In the local coordinates $f(u, v) = (u, v, h(u, v))$ write down the geodesic equations.

The definition I have for geodesic is:

Let $\alpha : [a, b] \to \Sigma$ be a regular parameterized curve then we call it geodesic if its tangent vector is parallel along $\alpha$, i.e $\nabla_{\alpha '} \alpha'=0.$

So if $\alpha (t) = f(a(t),b(t))$ is a geodesic then I must find what equations $a$ and $b$ must satisfy. How can I do that?

  • 1
    $\begingroup$ $\alpha''(t)\cdot f_u=\alpha''(t)\cdot f_v =0$ $\endgroup$ – HK Lee Dec 2 '13 at 3:02
  • $\begingroup$ @HKLee how to get $\alpha''(t)\cdot f_u=\alpha''(t)\cdot f_v =0$ from $\nabla_{\alpha '} \alpha'=0$ ? $\endgroup$ – lanse7pty Sep 24 '16 at 8:13
  • $\begingroup$ $f_u,\ f_v$ are coordinate vector fields That is they are basis of tangent space of the surface That is the condition is definition of $\nabla_{\alpha'}\alpha'=0$ $\endgroup$ – HK Lee Sep 24 '16 at 9:10

$$\alpha'(t) = (u',v',h_uu' + h_vv')$$ so that $$\alpha''(t)=(u'',v'',h_{uu}( u')^2+h_{vv}(v')^2 +2h_{uv}u'v' + h_uu'' + h_vv'')$$

And $$f_u=(1,0,h_u),\ f_v=(0,1,h_v)$$ so that we have two equalities $$ u''+h_u[h_{uu}( u')^2+h_{vv}(v')^2 +2h_{uv}u'v' + h_uu'' + h_vv'']=0 $$ $$ v''+h_v[h_{uu}( u')^2+h_{vv}(v')^2 +2h_{uv}u'v' + h_uu'' + h_vv'']=0 $$


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.