I'm trying to understand Riemann normal coordinates. This "simple" example using the surface of a unit sphere is from http://www.maths.bris.ac.uk/~macpd/gen_rel/snotes.pdf (p26). The “north pole” $\theta=0$ is the initial point. The geodesics radiating from the north pole are lines of constant $\phi$. The new coordinates $\xi$ and $\eta$ are related to the old (ordinary spherical) coordinates by $$\xi=\theta\cos\phi$$ and$$\eta=\theta\sin\phi.$$
I'm assuming (quite possibly incorrectly) that these are arbitrary definitions. They then say the metric is$$ds^{2}=d\theta^{2}+\sin^{2}\theta d\phi^{2}$$ $$=\frac{d\xi^{2}}{\theta^{4}}\left(\xi^{2}\theta^{2}+\eta^{2}\sin^{2}\theta\right)+\frac{d\eta^{2}}{\theta^{4}}\left(\eta^{2}\theta^{2}+\xi^{2}\sin^{2}\theta\right)$$ where $\theta=\sqrt{\xi^{2}+\eta^{2}}.$
Can anyone please explain where the second metric comes from and also why $\theta=\sqrt{\xi^{2}+\eta^{2}}$? Thank you.
This edit added 18 Sept 2014
Right, I'm going to wander off on my own with this and see where I get. Substituting $\eta$ and $\xi$ into the metric gives
$$ds^{2}=\frac{d\xi^{2}}{\theta^{4}}\left(\xi^{2}\theta^{2}+\theta^{2}\sin^{2}\phi\sin^{2}\theta\right)+\frac{d\eta^{2}}{\theta^{4}}\left(\eta^{2}\theta^{2}+\theta^{2}\cos^{2}\phi\sin^{2}\theta\right).$$
Note that for a unit sphere (with the $x,y$ plane passing through the equator) $$x=\sin\theta\cos\phi,$$ $$y=\sin\theta\sin\phi.$$ This gives$$ds^{2}=\frac{d\xi^{2}}{\theta^{4}}\left(\xi^{2}\theta^{2}+\theta^{2}y^{2}\right)+\frac{d\eta^{2}}{\theta^{4}}\left(\eta^{2}\theta^{2}+\theta^{2}x^{2}\right)$$
$$ds^{2}=\frac{d\xi^{2}}{\theta^{2}}\left(\xi^{2}+y^{2}\right)+\frac{d\eta^{2}}{\theta^{2}}\left(\eta^{2}+x^{2}\right)$$ $$ds^{2}=d\xi^{2}\frac{\left(\xi^{2}+y^{2}\right)}{\left(\xi^{2}+\eta^{2}\right)}+d\eta^{2}\frac{\left(\eta^{2}+x^{2}\right)}{\left(\xi^{2}+\eta^{2}\right)}.$$ And then I grind to a halt. The metric is supposed to reduce to $\delta_{\alpha\beta}$, but I can't see how it does.
Just a thought, but is it useful to point out that for small angles (where $\theta=\sin\theta)$, $\eta\approx y$ and $\xi\approx x$? The above metric then reduces to $$ds^{2}\approx d\xi^{2}+d\eta^{2}.$$
Is this what should happen for Riemann normal coordinates?
This edit added 21 September 2014
Following the latest hint from @Semsen $$ds^{2}=\frac{\zeta^{2}}{\zeta^{2}+\eta^{2}}d\zeta^{2}+\frac{2\eta\zeta}{\zeta^{2}+\eta^{2}}d\zeta d\eta+\frac{\eta^{2}}{\zeta^{2}+\eta^{2}}d\eta^{2}+\sin^{2}\theta\cos^{4}\phi\{\frac{1}{\zeta^{2}}d\eta^{2}-2\frac{\eta}{\zeta^{3}}d\eta d\zeta+\frac{\eta^{2}}{\zeta^{4}}d\zeta^{2}\}$$
$$ds^{2}=\frac{\zeta^{2}}{\zeta^{2}+\eta^{2}}d\zeta^{2}+\frac{2\eta\zeta}{\zeta^{2}+\eta^{2}}d\zeta d\eta+\frac{\eta^{2}}{\zeta^{2}+\eta^{2}}d\eta^{2}+\sin^{2}\theta\frac{\zeta^{4}}{\theta^{4}}\{\frac{1}{\zeta^{2}}d\eta^{2}-2\frac{\eta}{\zeta^{3}}d\eta d\zeta+\frac{\eta^{2}}{\zeta^{4}}d\zeta^{2}\}$$
$$ds^{2}=\frac{\zeta^{2}}{\theta^{2}}d\zeta^{2}+\frac{2\eta\zeta}{\theta^{2}}d\zeta d\eta+\frac{\eta^{2}}{\theta^{2}}d\eta^{2}+\sin^{2}\theta\frac{1}{\theta^{4}}\{\zeta^{2}d\eta^{2}-2\eta\zeta d\eta d\zeta+\eta^{2}d\zeta^{2}\}$$ $$ds^{2}=\frac{1}{\theta^{4}}\left(\theta^{2}\zeta^{2}d\zeta^{2}+\theta^{2}2\eta\zeta d\zeta d\eta+\theta^{2}\eta^{2}d\eta^{2}+\sin^{2}\theta\{\zeta^{2}d\eta^{2}-2\eta\zeta d\eta d\zeta+\eta^{2}d\zeta^{2}\}\right)$$ $$ds^{2}=\frac{1}{\theta^{4}}\left(\theta^{2}\zeta^{2}d\zeta^{2}+\theta^{2}2\eta\zeta d\zeta d\eta+\theta^{2}\eta^{2}d\eta^{2}+\sin^{2}\theta\zeta^{2}d\eta^{2}-\sin^{2}\theta2\eta\zeta d\eta d\zeta+\sin^{2}\theta\eta^{2}d\zeta^{2}\right)$$
$$ds^{2}=\frac{d\zeta^{2}}{\theta^{4}}\left(\theta^{2}\zeta^{2}+\eta^{2}\sin^{2}\theta\right)+\frac{d\eta^{2}}{\theta^{4}}\left(\theta^{2}\eta^{2}+\zeta^{2}\sin^{2}\theta\right)+\frac{2\eta\zeta d\zeta d\eta}{\theta^{4}}\left(\theta^{2}-\sin^{2}\theta\right)$$
If I could only now get rid of the right-hand term I'd have the original metric. But how to get rid of this term?