Einstein–Hilbert action

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The Einstein-Hilbert action in general relativity is the action that yields the Einstein's field equations when varied to obtain equations of motion for the spacetime metric. The action was first proposed by David Hilbert in 1915 (Hilbert, 1915). The derivation of the Einstein equations from an action has several advantages. First of all, it allows for easy unification of general relativity with other classical fields theories (such as Maxwell theory), which are also formulated in terms of an action. In the process the derivation from an action identifies a natural candidate for the source term coupling the metric to matter fields. Moreover, the action allows for the easy identification of conserved quantities through Noether's theorem by studying symmetries of the action.

In general relativity, the action is usually assumed to be a functional of the metric (and matter fields), and the connection is given by the Levi-Civita connection. The Palatini formulation of general relativity assumes the metric and connection to be independent, and varies with respect to both independently, which makes it possible to include fermionic matter fields with non-integral spin.

The action $S [g]$ which gives rise to the vacuum Einstein equations is given by the following integral of the Lagrangian

$S[g]= \int kR \sqrt{-g} \, \mathrm{d}^4x$

where $g = \, \det \, (g_{ab})$ is the determinant of a spacetime Lorentz metric, R is the Ricci scalar, k is the constant c⁴/16πG, the Lagrangian being $R\sqrt{-g}$ , and the integral is taken over a region of spacetime. The Einstein equations in the presence of matter are given by adding the Lagrangian for the matter into the integral. (In this article "Lagrangian" means "Lagrangian density"; other authors occasionally use it to mean the integral of the Lagrangian density over space or over spacetime.)

Note that $\sqrt{-g} \, \mathrm{d}^4x$ is an invariant 4-volume element, so the action can be also written in the following (somewhat more elegant) fashion:

$S[g]= k \int R \, \mathrm{dV}$

1 Derivation of Einstein's field equations
2 Cosmological constant
3 See also
4 References

[edit] Derivation of Einstein's field equations

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

$S = \int \left[ k \, R + \mathcal{L}_\mathrm{M} \right] \sqrt{-g} \, \mathrm{d}^4x$

The action principle then tells us that the variation of this action with respect to the inverse metric is zero, yielding

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

Since this equation should hold for any variation $δ g μν$ , it implies that

$\frac{\delta R}{\delta g^{\mu\nu}} + \frac{R}{\sqrt{-g}} \frac{\delta \sqrt{-g}}{\delta g^{\mu\nu}} = -\frac{1}{k}\frac{1}{\sqrt{-g}}\frac{\delta (\sqrt{-g} \mathcal{L}_\mathrm{M})}{\delta g^{\mu\nu}},$

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

$T_{\mu\nu}:= -2 \frac{1}{\sqrt{-g}}\frac{\delta (\sqrt{-g} \mathcal{L}_\mathrm{M})}{\delta g^{\mu\nu}}.$

To calculate the left hand side of the equation we need the variations of the Ricci scalar R and the determinant of the metric. These can be obtained by standard text book calculations such as the one given below, which is strongly based on the one given in Carroll 2004.

[edit] Variation of the Riemann tensor, the Ricci tensor, and the Ricci scalar

To calculate the variation of the Ricci scalar we calculate first the variation of the Riemann curvature tensor, and then the variation of the Ricci tensor. So, the Riemann curvature tensor is defined as,

${R^\rho}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma},$

Since the Riemann curvature depends only on the Levi-Civita connection $\Gamma^\lambda_{\mu\nu}$ , the variation of the Riemann tensor can be calculated as,

$\delta{R^\rho}_{\sigma\mu\nu} = \partial_\mu\delta\Gamma^\rho_{\nu\sigma} - \partial_\nu\delta\Gamma^\rho_{\mu\sigma} + \delta\Gamma^\rho_{\mu\lambda} \Gamma^\lambda_{\nu\sigma} + \Gamma^\rho_{\mu\lambda} \delta\Gamma^\lambda_{\nu\sigma} - \delta\Gamma^\rho_{\nu\lambda} \Gamma^\lambda_{\mu\sigma} - \Gamma^\rho_{\nu\lambda} \delta\Gamma^\lambda_{\mu\sigma}.$

Now, since $\delta\Gamma^\rho_{\nu\mu}$ is the difference of two connections, it is a tensor and we can thus calculate its covariant derivative,

$\nabla_\lambda (\delta \Gamma^\rho_{\nu\mu} ) = \partial_\lambda (\delta \Gamma^\rho_{\nu\mu} ) + \Gamma^\rho_{\sigma\lambda} \delta\Gamma^\sigma_{\nu\mu} - \Gamma^\sigma_{\nu\lambda} \delta \Gamma^\rho_{\sigma\mu} - \Gamma^\sigma_{\mu\lambda} \delta \Gamma^\rho_{\nu\sigma}$

We can now cleverly observe that the expression for the variation of Riemann curvature tensor above is equal to the difference of two such terms,

$\delta R^\rho{}_{\sigma\mu\nu} = \nabla_\mu (\delta \Gamma^\rho_{\nu\sigma}) - \nabla_\nu (\delta \Gamma^\rho_{\mu\sigma}).$

We may now obtain the variation of the Ricci curvature tensor simply by contracting two indices of the variation of the Riemann tensor,

$\delta R_{\mu\nu} \equiv \delta R^\rho{}_{\mu\rho\nu} = \nabla_\rho (\delta \Gamma^\rho_{\nu\mu}) - \nabla_\nu (\delta \Gamma^\rho_{\rho\mu}).$

The Ricci scalar is defined as

$R = g^{\mu\nu} R_{\mu\nu}.\!$

Therefore, its variation with respect to the inverse metric $g μν$ is given by

$\begin{align} \delta R &= R_{\mu\nu} \delta g^{\mu\nu} + g^{\mu\nu} \delta R_{\mu\nu}\\ &= R_{\mu\nu} \delta g^{\mu\nu} + \nabla_\sigma \left( g^{\mu\nu} \delta\Gamma^\sigma_{\nu\mu} - g^{\mu\sigma}\delta\Gamma^\rho_{\rho\mu} \right) \end{align}$

In the second line we used the previously obtained result for the variation of the Ricci curvature and the metric compatibility of the covariant derivative, $\nabla_\sigma g^{\mu\nu} = 0$ .

The last term $\nabla_\sigma ( g^{\mu\nu} \delta\Gamma^\sigma_{\nu\mu} - g^{\mu\sigma}\delta\Gamma^\rho_{\rho\mu} )$ is a total derivative and thus by Stokes' theorem only yields a boundary term when integrated. Hence when the variation of the metric $δ g μν$ vanishes at infinity, this term does not contribute to the variation of the action. And we thus obtain,

$\frac{\delta R}{\delta g^{\mu\nu}} = R_{\mu\nu}.$

[edit] Variation of the determinant

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

$\,\! \delta g = g \, g^{\mu\nu} \delta g_{\mu\nu}$

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

Using this we get

$\begin{align} \delta \sqrt{-g} &= -\frac{1}{2\sqrt{-g}}\delta g &= \frac{1}{2} \sqrt{-g} (g^{\mu\nu} \delta g_{\mu\nu}) &= -\frac{1}{2} \sqrt{-g} (g_{\mu\nu} \delta g^{\mu\nu}),\end{align}$