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I have been reading about Eddington and his measurement of 1.75 arc seconds in 1919. What I am wondering is why was the measurement lateral? I picture a beam of star light entering a curvature of spacetime. The beam has no sense of up, down or left or right. So as it enters this curvature it seems arbitrary as to which curved path to follow. It could go up, down or sideways or anywhere in between. All paths seem equal. What determines which path the light will take? It would be very strange if it always took a path that considered the earth’s location.

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Maybe a picture will help. The big yellow circle is the sun, the small blue circle is the earth, and the black lines are light rays from the distant star (with the gravitational deflections greatly exaggerated):

enter image description here

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The starlight was slightly bent by the gravity of the Sun, in the direction of the Sun, as it passed near the Sun. The Earth had nothing to do with it.

There is nothing arbitrary about it. You can calculate the curvature of spacetime near the Sun, and you can calculate a null (light-like) geodesic through that curved spacetime. A geodesic grazing the Sun bends by 1.75 arc seconds.

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  • $\begingroup$ Thanks for the response. Why didn’t it go over the top and down or under, what we would call the bottom, and then up. Why around the side as seen from us? $\endgroup$ – Lambda Jan 6 at 19:34
  • $\begingroup$ That’s where the particular star Eddington was observing happened to be... in some direction just outside the eclipsed Sun. $\endgroup$ – G. Smith Jan 6 at 19:38
  • $\begingroup$ The star wasn’t directly behind the center of the eclipsed Sun. That would require a much larger deflection. $\endgroup$ – G. Smith Jan 6 at 19:42
  • $\begingroup$ What I’m wonder is this: it wasn’t really just outside. It was also just above or just below the sun. So Eddington could have just have easily seen a shift up or down in the apparent location. $\endgroup$ – Lambda Jan 6 at 19:44
  • $\begingroup$ In the tangential direction? No, because the gravity of the Sun is radial and attractive. $\endgroup$ – G. Smith Jan 6 at 19:50

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