# Stress-energy tensor explicitly in terms of the metric tensor

I am trying to write the Einstein field equations $$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu} R=\frac{8\pi G}{c^4}T_{\mu\nu}$$ in such a way that the Ricci curvature tensor $R_{\mu\nu}$ and scalar curvature $R$ are replaced with an explicit expression involving the metric tensor $g_{\mu\nu}$. Using the equations $R=g^{\mu\nu}R_{\mu\nu}$ (relating the scalar curvature to the trace of the Ricci curvature tensor) and $R_{\mu\nu}=R^\lambda_{\mu\lambda\nu}$ (relating the Ricci curvature tensor to the trace of the Riemann curvature tensor), would anyone be willing to give recommendations on how to proceed, or already know the equation?

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I can tell you right now they would be enormously complicated for non zero $T_{\mu\nu}$. The Riemann tensor is a series of derivatives and products of Christoffel symbols, which in turn are also a series of derivatives and products of the metric. – TeeJay Feb 27 '14 at 2:27
In fact, the exercise is rather trivial but just utterly boring and unenlightening. – Danu Feb 27 '14 at 2:35

The Einstein equations are some of the most complicated PDE's people study. There is no shortcut for this, you just have to do all the horrific algebra. Start with the trace-reversed Einstein equation $$R_{\mu \nu}=8\pi G(T_{\mu \nu}-\frac{1}{2}Tg_{\mu \nu})$$ Use the equation for Ricci in terms of the Christoffel Connection $$R_{\mu \nu}=2\Gamma^\alpha_{\mu [\nu,\alpha]}+2\Gamma^{\alpha}_{\lambda[\alpha}\Gamma^\lambda_{\nu]\mu}$$ Plug in $$\Gamma^i_{kl}=\frac{1}{2}g^{im}(g_{mk,l}+g_{ml,k}-g_{kl,m})$$ In the end you will just have lots of terms with up to two derivatives of the metric.