One question that popped up during the studies of special and general relativity (which I am forced to take unfortunately) is the following:

Bending of space-time

How do we know that this is due to the bending of space-time and not just plain old 3rd semester diffraction. If you find this a silly question, downvote, otherwise consider the following picture.Razor

The Sun is the straightedge, the screen is the earth and the star is the point source.
This is simplified to a great extent but the idea still holds (I think)

  • $\begingroup$ I don't know enough about this, but the 1973 eclipse experiment measured the deflection of light of stars that appear a few solar radii away from the sun, rather than at the edge. The deflection gets smaller for the images further away from the sun, but it still follows the right curve. (Also, the Hippacros mission looked at the deflection curve all the way to stars that are 90 degrees away from the sun, so while there may be some other explanation, it definitely doesn't have to do with the sun's edge, I'd say.) I apologize if I misinterpreted anything or if this isn't relevant. $\endgroup$ – Maximal Ideal Apr 23 '18 at 0:25
  • $\begingroup$ The only way I figure that could happen is if gravity was a pseudo-force due to the rotation of the whole system (universe). Also, I've seen the measurements from the theoreticians at the neighboring Institute on how they calculate the correct position and I must say I am not convinced :D. I wouldn't be asking this question since I have no desire to pursue it but it did pop up as I was studying so it made me wonder. Maybe it is a silly question, I really don't know. $\endgroup$ – Dominik Car Apr 23 '18 at 9:35
  • $\begingroup$ You might find this interesting: spiff.rit.edu/richmond/occult/bessel/bessel.html also you can read my derivation of it by clicking on my website link at the top of my page. $\endgroup$ – Bill Alsept Apr 27 '18 at 17:01
  • $\begingroup$ In my opinion the diffraction of individual photons explains this phenomenon just fine. And there should not be a sharp divide at the shadow. It would be more spread out. $\endgroup$ – Bill Alsept Apr 27 '18 at 17:17
  • $\begingroup$ Dominik, how do you know they’re not the same thing? Light diffracting or curving around a straight edge, the sun, the moon or anything is probably caused by the same thing, gravity. What causes gravity is another thing. $\endgroup$ – Bill Alsept Apr 27 '18 at 21:12
  1. We know this because the position of the apparent star is perfectly matching the GR calculations about bent spacetime, depending on a few things including the mass of the star (the one in between that bends spacetime, in your case the Sun).

  2. What you are describing, interference, would not depend on the same way on the mass, the density, stress-energy and a few more things as GR describes bent spacetime.

  3. There were numerous calculations and experiments like the Shapiro test and they all perfectly gave the matching numbers according to GR.

  4. Interference would not depend on the same things, for example interference would react differently on the size/mass ratio or density of the star, whereas in GR it really matters what your star's energy density, for example, is compared to its size, for example, a black hole in your case would have an interference of what? I believe that interference would not even work with a black hole.

  • $\begingroup$ One thing I’ve never understood is how we view the starlight during an eclipse. How and where do the calculations deal with the light also passing the edge of the moon before we see it? $\endgroup$ – Bill Alsept Apr 27 '18 at 18:07
  • $\begingroup$ Dear Bill Alsept, the moon would distort the position of behind stars so small, it is not really measurable. $\endgroup$ – Árpád Szendrei Apr 27 '18 at 18:50
  • $\begingroup$ @ÁrpádSzendrei it is very measurable, I have done it. See here: spiff.rit.edu/richmond/occult/bessel/bessel.html $\endgroup$ – Bill Alsept Apr 27 '18 at 20:02
  • $\begingroup$ note that the effect (esp. Shaprio, which is a little different) is independent of wavelength. $\endgroup$ – JEB Mar 31 '19 at 4:07

On one hand, a typical diffraction angle $\theta$ for light with wavelength $\lambda$ by a spherical obstacle with the radius $R$ of the Sun is $$\theta~\sim~\frac{\lambda}{R}~\sim~~\frac{10^{-6} \text{ m} }{10^{9} \text{ m}}~\sim~10^{-15}\text{ rad};$$ while on the other hand, the gravitational bending/deflection of light by the Sun $$\theta~=~\frac{2r_s}{R}~\approx~\frac{2\cdot 3 \text{ km} }{7 \cdot 10^5\text{ km}}~\sim~10^{-5}\text{ rad}$$ is a roughly 10 orders of magnitude bigger!


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