If F is a regular surface, $c: I \rightarrow F $ a geodesic and $\phi: J \rightarrow I$ a reparametrization, why does it hold that $c \circ \phi: J \rightarrow F$ is a geodesic if and only if $\phi(t) = mt + n$ for some $m, n \in \mathbb R$?
We had this theorem today in class, but unfortunately without any proof.
If you write down the definition of a geodesic, you'll see that a geodesic must have constant speed. That is, if $c \colon I \rightarrow F$ is a geodesic, we must have $ \left< c'(t), c'(t) \right> \equiv C$ for some $C \geq 0$ which is the speed of the geodesic. Thus, if $\phi \colon J \rightarrow I$ is a reparametrization and you want $c \circ \phi$ to be a geodesic, you must have
$$ \left< (c \circ \phi)'(t), (c \circ \phi)'(t) \right> = \left< c'(\phi(t)) \phi'(t), c'(\phi(t)) \phi'(t) \right> = (\phi'(t))^2 \left< c'(\phi(t)), c'(\phi(t) \right> \equiv D. $$
If $C = 0$ then $c$ is a constant curve and so any reparametrization of $c$ will also be constant and a geodesic. If $C > 0$, then we see that we must have $\phi'(t)^2 = \frac{D}{C}$ and since $\phi$ is regular ($\phi'(t) \neq 0$ for all $t \in J$), we must have
$$ \phi'(t) \equiv \sqrt{\frac{D}{C}} \implies \phi(t) = \sqrt{\frac{D}{C}}t + n $$
for some $n \in \mathbb{R}$ or
$$ \phi'(t) \equiv -\sqrt{\frac{D}{C}} \implies \phi(t) = -\sqrt{\frac{D}{C}}t + n $$
for some $n \in \mathbb{R}$.
The other direction can be checked directly using the definition of a geodesic.