For the Newtonian or Special Relativistic calculation, it's easiest to treat the light as a particle, although a wave description would also work. The light is traveling at speed c. Say that the gravitational field g
is at right angles to the velocity. (That's the part of g
that gives curvature.) The radius of curvature (this is just simple geometry) is then c2
/g.In other words, for each little distance dx traveled, the direction changes by an angle dθ = gdx/c2
To get the total curvature as the light goes by the sun (say with closest approach distance L from the center) we have to integrate dθ over the whole path, taking into account the changing distance from the sun and the changing angle between the acceleration and the velocity. So long as you make the (excellent) approximation that the total curvature is very small you can just add the curvatures on the nearly uncurved path. If I've done the integral correctly, that gives θ=2GM/Lc2
for the total curvature.
Now what happens in GR? I'm a little more comfortable with a wave picture here. That part of the curvature we just calculated basically corresponds to the change in kinetic energy when something falls in a field. Energy is just the same thing as frequency, quantum mechanically. So what that part consists of is just that the part of the wave closer to the sun is higher frequency, with the wavefronts spaced closer together. Try drawing some wavefronts spaced a little closer together on one side and farther apart on the other, remembering that the wave is propagating at right angles to the fronts. You'll see the wave curve.
Why did I go through that wave exercise? In GR, the gravitational distortion affects not only the time part of spacetime (the part we just treated) but also the space part. In a standard coordinate choice, there's extra space, more path length, for the parts of the wave nearer the sun. Projection of that curved space onto a flat plane makes waves which have equal spacing nearer and farther from the sun look like they are more tightly spaced near the sun, since the flat space doesn't have that extra stretch in there. So that's the source of the extra bending. This second bending is just as big for massive particles as for light, but for particles moving slowly compared to c, the first component of the bending is so much bigger that it's hard to see this second part.
Why are the two components of equal importance for light? In dividing the effect up into the two parts, we've implicitly assumed a conventional coordinate system in which the speed of light is isotropic and everywhere constant. [The previous answer I'd given here was messed up./mw] I can now do the calculation, and it works out. It looks like the time-like part of the effect comes mainly from the region where the light gets closest to the sun. The space-like part seems to come mostly from two regions, where the light is approaching and leaving that region of closest approach. The reason for that is that the peculiar lengths (in this choice of coordinates) are only along the radial direction, not along the tangential direction. Unfortunately I don't yet have a deep enough feel for the calculation to summarize simply why the two parts have to come out equal.
As for the observational results, there are many references in http://en.wikipedia.org/wiki/Tests_of_general_relativity
(published on 08/28/11)