advanced30 min studyLesson 811 in Physics

Einstein–Hilbert action and boundary terms

Derive the field equations with explicit curvature conventions and identify what a fixed boundary metric actually fixes.

Before you begin

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The question for today

What you’ll understand

Declare curvature, signature, and action units consistently.
Explain why the Dirichlet gravitational action includes a boundary term.
Separate initial constraints from equations that evolve admissible data.

An action principle specifies both bulk dynamics and admissible variations at the edge of a region. Ignoring the second task can produce the correct Einstein equation while leaving the variational problem incomplete. Throughout this module spacetime has four dimensions and signature (−,+,+,+). We define curvature by $[\nabla_\mu,\nabla_\nu]V^\rho=R^\rho{}_{\sigma\mu\nu}V^\sigma$ and $R_{\sigma\nu}=R^\rho{}_{\sigma\rho\nu}$ . The Einstein tensor is $G_{\mu\nu}=R_{\mu\nu}-Rg_{\mu\nu}/2$ . These conventions fix the signs of the cosmological and source terms rather than leaving them implicit.

An invariant action with declared units

Take $x^0=ct$ and the other coordinates to have length units, with dimensionless metric components in such a chart. The scalar curvature and cosmological constant have inverse square length units. The factor multiplying the bulk integral then gives action units, joule seconds. Vary the inverse metric while holding the matter fields independently variable. With stress-energy expressed as energy density and pressure in SI units, the matter variation includes an inverse factor of c. This convention produces the familiar source coefficient without hiding a time-coordinate conversion.

The determinant variation supplies a term proportional to the metric; varying the Ricci scalar supplies the Ricci tensor plus a total derivative. The interior coefficient of an arbitrary metric variation therefore contains the Einstein tensor. This calculation assumes the Levi-Civita connection, so the metric determines a torsion-free, metric-compatible connection. An independent-connection variational principle, higher-curvature action, or nonminimal matter coupling requires its own hypotheses. The cosmological term remains present even when all nongravitational fields vanish.

S_{\mathrm{EH}}=\frac{c^3}{16\pi G}\int_Md^4x\sqrt{-g}\,(R-2\Lambda),\qquad \delta S_m=-\frac1{2c}\int_Md^4x\sqrt{-g}\,T_{\mu\nu}\delta g^{\mu\nu}

G is Newton’s constant, c is the speed of light, and T has energy-density units in an orthonormal frame. The displayed matter variation defines the metric stress-energy convention with length-valued coordinates.↔ Scroll wide equations horizontally.

Fixed induced geometry does not fix normal derivatives

For a smooth non-null boundary, let the outward unit normal satisfy $n^\mu n_\mu=\epsilon$ , where epsilon is plus one for a timelike boundary and minus one for a spacelike boundary. Write h for the induced metric and define extrinsic curvature by projecting the covariant derivative of the normal onto the boundary. The trace K records how that boundary sits in the bulk. Its sign changes when the normal reverses, which makes an orientation convention indispensable.

The Einstein–Hilbert variation contains normal derivatives of the metric variation. Holding the induced metric fixed does not make those terms disappear. Adding the Gibbons–Hawking–York term cancels them, leaving a boundary variation proportional to the variation of h itself. This is a Dirichlet problem, not a demand to freeze both position-like and derivative-like data. Piecewise smooth boundaries need joint terms; null boundaries require another formulation. Compactly supported variations avoid these edge issues only for the restricted purpose of deriving interior equations.

S_{\mathrm{GHY}}=\frac{c^3}{8\pi G}\int_{\partial M}d^3y\sqrt{|h|}\,\epsilon K,\qquad K_{ab}=h_a{}^\mu h_b{}^\nu\nabla_\mu n_\nu

This expression applies to smooth non-null boundary pieces and fixed induced metric. It is not a complete action prescription for corners or null edges.↔ Scroll wide equations horizontally.

