 Solution

For radial motion, use the geodesic LagrangianThe Euler-Lagrange equation for $t$ gives $c^2\ddot{t}+a(da/dt){\dot{\chi }}^2=0$. Since $\chi$ is a cyclic coordinate, its conjugate momentum is constant. HenceFor a massive particle, take $\tau$ as proper time; the four-velocity normalization is $c^2{\dot{t}}^2-a^2{\dot{\chi }}^2=c^2$. For a photon, use an affine parameter rather than proper time, and the corresponding tangent is a null vector. ThereforeThe normalization of the null affine parameter remains arbitrary.