Proving a curve is a geodesic. I am really stuck on the following question.
Let $ \ \gamma : I \longrightarrow M \ $ be a non-constant (i.e  $ \ \gamma'\ $  is not identically zero) geodesic. Show that a reparametrization $\ \gamma \circ h : J \longrightarrow M \ $ is a geodesic if and only if $ \ h: J \longrightarrow I \ $ is of the form $h(t) = at+b \ $ with $ \ a, b \in \mathbb{R}$.  
Does anyone have any idea how to prove this?
 A: For $h: J\rightarrow I$ such that $s=h(t)$, we use $'$ to denote $\frac{d}{dt}$ and $\cdot$ to denote $\frac{d}{ds}$. Then  we have
$$(\gamma\circ h)'(t)=h'(t)\dot{\gamma}(s)$$
by chain rule.
Therefore, we have
$$\nabla_{(\gamma\circ h)'(t)}(\gamma\circ h)'(t)=
\nabla_{h'(t)\dot{\gamma}(s)}\Big(h'(t)\dot{\gamma}(s)\Big)$$
$$=
h'(t)\nabla_{\dot{\gamma}(s)}\Big(h'(t)\dot{\gamma}(s)\Big)=
h'(t)^2\nabla_{\dot{\gamma}(s)}\dot{\gamma}(s)+\frac{d}{ds}\big(h'(t)\big)\dot{\gamma}(s).$$ Since $\gamma(s)$ is geodesic, $\nabla_{\dot{\gamma}(s)}\dot{\gamma}(s)=0$, which implies that
$$\tag{1}\nabla_{(\gamma\circ h)'(t)}(\gamma\circ h)'(t)=\frac{d}{ds}\big(h'(t)\big)\dot{\gamma}(s).$$
Therefore, if $(\gamma\circ h)(t)$ is geodesic, i.e. $\nabla_{(\gamma\circ h)'(t)}(\gamma\circ h)'(t)=0$  if and only if 
$$\tag{2}\frac{d}{ds}\big(h'(t)\big)=0.$$
Integrating it, we have
$h'(t)=a$, which implies that 
$h(t)=at+b$ for some constant $a, b\in\mathbb{R}$. Conversely, if $h(t)=at+b$ for some constant $a, b\in\mathbb{R}$, then $(2)$ is satisfied, which implies that the expression in $(1)$ is zero, i.e. $(\gamma\circ h)(t)$ is geodesic.