Equations of motion include restrictions on initial data

Stationarity of the complete action gives the field equation below. Its ten component equations are linked by differential identities and coordinate freedom. On a spacelike initial hypersurface, the normal-normal and normal-tangent projections impose the Hamiltonian and momentum constraints on its intrinsic metric and extrinsic curvature. The remaining equations evolve admissible data after a choice of lapse, shift, and an appropriate evolution formulation. One cannot assign every metric component and its time derivative arbitrarily and expect a solution.

For smooth vacuum data satisfying the constraints, the Cauchy theory provides a local spacetime development, unique in the appropriate sense up to diffeomorphism. Suitable matter models need their own well-posed equations and compatibility assumptions. Local existence does not promise geodesic completeness or absence of later singularities. Exact metrics in the following lessons are special analytic solutions with stated domains, not a replacement for this initial-value problem. The small derivative toy below isolates the boundary issue without claiming to reproduce gravitational dynamics.

G_{\mu\nu}+\Lambda g_{\mu\nu}=\frac{8\pi G}{c^4}T_{\mu\nu}

Both sides have inverse square length units in the declared chart. The constraint/evolution distinction refers to projections relative to a chosen spacelike foliation, not a division into different physical laws.↔ Scroll wide equations horizontally.

Work it through

Endpoint values leave derivative variations free

Use the dimensionless toy functional $I[q]=\int_0^1q''(x)\,dx$ . Let $\delta q=\eta x(1-x)$ , with constant infinitesimal eta. Check whether fixed endpoint values make its variation zero, then find a boundary term that cancels the variation.

The variation vanishes at both endpoints, so the permitted change preserves q there. Its derivative is $\delta q'=\eta(1-2x)$ , giving endpoint derivative variations eta and minus eta.
Integrating the second derivative gives $\delta I=[\delta q']_0^1=-2\eta$ . Fixed values alone therefore do not make this functional stationary.
Add $B=-[q']_0^1$ . Its variation is plus twice eta and cancels the bulk total derivative exactly. This deliberately simple toy removes the entire functional, whereas the gravitational boundary term leaves nontrivial bulk equations.

The result

The fixed-endpoint variation gives −2η; adding the stated derivative boundary term cancels it and illustrates why endpoint values and derivatives differ.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Read a value and its slope separately

Reset and set coefficient $a=1$ , position $x=0$ ; predict height zero and slope zero. Repeat from Reset with both controls $a=1$ , $x=1$ ; predict height one and slope two. The lab evaluates only $y=ax^2$ , its slope, and ordinary area. It does not vary a gravitational action or impose a boundary condition; the two readouts simply distinguish values from derivatives.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA boundary prescription must specify which quantity is fixed; a function value and its derivative carry different information.

Pause & reason

Check your understanding

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Read the checkpoint explanation

Fixing the induced metric makes the remaining boundary variation vanish after the GHY term is included; fixing its normal derivative as well would specify a different, generally excessive prescription.

Your turn

Try it yourself

In four-dimensional vacuum with cosmological constant, trace the Einstein equation and determine both R and the Ricci tensor. Does vacuum force Minkowski geometry?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Contract with the inverse metric; its contraction with itself gives the spacetime dimension.

Compare with the solution

The trace of the Einstein tensor is minus R in four dimensions. With no matter, $-R+4\Lambda=0$ , so $R=4\Lambda$ . Substitution gives $R_{\mu\nu}=\Lambda g_{\mu\nu}$ . Nonzero Lambda therefore excludes Minkowski space as a vacuum solution. Even when Lambda vanishes, Ricci-flatness alone does not imply zero Riemann curvature: Schwarzschild provides a later counterexample.

Keep the idea

Take it with you

Action units and the definition of stress-energy determine the powers of c.
The GHY term completes a smooth non-null Dirichlet variational problem.
Constraint-satisfying initial data admit local evolution under precise hypotheses, not a universal global regularity guarantee.

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