Regarding the shortest path connecting two points on a sphere of radius $R$

Question

Consider a path on the surface of the sphere given by a function $\phi(\theta)$. The line element along the path $\phi$ is given by $$ \mathrm{d}s=R\mathrm{d}\theta\sqrt{1+\left (\frac{ \mathrm{d} \phi}{\mathrm{d} \theta}\right )^2\sin^2\theta}. $$ Deriving the Euler-Lagrange equation for $\phi(\theta)$, they lead to the differential equation $$ \frac{\mathrm{d} \phi}{\mathrm{d} \theta}=\frac{c}{\sin\theta\sqrt{\sin^2\theta-c^2}}\tag{1} $$ with $c$ some constant. Now, there are two things I'd like to figure out. How do I show that Euler-Lagrange equation is solved by $$ \phi=\arcsin(C_1\cot\theta), $$ where $C_1$ is constant? Determining such an integral is difficult for me. I think I should apply the identites identites $(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}$ and $(\cot x)'=-\frac{1}{\sin^2(x)}$. Secondly, I want to transform $(1)$ into cartesian coordinates, but I am not sure how to do it. Should I use the trigonometric addition formula here?

Answer 1

for your line element:

$\left(\frac{ds}{dt}\right)^2=R^2\,\left(\dot{\vartheta}\right)^2+ R^2\,\sin^2(\vartheta)\left(\dot{\varphi}\right)^2 $

I get this differential equations :

$ \ddot{\vartheta}-\frac{c^2\cos(\vartheta)}{R^4\,m^2\,\sin^3(\vartheta)}=0 $

and $\dot{\varphi}=\frac{c}{R^2\,m\,\sin^2(\vartheta)}$

There is no analytical solution ?