Einstein-Hilbert Action
Get Einstein%E2%80%93Hilbert Action essential facts below. View Videos or join the Einstein%E2%80%93Hilbert Action discussion. Add Einstein%E2%80%93Hilbert Action to your PopFlock.com topic list for future reference or share this resource on social media.
Einstein%E2%80%93Hilbert Action

The Einstein-Hilbert action (also referred to as Hilbert action[1]) in general relativity is the action that yields the Einstein field equations through the principle of least action. With the metric signature, the gravitational part of the action is given as[2]

${\displaystyle S={1 \over 2\kappa }\int R{\sqrt {-g}}\,\mathrm {d} ^{4}x,}$

where ${\displaystyle g=\det(g_{\mu \nu })}$ is the determinant of the metric tensor matrix, ${\displaystyle R}$ is the Ricci scalar, and ${\displaystyle \kappa =8\pi Gc^{-4}}$ is Einstein's constant (${\displaystyle G}$ is the gravitational constant and ${\displaystyle c}$ is the speed of light in vacuum). If it converges, the integral is taken over the whole spacetime. If it does not converge, ${\displaystyle S}$ is no longer well-defined, but a modified definition where one integrates over arbitrarily large, relatively compact domains, still yields the Einstein equation as the Euler-Lagrange equation of the Einstein-Hilbert action.

The action was first proposed by David Hilbert in 1915.

## Discussion

The derivation of equations from an action has several advantages. First of all, it allows for easy unification of general relativity with other classical field 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 Einstein equations in the presence of matter are given by adding the matter action to the Einstein-Hilbert action.

## Derivation of Einstein's 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.

The action principle then tells us that to recover a physical law, we must demand that the variation of this action with respect to the inverse metric be zero, yielding

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

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 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

${\displaystyle {R^{\rho }}_{\sigma \mu \nu }=\partial _{\mu }\Gamma _{\nu \sigma }^{\rho }-\partial _{\nu }\Gamma _{\mu \sigma }^{\rho }+\Gamma _{\mu \lambda }^{\rho }\Gamma _{\nu \sigma }^{\lambda }-\Gamma _{\nu \lambda }^{\rho }\Gamma _{\mu \sigma }^{\lambda }}$.

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

${\displaystyle \delta {R^{\rho }}_{\sigma \mu \nu }=\partial _{\mu }\delta \Gamma _{\nu \sigma }^{\rho }-\partial _{\nu }\delta \Gamma _{\mu \sigma }^{\rho }+\delta \Gamma _{\mu \lambda }^{\rho }\Gamma _{\nu \sigma }^{\lambda }+\Gamma _{\mu \lambda }^{\rho }\delta \Gamma _{\nu \sigma }^{\lambda }-\delta \Gamma _{\nu \lambda }^{\rho }\Gamma _{\mu \sigma }^{\lambda }-\Gamma _{\nu \lambda }^{\rho }\delta \Gamma _{\mu \sigma }^{\lambda }}$.

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

${\displaystyle \nabla _{\mu }\left(\delta \Gamma _{\nu \sigma }^{\rho }\right)=\partial _{\mu }(\delta \Gamma _{\nu \sigma }^{\rho })+\Gamma _{\mu \lambda }^{\rho }\delta \Gamma _{\nu \sigma }^{\lambda }-\Gamma _{\mu \nu }^{\lambda }\delta \Gamma _{\lambda \sigma }^{\rho }-\Gamma _{\mu \sigma }^{\lambda }\delta \Gamma _{\nu \lambda }^{\rho }}$.

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

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

We may now obtain the variation of the Ricci curvature tensor simply by contracting two indices of the variation of the Riemann tensor, and get the Palatini identity:

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

The Ricci scalar is defined as

${\displaystyle R=g^{\sigma \nu }R_{\sigma \nu }}$.

Therefore, its variation with respect to the inverse metric ${\displaystyle g^{\sigma \nu }}$ is given by

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

In the second line we used the metric compatibility of the covariant derivative, ${\displaystyle \nabla _{\sigma }g^{\mu \nu }=0}$, and the previously obtained result for the variation of the Ricci curvature (in the second term, renaming the dummy indices ${\displaystyle \rho }$ and ${\displaystyle \nu }$ to ${\displaystyle \mu }$ and ${\displaystyle \rho }$ respectively).

The last term,

${\displaystyle \nabla _{\rho }\left(g^{\sigma \nu }\delta \Gamma _{\nu \sigma }^{\rho }-g^{\sigma \rho }\delta \Gamma _{\mu \sigma }^{\mu }\right)}$, i.e. ${\displaystyle \nabla _{\rho }A^{\rho }\equiv A^{\lambda }{}_{;\lambda }}$ with ${\displaystyle A^{\rho }=g^{\sigma \nu }\delta \Gamma _{\nu \sigma }^{\rho }-g^{\sigma \rho }\delta \Gamma _{\mu \sigma }^{\mu }}$,

multiplied by ${\displaystyle {\sqrt {-g}}}$, becomes a total derivative, since for any vector ${\displaystyle A^{\lambda }}$ and any tensor density ${\displaystyle {\sqrt {-g}}\,A^{\lambda }}$ we have:

${\displaystyle {\sqrt {-g}}\,A_{;\lambda }^{\lambda }=({\sqrt {-g}}\,A^{\lambda })_{;\lambda }=({\sqrt {-g}}\,A^{\lambda })_{,\lambda }}$ or ${\displaystyle {\sqrt {-g}}\,\nabla _{\mu }A^{\mu }=\nabla _{\mu }\left({\sqrt {-g}}\,A^{\mu }\right)=\partial _{\mu }\left({\sqrt {-g}}\,A^{\mu }\right)}$

and thus by Stokes' theorem 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 ${\displaystyle \partial _{\lambda }\,\delta g^{\mu \nu }\equiv \delta \,\partial _{\lambda }g^{\mu \nu }}$; 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. And we thus obtain

at events not in the closure of the boundary.

### Variation of the determinant

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

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

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 Einstein's field equation, 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 gravity law, where ${\displaystyle G}$ is the gravitational constant (see here for details).

## Cosmological constant

When a cosmological constant ? is included in the Lagrangian, the action:

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

Taking variations with respect to the inverse metric:

{\displaystyle {\begin{aligned}&\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 {\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 {\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{aligned}}}

Using the action principle:

{\displaystyle {\begin{aligned}&\delta S=0\\&{\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 {\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 }}}=0\\\end{aligned}}}

Combining this expression with the results obtained before:

{\displaystyle {\begin{aligned}&{\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 {{\mathcal {L}}_{\mathrm {M} }}{\delta g^{\mu \nu }}}\\\end{aligned}}}

We can obtain:

{\displaystyle {\begin{aligned}&{\frac {1}{2\kappa }}R_{\mu \nu }+{\frac {R}{2\kappa }}{\frac {-g_{\mu \nu }}{2}}-{\frac {\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 }+\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 }+\Lambda g_{\mu \nu }-\kappa T_{\mu \nu }=0\end{aligned}}}

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

${\displaystyle R_{\mu \nu }-{\frac {1}{2}}g_{\mu \nu }R+\Lambda g_{\mu \nu }={\frac {8\pi G}{c^{4}}}T_{\mu \nu }.}$