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Is there a fundamental misunderstanding here or have I made an algebraic slip?

I have a Riemannian metric of the form $ds^2={du^2+dv^2\over 1-u^2-v^2}$ on an open disc and I want to prove that radial curve $(r(t), 0)$ is a geodesic.

I wrote the metric in polar form -- $ds^2={dr^2+r^2d\theta^2\over 1-r^2}$, so the coefficients of the first fundamental form are $E={1\over 1-r^2}, F=0, G={r^2\over 1-r^2}$.

So the geodesic equations become ${d\over dt} (E\dot{r})={1\over 2}{\partial \over \partial r} (E\dot{r}^2)$ and (another equation which works).

I can always reparametrise $r(t)$ so that it has unit speed.

And I get $2r\over (1-r^2)^2$= $r\over (1-r^2)^2$, which is clearly wrong! Why is there an extra factor of 2?

Thank you.

ADDED: Perhaps it would be helpful for me to point out that generally, for a Riemannian metric of the form $Edu^2+2Fdudv+Gdv^2$ and geodesic $g(t)=(a(t),b(t))$, the Euler Lagrange equations are

${d\over dt} (E\dot{a}+F\dot{b})={1\over 2} (E_u\dot{a}^2+2F_u\dot{a}\dot{b}+G_u\dot{b}^2)$ and

${d\over dt} (F\dot{a}+G\dot{b})={1\over 2} (E_v\dot{a}^2+2F_v\dot{a}\dot{b}+G_v\dot{b}^2)$

where $\dot{a}={d\over dt}a,\,\,\,E_u={\partial\over\partial u}E$ and so on.

EUREKA! Ooh, I think I know what the bug is, The geodesic has constant speed, but the inner product here is not simply the Euclidean one which is what I assumed when I set $\dot{r}=1$! I think it works fine now. :) -- I write it here because I can't post it as an answer...

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    Anyone, please?2012-04-30
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    I'm not familiar with this form of the Euler equations. Do you have a reference for them?2012-04-30
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    @JasonDeVito: thanks for commenting! http://en.wikipedia.org/wiki/Differential_geometry_of_surfaces#Geodesics offers a bit of info. But generally, for a Riemannian metric of the form $Edu^2+2Fdudv+Gdv^2$ and a geodesic $g(t)=(a(t),b(t))$, the Euler Lagrange equations are ${d\over dt} (E\dot{a}+F\dot{b})={1\over 2} (E_u\dot{a}^2+2F_u\dot{a}\dot{b}+G_u\dot{b}^2)$ and ${d\over dt} (F\dot{a}+G\dot{b})={1\over 2} (E_v\dot{a}^2+2F_v\dot{a}\dot{b}+G_v\dot{b}^2)$2012-04-30
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    You actually _can_ post it as an answer.2012-04-30
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    @FernandoMartin: But I have to wait a further 3 hours since it says that I have too few "reputation points" so I can't post an answer within 8 hours of my asking the question...2012-04-30
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    I'm glad you figured it out. I'm not sure how the metric reenters the picture, so I'd very much like it if you posted your answer as soon as the system allows.2012-04-30
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    @JasonDeVito: I have added an answer, as per requested.2012-04-30

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