Geodesics on $S^2$ with specific Riemannian metric This is a revision of the statement in the question Geodesics on $S^2$

Suppose a Riemannian metric on $S^2$ satisfies the condition that there exist points $N,S \in S^2$ such that all geodesics originating from $N$ pass through $S$. Then the arc-lengths of all these geodesics are the same, assuming each geodesic connects $N$ to $S$ and does not pass through $S$ in the middle. 

I started with picking a $v_0 \in T_NS^2$ such that $\exp_N(v_0) = S$, and let $\gamma_0 \colon [0,1] \to S^2$ be defined by $\gamma_0(t) = \exp_N(tv_0)$. 
Then let $v \colon (-\epsilon,\epsilon) \to T_nS^2$ be a smooth map such that $v(0) = v(0)$, and $\exp_N(v(u)) = S$. 
Now we have a one-parameter family of geodesics $\gamma_u(t) = \exp_N(tv(u))$. 
If we define $\alpha \colon (-\epsilon,\epsilon) \times [0,1] \to S^2$ by $\alpha(u,t) = \gamma_u(t) = \exp_N(tv(u))$, we get a variation of $\gamma_0$ through geodesics, and this variation fixes endpoints $N$ and $S$. 
So we have the following facts: 
(1) $\displaystyle v(u) = \frac{d}{dt} \gamma_u(0) = \frac{\partial}{\partial t} \alpha(u,t) \Big|_{t=0}$
(2) $\displaystyle \frac{D}{\partial t} \frac{\partial}{\partial t} \alpha(u,t) = 0$
(3) $\displaystyle \frac{\partial}{\partial u} \alpha(u,0) = 0$, and $\displaystyle \frac{\partial}{\partial u} \alpha(u,1) = 0$.
If we can show that $\displaystyle \frac{d}{du} \|v(u)\| = 0$, we can use a compactness argument on $S^1$ to say that the arc-length of $\gamma_u$ does not depend on $u$. 
My trouble is in the following argument:

\begin{align*}
\frac{d}{du} \langle v(u),v(u) \rangle 
&= \frac{\partial}{\partial u} \langle \frac{\partial}{\partial t} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \Big|_{t=0} 
= 2 \langle \frac{D}{\partial u} \frac{\partial}{\partial t} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t)\rangle\Big|_{t=0} \\
&= 2 \langle \frac{D}{\partial t} \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t)\rangle\Big|_{t=0} \\
&= 2 \frac{\partial}{\partial t} \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \Big|_{t=0} - 2 \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{D}{\partial t} \frac{\partial}{\partial t} \alpha(u,t) \rangle \Big|_{t=0} \\
&= 2 \frac{\partial}{\partial t} \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \Big|_{t=0} 
\end{align*}

I am not sure how to conclude that the expression evaluates to be zero. But I assume I need to make use of fact (3) afterwards, since I have not used it anywhere in my argument. Any help is appreciated! Thanks!
 A: I think I have figured it out. 
To show that $\displaystyle \frac{\partial}{\partial t} \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \Big|_{t=0} = 0$, I only need to show that $\displaystyle \frac{\partial}{\partial t} \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle$ is constant in $t$, i.e., $\displaystyle \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle$ is linear in $t$, and conclude that it must be $0$ by fact (3).
Now,
\begin{align*}
\frac{\partial}{\partial t}\frac{\partial}{\partial t} \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle
=& \frac{\partial}{\partial t} \langle \frac{D}{\partial t}\frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \\
=& \langle \frac{D}{\partial t}\frac{D}{\partial t}\frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \\
=& \langle \frac{D}{\partial t}\frac{D}{\partial u}\frac{\partial}{\partial t} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \\
=& \langle \frac{D}{\partial u}\frac{D}{\partial t}\frac{\partial}{\partial t} \alpha(u,t) + R(\frac{\partial\alpha}{\partial u},\frac{\partial\alpha}{\partial t})\frac{\partial \alpha}{\partial t}, \frac{\partial}{\partial t} \alpha(u,t) \rangle \\
=& \langle R(\frac{\partial\alpha}{\partial u},\frac{\partial\alpha}{\partial t})\frac{\partial \alpha}{\partial t}, \frac{\partial \alpha}{\partial t} \rangle = 0
\end{align*}
