Editing Einstein–Hilbert action (section)

==Derivation of Einstein field equations==
Suppose that the full action of the theory is given by the Einstein–Hilbert term plus a term <math>\mathcal{L}_\mathrm{M}</math> describing any matter fields appearing in the theory.

{{NumBlk|:|<math>S = \int \left[ \frac{1}{2\kappa} R + \mathcal{L}_\mathrm{M} \right] \sqrt{-g} \, \mathrm{d}^4 x </math>.|{{EquationRef|1}}}}

The [[stationary-action principle]] then tells us that to recover a physical law, we must demand that the variation of this [[Action (physics)|action]] with respect to the inverse metric be zero, yielding

:<math>\begin{align}
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}^4x \\
  &= \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}^4x
\end{align}</math>.

Since this equation should hold for any variation <math>\delta g^{\mu\nu}</math>, it implies that

{{NumBlk|:|<math>\frac{\delta R}{\delta g^{\mu\nu}} + \frac{R}{\sqrt{-g}} \frac{\delta \sqrt{-g}}{\delta g^{\mu\nu}} = -2\kappa \frac{1}{\sqrt{-g}} \frac{\delta (\sqrt{-g} \mathcal{L}_\mathrm{M})}{\delta g^{\mu\nu}}</math>|{{EquationRef|2}}}}

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]],<ref>{{Citation
 | last = Blau
 | first = Matthias
 | title = Lecture Notes on General Relativity
 | journal = 
 | volume = 
 | url = http://www.blau.itp.unibe.ch/newlecturesGR.pdf
 | page = 207 
 | date = August 27, 2024
}}</ref>

:<math>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}</math>.

To calculate the left hand side of the equation we need the variations of the Ricci scalar <math>R</math> 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).<ref>{{Citation|author=Carroll, Sean M. |author-link=Sean M. Carroll |title=Spacetime and Geometry: An Introduction to General Relativity |location=San Francisco |publisher=Addison-Wesley |date=2004 |isbn=978-0-8053-8732-2}}</ref>

===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]]

:<math>
  \delta R_{\sigma\nu} \equiv \delta {R^\rho}_{\sigma\rho\nu} =
  \nabla_\rho \left( \delta \Gamma^\rho_{\nu\sigma} \right) - \nabla_\nu \left( \delta \Gamma^\rho_{\rho\sigma} \right)</math>.

Using the [[product rule]], the variation of the Ricci scalar <math>R = g^{\sigma\nu} R_{\sigma\nu}</math> then becomes

:<math>\begin{align}
\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^\rho_{\nu\sigma} - g^{\sigma\rho} \delta \Gamma^\mu_{\mu\sigma} \right),
\end{align}</math>

where we also used the [[Metric connection#Riemannian connection|metric compatibility]] <math>\nabla_\sigma g^{\mu\nu} = 0</math>, and renamed the summation indices <math>(\rho,\nu) \rightarrow (\mu,\rho)</math> in the last term.

When multiplied by <math>\sqrt{-g}</math>, the term <math>\nabla_\rho \left( g^{\sigma\nu} \delta\Gamma^\rho_{\nu\sigma} - g^{\sigma\rho}\delta\Gamma^\mu_{\mu\sigma} \right)</math> becomes a [[total derivative]], since for any [[Ricci calculus|vector]] <math>A^\lambda</math> and any [[tensor density]] <math>\sqrt{-g}\,A^\lambda</math>, we have 
:<math>
        \sqrt{-g} \, A^\lambda_{;\lambda} =
  \left(\sqrt{-g} \, A^\lambda\right)_{;\lambda} =
  \left(\sqrt{-g} \, A^\lambda\right)_{,\lambda}
</math> or <math>
  \sqrt{-g} \, \nabla_\mu A^\mu =
    \nabla_\mu\left(\sqrt{-g} \, A^\mu\right) =
  \partial_\mu\left(\sqrt{-g} \, A^\mu\right)
</math>.

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 <math>\delta g^{\mu\nu},</math> but also on its partial derivatives <math>\partial_\lambda\, \delta g^{\mu\nu} \equiv \delta\, \partial_\lambda g^{\mu\nu}</math>; see the article [[Gibbons–Hawking–York boundary term]] for details. However, when the variation of the metric <math>\delta g^{\mu\nu}</math> 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

{{NumBlk|:|<math>\frac{\delta R}{\delta g^{\mu\nu}} = R_{\mu\nu}</math>.|{{EquationRef|3}}}}

at [[event (relativity)|events]] not in the [[closure (topology)|closure]] of the boundary.

===Variation of the determinant===
[[Jacobi's formula]], the rule for differentiating a [[determinant#Derivative|determinant]], gives:

:<math>\delta g = \delta \det(g_{\mu\nu}) = g g^{\mu\nu} \delta g_{\mu\nu}</math>,

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

:<math>\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)</math>

In the last equality we used the fact that

:<math>g_{\mu\nu}\delta g^{\mu\nu} = -g^{\mu\nu} \delta g_{\mu\nu}</math>

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

:<math>\delta g^{\mu\nu} = - g^{\mu\alpha} \left( \delta g_{\alpha\beta} \right) g^{\beta\nu}</math>.

Thus we conclude that

{{NumBlk|:|<math>\frac{1}{\sqrt{-g}} \frac{\delta \sqrt{-g}}{\delta g^{\mu\nu} } = -\frac{1}{2} g_{\mu\nu}</math>.|{{EquationRef|4}}}}

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

{{NumBlk|:|<math>R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = \frac{8 \pi G}{c^4} T_{\mu\nu}</math>,|{{EquationRef|5}}}}

which is the [[Einstein field equations]], and

:<math>\kappa = \frac{8\pi G}{c^4}</math>

has been chosen such that the non-relativistic limit yields [[Newton's law of universal gravitation|the usual form of Newton's gravity law]], where <math>G</math> is the [[gravitational constant]] (see [[Einstein field equations#The correspondence principle|here]] for details).