Einstein–Hilbert action

Derivation of Einstein field equations

Suppose that the full action of the theory is given by the Einstein–Hilbert term plus a term ${\displaystyle {\mathcal{L}}_{\mathrm{M}}}$ describing any matter fields appearing in the theory,

Requiring the variation of this action with respect to the inverse metric be zero yields

${\displaystyle \begin{array}{cc}0& =\delta S\\ & =\int \left[\frac{1}{2\kappa }\frac{\delta \left(\sqrt{-g}R\right)}{\delta g^{\mu \nu }}+\frac{\delta \left(\sqrt{-g}{\mathcal{L}}_{\mathrm{M}}\right)}{\delta g^{\mu \nu }}\right]\delta g^{\mu \nu }{\mathrm{d}}^{4}x\\ & =\int \left[\frac{1}{2\kappa }\left(\frac{\delta R}{\delta g^{\mu \nu }}+\frac{R}{\sqrt{-g}}\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu }}\right)+\frac{1}{\sqrt{-g}}\frac{\delta \left(\sqrt{-g}{\mathcal{L}}_{\mathrm{M}}\right)}{\delta g^{\mu \nu }}\right]\delta g^{\mu \nu }\sqrt{-g}{\mathrm{d}}^{4}x.\end{array}}$.

Since this equation should hold for any variation ${\displaystyle \delta g^{\mu \nu }}$, it implies that

is the equation of motion for the metric field. The right hand side of this equation is (by definition) proportional to the stress–energy tensor,

${\displaystyle T_{\mu \nu }:=\frac{-2}{\sqrt{-g}}\frac{\delta (\sqrt{-g}{\mathcal{L}}_{\mathrm{M}})}{\delta g^{\mu \nu }}=-2\frac{\delta {\mathcal{L}}_{\mathrm{M}}}{\delta g^{\mu \nu }}+g_{\mu \nu }{\mathcal{L}}_{\mathrm{M}}}$.

To calculate the left hand side of the equation we need the variations of the Ricci scalar ${\displaystyle R}$ and the determinant of the metric. These can be obtained by standard textbook calculations such as the one given below, which is strongly based on the one given in Carroll (2004).

Variation of the Ricci scalar

The variation of the Ricci scalar follows from varying the Riemann curvature tensor, and then the Ricci curvature tensor.

The first step is captured by the Palatini identity

${\displaystyle \delta R_{\sigma \nu }\equiv \delta {R^{\rho }}_{\sigma \rho \nu }={\mathrm{\nabla }}_{\rho }\left(\delta {\mathrm{\Gamma }}_{\nu \sigma }^{\rho }\right)-{\mathrm{\nabla }}_{\nu }\left(\delta {\mathrm{\Gamma }}_{\rho \sigma }^{\rho }\right)}$.

Using the product rule, the variation of the Ricci scalar ${\displaystyle R=g^{\sigma \nu }{R}_{\sigma \nu }}$ then becomes

${\displaystyle \begin{array}{cc}\delta R& =R_{\sigma \nu }\delta g^{\sigma \nu }+g^{\sigma \nu }\delta R_{\sigma \nu }\\ & =R_{\sigma \nu }\delta g^{\sigma \nu }+{\mathrm{\nabla }}_{\rho }\left(g^{\sigma \nu }\delta {\mathrm{\Gamma }}_{\nu \sigma }^{\rho }-g^{\sigma \rho }\delta {\mathrm{\Gamma }}_{\mu \sigma }^{\mu }\right),\end{array}}$

where we also used the metric compatibility ${\displaystyle {\mathrm{\nabla }}_{\sigma }{g}^{\mu \nu }=0}$, and renamed the summation indices ${\displaystyle (\rho ,\nu )\to (\mu ,\rho )}$ in the last term.

When multiplied by ${\displaystyle \sqrt{-g}}$, the term ${\displaystyle {\mathrm{\nabla }}_{\rho }\left(g^{\sigma \nu }\delta {\mathrm{\Gamma }}_{\nu \sigma }^{\rho }-g^{\sigma \rho }\delta {\mathrm{\Gamma }}_{\mu \sigma }^{\mu }\right)}$ becomes a total derivative, since for any vector ${\displaystyle A^{\lambda }}$ and any tensor density ${\displaystyle \sqrt{-g}{A}^{\lambda }}$, we have

${\displaystyle \sqrt{-g}{\mathrm{\nabla }}_{\mu }{A}^{\mu }={\mathrm{\nabla }}_{\mu }\left(\sqrt{-g}{A}^{\mu }\right)={\mathrm{\partial }}_{\mu }\left(\sqrt{-g}{A}^{\mu }\right)}$.

By Stokes' theorem, this only yields a boundary term when integrated. The boundary term is in general non-zero, because the integrand depends not only on ${\displaystyle \delta g^{\mu \nu },}$ but also on its partial derivatives ${\displaystyle {\mathrm{\partial }}_{\lambda }\delta g^{\mu \nu }\equiv \delta {\mathrm{\partial }}_{\lambda }{g}^{\mu \nu }}$; see the article Gibbons–Hawking–York boundary term for details. However, when the variation of the metric ${\displaystyle \delta g^{\mu \nu }}$ vanishes in a neighbourhood of the boundary or when there is no boundary, this term does not contribute to the variation of the action. Thus, we can forget about this term and simply obtain

at events not in the closure of the boundary.

