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.


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.

  • $\begingroup$ 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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