I think there is something wrong with the proof this text gives of Lemma $2.1.5$, in pages $19$ and $20$, for timelike curves. I used another function, and it seems to work. Either I'm wrong, or he is. Can someone check, please? I'm using the convention $$\langle (x_1,y_1,z_1),(x_2,y_2,z_2) \rangle = x_1x_2+y_1y_2 - z_1z_2$$
and, just to remind, $\|(x,y,z)\| = \sqrt{|\langle (x,y,z),(x,y,z)\rangle|}$.
My work: We consider $\alpha: I \rightarrow \mathbb{L}^3$ a timelike curve. Fix $t_0 \in I$ and consider the function $$s(t) = \int_{t_0}^t \sqrt{- \langle \alpha ' (\xi), \alpha'(\xi) \rangle} \mathrm{d}\xi$$
Then we have $s'(t) = \sqrt{- \langle \alpha ' (t), \alpha'(t) \rangle} > 0$, then $s$ is strictly increasing, so we have an inverse $h$. Since $h(s(t)) = t$ for all $t \in I$, by differentiation, we have $$\begin{align} h'(s(t)) \cdot s'(t) &= 1 \\ h'(s(t)) &= \frac{1}{\sqrt{- \langle \alpha ' (t), \alpha'(t) \rangle}}\end{align}$$
and so $h'(t) = \dfrac{1}{\sqrt{- \langle \alpha ' (h(t)), \alpha'(h(t)) \rangle}}$. To finish it, I just have to check that the parametrization $\alpha \circ h$ satisfies $\| (\alpha \circ h)'(t) \| = 1$. So: $$\begin{align}\| (\alpha \circ h)'(t) \| &= \| \alpha'(h(t)) \cdot h'(t) \| \\ &= \left\|\alpha'(h(t)) \cdot \frac{1}{\sqrt{- \langle \alpha ' (h(t)), \alpha'(h(t)) \rangle}} \right\| \\[2pt] &=\sqrt{\left|\left\langle \frac{\alpha'(h(t))}{\sqrt{- \langle \alpha ' (h(t)), \alpha'(h(t)) \rangle}}, \frac{\alpha'(h(t))}{\sqrt{- \langle \alpha ' (h(t)), \alpha'(h(t)) \rangle}} \right\rangle\right|} \\[2pt] &= \sqrt{\left| \frac{1}{-\langle \alpha'(h(t)), \alpha'(h(t))\rangle} \cdot \langle \alpha'(h(t)), \alpha'(h(t)) \rangle\right|} \\ &=\sqrt{\left| \frac{1}{-1}\right|} = 1\end{align}$$ $\hspace{15 cm}\square$
Thank you!
You are perfectly right.
If $s(t)$ were $-\int_{t_0}^t\langle\alpha',\alpha'\rangle$ without the square root, we would get $\frac1{\|\alpha'\|}$ for the norm of $\alpha\circ s^{-1}$ instead of $1$.
Assume that the reparametrized curve is $\beta$, so $\beta\circ s=\alpha$ for some diffeomorphism $s$ (at least locally around $t_0$). Then we have $$\alpha'=(\beta\circ s)'=(\beta'\circ s)\cdot s'$$ consequently, $\|\beta'(s(t))\|=\displaystyle\frac{\|\alpha'(t)\|}{|s'(t)|}$, so what we need is $|s'(t)|=\|\alpha'(t)\|$, yielding to your solution.