How do I calculate scalar curvature? How can I calculate the scalar curvature of an object? Maybe a tennis ball?
I'm quite new to this stuff. And I have a problem understanding things I believe maybe complex...
Like, when I looked up how can I calculate scalar curvature, one question said how to calculate scalar curvature in a local chart, I tried reading through it to find my answer.  I wish I could understand everything, or even 1% of it, or maybe know a little of it, because I took Intermediate Algebra(even though I got a C-,) I really do have quite a passion for mathematics.
I am going to be taking CS101, next week.
I want to ask more questions like these, like how to calculate Stress-energy-tensor.
Thank you so much!
 A: I don't know if this is going to be what you're asking for, but it's what I thought it could have been so.. take it or leave :)
It's about curvature scalar and Ricci tensor, then. We will find them for the $2-D$ surface of a sphere, that is your tennis ball (a good sphere). The Ricci tensor is calculated from the Riemann tensor, and that in turn depends on the Christoffel symbols, so we need them first. It’s easiest to find them from the geodesic equation. I hope you know a bit of General Relativity or at least Differential Geometry.
$$\displaystyle  g_{aj}\ddot{x}^{\ j}+\left(\partial_{i}g_{aj}-\frac{1}{2}\partial_{a}g_{ij}\right)\dot{x}^{\ j}\dot{x}^{\ i}=0 \ \ \ \ \ (1)$$
which is formally equivalent to
$$ \displaystyle  \ddot{x}^{\ m}+\Gamma_{\; ij}^{m}\dot{x}^{\ j}\dot{x}^{\ i}=0 \ \ \ \ \ (2)$$
The metric for a sphere is
$$\displaystyle  ds^{2}=r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (3)$$
where ${r}$ is the constant radius of the sphere, so
$$\displaystyle   g_{\theta\theta}  \displaystyle  =  \displaystyle  r^{2}\ \ \ \ \ (4)$$
$$\displaystyle  g_{\phi\phi}  \displaystyle  =  \displaystyle  r^{2}\sin^{2}\theta \ \ \ \ \ (5)$$
We have two equations arising from $1$. For ${a=\theta}$
$$\displaystyle  r^{2}\ddot{\theta}-r^{2}\sin\theta\cos\theta\dot{\phi}^{2}=0 \ \ \ \ \ (6)$$
Comparing with $2$ we get, after dividing out the ${r^{2}}$:
$$\displaystyle  \Gamma_{\;\phi\phi}^{\theta}=-\sin\theta\cos\theta=-\frac{1}{2}\sin2\theta \ \ \ \ \ (7)$$
For ${a=\phi}$:
$$\displaystyle  r^{2}\sin^{2}\theta\ddot{\phi}^{2}+2r^{2}\sin\theta\cos\theta\dot{\theta}\dot{\phi}=0 \ \ \ \ \ (8)$$
Dividing through by ${r^{2}\sin^{2}\theta}$ and comparing with $2$ we get (remember that the second term is ${\left(\Gamma_{\;\theta\phi}^{\phi}+\Gamma_{\;\phi\theta}^{\phi}\right)\dot{\theta}\dot{\phi}}$ and that ${\Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}})$:
$$\displaystyle  \Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}=\cot\theta \ \ \ \ \ (9)$$
All other Christoffel symbols are zero.
In $2D$, the Riemann tensor has only one independent component, which we can take to be ${R_{\theta\phi\theta\phi}}$, which can be calculated from
$$\displaystyle  R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i}+\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}-\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i} \ \ \ \ \ (10)$$
Lowering the first index, we have
$$\displaystyle   R_{\theta\phi\theta\phi}  \displaystyle  =  \displaystyle  g_{\theta i}R_{\;\phi\theta\phi}^{i}\ \ \ \ \ (11)$$
$$\displaystyle   \displaystyle  =  \displaystyle  g_{\theta\theta}R_{\;\phi\theta\phi}^{\theta}\ \ \ \ \ (12)$$
$$\displaystyle   \displaystyle  =  \displaystyle  r^{2}\left(\partial_{\theta}\Gamma_{\;\phi\phi}^{\theta}-\partial_{\phi}\Gamma_{\;\theta\phi}^{\theta}+\Gamma_{\;\phi\phi}^{k}\Gamma_{\;\theta k}^{\theta}-\Gamma_{\;\theta\phi}^{k}\Gamma_{\; k\phi}^{\theta}\right)\ \ \ \ \ (13)$$
$$\displaystyle   \displaystyle  =  \displaystyle  r^{2}\left(-\cos2\theta-0+0-\Gamma_{\;\theta\phi}^{\phi}\Gamma_{\;\phi\phi}^{\theta}\right)\ \ \ \ \ (14)$$
$$\displaystyle   \displaystyle  =  \displaystyle  r^{2}\left(\sin^{2}\theta-\cos^{2}\theta+\cos^{2}\theta\right)\ \ \ \ \ (15)$$
$$\displaystyle   \displaystyle  =  \displaystyle  r^{2}\sin^{2}\theta \ \ \ \ \ (16)$$
We can now find the Ricci tensor.
