Question on Shortest Path on the Sphere

Question

Consider the following parameterization of the unit sphere: $$X(u,v)=(\sin v \cos u , \sin v \sin u, \cos v)$$ where $u \in (-\pi,\pi),v \in (0,\pi)$. First of all, I am told to find the length of the curve $u=u_0, t \in [a,b]$, where $u_0$ is a constant and $0<a<b<\pi$ (I call this curve $\alpha$). It is easy to see that it is given by $b-a$.

After that, I am told to show that if $(u(t),v(t)), t \in [a,b]$ is a curve on $(-\pi,\pi) \times (0,\pi)$, then the curve on the surface $\beta (t)=X(u(t),v(t))$ which satsifies the $\alpha(a)=\beta(a)$ and $\alpha(b)=\beta(b)$ has a length $\geq b-a$.

I know there is something called "geodesic", but I would to like to show it explicitly without evoking any large theorem.

I tried to compute the length of $X(a(t),b(t))$, which is given by $$L=\int_a^b \sqrt{\sin ^2 (v(t)) \left(\frac{du}{dt}\right)^2+\left(\frac{dv}{dt}\right)^2}dt$$ I am stuck here. Can anyone give me some hint?

Answer 1

Well, \begin{align} L &=\int_a^b \sqrt{\sin ^2 (v(t)) (\frac{du}{dt})^2+(\frac{dv}{dt})^2}dt \\ &= \int_a^b \sqrt{\sin ^2 v du^2+dv^2} \\ &=\int_a^b \sqrt{(1+\sin ^2 v(u))u'^2(v)}dv \\ &\ge \int_a^b dv \end{align} The lower bound is achieved when $u'(v)=0$, i.e. $u(v)=\text{Constant}=0$. The length then is $b-a$.

Answer 2

\begin{align} L &=\int_a^b \sqrt{\sin ^2 (v(t)) (\frac{du}{dt})^2+(\frac{dv}{dt})^2}\space dt \\ &\ge \int_a^b \sqrt{(\frac{dv}{dt})^2} \space dt\\ &= \int_a^b |\frac{dv}{dt}|\space dt \\ &\ge \int_a^b \frac{dv}{dt}\space dt \\ &= v(b)-v(a) \\ &= b-a \\ \end{align} where $v(a)=a$ and $v(b)=b$ because the two curves have the same starting and ending point.