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A simple Riemannian manifold has strictly convex boundary, no conjugate points, and a unique connecting geodesic depending smoothly on its endpoints. In particular its unit-speed geodesics are nontrapping. Write $\nu$ for the inward unit normal, and identify inward and outward parts of its boundary unit tangent bundle byThe signs are a convention; here $+$ means inward. If $(x,v)$ is strictly inward, let $\tau (x,v)>0$ be its first exit time. The geodesic scattering relation isThus it records both the exit point and the outward unit velocity, not just the exit point. On strictly outward vectors extend $\alpha$ by following the same geodesic flow backwards to its previous entry point, retaining the velocity orientation. On tangential vectors set $\alpha (x,v)=(x,v)$. The extended scattering relation is an involution on the full boundary unit bundle, since its forward and backward constructions undo one another. On a simple Riemannian manifold, the travel time equals the boundary distance function between the two endpoints; travel time is often recorded separately as lens data.

A simple Riemannian manifold has strictly convex boundary and a unique geodesic between each pair of points, smoothly dependent on its endpoints; there are no conjugate points and no trapped geodesics. Let $\nu$ denote the outward unit normal. For a boundary unit vector with $⟨v,\nu ⟩<0$, follow its unit-speed geodesic ${\gamma }_{x,v}$ until the next boundary intersection at time $\tau (x,v)>0$. Its geodesic scattering relation isThus it maps incoming boundary unit vectors to outgoing ones, recording both exit point and direction.

If the scattering relation is defined on the full boundary unit bundle, extend it on outgoing vectors by following the same oriented geodesic backwards to its previous boundary intersection, and fix tangential vectors. This extension is an involution, because the second application follows the same segment back to its starting point with the same velocity. The exit time is not a component of the scattering relation itself; together with it one obtains lens data. For a simple Riemannian metric, $\tau$ is the boundary distance between the endpoints of the segment.

Full equality of the Riemannian metrics at the boundary identifies their boundary unit bundles, outward normals and induced boundary Riemannian metrics. We show directly that the common boundary distance function determines both endpoint directions.

For distinct boundary points $x,y$, let $\gamma$ be the unique joining unit-speed geodesic, with initial velocity $v$ and final velocity $w$. The first variation of its length gives, for $a\in T_x\partial M$ and $b\in T_y\partial M$,Indeed, integration by parts in the length variation leaves just the endpoint terms, because the interior term contains ${\nabla }_{\dot{\gamma }}\dot{\gamma }=0$. The Riemannian metric is simple, so the off-diagonal endpoint dependence is smooth and these derivatives are valid.

Thus the tangential components are $v_T=-{\mathrm{grad}}_x^{\partial M}{d}_g(x,y)$ and $w_T={\mathrm{grad}}_y^{\partial M}{d}_g(x,y)$. Unit speed and the incoming/outgoing signs supply the missing normal components:All quantities on the right use only boundary distance and the common boundary Riemannian metric. Every strictly incoming vector for $g_1$ has some exit point $y$; the geodesic joining the same pair for $g_2$ has exactly the same reconstructed initial and final vectors. Hence both geodesic scattering relations give the same exit for that input. Tangential vectors are fixed, and the outgoing extensions agree as inverses. ThereforeIn fact their exit times also agree, since each is $d_g(x,y)$.

A compact manifold with boundary is simple when the boundary is strictly convex and any two points are joined by a unique geodesic depending smoothly on the endpoints, without conjugate points. Geodesics exit in finite time. These properties make off-diagonal boundary distance derivatives and the geodesic scattering relation well behaved.