# 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? – Žarko Tomičić Jan 27 '19 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 $R_{ik}-0.5g_{ik}R=0$ – Frederic Thomas Jan 27 '19 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? – Žarko Tomičić Jan 28 '19 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. – Qmechanic Jan 28 '19 at 13:13
• Yes yes...I see. I was just, you know, sayin.. – Žarko Tomičić Jan 28 '19 at 13:15
• That is a great way to think of it. – R. Rankin Jan 28 '19 at 19:15