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For $s(t)$ the geodesic confined to the surface of a (3D) sphere, how does one get $\|\dot{s}\|^2 s + \ddot{s} = 0$ by setting $\frac{d}{d\delta} \left( \int \|\frac{d}{dt} \frac{s(t)+\delta h(t)}{\|s(t)+\delta h(t) \|}\|^2 dt \right) = 0$ at $\delta = 0$ for any arbitrary (but smooth) $h(t)$?

Is brute-force differentiation of the expression (horribly ugly and I have not succeeded in arriving at the desired expression) the only way to do it?


Edit: Just wanted to clarify that I am asking for how to carry out the calculation using, in particular, the $\frac{d}{d\delta} (...) = 0$ method. (Thanks to all who have pointed out the ambiguities.)

Oh no! I have just realized that I foolishly left out bits of the derivand! (Was too bogged down by the formatting and clearly wasn't checking properly.) Would the question make more sense now? (Sorry :S)

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Are you asking for an explanation of why this procedure yields geodesics, or asking how to carry out the calculation? – Zhen Lin Aug 11 '11 at 10:46
I know that this post should really be a comment, but someone logged me out of my unregistered account ("Hitchhiker"), which I believe is the reason why I don't see a "Comment" (or something of the sort) button. Anyway, @Zhen Lin, I am asking for the latter -- how to carry out the calculation. Thanks! --Hitchhiker – Hitchhiker Aug 11 '11 at 13:47
Hitchhiker, you might want to register to avoid these troubles. – J. M. Aug 11 '11 at 14:27
@Hitchhiker: this is my understanding. As an unregistered user, you only stay logged in as long as a particular cookie remains on your system, so it's easy to lose access. This will stop being an issue if you register. – Qiaochu Yuan Aug 11 '11 at 15:19
@Hitchhiker: I recommend trickery. $\left\| \mathord{-} \right\|^2$ is a lot easier to work with than $\left\| \mathord{-} \right\|$, so we should replace the derivand $y$ by its square. Note that $\frac{d}{dx}\left[y^2\right] = 2y \frac{dy}{dx}$, and so if $y$ is nowhere zero, $\frac{d}{dx}\left[y^2\right] = 0$ if and only if $\frac{dy}{dx} = 0$. – Zhen Lin Aug 12 '11 at 8:56
up vote 1 down vote accepted

First a general piece of advice in response to your assessment of the derivative as "horribly ugly": In this sort of calculation, it simplifies things quite considerably if you keep in mind while differentiating that you're about to set $\delta=0$; then you can immediately drop everything proportional to $\delta$ that you're not differentiating.

Keeping that in mind, taking the derivative with respect to $\delta$ and setting $\delta=0$ yields:

$$ \begin{eqnarray} \left.\frac{\mathrm d}{\mathrm d\delta}\left(\frac{\mathrm d}{\mathrm dt}\frac{s+\delta h}{|s+\delta h|}\right)^2\right|_{\delta=0} &=& \left.2\left(\frac{\mathrm d}{\mathrm dt}\frac{s+\delta h}{|s+\delta h|}\right)\frac{\mathrm d}{\mathrm d\delta}\left(\frac{\mathrm d}{\mathrm dt}\frac{s+\delta h}{|s+\delta h|}\right)\right|_{\delta=0} \\ &=& 2\left(\frac{\mathrm d}{\mathrm dt}\frac{s}{|s|}\right)\frac{\mathrm d}{\mathrm dt}\left(\frac{h}{|s|}-\frac{(h\cdot s)s}{|s|^3}\right) \\ &=& 2\dot s\frac{\mathrm d}{\mathrm dt}\left(h-(h\cdot s)s\right)\;, \end{eqnarray}$$

where in the end I used $|s|=1$. We can use integration by parts to move the derivative to the first factor to conclude that $\ddot s\cdot (h-(h\cdot s)s)$ must vanish. The second factor is the component of $h$ orthogonal to $s$, so this implies $\ddot s=\lambda s$. We can find $\lambda$ by differentiating $0=s\cdot\dot s$, which yields $0=s\cdot\ddot s+\dot s\cdot\dot s=\lambda s\cdot s+\dot s\cdot\dot s=\lambda+\dot s\cdot\dot s$, so $\lambda=-\dot s\cdot\dot s$, and thus $\ddot s=-(\dot s\cdot\dot s)s$.

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Thanks, joriki! :) – Hitchhiker Aug 13 '11 at 9:22

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