You must have responded in hurry here. For non euclidean surface, I feel motion of a particle can be on geodesic even in presence of external force. So it cannot be said that all such loci are non-straightline.
Only if external forces sum to zero will the trajectory of a particle be on a geodesic.
That's a direct translation of the concept of Newton's $$\sum_i \vec{f}_i = m \vec{a}$$ into general relativistic terms.
1. So in principle you are saying that two photons, initially prallel will not remain parallel.
Actually I wanted to say about instersection of the photons. But still ok.
As per below link, atleast Photons will not loose parallel aspect, due to self gravitation.
http://www.ptep-online.com/index_files/2013/PP-35-L3.PDF
Your source is talking about a universe devoid of all matter, energy and cosmological constant other that two parallel photons. No one who read your post assumed that was the case being considered. I was talking about two photons in our universe.
Here is the paper (ref 1 of your source) from 1931 which supports your claim at the end of §6 and also the end of §9:
Richard C. Tolman, Paul Ehrenfest, and Boris Podolsky. "
On the gravitational field produced by light."
Physical Review 37 (5) : 602. (March 1, 1931)
In order to reach that conclusion they used approximation techniques to calculate how spacetime curves about beams and pulses of light and then asked about the directional dependence of how that curvature changes the description of speed of light and geodesics of particles. For example in section §7 they show that a pencil of light has twice the gravitational attraction on stationary particles than expected from naively considering $$E=mc^2$$ to give the "mass of the photon".
Now the issue is will the photons (initially parallel) on two geodesics (one at r and another at r+dr) outside a spherical symmetry massive source, will they diverge as Farsight is stating, meaning inner geodesic curving more than the outer geodesic ? Intuitively it does not appear so, but then you know the maths.
I know the maths are quite rough, but here is a case where an exact solution is known.
If the at the event given by Schwarzschild coordinates $$(t,r,\theta,\phi)$$ we choose $$T^{\mu} d\lambda$$ to be the tangential light-like direction such that:
$$dt = T^{t} d \lambda = d \lambda , \; dr = T^{r} d \lambda = 0, \; d\theta = T^{\theta} d \lambda = \frac{c}{r} \sqrt{1 - \frac{r_s}{r} } d \lambda, \; dr = T^{\phi} d \lambda = 0 $$ (here I choose a dimensional $$\lambda$$ with units of time)
then it follows that if there are two nearby geodesics separated by infinitesimal bit $$X^{\mu}$$ that the geodesics diverge with an acceleration of:
$$R_{\beta \gamma \delta}^{\alpha} T^{\beta} T^{\gamma} X^{\delta} = \frac{ c^2 r_{s} \left(r - r_{s}\right)}{2 r^{4}} \begin{pmatrix} X^{t} - c^{-1} \sqrt{\frac{r^{3}}{r - r_{s}} } X^{\theta}
\\ 3 X^{r}
\\ c \sqrt{\frac{r - r_{s}}{r^{3}} } X^{t} - X^{\theta}
\\ - 3 X^{\phi} \end{pmatrix} $$
which is never zero except when $$X^{\mu} \propto T^{\mu}$$, i.e. when the "two infintesimally separated" geodesics are part of the same space-time geodesic (or when $$r_s = 0$$ which is flat space-time).