Deriving the Einstein Equation from the Action

The Einstein equation follows from a single stationary-action calculation only when the metric variable, curvature sign, matter stress tensor, normalization, and boundary assumptions are kept synchronized. The derivation below varies the inverse metric, retains the surface divergence until its treatment is declared, and then checks the factor $8\pi G$ against the Newtonian limit.

Required background. We use variational and presymplectic geometry, curvature, and hypersurface data.

Helpful background. Special relativity, tensor calculus, and classical field actions provide the limiting and matter-sector checks.

Total action and declared variation

Take

$$S[g,\psi] =\frac{1}{16\pi G}\int_{\mathcal M}d^4x\,\sqrt{-g}(R-2\Lambda) +S_{\mathrm m}[g,\psi]+S_{\partial\mathcal M}.$$

Either let $\delta g^{ab}$ have compact support in the interior, or choose $S_{\partial\mathcal M}$ and boundary data so that the total surface variation vanishes. On a smooth non-null Dirichlet boundary, the appropriate completion is the Gibbons–Hawking–York term. The bulk derivation is the same in both cases.

Define the matter stress tensor by

$$T_{ab}=-\frac{2}{\sqrt{-g}} \frac{\delta S_{\mathrm m}}{\delta g^{ab}}.$$

On the matter equations of motion and up to matter boundary terms,

$$\delta S_{\mathrm m} =-\frac12\int_{\mathcal M}d^4x\,\sqrt{-g}\,T_{ab}\delta g^{ab}.$$

This sign belongs to inverse-metric variation with signature $(-,+,+,+)$; changing variables requires translating it rather than copying the formula unchanged.

Gravitational and cosmological variations

The curvature identity is

$$\delta(\sqrt{-g}R) =\sqrt{-g}\,G_{ab}\delta g^{ab} +\sqrt{-g}\,\nabla_a v^a.$$

The cosmological density gives

$$\delta(-2\Lambda\sqrt{-g}) =\Lambda\sqrt{-g}\,g_{ab}\delta g^{ab}.$$

Therefore, after the declared boundary treatment,

$$\delta S =\int_{\mathcal M}d^4x\,\sqrt{-g} \left[ \frac{G_{ab}+\Lambda g_{ab}}{16\pi G} -\frac12T_{ab} \right]\delta g^{ab}.$$

Arbitrariness of the interior symmetric variation gives

$$G_{ab}+\Lambda g_{ab}=8\pi G\,T_{ab}.$$

The factor $8\pi G$ is the ratio of $1/2$ in the stress-tensor definition to $1/(16\pi G)$ in the gravitational action. Reall carries out the curvature variation, defines the matter stress tensor with the same inverse-metric sign, and obtains the normalized field equation in Reall 2013, §§ 10.2–10.3, pp. 116–120, Eqs. (10.14)–(10.34), PDF.

Repeating the calculation with the direct metric

If $g_{ab}$ is the independent variable,

$$\delta g^{ab}=-g^{ac}g^{bd}\delta g_{cd}.$$

The matter variation becomes

$$\delta S_{\mathrm m} =\frac12\int\sqrt{-g}\,T^{ab}\delta g_{ab},$$

while the gravitational bulk term becomes

$$\delta S_{\mathrm{grav}} =-\frac{1}{16\pi G}\int\sqrt{-g} (G^{ab}+\Lambda g^{ab})\delta g_{ab}.$$

Both terms reverse their relative presentation, and stationarity gives the identical raised-index field equation. This second derivation is an efficient check on every sign.

Newtonian normalization check

Set $\Lambda=0$ and consider a weak stationary field with

$$g_{00}=-(1+2\Phi), \qquad |\Phi|\ll1,$$

and slowly moving matter with $T_{00}\simeq\rho$ and $T\simeq-\rho$. It is safer not to assert a formula for $G_{00}$ from $g_{00}$ alone: at linear order $G_{00}$ also depends on the spatial metric perturbation and the gauge choice. Instead, trace-reverse the four-dimensional field equation,

$$R_{ab}=8\pi G\left(T_{ab}-\frac12g_{ab}T\right).$$

Stationarity makes the leading curvature component

$$R_{00}\simeq\nabla^2\Phi,$$

while the source is

$$8\pi G\left(T_{00}-\frac12g_{00}T\right) \simeq4\pi G\rho.$$

Consequently,

$$\nabla^2\Phi=4\pi G\rho.$$

The slow-particle geodesic equation gives $d^2\mathbf x/dt^2=-\boldsymbol\nabla\Phi$. Thus the same normalization reproduces both Poisson’s equation and Newtonian free fall without silently choosing the spatial perturbation. The weak-field curvature and trace-reversal steps are worked out in Reall 2013, § 8.2, pp. 80–82, Eqs. (8.19)–(8.35), PDF. A prefactor $1/(8\pi G)$ in the action would miss this matching by a factor of two under the declared stress definition.

Identities and what stationarity does not prove

The contracted Bianchi identity makes $\nabla^a(G_{ab}+\Lambda g_{ab})=0$. Diffeomorphism invariance of the matter action implies $\nabla^aT_{ab}=0$ when matter equations hold. This compatibility is a check, not an additional metric equation.

Tracing the vacuum equation supplies another quick diagnostic. With $T_{ab}=0$ in four dimensions, one obtains $R=4\Lambda$ and hence $R_{ab}=\Lambda g_{ab}$. Substituting that relation back gives zero on both sides. If a proposed action variation instead yields $R_{ab}=-\Lambda g_{ab}$ for the displayed $R-2\Lambda$ convention, the determinant variation or cosmological sign has been mishandled.

For matter, taking the divergence before imposing the matter equations shows why stationarity must include every varied field. The metric equation alone would demand a conserved source, but a trial matter configuration off its equations of motion need not have conserved stress-energy. The full diffeomorphism identity relates its divergence to the matter Euler–Lagrange expressions; conservation emerges on shell.

Stationarity supplies local field equations within the chosen class of variations. It does not determine boundary conditions, guarantee a well-posed initial-boundary problem, choose a global topology, or assert that an extremum is a minimum. In Lorentzian field theory “least action” is generally misleading; stationary action is the correct statement.

Exercises

Suppose the matter variation were mistakenly written with a plus sign while keeping inverse-metric variation. What field equation would result, and which limit exposes the error?

Solution

Using $\delta S_{\mathrm m}=+(1/2)\int\sqrt{-g}T_{ab}\delta g^{ab}$ would give $G_{ab}+\Lambda g_{ab}=-8\pi GT_{ab}$. With $\Lambda=0$, trace reversal then gives $R_{00}\simeq-4\pi G\rho$ instead of $+4\pi G\rho$. Since $R_{00}\simeq\nabla^2\Phi$ in the weak stationary limit, this predicts $\nabla^2\Phi=-4\pi G\rho$, the wrong sign for the potential of positive mass.

References

• Reall, Harvey S. General Relativity. Cambridge Part III lecture notes (2013). PDF.