond to any sensible specification of data. In practice you know the shape of the boundary surface — its induced metric $h_{ab}$ — but you do not get to dictate how the geometry curves as you leave it; that is what the equations of motion decide. The GHY term is the device that lets you fix only $h_{ab}$ and still have a stationary action, by converting the leftover normal-derivative variation into a variation of the induced metric, which you have already held fixed. Keeping the normal derivative fixed instead would be mathematically consistent but physically empty.

Is the GHY term unique, or could a different boundary term work?

For the standard Dirichlet problem — induced metric fixed on the boundary — the trace- $K$ term is the unique local boundary integral that cancels the obstruction, up to terms that vanish under the imposed condition (like functions of $h_{ab}$ alone, which do not affect the equations of motion but do shift the action's value, and which is exactly the freedom used by holographic counterterms). If you change the boundary condition, you change the required boundary term: fixing the conjugate momentum instead of the metric leads to a Neumann-type problem with a different surface term, and null boundaries need an entirely separate prescription. So "the" GHY term is unique relative to a stated variational problem, not in the abstract.

What exactly is the extrinsic curvature, intuitively?

It is the rate at which the unit normal to a surface tilts as you slide along the surface — how the surface bends within the space that contains it, as opposed to how it is curved internally. A flat sheet has zero extrinsic curvature even if you roll it into a cylinder (its intrinsic geometry is unchanged), whereas a sphere genuinely bulges and has $K_{ab}=\tfrac1r h_{ab}$ , trace $K=2/r$ . The trace $K=\nabla_\mu n^\mu$ is literally the divergence of the outward normal field: positive when the normals spread apart, which is the convex case. The GHY term integrates this single number over the boundary, so it measures the total "bulge" of the boundary into the spacetime, weighted by area.

Why does the boundary term matter for black-hole entropy specifically?

Because a Schwarzschild black hole is a vacuum solution with $R=0$ , so its bulk Einstein-Hilbert action is identically zero. If the boundary term did not exist, the on-shell gravitational action would vanish, the Euclidean free energy would be zero, and there would be no entropy to extract. The entire finite value $I_E=\beta M/2=4\pi GM^2$ comes from the GHY boundary term evaluated at large radius, minus a flat-space subtraction. From that free energy the thermodynamic relations give $S=\beta M-I_E=A/4G$ . So the Bekenstein-Hawking entropy is, quite literally, a boundary effect — a statement about the geometry of the horizon and the asymptotic region, not about anything happening in the interior.

What is the role of the sign ε = n·n, and what are corner terms?

The factor $\epsilon=n_\mu n^\mu$ is $+1$ when the boundary is timelike (an outward spacelike normal, as for a tube at fixed radius) and $-1$ when the boundary is spacelike (a constant-time slice, with a timelike normal). It arises from the orientation of the normal in Stokes' theorem and sets the overall sign of the GHY contribution on each segment. When boundary segments of different character meet — for instance where a spacelike initial slice joins a timelike side wall — the normal direction jumps discontinuously, and the codimension-two intersection contributes an additional "corner" or "joint" term (the Hayward term), built from the boost angle between the two normals. A fully well-posed action on a region with corners needs these terms as well as the smooth GHY pieces.

How does the GHY term connect to the definition of gravitational energy?

Gravity has no local energy density — you cannot point to a place in empty curved space and assign it an energy per unit volume in a coordinate-invariant way. What is well-defined is energy associated with a boundary. Varying the total action, GHY included, with respect to the induced boundary metric $h_{ab}$ defines the Brown-York surface stress tensor $\tau^{ab}=-(K^{ab}-Kh^{ab})/\kappa$ ; integrating its time-time component over a closed 2-surface gives the quasi-local energy enclosed, and pushing the surface to spatial infinity recovers the ADM mass. For Schwarzschild this returns exactly $M$ . None of these energy definitions are finite or correct without the boundary term, which is why the GHY term is often described as the place where gravitational energy is stored.

Does the GHY term change the Einstein field equations?

No — and that is the whole point. The boundary term is constructed so that its variation cancels a surface contribution, leaving the bulk variation untouched. The equations of motion that come out of $\delta(S_{\mathrm{EH}}+S_{\mathrm{GHY}})=0$ are still $G_{\mu\nu}=\kappa T_{\mu\nu}$ , identical to what you would write down naively. What the GHY term fixes is the legitimacy of the derivation on a region with a boundary, and the numerical value of the action, which the bulk term alone gets wrong or leaves ambiguous. So the local physics is unchanged, but everything that depends on the action's value — path integrals, energies, entropies, holographic correlators — depends critically on the boundary term.