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How to calculate the first variation of length of a curve $\gamma?$ The length is defined as $$L(\gamma) = \int_{\gamma}ds.$$ So the first variation is $$\frac{d}{dc}L(\gamma + c\phi)|_{c=0} = \frac{d}{dc}\int_{\gamma + c\phi}ds\bigg|_{c=0} = \int_{\gamma +c\phi}\nabla_{\gamma + c\phi}\cdot v_c\bigg|_{c=0}$$ (where $v_c$ is the velocity of the curve $\gamma + c\phi$) $$= \int_{\gamma +c\phi}v_c \cdot \kappa_c \nu_c\bigg|_{c=0}$$ where $\kappa_c$ and $\nu_c$ are the mean curvature and unit normal vector on $\gamma + c\phi.$

How can I relate the quantities [$v_c$, $\kappa_c$ and $\nu_c$] to [$v$, $\kappa$ and $\nu$] (the respective quantities on $\gamma$)? I don't think I can just evaluate the above on $c=0$ and remove the subscript $c$'s.

Edit: I'd prefer answers not to have any heavy Riemannian geometry stuff (so tensors and g^{ij} and stuff like that). Thanks. Also the reason I'm asking is to show that mean curvature is the gradient flow of the length integral so I want to see how they end up with $\int_\gamma{\kappa \phi}$ at the end of the above calculation.

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I don't know how you can get away from Riemannian geometry. You need the covariant derivative for the first variation. Also your life might be easier if you look at energy instead of arc length, $E(\gamma) = \frac{1}{2}\int_\gamma \langle \gamma',\gamma'\rangle$, since (loosely speaking) you won't have to deal with a square root. –  Neal Jun 19 '12 at 13:25
@Neal maybe a little bit of RG would be OK. I can handle the $g^{ij}$ stuff... –  soup Jun 22 '12 at 9:31

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