these points. Let X G Ft and let u be the Finsler unit

to the trajectory of light from A to X. An infinitesimal

Huygens principle states that the tangent space to the

parallel to the tangent space to the indicatrix TUS(X)

figure 1.11.

Figure 1.11. Huygens principle

We are in a position to deduce the billiard reflection

geometry. To fix ideas, let us consider the two-dimensi

Let I be a smooth curved mirror (or the boundary of a

and AXB the trajectory of light from A to B. As us

that point X extremizes the Finsler length of the brok

Theorem 1.14. Let u and v be the Finsler unit vectors

incoming and outgoing rays. Then the tangent lines to

S(X) at points u and v intersect at a point on the ta

at X; see figure 1.12 featuring the tangent space at po

Proof. We repeat, with appropriate modifications, th

the Euclidean case. Consider the functions f{X) = \AX

\BX\ where the distances are understood in the Finsl

and r\ be tangent vectors to the indicatrix S(X) at p

One has, for the directional derivative, Du(f) = 1 sin

to the trajectory of light from A to X. On the othe

Huygens principle, £ is tangent to the front of point

through point X. This front is a level curve of the fun

Dz(f) = 0. Likewise, Dv(g) = 0 and Dv(g) = - 1 .