# Riemann curvature scalar, Ricci Scalar.What does they measure ?

1. Jun 25, 2013

### zn5252

hello
Can you perhaps explain what does the Riemann curvature scalar R measure? or is just an abstract entity ?
What does the Ricci tensor measure ?
I just want to grasp this and understand what they do.
cheers,

typo: What DO they measure in the title.

2. Jun 25, 2013

### WannabeNewton

Let $M$ be a Riemannian manifold and let $p \in M$. Given any unit vector $X\in T_p M$, pick an orthonormal basis $\{E_{i}\}$ for $T_p M$ such that $E_1 = X$. It can be shown that $\text{Ricci}(X,X) = \sum _{k = 2}^{n}K(E_1,E_k)$ where $K(E_1,E_i)$ is the sectional curvature of the plane spanned by $E_1 = X$ and $E_i$. So the Ricci tensor can be interpreted as the sum of sectional curvatures of planes spanned by a unit vector $X$ in the tangent space and other elements of an orthonormal basis to which $X$ belongs.

Then, $R = R_{j}{}{}^{j} = \sum_{j \neq k} (E_j,E_k)$ i.e. the Ricci scalar measures the sum of all sectional curvatures of planes spanned by distinct pairs of elements in a given orthonormal basis.

There are many other ways to interpret it as well. See for example: http://en.wikipedia.org/wiki/Ricci_curvature

Also see problem 9 of chapter 4 in Do Carmo "Riemannian Geometry" for an absolutely beautiful relationship between the Ricci scalar and the Ricci tensor in terms of the area of a sphere in the tangent space and the integral of the Ricci tensor over that sphere.

3. Jun 25, 2013

### Staff: Mentor

There are pretty decent intuitive definitions at math.ucr.edu/home/baez/gr/

4. Jun 25, 2013

### zn5252

Thanks to all.
In wikipedia it is mentioned :
"Specifically, the scalar curvature represents the amount by which the volume of a geodesic ball in a curved Riemannian manifold deviates from that of the standard ball in Euclidean space"

Does this mean that the volume of the sphere in the curved space which is 4/3Pi*r*r*r , which is also the same in eucledian space yields a null value for the Scalar curvature ?

5. Jun 25, 2013

### WannabeNewton

Let $\epsilon > 0$ be sufficiently small so that for $p \in M$, $B_{\epsilon}(p)\subset M$ is a geodesic ball under the exponential map. It can be shown then that $\frac{\text{Vol}(B_{\epsilon}(p))}{\text{Vol}(B_{\epsilon}(0)\subset \mathbb{R}^{n})} = 1 - \frac{R}{6(n + 2)}\epsilon^{2}+ O(\epsilon^{4})$. If the two volumes agree then $R = 0$ to fourth order in $\epsilon$.