Confusion regarding Riemann normal coordinates

Let's reformulate the argument of the problem in the following equivalent way first (just making it mathematically a bit more accurate):

Denote the north pole by $o$. Then we are given the map (normal coordinates) $$\mathrm{exp}_{\ o}^{\ -1}: \mathcal{U}\subset S^2 \longrightarrow \mathcal{U}^\prime \subset T_{o}(S^2)$$ $$\left( \theta, \phi \right) \longmapsto \left( \eta, \xi \right)$$

(where $\mathcal{U}^\prime$ is the neighborhood around $0$ in $T_{o}(S^2)$ on which the exponential map $\mathrm{exp}_{o}$ is a diffeomorphism onto a neighborhood $\mathcal{U}$ of $o$ in $S^2$). If $g = d\ \theta ^ {\ 2} + sin^2 (\theta )\ d\ \phi^{\ 2} $ is the metric tensor on $S^2$ then our goal is to show

$$g^\prime =\mathrm{exp}_{o}^{\ast} (g) = \frac {{\xi}^{2}{\theta}^{2}+{\eta}^{2} \sin ^2 \left( \theta \right)}{{\theta}^{4}}\ d \xi ^2 + \frac {{\eta}^{2}{\theta}^{2}+{\xi}^{2} \sin ^2 \left( \theta \right)}{{\theta}^{4}}\ d \eta ^2$$

on $T_{o}(S^2)$ where $\theta = \sqrt{\eta^2 + \xi^2}$.

Using the "matrix of the corresponding metric tensor with respect to coordinate neighborhood basis" of $S^2$, we have:

$$g = \left( \begin {array}{cc} 1&0\\ 0& \sin ^2 \left( \theta \right) \end {array} \right)$$

Then showing the matrix of $d\ \mathrm{exp}_{o}$ by $J$ we have (note that since $\mathrm{exp}$ is diffeomorphism over $\mathcal{U}^\prime$, we can obtain $J$ by simply computing the inverse of $d\ \mathrm{exp}_{o}^{-1}$) (perhaps you might want to take a look at Chapter 2, Problem 8, Semi-Riemannian geometry with applications to relativity, O'Neil) :

$$g^\prime = \left( J \right)^{T} g\ J = \left( \begin {array}{cc} {\frac { \cos ^2 \left( \theta \right) \cos ^2 \left( \phi \right) + \cos ^2 \left( \phi \right) {\theta}^{2}- \cos ^2 \left( \theta \right) - \cos ^2 \left( \phi \right) +1}{{\theta}^{2}}}&{\frac {\cos \left( \phi \right) \sin \left( \phi \right) \left( \left( \cos \left( \theta \right) \right) ^{2}+{\theta}^{2}-1 \right) }{{\theta}^{2}}} \\ {\frac {\cos \left( \phi \right) \sin \left( \phi \right) \left( \left( \cos \left( \theta \right) \right) ^{2}+{ \theta}^{2}-1 \right) }{{\theta}^{2}}}&{\frac {- \cos ^2 \left( \theta \right) \cos ^2 \left( \phi \right)- \cos ^2 \left( \phi \right) {\theta}^{2}+ \cos ^2 \left( \phi \right) +{\theta}^{2}}{{\theta}^ {2}}}\end {array} \right) $$ which upon simplification gives $$\left( \begin {array}{cc} {\frac {{\xi}^{2}{\theta}^{2}+{\eta}^{2} \sin ^2 \left( \theta \right) }{{\theta}^{4}}} &{\frac {\cos \left( \phi \right) \sin \left( \phi \right) \left( {\theta}^{2}- \sin ^2 (\theta )\right) }{{\theta}^{2}}} \\ {\frac {\cos \left( \phi \right) \sin \left( \phi \right) \left( {\theta}^{2}- \sin ^2 (\theta ) \right) }{{\theta}^{2}}} &{\frac {{\eta}^{2}{\theta}^{2}+{\xi}^{2} \sin ^2 \left( \theta \right)}{{\theta}^{4}}} \end {array} \right) $$ Note that this is exactly the matrix of the metric tensor that you have obtained on 21 September 2014 edit in a more "compact" formulation. Note that at the north pole ($\eta = 0$ and $\xi = 0$), $g^\prime (1,2) = g ^ \prime (2,1) = 0$ and hence we have obtained the metric tensor (note that around $\theta = 0$, $sin (\theta) \approx \theta$ ).


Things I didn't realise when I first posted this question.

  1. The Riemann Normal Coordinates sit like a little cross centred on the north pole of the two-sphere. One geodesic, the coordinate axis $\xi$, lies along $\phi=0$. The other geodesic, the coordinate axis $\eta$, lies along $\phi=\pi/2$. For small $\theta$ these coordinates are approximately flat (ie Cartesian).

  2. So $\xi=\theta\cos\phi$ and $\eta=\theta\sin\phi$ are not, as I suggested, arbitrary definitions.

  3. From the definitions of $\xi$ and $\eta$ we can derive $\theta=\sqrt{\xi^{2}+\eta^{2}}$.

  4. I don't understand Arvin's higher level explanation, so it took me a while to figure out his answer. He is using$$g=J^{T}g^{\prime}J$$ $$g=\left[\begin{array}{cc} \frac{\xi}{\sqrt{\xi^{2}+\eta^{2}}} & \frac{-\eta}{\xi^{2}+\eta^{2}}\\ \frac{\eta}{\sqrt{\xi^{2}+\eta^{2}}} & \frac{\xi}{\xi^{2}+\eta^{2}} \end{array}\right].\left[\begin{array}{cc} 1 & 0\\ 0 & \sin^{2}\left(\theta\right) \end{array}\right].\left[\begin{array}{cc} \frac{\xi}{\sqrt{\xi^{2}+\eta^{2}}} & \frac{\eta}{\sqrt{\xi^{2}+\eta^{2}}}\\ \frac{-\eta}{\xi^{2}+\eta^{2}} & \frac{\xi}{\xi^{2}+\eta^{2}} \end{array}\right]=\left[\begin{array}{cc} \frac{\xi^{2}}{\xi^{2}+\eta^{2}}+\frac{\eta^{2}\sin^{2}\theta}{\left(\xi^{2}+\eta^{2}\right)^{2}} & \frac{\eta\xi}{\xi^{2}+\eta^{2}}-\frac{\eta\xi\sin^{2}\theta}{\left(\xi^{2}+\eta^{2}\right)^{2}}\\ \frac{\eta\xi}{\xi^{2}+\eta^{2}}-\frac{\eta\xi\sin^{2}\theta}{\left(\xi^{2}+\eta^{2}\right)^{2}} & \frac{\eta^{2}}{\xi^{2}+\eta^{2}}+\frac{\xi^{2}\sin^{2}\theta}{\left(\xi^{2}+\eta^{2}\right)^{2}} \end{array}\right].$$ For small angles, $\theta^{2}\approx\sin^{2}\theta$, $g_{21}=g_{12}=0$ and we have the desired line element $$ds^{2}=\frac{d\xi^{2}}{\theta^{4}}\left(\xi^{2}\theta^{2}+\eta^{2}\sin^{2}\theta\right)+\frac{d\eta^{2}}{\theta^{4}}\left(\eta^{2}\theta^{2}+\xi^{2}\sin^{2}\theta\right).$$All this only took me four and half years to figure out.