Einstein-Hilbert action

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The Einstein-Hilbert action is a mathematical object (an action) that is used to derive Einstein's field equations of general relativity.

In 1913, Albert Einstein was working on the equivalence of gravity and acceleration. He realized the geometry would work if spacetime was curved, and not flat as had always been the classical assumption. Einstein proposed that mass and energy would warp spacetime and that gravitational force was merely an expression of the spacetime curvature. It has been said that Hilbert independently found the equations a few days before Einstein (discussion). These equations were based on the Einstein-Hilbert action.

In general relativity, the action is assumed to be a functional only of the metric, i.e. the connection is given by the Levi-Civita connection. Some extensions of general relativity assume the metric and connection to be independent, however, and vary with respect to both independently.

The action $S [g]$ which gives rise to the vacuum Einstein equations is given by the following integral of the Lagrangian density (sometimes just called the Lagrangian)

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

where $g = \, \det \, (g_{ab})$ is the determinant of a spacetime Lorentz metric, R is the Ricci scalar, k is a constant which depends on the units chosen (see below), the Lagrangian being $R\sqrt{-g}$ , and the integral is taken over a region of spacetime. In the rival field theory of gravitation called Brans-Dicke theory, k is replaced by a scalar field.

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

[edit] Derivation of Einstein's field equations

To derive the full field equations, it is natural to assume that an extra term - a matter Lagrangian $\mathcal{L}_\mathrm{M}$ be added:

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

The variation with respect to the reciprocal of the metric yields

$0 = \delta S = \int k \delta (R \sqrt{-g}) + \delta (\sqrt{-g} \mathcal{L}_\mathrm{M}) \mathrm{d}^4x$

$= \int \left[ k \left( \frac{\delta R}{\delta g^{mn}} + R \frac{\delta \sqrt{-g}}{\sqrt{-g}\delta g^{mn} } \right) + \frac{1}{\sqrt{-g}} \frac{\delta (\sqrt{-g} \mathcal{L}_\mathrm{M})}{\delta g^{mn}} \right] (\delta g^{mn}) \sqrt{-g}\, \mathrm{d}^4x$

The following are standard text book calculations which have in part been taken from Carroll (see References).

[edit] Variation of the Ricci scalar

Recall that the Riemann curvature tensor is

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

If we calculate its variation, the first two terms on the right side become

$\partial_\mu\delta\Gamma^\rho_{\nu\sigma} - \partial_\nu\delta\Gamma^\rho_{\mu\sigma}$

Notice that $\delta\Gamma^\rho_{\nu\sigma}$ is a tensor. So we can transform to a coordinate system where $\Gamma^\rho_{\nu\sigma}$ is zero. This causes the terms without partial derivatives to vanish. Also, in the terms with partial derivatives, they become covariant derivatives. So we get

$\delta R^r{}_{mln} = \nabla_l (\delta \Gamma^r_{nm}) - \nabla_n (\delta \Gamma^r_{lm})$

where δΓ is the variation of the Levi-Civita connection (which is not expanded in terms of δg because those details are not required subsequently).

Due to R=g^mnR_mn and R_mn=R^r_{mr n} the variation of the scalar curvature is

$\delta R = R_{mn} \delta g^{mn} + \nabla_s ( g^{mn} \delta\Gamma^s_{nm} - g^{ms}\delta\Gamma^r_{rm} )$

where the second term yields a surface term by Stokes' theorem as long as k is a constant and does not contribute when the variation δg^mn is supposed to vanish at infinity. So we get

$\frac{\delta R}{\delta g^{mn}} = R_{mn}$

[edit] Variation of the determinant

The following property of a determinant

$\,\! g = \exp \mathrm{tr} \ln \, (g_{mn})$

is used to determine the variation:

$\,\! \delta g=g (g^{mn} \delta g_{mn})$

or one can tranform to a coordinate system where $g_{mn}\!$ is diagonal and then apply the product rule to differentiate the product of factors on the main diagonal.

Using this we get

$\delta \sqrt{-g} = \frac{1}{2\sqrt{-g}}\cdot -\delta g = \frac{1}{2} \sqrt{-g} (g^{mn} \delta g_{mn})$

Now use $0=\delta(4)=\delta(g^{mn}g_{mn})=(\delta g^{mn})g_{mn}+g^{mn}\delta g_{mn}\!$ to infer

$\delta \sqrt{-g} = \frac{1}{2} \sqrt{-g} (-g_{mn} \delta g^{mn})$

So we conclude

$\frac{\delta \sqrt{-g}}{\sqrt{-g}\delta g^{mn} } = -\frac{1}{2}\, g_{mn}$

[edit] Variation of matter term

The matter term is used to define the stress-energy tensor T_mn.

$-\frac{1}{2}\, T_{mn} := \frac{1}{\sqrt{-g}} \frac{\delta (\sqrt{-g} \mathcal{L}_\mathrm{M})}{\delta g^{mn}} = -\frac{1}{2}g_{mn}\mathcal{L}_\mathrm{M} + \frac{\delta \mathcal{L}_\mathrm{M}}{\delta g^{mn}}$

Multiplying by minus two, we get that the stress-energy tensor may be written as

$T_{mn} = g_{mn} \mathcal{L}_\mathrm{M} - 2 \frac{\delta \mathcal{L}_\mathrm{M}}{\delta g^{mn}}$

where the functional derivative can be replaced by a partial derivative, if the matter Lagrangian does not depend on derivatives of the metric (as is common in general relativity). Also the derivative of the matter Lagrangian must be made symmetric in its indices m and n, if it is not already so, because the variation of g is symmetric since g itself is constrained to be symmetric.

Note that this is the conventional definition in general relativity, although there are several inequivalent definitions, in particular the canonical stress-energy tensor.