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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?

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    $\begingroup$ $\alpha''(t)\cdot f_u=\alpha''(t)\cdot f_v =0$ $\endgroup$ – 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
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$$\alpha'(t) = (u',v',h_uu' + h_vv')$$ 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 $$ u''+h_u[h_{uu}( u')^2+h_{vv}(v')^2 +2h_{uv}u'v' + h_uu'' + h_vv'']=0 $$ $$ v''+h_v[h_{uu}( u')^2+h_{vv}(v')^2 +2h_{uv}u'v' + h_uu'' + h_vv'']=0 $$

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