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?


2 Answers 2


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 $R_{ik}:=g^{lm} R_{limk}$

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

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

  • $\begingroup$ Yes, I know, but the key word is "believe" . Why would anyone choose that action? $\endgroup$ Commented Jan 28, 2019 at 12:59
  • $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Commented Jan 28, 2019 at 13:13
  • $\begingroup$ Yes yes...I see. I was just, you know, sayin.. $\endgroup$ Commented Jan 28, 2019 at 13:15
  • $\begingroup$ That is a great way to think of it. $\endgroup$
    – R. Rankin
    Commented Jan 28, 2019 at 19:15

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