1
$\begingroup$

The two geodesic equations are:

$$\frac d{ds} (Eu' + Fv') = 1/2(E_u {u'}^2 + 2F_u u'v' + G_u {v'}^2) $$

$$\frac d{ds} (Fu' + Gv') = 1/2(Ev{u'}^2 + 2F_v u'v' + G_v{v'}^2) $$

And in the hyperbolic plane $H^2$, $E=G=1/y^2$, $F=0$. Put them into the first geodesic equation, I can get two solutions: $x=\text{const}$ and $(x-a)^2+y^2=c$, which are just the geodesic in $H^2$. But is that true that we can get the geodesic only by the first geodesic equation? Thank you.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.