Metric Tensors in Geodesic Normal Coordinates

Question

Consider the unit sphere and its north pole $(0, 0, 1)$. My question is how to write the metric tensor $g_{ij}$ in geodesic normal coordinates. I know that metric tensors are defined as inner product $\langle \partial_i, \partial_j\rangle$. However, how do I find $\partial_i$ and $\partial_j$ in normal coordinates, please? Thank you!

Answer 1

The exponential map is given by

$$ \exp_p : T_pS \to S,\ \ v \mapsto \gamma_v(1),$$

where $\gamma_v$ is a geodesic with $\gamma(0)= p = (0,0,1)$ and $\gamma'(0) = v$. Now we identify $T_pS$ with the $(x, y)$-plane. We know that all geodesics are great circles, so for any $v = (r\cos\theta, r\sin\theta)$, then $\gamma_v$ will be a geodesic in the great circle $S \cap L_v$, where $L_v$ is the plane spanned by $v$ and $(0,0,1)$ in $\mathbb R^3$. So you need to find the point $q$ in $S\cap L_v$ such that the $d(p, q) = r = |v|$. Then $q = \gamma_v(1) = \exp_p (v)$.

So the map can be written down explicitly:

$$\exp_p(r, \theta) = (\sin r\cos\theta, \sin r \sin\theta , \cos r)$$

and now everything is standard:

$$\partial_r =(\cos r\cos \theta, \cos r \sin\theta, -\sin r), $$

$$\partial_\theta = (-\sin r\sin \theta, \sin r\cos\theta, 0),$$

$$g_{rr} = 1,\ g_{r\theta} = 0, \ g_{\theta\theta} = \sin ^2 r.$$