Einstein Field Equations

Bianchi identity

Previously, we have defined the Riemann curvture tensor as $$R^\alpha_{\beta\mu\nu} := \Gamma^\alpha_{\beta\nu,\mu} - \Gamma^\alpha_{\beta\mu,\nu} + \Gamma^\alpha_{\sigma\mu}\Gamma^\sigma_{\beta\nu} - \Gamma^\alpha_{\sigma\nu}\Gamma^\sigma_{\beta\nu}.$$ Substitute $$\Gamma^\alpha_{\mu\nu,\sigma} = \frac{1}{2}g^{\alpha\beta}(g_{\beta\mu,\nu\sigma}+g_{\beta\nu,\mu\sigma}-g_{\mu\nu,\beta\sigma}),$$ we have $$R^\alpha_{\beta\mu\nu} = \frac{1}{2}g^{\alpha\sigma}(g_{\sigma\nu,\beta\mu}-g_{\sigma\mu,\beta\nu}+g_{\beta\mu,\sigma\nu}-g_{\beta\nu,\sigma\mu}).$$ Lowering an index, $$R_{\alpha\beta\mu\nu}=g_{\alpha\lambda}R^\lambda_{\beta\mu\nu} = \frac{1}{2}(g_{\sigma\nu,\beta\mu}-g_{\alpha\mu,\beta\nu}+g_{\beta\mu,\sigma\nu}-g_{\beta\nu,\sigma\mu}).$$ Since the metric tensor is symmetric and partial derivatives commute, $$R_{\alpha\beta\mu\nu,\lambda} + R_{\alpha\beta\lambda\mu,\nu} + R_{\alpha\beta\nu\lambda,\mu} = 0.$$ Since a manifold is locally flat, if locally the partial derivative is zero, then so is the covariant derivative, i.e. $$R^\gamma_{\beta\gamma\delta;\epsilon} + R^\gamma_{\beta\epsilon\gamma;\delta} + R^\gamma_{\beta\delta\epsilon;\gamma} = 0$$

Contraction of Bianchi Identity

Since Riemann tensor is antisymmetric, $$\begin{align} R^\gamma_{\beta\gamma\delta;\epsilon} - R^\gamma_{\beta\gamma\epsilon;\delta} + R^\gamma_{\beta\delta\epsilon;\gamma} &= 0 \\ R_{\beta\delta;\epsilon} - R_{\beta\epsilon;\delta} + R^\gamma_{\beta\delta\epsilon;\gamma} &= 0\\ g^{\beta\delta}(R_{\beta\delta;\epsilon} - R_{\beta\epsilon;\delta} + R^\gamma_{\beta\delta\epsilon;\gamma}) &= 0\\ R^\delta_{\delta;\epsilon} - R^\delta_{\epsilon;\delta} - R^{\gamma\delta}_{\delta\epsilon;\gamma} &= 0 \\ R^\delta_{\delta;\epsilon} - 2R^\gamma_{\epsilon;\gamma} &= 0\\ \left(R^\gamma_\epsilon - \frac{1}{2}g^\gamma_\epsilon R\right)_{;\gamma} &= 0\\ g^{\beta\delta}\left(R^\gamma_\epsilon - \frac{1}{2}g^\gamma_\epsilon R\right)_{;\gamma} &= 0\\ \left(R^{\gamma\delta}-\frac{1}{2}g^{\gamma\delta}R\right)_{;\gamma} &= 0.\\ \end{align}$$ This is the "equation of motion" for a locally flat manifold.

Hilbert Action

Write down the action as $$S = \int_V R\sqrt{-g}\,d^4x.$$ Under general coordinate transformation, $$\delta S = \int_V \delta\left(R\sqrt{-g}\right)d^4x$$

