4

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.

Qmechanic
  • 201,751
user1157
  • 155

1 Answers1

1

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 $P$.

Lee Mosher
  • 269
  • Can we modify the Gaussian curvature to become the Ricci scalar curvature ? – user1157 Apr 02 '18 at 08:51
  • 1
    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. – Lee Mosher Apr 02 '18 at 11:51