How did Einstein arrive at the right hand side of his general relativity tensor equation?

It seems Einstein postulated the right hand side of his field equations… I see in books that tell the story of how he arrived at what to put on the right hand side of his tensor equations, the physics that causes the curvature tensor on the left hand side, that he tried this and that while trying to satisfy the requirements of having a valid tensor equation, not breaking known physical laws, and collapsing back to Newtonian mechanics for conditions of low speed and weak gravity. Is there more than one form that satisfies these requirements? It seems perhaps yes… If so, how did he know to choose a form that receives empirical support?

• The RHS contains no information until you specify $T^{\mu\nu}$. One can use the Newtonian limit to help with that. Jan 11, 2015 at 1:39
• Jan 11, 2015 at 2:47
• Comment to the post (v2): What is left and right hand side of EFE according to your convention? Sep 26, 2016 at 21:25

First let's consider the geodesic equation.

$$\frac{d^2x^\mu}{d\tau^2}=-\Gamma_{\alpha\beta}^\mu \frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}$$

In the Newtonian limit, the t-t component of $g^{\mu\nu}$ plays the role of the gravitational potential $\Phi$ in classical mechanics. The right hand side of the Einstein field equations does make sense. He wanted a theory to describe how spacetime is affected by matter and energy, and so it seemed logical to chose the stress energy tensor. In the Newtonian limit, he wanted the equation to reduce to

$$\nabla^2 \Phi = 4\pi \rho G.$$

If the t-t component of the stress energy tensor played the role of $\rho$, the left side was to be made up of second derivatives of the metric tensor. Initially, he tried using the Ricci tensor, but it didn't work out. The field equations had to have the property that covariant derivative of both sides had to be 0. This is true because $\nabla_\mu T_{\mu \nu}= 0$ (The continuity equation.) Since using the Ricci tensor failed, he tried constructed a tensor and got the equation:

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

The constant in front of the stress energy tensor was obtained by in depth analysis of the conditions at the Newtonian limit.