$$\begin{align} \delta\left(R\sqrt{-g}\right) &= \delta\left(g^{\mu\nu}R_{\mu\nu}\sqrt{-g}\right)\\ &= \left(\delta g^{\mu\nu}\right)R_{\mu\nu}\sqrt{-g} + g^{\mu\nu}\left(\delta R_{\mu\nu}\right)\sqrt{-g} + g^{\mu\nu}R_{\mu\nu}\left(\delta\sqrt{-g}\right)\\ &= \left(\delta g^{\mu\nu}\right)R_{\mu\nu}\sqrt{-g} + g^{\mu\nu}\left(\delta R_{\mu\nu}\right)\sqrt{-g} + R\left(\delta\sqrt{-g}\right).\\ \end{align}$$ By $\delta\sqrt{-g}=-\frac{1}{2}\sqrt{g}g_{\mu\nu}\delta g^{\mu\nu}$, $$\delta\left(R\sqrt{-g}\right)=\left(R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R\right)\sqrt{-g}(\delta g^{\mu\nu}) + g^{\mu\nu}\sqrt{-g}\left(\delta R_{\mu\nu}\right)$$

By the definitions, the Riemann Tensor is $$R^\alpha_{\mu\sigma\nu} = \left(\Gamma^\alpha_{\mu\nu,\sigma}-\Gamma^\alpha_{\mu\sigma,\nu}+\Gamma^\alpha_{\sigma\gamma}\Gamma^\gamma_{\mu\nu}-\Gamma^\alpha_{\nu\gamma}\Gamma^\gamma_{\mu\sigma}\right)$$ and the Ricci Tensor is $$R_{\mu\nu}=R^\alpha_{\mu\alpha\nu}=\Gamma^\alpha_{\mu\nu,\alpha}-\Gamma^\alpha_{\mu\alpha,\nu}+\Gamma^\alpha_{\alpha\gamma}\Gamma^\gamma_{\mu\nu}-\Gamma^\alpha_{\nu\gamma}\Gamma^\gamma_{\mu\alpha}.$$ So, $$\delta R_{\mu\nu} = \delta\Gamma^\alpha_{\mu\nu,\alpha}-\delta\Gamma^\alpha_{\mu\alpha,\nu}+\delta\Gamma^\alpha_{\alpha\gamma}\Gamma^\gamma_{\mu\nu} + \Gamma^\alpha_{\alpha\gamma}\delta\Gamma^\gamma_{\mu\nu} - \delta\Gamma^\alpha_{\nu\gamma}\Gamma^\gamma_{\mu\alpha} - \Gamma^\alpha_{\nu\gamma}\left(\delta\Gamma^\gamma_{\mu\alpha}\right). \tag{1}$$ The covariant derivative of the delta Christoffel symbol is given by $$\left(\delta\Gamma^\rho_{\nu\sigma}\right)_{;\mu}=\left(\delta\Gamma^{\rho}_{\nu\sigma}\right)_{,\mu}-\left(\delta\Gamma^\rho_{\lambda\sigma}\right)\Gamma^\lambda_{\nu\mu} - \Gamma^\gamma_{\sigma\mu}\left(\delta\Gamma^\rho_{\nu\lambda}\right) + \left(\delta\Gamma^\lambda_{\nu\sigma}\right)\Gamma^\rho_{\lambda\mu}.$$ Therefore, $$\left(\delta\Gamma^\alpha_{\mu\alpha}\right)_{;\nu} = \left(\delta\Gamma^\alpha_{\mu\alpha}\right)_{,\nu}-\left(\delta\Gamma^\alpha_{\gamma\alpha}\right)\Gamma^\gamma_{\mu\nu} - \Gamma^\gamma_{\alpha\nu}\left(\delta\Gamma^\alpha_{\mu\gamma}\right) + \left(\delta\Gamma^\gamma_{\mu\alpha}\right)\Gamma^\alpha_{\gamma\nu}, \tag{2}$$ where the first, second and fourth term is the negation of second, third and sixth term of (1) respectively, and $$\left(\delta\Gamma^\alpha_{\mu\nu}\right)_{;\alpha}=\left(\delta\Gamma^\alpha_{\mu\nu}\right)_{,\alpha}-\left(\delta\Gamma^\alpha_{\gamma\nu}\right)\Gamma^\gamma_{\mu\alpha}-\Gamma^\gamma_{\nu\alpha}\left(\delta\Gamma^\alpha_{\mu\gamma}\right) + \left(\delta\Gamma^\gamma_{\mu\nu}\right)\Gamma^\alpha_{\gamma\alpha},$$ where the first, second and fourth term is the first, fifth and fourth term of (1) respectively. Also, the third term of (2) and (3) are the same. Thus, (3) minus (2) gives (1) $$\delta R_{\mu\nu} = \left(\delta\Gamma^\alpha_{\mu\nu}\right)_{;\alpha} - \left(\delta\Gamma^\alpha_{\mu\alpha}\right)_{;\nu}.$$ Hence, $$\begin{align} g^{\mu\nu}\delta R_{\mu\nu}\sqrt{-g} &= g^{\mu\nu}\left(\left(\delta\Gamma^\alpha_{\mu\nu}\right)_{;\alpha}-\left(\delta\Gamma^\alpha_{\mu\alpha}\right)_{;\nu}\right)\sqrt{-g}\\ &= \left[\left(g^{\mu\nu}\delta\Gamma^\alpha_{\mu\nu}\right)_{;\alpha} - \left(g^{\mu\nu}\delta\Gamma^\alpha_{\mu\alpha}\right)_{;\nu}\right]\sqrt{-g} \\ &= \left(g^{\mu\nu}\delta\Gamma^\alpha_{\mu\nu}-g^{\mu\alpha}\delta\Gamma^\nu_{\mu\nu}\right)_{;\alpha}\sqrt{-g}, \end{align}$$ in which dummy variables $\alpha \leftrightarrow \nu$ changed on last line. Define $$A^\alpha := g^{\mu\nu}\delta\Gamma^\alpha_{\mu\nu} - g^{\mu\alpha}\delta\Gamma^\nu_{\mu\nu}.$$ Then, the integral becomes $$\int_Vg^{\mu\nu}\delta R_{\mu\nu}\sqrt{-g}d^4x = \int_V A^\alpha_{;\alpha}\sqrt{-g}d^4x.$$ By divergence theorem, $$\int_Vg^{\mu\nu}\delta R_{\mu\nu}\sqrt{-g}d^4x = \oint A^\alpha n_\alpha \sqrt{-g}d^3S.$$ As boundary terms can be eliminated by redefining the Hilbert action, we can take this term to be zero $$\int g^{\mu\nu}\delta R_{\mu\nu}\sqrt{-g}d^4x = 0.$$ So, the invariance of the action yields $$\left(R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R\right) = 0,$$ which is definitely zero upon convariant derivative.

