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I'm trying to derive the extremal solutions to the Lagrangian for arclength on the unit sphere by setting up the Euler-Lagrange equations.

Starting from

$$L = \sqrt{\dot \theta^2 + \sin^2 \theta \ \dot \phi^2} \ $$

I see not long after that

$$\frac{\partial L}{\partial \dot\theta}=\frac{\dot\theta}{\sqrt{\dot \theta^2 + \sin^2 \theta \ \dot \phi^2}}\\ \frac{\partial L}{\partial\theta}=\frac{\dot\phi^2\sin(\theta)\cos(\theta)}{\sqrt{\dot \theta^2 + \sin^2 \theta \ \dot \phi^2}}\\ \frac{\partial L}{\partial\dot\phi}=\frac{\sin^2(\theta)\dot\phi}{\sqrt{\dot \theta^2 + \sin^2 \theta \ \dot \phi^2}}\\ \frac{\partial L}{\partial\phi}=0$$

I'm pretty sure I've understood the planar case, and that had taught me to recognize that in the last two equations, I could then say that $$\frac{\partial L}{\partial\dot\phi}=C_1$$ a constant $C_1$.

Then one can rewrite

$$\frac{\partial L}{\partial\theta}=C_1\dot\phi\cot(\theta)\\ \frac{\partial L}{\partial\dot\theta}=C_1\frac{\dot\theta}{\sin^2(\theta)\dot\phi}$$

but then $$\frac{d}{dt}\frac{\partial L}{\partial\dot\theta}=\frac{\partial L}{\partial \theta}$$ still seems to be a second order differential equation in both variables, and I feel like I've run aground.

How do you go from here? Maybe I'm supposed to do more at the step where I found $C_1$ to make some reductions?

The final goal is to see concretely that the answer is going to yield a single solution when I'm going from $(0,0)$ and $(\pi/2, \pi/2)$, say, and that there are infinitely many extremals when going between antipodal points.

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When minimizing length, there are always infinitely many extremals because reparameterizations do not change the length. For this reason, one usually studies the better behaved extremal problem $\int |\dot \gamma|^2\to \min$ (rather than $\int |\dot \gamma|\to \min$). The extremals are geometrically the same, since the Cauchy-Schwarz inequality $$ \left(\int_0^1 |\dot \gamma| \right)^2 \le \int_0^1 |\dot \gamma|^2 $$ turns into equality for constant-speed parameterizations. And without the square root we get a lot simpler equations. So, $L=\dot\theta^2 + \sin^2\theta \,\dot \phi^2$ leads to $$ \frac{d}{dt}\frac{\partial L}{\partial\dot\theta} = \frac{d}{dt}(2\dot\theta) = 2\ddot\theta $$ $$ \frac{\partial L}{\partial\theta} = 2\sin\theta\cos\theta \, \dot\phi^2 $$ hence $$ \ddot\theta = \sin\theta\cos\theta \, \dot\phi^2 \tag1 $$ Also, $$ \frac{\partial L}{\partial\dot\phi} = 2\sin^2\theta \, \dot\phi $$ $$ \frac{\partial L}{\partial\phi} = 0 $$ hence $$ \sin^2\theta \, \dot\phi = C \tag2 $$ From (1) and (2) we get $$ \ddot\theta = C^2 \frac{\cos\theta}{\sin^3\theta} \tag 3$$ which is still unpleasant but is somewhat solvable for the inverse function $t(\theta)$. When working with (3) by hand, it seems advisable to let $y=\cot \theta$ and rewrite the equation in terms of $y$.

I somehow doubt that this approach will shed any light on the (non)-uniqueness of geodesics.

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  • $\begingroup$ Ah! A huge simplification from the get-go. But surely people talk counting minima and maxima up to some reasonable equivalence. Blindly putting reparametrizations on an equal footing with each other does not seem useful. $\endgroup$ – rschwieb Aug 15 '18 at 2:38
  • $\begingroup$ The solutions do seem rather horrendous. And here I thought this and curve length in the plane were going to be baby examples :/ $\endgroup$ – rschwieb Aug 15 '18 at 2:50
  • $\begingroup$ One can still use variational method here, just not by writing everything so concretely from the beginning. Length-minimizing curves are shown to have geodesic curvature zero by a variational argument (e.g. Lee's Riemannian Manifolds, pp 99-100), which on the sphere leads to their characterization as arcs of great circles without much difficulty. $\endgroup$ – user357151 Aug 15 '18 at 3:11
  • $\begingroup$ Sure... I'm just trying to get my hands dirty with examples I know should work out and I already know the answers to. Doubtless there are more economical solutions with more advanced concepts. Even if I don't work out every little detail, seeing the E-L equations in action is my goal. $\endgroup$ – rschwieb Aug 15 '18 at 11:07
  • $\begingroup$ @user357151: The step to obtain (2) is a bit obscure. You appear to have integrated the second Euler-Lagrange equation w.r.t. time to get this (i.e. integrated 0 instead of differentiating the other side). It would be nice if this was stated. $\endgroup$ – qman Aug 16 '18 at 3:11

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