Einstein uses Lorentz-Heaviside units!

(If he just set c=1 without further reasons, c wouldn't show up in the expression for $$\chi$$ at all.)

To be honest I have my doubt about that statement of a Wikipedia editor. It seems to me like Einstein also uses energy density but uses units of light meters instead of seconds for time, such that he can write E=m like he does earlier in the book...

Also Einstein uses covariant indices at the x for the geodesic.

(I'm comparing with my earlier result here, which in turn came form comparing Wikipedia with Einstein's "The Meaning of Relativity" coursebook.(German title: "Grundzüge der Relativitätstheorie") )

I assume the sign for $$\chi$$ is convention?

Wikipidia says it has chosen to write the energy-stress tensor in units of energy density, while Einstein in his original publication used units of mass density.

$$\chi = \frac{8\pi K}{c^2} ?!, K=G$$

$$R_{\mu\nu}-\frac12g_{\mu\nu}R= \kappa T_{\mu\nu}$$

$$G_{\mu\nu} = \kappa T_{\mu\nu}$$

$$\frac{d^2x^\mu}{ds^2} + \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{ds} \frac{dx^\beta}{ds} = 0$$

$$\kappa = - \chi$$

$$G_{\mu\nu} := {R_{\mu\nu}-\frac12g_{\mu\nu}R }$$

$$\frac{d^2x^\mu}{ds^2} + \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{ds} \frac{dx^\beta}{ds} = 0$$

$$G_{\mu\nu} = \kappa T_{\mu\nu}$$

$$c=1 \implies R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = 8\pi G T_{\mu\nu}$$

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

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