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Light bends twice as much due to gravity as Newtonian theory predicts as discussed in this related question. The argument couched in general relativity, quote from here.

Since most objects move much slower than the speed of light, meaning that they travel much farther in time than in space, they feel mostly the time curvature. The Newtonian analysis is fine for those objects. Since light moves at the speed of light, it sees equal amounts of space and time curvature, so it bends twice as far as the Newtonian theory would predict.

But is it possible to show this without general relativity, by considering a uniformly accelerated rocket and looking at how much light bends? Under the equivalence principle this situation should be the same as light in a uniform gravitational field.

A simple argument for the rocket example is that light goes in a straight line while the rest of the rocket accelerates at 1g. Suppose there are is a series of rulers that measure the height of projectiles as they pass by (the rocket is way too slow to reach relativistic speeds in the time it takes to run these experiments). Let t be the elapsed time after any projectile hits the "top" of it's trajectory (highest "height" as read by the rulers). The rocket has moved ahead by (1/2)9.81 t^2 m. Regardless whether it's a thrown stone or a photon, the projectile is travels in a straight line and thus has fallen behind in our accelerated frame by the same amount at any given t. There seems to be a flaw in this reasoning because it not giving us the needed factor of 2. What is the "catch"?

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    $\begingroup$ Newtonian gravity has nothing to do with bending space, so I don’t see how bending spacetime explains the factor of two. (Not to mention that there are three spatial dimensions and only one time dimension, so it seems like this “argument” would give a factor of 4/3.) Where did you get that argument from? $\endgroup$ – G. Smith May 13 at 0:04
  • $\begingroup$ @ G. Smith: Added the argument and it's source. $\endgroup$ – Kevin Kostlan May 13 at 0:14
  • $\begingroup$ Thanks! I see that it is from an MIT electrical engineering course. Although MIT is prestigious, I don’t consider it a valid physics argument. (The author calls it a “rough explanation”.) I will be interested to see what other physicists think. The point that you can derive the value up to a constant factor by dimensional considerations is completely valid. $\endgroup$ – G. Smith May 13 at 0:38
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But is it possible to show this without general relativity, by considering a uniformly accelerated rocket and looking at how much light bends?

No. The factor of 2 only comes out of the full formalism of General Relativity, in which there $g_{tt}$ term in the metric is not constant. Had there been a semi-Newtonian explanation of the factor of two, Eddington's confirmation of the GR prediction would not necessarily have been seen as a vindication of the whole GR theory.

Your seem to expect that the equivalence principle on its own should work, but that is not sufficient, because the bending of light involves the motion of a ray through a inhomogeneous gravitational field. The central field (Schwarzschild solution) is not spatially uniform, so its effects are not the same as the effects of an accelerating frame of reference.

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