OK, let me give you guys my approach to the problem; it's long so be patient.

First, I assume a uniform gravitational field. Uniform gravitational fields have the property that the force is directly proportional to the mass of an object by a constant a. (This is a property of all fictitious forces.)

F=ma

We also have that

F=dp/dt=d(γmv)/dt=γ^{3}mdv/dt.

(by the chain rule)

Mass cancels so we have the differential equation

a=γ^{3}dv/dt.

Solving this differential equation for v with the initial condition of v(0)=0, we have

v(t)=at/√(1 + a^{2}t^{2}/c^{2}).

With our uniform gravitational field, we are working in a non-inertial frame. We want to work in an inertial frame because that's where light travels in straight lines. So let's move to the frame where objects fall freely, that is, the frame that moves with speed v(t) with respect to our non-inertial rest frame. To do this we need to perform a Lorentz transformation.

First we need to find the position of the non-inertial frame in our inertial frame as a function of time, x(t). To find this we simply integrate -v(t) and obtain

x(t)= -c^{2}/a (√(1 + a^{2}t^{2}/c^{2}) - 1)

with the initial condition that x(0)=0. We can now apply the Lorentz transform on the x coordinate.

x'=γ(x - x(t))

x is the position in the inertial frame, and x' is the position in our non-inertial rest frame. In addition,

γ=1/√(1 - v(t)^{2}/c^{2})=√(1+a^{2}t^{2}/c^{2}),

so we have

x'=√(1 + a^{2}t^{2}/c^{2})(x + c^{2}/a(√(1 + a^{2}t^{2}/c^{2}) - 1))),

which simplifies to

x'=√(1 + a^{2}t^{2}/c^{2})(x - c^{2}/a) + at^{2} + c^{2}/a

.

Now consider a light beam which travels in a straight line from a height of h to the ground at some angle θ from the vertical. Its x coordinate is given by

x=h - cos(θ)ct

and y coordinate is given by

y=sin(θ)ct.

We notice that the y component is perpendicular to the direction of motion so it is unaffected by Lorentz contraction and y=y'. We solve this relationship for t,

t=y'/(sin(θ)c),

and plug in to the Lorentz transformation equation along with the previous equation for x. We now have an equation for the path of light in terms of x' (the vertical height), y' (the horizontal distance), θ (the angle from which the light was emitted from the sun), h (the height of the sun) and a (the constant acceleration).

x'=√(1+a^{2}y'^{2}/(c^{4}sin^{2}(θ)))(h - y'/tan(θ) - c^{2}/a) + ay'^{2}/(c^{2}sin^{2}(θ)) + c^{2}/a

In the more familiar xy coordinates where y is vertical and x is horizontal this is

y=√(1+a^{2}x^{2}/(c^{4}sin^{2}(θ)))(h - x/tan(θ) - c^{2}/a) + ax^{2}/(c^{2}sin^{2}(θ)) + c^{2}/a

.

Clearly, this path is not parabolic, but is nearly so, as can be seen if one graphs it. It only remains to choose the free parameters a, h, and θ to match observation.