Since the g force of earth is 9.8 m/s*2 does that mean light 'drops' at that rate as it travels past earth? Or is general relativity a lot more complex than that?


If you assume that light "drops" at the gravitational acceleration $g$, and calculate the angle by which light is deflected using Newtonian mechanics, you end up with the formula:

$$ \theta = \frac{2GM}{c^2r} $$

where $M$ is the mass of the deflecting object and $r$ is the distance of closest approach. A quick Google should find the calculation, or see this paper on the Arxiv. However when you do the calculation using general relativity you get the result:

$$ \theta = \frac{4GM}{c^2r} $$

So GR predicts an angle of deflection twice as great as predicted by Newtonian mechanics. This was the basis of the Eddington measurement of bending by the Sun. The was not just to observe bending, since Newtonian mechanics predicted that. The aim was to show the bending was twice as large as the Newtonian prediction.

  • $\begingroup$ Wait did Newton know that light would bend due to gravity? $\endgroup$ – Ray Kay Jan 25 '15 at 1:04
  • 3
    $\begingroup$ It is not necessary that Newton knew it, you can still use his mechanics to make a prediction. $\endgroup$ – dmckee --- ex-moderator kitten Jan 25 '15 at 5:26

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