I know about curvature by this notation $$\tau=\frac{dt}{ds}$$ the change of tangent vector with respect to arc length $s$ .

I also know about Ricci scalar curvature is $$g^{ij}R_{ij}=R$$

I know the formulas. But i want to really know about their differences and some helpful geometrical interpretations.


Your formula $\tau = \frac{dt}{ds}$ requires a correction, namely $t$ is not just any old tangent vector but is instead the unit length tangent vector. Also, that notion of curvature, which is known as "geodesic curvature", applies only to curves (1-dimensional objects) in space, the one parameter being $s$.

On the other hand Ricci curvature applies only to 2-or-higher dimensional objects in space.

So, you're not going to find much in the way of a direct comparison between those two types of curvature.

Still, though, there are some indirect comparisons in some limited situations. One particularly close connection occurs for a 2-dimensional surface $S$ in 3-dimensional space. The Ricci curvature at a point $P \in S$ is equal to the Gaussian curvature (because in 2 dimensions there's nothing to contract in the contraction formula that you give). And the Gaussian curvature is equal to the product of two different geodesic curvatures, namely the so-called "principle curvatures" which are the maximum and minimum values of $\tau$ for curves passing through $P$.

  • $\begingroup$ Can we modify the Gaussian curvature to become the Ricci scalar curvature ? $\endgroup$ – user1157 Apr 2 '18 at 8:51
  • 1
    $\begingroup$ As I said, in dimension 2 they are the same. In higher dimensions, there is still a relation between Gaussian curvature and Ricci curvature, the intermediary between the two being the Riemann curvature tensor. $\endgroup$ – Lee Mosher Apr 2 '18 at 11:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.