$$\displaystyle  R_{ij}=g^{ab}R_{aibj} \ \ \ \ \ (17)$$
Since the metric is diagonal and ${g^{ab}}$ is the inverse of ${g_{ab}}$, we have
$$\displaystyle   g^{\theta\theta}  \displaystyle  =  \displaystyle  \frac{1}{g^{\theta\theta}}=\frac{1}{r^{2}}\ \ \ \ \ (18)$$
$$\displaystyle  g^{\phi\phi}  \displaystyle  =  \displaystyle  \frac{1}{g^{\phi\phi}}=\frac{1}{r^{2}\sin^{2}\theta} \ \ \ \ \ (19)$$
so, using the symmetries of the Riemann tensor,
$$\displaystyle   R_{\theta\theta}  \displaystyle  =  \displaystyle  g^{ab}R_{a\theta b\theta}\ \ \ \ \ (20)$$
$$\displaystyle   \displaystyle  =  \displaystyle  g^{\phi\phi}R_{\phi\theta\phi\theta}\ \ \ \ \ (21)$$
$$\displaystyle   \displaystyle  =  \displaystyle  \frac{1}{r^{2}\sin^{2}\theta}r^{2}\sin^{2}\theta\ \ \ \ \ (22)$$
$$\displaystyle   \displaystyle  =  \displaystyle  1\ \ \ \ \ (23)$$
$$\displaystyle  R_{\phi\phi}  \displaystyle  =  \displaystyle  g^{ab}R_{a\phi b\phi}\ \ \ \ \ (24)$$
$$\displaystyle   \displaystyle  =  \displaystyle  g^{\theta\theta}R_{\theta\phi\theta\phi}\ \ \ \ \ (25)$$
$$\displaystyle   \displaystyle  =  \displaystyle  \sin^{2}\theta\ \ \ \ \ (26)$$
$$\displaystyle  R_{\theta\phi}  \displaystyle  =  \displaystyle  R_{\phi\theta}=0 \ \ \ \ \ (27)$$
We can get the upstairs version of the Ricci tensor as well:
$$\displaystyle   R^{\theta\theta}  \displaystyle  =  \displaystyle  g^{\theta i}g^{\theta j}R_{ij}\ \ \ \ \ (28)$$
$$\displaystyle   \displaystyle  =  \displaystyle  g^{\theta\theta}g^{\theta\theta}R_{\theta\theta}\ \ \ \ \ (29)$$
$$\displaystyle   \displaystyle  =  \displaystyle  \frac{1}{r^{4}}\ \ \ \ \ (30)$$
$$\displaystyle  R^{\phi\phi}  \displaystyle  =  \displaystyle  g^{\phi\phi}g^{\phi\phi}R_{\phi\phi}\ \ \ \ \ (31)$$
$$\displaystyle   \displaystyle  =  \displaystyle  \frac{1}{r^{4}\sin^{2}\theta} \ \ \ \ \ (32)$$
The curvature scalar is
$$\displaystyle   R  \displaystyle  =  \displaystyle  g_{ij}R^{ij}\ \ \ \ \ (33)$$
$$\displaystyle   \displaystyle  =  \displaystyle  r^{2}\frac{1}{r^{4}}+r^{2}\sin^{2}\theta\frac{1}{r^{4}\sin^{2}\theta}\ \ \ \ \ (34)$$
$$\displaystyle   \displaystyle  =  \displaystyle  \frac{2}{r^{2}} \ \ \ \ \ (35)$$
As we would expect, the curvature of a sphere decreases as its radius gets larger.
