# Einstein and Riemann curvature tensor

Riemann curvature tensor is dirrectly related to a path dependence of parallel transport. I read that Einstein first thought of this tensor to be the one that goes into his field equation but it didn't fit. Now, I know why this was problematic. There is a clear explanation of that. What I am asking is, is there a formulation of theory of gravity where equations give us directly $$R$$ and not $$G$$? $$G$$ is Einsteins tensor and $$R$$ Riemanns. Because, non-relativistic equation was used and directly translated, with energy-momentum tensor. Is there some other tensor related to the physics which would give $$R$$ instead of $$G$$? $$G$$ itself is not curvature...but from $$G$$ I guess we could calculate $$R$$ so there should be some equation that gives $$R$$ directly?

The formulation you are looking for exists. Starting off from Einstein's field equations: ($$R_{ik}$$ Ricci tensor, $$R$$ curvature scalar and $$T_{ik}$$ the stress tensor and $$g_{ik}$$ the metric tensor, $$\kappa=\frac{8\pi G}{c^4}$$ with $$G$$ gravitational constant)

$$R_{ik} -\frac{1}{2}g_{ik} R = \kappa T_{ik}$$

One can alternatively move one index up in all tensors:

$$R_i^k -\frac{1}{2}\delta_i^k R = \kappa T_i^k$$

and then taking the trace (we put the indices $$i$$ and $$k$$ equal and sum over it according to Einstein's summation convention):

$$R = - \kappa T$$

with $$T = T_i^i\equiv\sum^3_{i=0} T_i^i$$. We substitute this expression for $$R$$ in Einstein's original field equations and bring the term with $$T$$ on the right side:

$$R_{ik} = \kappa\left(T_{ik} -\frac{1}{2} g_{ik} T\right)$$

and if you like with the Riemann curvature tensor according to the definition of the Ricci tensor $$R_{ik}:=g^{lm} R_{limk}$$

$$g^{lm} R_{limk} = \kappa\left(T_{ik} -\frac{1}{2} g_{ik} T\right)$$

• Great! Isnt this then better? From geometrical point of view? Jan 27, 2019 at 21:45
• When one searches for a vacuum solution of the EFEs, one can immediately start off with $R_{ik}=0$ which is easier than $R_{ik}-0.5g_{ik}R=0$ Jan 27, 2019 at 22:12

Here is one difference: If you believe that the EFEs should follow from a stationary action principle (as Hilbert did), then the Einstein tensor $$G_{\mu\nu}$$ is preferred over the Ricci curvature tensor $$R_{\mu\nu}$$, as only the former has the form of a functional derivative $$\frac{1}{\sqrt{|g|}}\frac{\delta S}{\delta g^{\mu\nu}}$$ of some action functional $$S$$. (For the Einstein tensor $$G_{\mu\nu}$$ this is the Einstein-Hilbert action.)

• Yes, I know, but the key word is "believe" . Why would anyone choose that action? Jan 28, 2019 at 12:59
• The point is whether an action exists or not. Its specific form is irrelevant for the argument presented here. Actions play an important role in modern physics. Jan 28, 2019 at 13:13
• Yes yes...I see. I was just, you know, sayin.. Jan 28, 2019 at 13:15
• That is a great way to think of it. Jan 28, 2019 at 19:15