The action is called the Hilbert action. We see that Hilbert action defined in this way reproduces the contracted Bianchi identiy, the "equation of motion" of a locally flat manifold.

Einstein-Hilbert Action

We write the whole action together with the matter field Lagrangian, $$S = \int_V \left[\frac{1}{2K}R+L_M\right]\sqrt{-g}\,d^4x,$$ where $L_M$ is some matter fields and $K$ is some constant to make the dimension right. Under general coordinate transformation, $$\delta S = \int_V \delta\left(\left[\frac{1}{2K}R+L_M\right]\sqrt{-g}\right)d^4x.$$

$$\delta\left(L_M\sqrt{-g}\right)=\frac{1}{\sqrt{-g}}\frac{\delta\left(\sqrt{-g}L_M\right)}{\delta g_{\mu\nu}} \left(\delta g^{\mu\nu}\sqrt{-g}\right)$$ If we define the stress energy tensor as $$T_{\mu\nu}:=\frac{-2}{\sqrt{-g}}\frac{\delta \left(\sqrt{-g}L_M\right)}{\delta g^{\mu\nu}},$$ the invariance of the action under general coordinate transformation yields, $$\bbox[5px, border: 2px solid #666]{ R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R - KT^{\mu\nu} = 0. }$$ This is called the Einstein field equations (EFE). Then, by conservation of energy-momentum, $\nabla T^{\mu\nu}_{;\nu} = 0$, $$\left(R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R\right)_{;\nu} = KT^{\mu\nu}_{;\nu} = 0.$$ So, defining the stress energy tensor in this way yields the "equation of motion" of a locally smooth manifold. EFE relates how energy-momentum curves the manifold. We will find $K$ later using Newtonian limit.