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Concerning the using covariant derivative along a curve to find its geodesic curvature.

Let S be an oriented surface, and $\alpha(u^1, u^2) =(\alpha^1(u^1, u^2), \alpha^2(u^1, u^2))$ a curve in S parametrized by arclength. Then the unit tangent vectors $\alpha'$ is a vectorfield along $\alpha$. Since I'm trying to find the geodesic curvature of $\alpha$, I need (using einstein summation here): $\nabla_{\alpha'}\alpha' = {\partial \alpha'^i \over \partial u^k} + u^j\Gamma^i_{kj}$ where the $\Gamma$s are christoffel symbols.

What I'm not seeing is where/how this depends on arclength parametrization? $\alpha'$ could be normalized if need be.

I have tried to find a post that answers this question, but nothing I find seems to quite cover it.

I will add that I'd prefer hints! I'm trying to get a feel for the covariant derivative and it's kicking my butt thus far.

Thank you

edit: oh, and my level is ~3 years worth haphazard university math

Gustav Franklin
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  • curves depend only on one parameter – janmarqz Dec 15 '16 at 23:23
  • But does the covariant derivative care? Does it still view the $u_i$s in $\alpha$ as just coordinates in S, or as dependent on a parameter, say t? – Gustav Franklin Dec 16 '16 at 08:14
  • to reach a curve over a surface you need $t\mapsto (v(t),w(t))\mapsto (x(v(t),w(t)),y(v(t),w(t)),z(v(t),w(t)))$ which is a composition $\Bbb R\to\Bbb R^2\to\Bbb R^3$ – janmarqz Dec 16 '16 at 15:02

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