#2 – Hilbert – Einstein action – Part 1

The Hilbert-Einstein action is:

$S_{HE}=\int\left(\frac{1}{2\kappa}R+\mathcal{L}_{M}\right)\sqrt{-g}\ d^{4}x.$

Einstein equations (the evolution equations of the metrig $\mathbf{g}$) can be found if we set $\delta S_{HE}=0$.

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

Rearranging:

$\int\left[\frac{1}{2\kappa}\left(\frac{\delta R}{\delta g_{\mu\nu}}+\frac{R}{\sqrt{-g}}\frac{\delta\sqrt{-g}}{\delta g_{\mu\nu}}\right)+\frac{1}{\sqrt{-g}}\frac{\delta (\mathcal{L}_{M}\sqrt{-g})}{\delta g_{\mu\nu}}\right]\sqrt{-g}\ \delta g_{\mu\nu}\ d^{4}x = 0$

Since this must be true for any variation $\delta g_{\mu\nu}$, what’s inside the brackets must be zero, so:

$\frac{1}{2\kappa}\left(\frac{\delta R}{\delta g_{\mu\nu}}+\frac{R}{\sqrt{-g}}\frac{\delta\sqrt{-g}}{\delta g_{\mu\nu}}\right)+\frac{1}{\sqrt{-g}}\frac{\delta (\mathcal{L}_{M}\sqrt{-g})}{\delta g_{\mu\nu}}= 0$

or

$\frac{1}{\kappa}\left(\frac{\delta R}{\delta g_{\mu\nu}}+\frac{R}{\sqrt{-g}}\frac{\delta\sqrt{-g}}{\delta g_{\mu\nu}}\right)=-\frac{2}{\sqrt{-g}}\frac{\delta (\mathcal{L}_{M}\sqrt{-g})}{\delta g_{\mu\nu}}.$

The r.h.s. is defined as the energy-stress tensor:

$T^{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta (\mathcal{L}_{M}\sqrt{-g})}{\delta g_{\mu\nu}}=-\frac{2}{\sqrt{-g}}\left(\frac{\delta \sqrt{-g}}{\delta g_{\mu\nu}}\mathcal{L}_{M}+\sqrt{-g}\frac{\delta \mathcal{L}_{M}}{\delta g_{\mu\nu}}\right)= -\frac{2}{\sqrt{-g}}\left(-\frac{1}{2\sqrt{-g}}g\ g^{\mu\nu}\mathcal{L}_{M}+\sqrt{-g}\frac{\delta \mathcal{L}_{M}}{\delta g_{\mu\nu}}\right)=-2\frac{\delta \mathcal{L}_{M}}{\delta g_{\mu\nu}}+g^{\mu\nu}\mathcal{L}_{M},$

where we used the result

$\frac{\delta \sqrt{-g}}{\delta g_{\mu\nu}}=\frac{\partial\sqrt{-g}}{\partial g}\frac{\delta g}{\delta g_{\mu\nu}}=-\frac{1}{2}\sqrt{-g}\ g^{\mu\nu}.$

Then

$T^{\mu\nu}=\frac{1}{\kappa}\left(\frac{\delta R}{\delta g_{\mu\nu}}+\frac{R}{\sqrt{-g}}\frac{\delta\sqrt{-g}}{\delta g_{\mu\nu}}\right)=\frac{1}{\kappa}\left(\frac{\delta R}{\delta g_{\mu\nu}}-\frac{1}{2}Rg^{\mu\nu}\right).$