 Solution

A photon follows a null geodesic, so the parameter called proper time in the question must instead be an affine parameter: proper time is identically zero on a null worldline. Denote this parameter by $\tau$ for continuity with the displayed formulas.

The geodesic Lagrangian in the equatorial plane is $\frac{1}{2}(-F{\dot{t}}^2+F^{-1}{\dot{r}}^2+r^2{\dot{\varphi }}^2)$. The cyclic coordinates give conserved quantities $E=F\dot{t}$ and $h=r^2\dot{\varphi }$, proportional to the energy measured at infinity and angular momentum. The null constraint then givesThe overall scale depends on the affine parameter, while $b=h/E$ is the invariant impact parameter. For $u=1/r$, we have $\dot{r}=-hdu/d\varphi$, giving

At zeroth order choose the direction of the orbit so that $u_0=\sin \varphi /b$. Then $r\sin \varphi =b$, a straight line at distance $b$ from the origin. Differentiating the orbit equation gives $u^{\prime \prime }+u=\frac{3}{2}{r}_s{u}^2$. To first order,solves this equation; the homogeneous first-order correction can be absorbed into the choice of asymptotic direction and impact parameter. It also satisfies the first integral to this order: the terms $2u_0^{\prime }{u}_1^{\prime }+2u_0{u}_1-u_0^3$ cancel.

Near the incoming asymptote its zero is at $\varphi =-r_s/b+O(r_s^2/b^2)$, and near the outgoing asymptote at $\varphi =\pi +r_s/b+O(r_s^2/b^2)$. Thus the excess angular change over a straight trajectory isto first order. Solar light-deflection measurements and gravitational lensing provide observational evidence for this effect. The perturbation requires $r_s/b\ll 1$.