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Let $\rho (x,y)$ be the common boundary distance function, with $x\ne y$. For each metric, the simple Riemannian manifold property gives a unique unit-speed geodesic from $x$ to $y$, smoothly depending on the endpoint pair. Denote its initial and terminal velocities by $v_i,w_i$.

The first variation of geodesic energy, or equivalently length variation at unit speed, gives the endpoint formulaFor completeness, if $J$ is an endpoint variation along a geodesic, differentiating its length and integrating the covariant derivative by parts givesThe integral vanishes by the geodesic equation, leaving precisely these two boundary terms and their opposite signs.

Because the full metrics agree on the boundary tangent spaces $T_{x}M$, the induced boundary metric $h$, unit spheres and inward unit normals agree there. The displayed covectors therefore determine the same tangential velocity components for the two metrics:Their normal components are fixed by unit length and the entry or exit sign. With inward unit normals ${\nu }_x,{\nu }_y$,Strict convexity makes these signs strict for distinct endpoints. Thus $v_1=v_2$ and $w_1=w_2$ for the same endpoint pair.

Now start with any strictly inward $(x,v)$ and let its $g_1$ exit point be $y$. The $g_2$ geodesic joining $x$ to this same $y$ has the same initial velocity $v$, so uniqueness of the geodesic equation identifies it with the $g_2$ trajectory launched from $(x,v)$. Its exit velocity is also the same. HenceThis is the boundary distance determines geodesic scattering argument. The extension to outward vectors follows by the inverse scattering relation, and tangential vectors are fixed for both metrics. Equality of the travel times follows as well, since both equal $\rho (x,y)$.