Variation of the determinant

Jacobi's formula, the rule for differentiating a determinant, gives:

${\displaystyle \delta g=\delta det(g_{\mu \nu })=g{g}^{\mu \nu }\delta g_{\mu \nu }}$,

or one could transform to a coordinate system where ${\displaystyle g_{\mu \nu }}$ is diagonal and then apply the product rule to differentiate the product of factors on the main diagonal. Using this we get

${\displaystyle \delta \sqrt{-g}=-\frac{1}{2\sqrt{-g}}\delta g=\frac{1}{2}\sqrt{-g}\left(g^{\mu \nu }\delta g_{\mu \nu }\right)=-\frac{1}{2}\sqrt{-g}\left(g_{\mu \nu }\delta g^{\mu \nu }\right)}$

In the last equality we used the fact that

${\displaystyle g_{\mu \nu }\delta g^{\mu \nu }=-g^{\mu \nu }\delta g_{\mu \nu }}$

which follows from the rule for differentiating the inverse of a matrix

${\displaystyle \delta g^{\mu \nu }=-g^{\mu \alpha }\left(\delta g_{\alpha \beta }\right)g^{\beta \nu }}$.

Thus we conclude that

Equation of motion

Now that we have all the necessary variations at our disposal, we can insert (3) and (4) into the equation of motion (2) for the metric field to obtain

which is the Einstein field equations, and

${\displaystyle \kappa =\frac{8\pi G}{c^4}}$

has been chosen such that the non-relativistic limit yields the usual form of Newton's law of gravity, where ${\displaystyle G}$ is the gravitational constant (see here for details).

Cosmological constant

When a cosmological constant ${\displaystyle \mathrm{\Lambda }}$ is included in the Lagrangian, the action becomes

${\displaystyle S=\int \left[\frac{1}{2\kappa }(R-2\mathrm{\Lambda })+{\mathcal{L}}_{\mathrm{M}}\right]\sqrt{-g}{\mathrm{d}}^{4}x.}$

Taking variations with respect to the inverse metric:

${\displaystyle \begin{array}{cc}\delta S& =\int \left[\frac{\sqrt{-g}}{2\kappa }\frac{\delta R}{\delta g^{\mu \nu }}+\frac{R}{2\kappa }\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu }}-\frac{\mathrm{\Lambda }}{\kappa }\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu }}+\sqrt{-g}\frac{\delta {\mathcal{L}}_{\mathrm{M}}}{\delta g^{\mu \nu }}+{\mathcal{L}}_{\mathrm{M}}\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu }}\right]\delta g^{\mu \nu }{\mathrm{d}}^{4}x\\ & =\int \left[\frac{1}{2\kappa }\frac{\delta R}{\delta g^{\mu \nu }}+\frac{R}{2\kappa }\frac{1}{\sqrt{-g}}\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu }}-\frac{\mathrm{\Lambda }}{\kappa }\frac{1}{\sqrt{-g}}\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu }}+\frac{\delta {\mathcal{L}}_{\mathrm{M}}}{\delta g^{\mu \nu }}+\frac{{\mathcal{L}}_{\mathrm{M}}}{\sqrt{-g}}\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu }}\right]\delta g^{\mu \nu }\sqrt{-g}{\mathrm{d}}^{4}x.\end{array}}$

Using the action principle:

${\displaystyle 0=\delta S=\frac{1}{2\kappa }\frac{\delta R}{\delta g^{\mu \nu }}+\frac{R}{2\kappa }\frac{1}{\sqrt{-g}}\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu }}-\frac{\mathrm{\Lambda }}{\kappa }\frac{1}{\sqrt{-g}}\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu }}+\frac{\delta {\mathcal{L}}_{\mathrm{M}}}{\delta g^{\mu \nu }}+\frac{{\mathcal{L}}_{\mathrm{M}}}{\sqrt{-g}}\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu }}}$

Combining this expression with the results obtained before:

${\displaystyle \begin{array}{cc}\frac{\delta R}{\delta g^{\mu \nu }}& =R_{\mu \nu }\\ \frac{1}{\sqrt{-g}}\frac{\delta \sqrt{-g}}{\delta g^{\mu \nu }}& =\frac{-g_{\mu \nu }}{2}\\ T_{\mu \nu }& ={\mathcal{L}}_{\mathrm{M}}{g}_{\mu \nu }-2\frac{\delta {\mathcal{L}}_{\mathrm{M}}}{\delta g^{\mu \nu }}\end{array}}$

We can obtain:

${\displaystyle \begin{array}{cc}\frac{1}{2\kappa }{R}_{\mu \nu }+\frac{R}{2\kappa }\frac{-g_{\mu \nu }}{2}-\frac{\mathrm{\Lambda }}{\kappa }\frac{-g_{\mu \nu }}{2}+\left(\frac{\delta {\mathcal{L}}_{\mathrm{M}}}{\delta g^{\mu \nu }}+{\mathcal{L}}_{\mathrm{M}}\frac{-g_{\mu \nu }}{2}\right)& =0\\ R_{\mu \nu }-\frac{R}{2}{g}_{\mu \nu }+\mathrm{\Lambda }{g}_{\mu \nu }+\kappa \left(2\frac{\delta {\mathcal{L}}_{\mathrm{M}}}{\delta g^{\mu \nu }}-{\mathcal{L}}_{\mathrm{M}}{g}_{\mu \nu }\right)& =0\\ R_{\mu \nu }-\frac{R}{2}{g}_{\mu \nu }+\mathrm{\Lambda }{g}_{\mu \nu }-\kappa T_{\mu \nu }& =0\end{array}}$

With $\kappa ={\displaystyle \frac{8\pi G}{c^4}}$, the expression becomes the field equations with a cosmological constant:

${\displaystyle R_{\mu \nu }-\frac{1}{2}{g}_{\mu \nu }R+\mathrm{\Lambda }{g}_{\mu \nu }=\frac{8\pi G}{c^4}{T}_{\mu \nu }.}$