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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? –  Sup Apr 30 '12 at 15:39
    
I'm not familiar with this form of the Euler equations. Do you have a reference for them? –  Jason DeVito Apr 30 '12 at 15:46
    
@JasonDeVito: thanks for commenting! en.wikipedia.org/wiki/… 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 ${d\over dt} (F\dot{a}+G\dot{b})={1\over 2} (E_v\dot{a}^2+2F_v\dot{a}\dot{b}+G_v\dot{b}^2)$ –  Sup Apr 30 '12 at 15:55
    
You actually can post it as an answer. –  Fernando Martin Apr 30 '12 at 17:53
    
@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... –  Sup Apr 30 '12 at 17:54
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1 Answer

This is for @JasonDeVito:

It is actually easier to use the original ("Cartesian") metric. Noting that for a curve $g(t)=(a(t),b(t)),\,\,\,||\dot{g}(t)||^2={\dot{a}^2+\dot{b}^2\over1-a^2-b^2}$ instead of simply $\dot{a}^2+\dot{b}^2$, then plugging and chugging gives the answer.

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Ok, apparently I cannot accept the answer within 2 days... –  Sup May 1 '12 at 0:00
    
I appreciate you typing this up (and I upvoted), but I'm still not seeing where you use $\| \dot{g}\|$ in your computations. In fact, I'm still having a lot of trouble getting my computations to work out. For example, in the $\frac{d}{dt}(E\dot{a})$ term, a $\ddot{a}$ appears. Where does it go? Do you get rid of it by differentiating the length = 1 equation? If you wouldn't mind, I'd really like to see a few more details (but please, feel free to ignore me as well!) –  Jason DeVito May 1 '12 at 2:11
    
@JasonDeVito : No probs, note that in the original coordinates, that curve is just $a=u, b=0$. ${d\over dt} ({1\over 1-u^2}\dot{u})={1\over 2}({\partial \over \partial u}({1\over 1-u^2})\dot{u}^2)$ –  Sup May 1 '12 at 6:50
    
@JasonDeVito : Also it is now ${\dot{u}^2\over 1-u^2}=1$ not simply $\dot{u}=1$ –  Sup May 1 '12 at 6:56
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