Let $(M, g)$ be a compact non-trapping manifold with strictly convex boundary. We have to show that $$ X \tilde{\tau}=-2 $$ where $\tilde{\tau}=\tau(x, v)-\tau(x,-v) .$ If $(x, v) \in S M$, let $\tau(x, v) \in[0, \infty]$ be the first time when $\gamma_{x, v}(t)$ exits $\boldsymbol{M}$ $$ \tau(x, v):=\sup \left\{t \geq 0 ; \gamma_{x, v}([0, t]) \subset M\right\} $$ We call $(M, g)$ non-trapping, if $\tau(x, v)<\infty$ for each $(x, v) \in S M$. Define geodesic flow on $S N$ as $$ \varphi_{t}: S N \rightarrow S N, \quad \varphi_{t}(x, v)=\left(\gamma_{x, v}(t), \dot{\gamma}_{x, v}(t)\right) $$ The geodesic vector field $X: C^{\infty}(S N) \rightarrow C^{\infty}(S N)$, which differentiates a function on $S N$ along geodesic flow: $$ X w(x, v)=\left.\frac{d}{d s} w\left(\varphi_{s}(x, v)\right)\right|_{s=0} $$
My attempt:
\begin{align*} X \tilde{\tau}(x,v)&=\left.\frac{d}{dt}\tilde {\tau}(\varphi_t(x,v))\right|_{t=0}\\ &=\left.\frac{d}{dt}\left(\tau(\varphi_t(x,v))-\tau(\varphi_t(x,-v)) \right) \right|_{t=0} \end{align*} I know that $\tau(\varphi_t(x,v))=\tau(x,v)-t$. And also $\tau(\varphi_t(x,-v)=\tau(x,-v)-t$. So this implies
\begin{align*} X \tilde{\tau}(x,v)&=\left.\frac{d}{dt}\tilde {\tau}(\varphi_t(x,v))\right|_{t=0}\\ &=\left.\frac{d}{dt}\left(\tau(\varphi_t(x,v))-\tau(\varphi_t(x,-v)) \right) \right|_{t=0}\\ &=\left.\frac{d}{dt}\left(\tau(x,v)-t-(\tau(x,-v)-t) \right) \right|_{t=0}\\ &=0. \end{align*}
Where I am missing I do not know. Any help or hint will be greatly appreciated.
Notice that: \begin{align} \varphi_t(x,-v) &= \left(\gamma_{x,-v}(t), \gamma'_{x,-v}(t) \right) \\ &= \left( \gamma_{x,v}(-t),-\gamma'_{x,v}(-t)\right). \end{align} Therefore: $$ \tau(\varphi_t(x,v)) = \tau (x,v)-t $$ and $$ \tau(\varphi_{-t}(x,v)) = \tau(\gamma_{x,v}(-t),-\gamma'_{x,v}(-t)) = \tau(x,-v) + t. $$ It follows that $\overline\tau(\varphi_t(x,v)) = -2t + \tau(x,v) - \tau(x,-v)$ and the result follows from a direct computation.