I note that there is a fair amount of light-bending questions, but I don’t really see the observer effect I am asking about. I apologize there is an answer already.

I wonder if the Earth’s gravity – at least in principle - were to enter the calculation of the effect of starlight bending noted by Eddington 1919.

Alternatively, one might wonder:

If the observer of starlight bending around the Sun, such as the case for Eddington’s measurements 1919, is hypothetically placed in a field of gravity equal to that of the Sun, i.e. if the Earth’s gravity during Eddington’s experiment were equal to that of the Sun, what would the observer see, in terms of apparent stellar displacement?

Would there be no effect at all, a classical Newtonian effect, i.e. the same as for an object with near light velocity in the same path, or something else again?

  • $\begingroup$ Related: physics.stackexchange.com/q/122319/2451 and links therein. $\endgroup$
    – Qmechanic
    Dec 6, 2020 at 16:07
  • $\begingroup$ @Qmechanic I can't see that is answers my question unless you know the details of the relativity equations. Two links down "observer" occurs in connection with the Newtonian mechanics. $\endgroup$ Dec 6, 2020 at 21:55
  • $\begingroup$ I actually found an answer in "The solar gravitational redshift from HARPS-LFC Moon spectra".: "A test of the general theory of relativity: "in accounting for the gravitational potential of the Earth, vGRS,⊕ = (GM⊕/cR⊕) = 0.21 m s−1 , where M⊕ and R⊕ are the nominal mass and radius of the Earth, we get a final value of: vGRS,theo = vGRS,1AU − vGRS,⊕ = 633.10 m s−1" . It is probably too obvious to experts. $\endgroup$ Dec 6, 2020 at 22:01
  • $\begingroup$ - the above effect is related to gravitational redshift where perhaps the issue is more evident than for angular reflection. $\endgroup$ Dec 6, 2020 at 22:28

1 Answer 1


Assuming spherical symmetry, the gravitational deflection of light from infinity that you see just on the horizon will be $2GM/c^2R$, because the path that it follows is half of the infinity-to-infinity geodesic that is used in the standard calculation. The deflection of light from straight overhead and from straight below your feet (imagining the planet/star to be transparent) will be zero by symmetry. So the deflection is probably a roughly sinusoidal function of the angle of elevation. At high elevation angles this would cause a roughly isotropic reduction of the apparent angular size of objects by $2GM/c^2R$, which is a few parts per million on the surface of the sun or about one part per billion on the surface of the earth.

The effect the Eddington experiment was looking for depends on the angular distance of the star from the sun in the sky, while this effect depends on the angle of the telescope, so I think it would be easy to disentangle them even if the earth had a much larger mass.


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