= Solution
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 $b\in T_y\partial M$,
$$
d_xd_g(x,y)(a)=-\langle v,a\rangle_g,
\qquad d_yd_g(x,y)(b)=\langle w,b\rangle_g.
$$
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 $v_T=-\operatorname{grad}_x^{\partial M}d_g(x,y)$ and $w_T=\operatorname{grad}_y^{\partial M}d_g(x,y)$. Unit speed and the incoming/outgoing signs supply the missing normal components:
$$
v=v_T-\sqrt{1-|v_T|^2}\,\nu_x,
\qquad w=w_T+\sqrt{1-|w_T|^2}\,\nu_y.
$$
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. Therefore
$$
\boxed{\alpha_{g_1}=\alpha_{g_2}}.
$$
In fact their exit times also agree, since each is $d_g(x,y)$.
Back to article page
