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When I see calculations of the Sun’s redshift as seen from Earth I believe, as I understand it, that 1) the gravity potential at the emission point near the Sun is taken into account, then 2) the Sun’s gravity potential at the Earth’s orbit and finally 3) the gravity potential at absorption, i.e. the Earth’s surface. I have at least seen all three components mentioned.

I am thinking about the acceleration that a point on Earth would have towards the Sun in its orbit. This is balanced against the Sun’s gravity (since the orbit is stable). I therefore wonder if that part, i.e. point 2), should be added at all, in the calculation.

Mikael Jensen
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There is a loss of frequency for a photon rising in the sun's gravitational field and a gain as it drops in the earth's field. Both of these are quite small. There would also be a Doppler shift for photons coming from various points on a rotating sun and being observed at various points on a rotating earth. (Though the gravitational shift is small from satellite orbits , both it and the Doppler shift must be taken into account for the accurate functioning of the GPS system.)

R.W. Bird
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  • I am mainly thinking about the component arising from the Sun's gravitation at the Earth's orbit, i.e. No 2 in my list. I am not shure you mentioned that, perhaps because it cancels out. The Earth is i) located in the Sun's gratitation field and also ii) accelerated by its orbit $v^2/r$, and since by the equivalence principle these are equal, I argue that this component should vanish. I disregard the rotation of both the Sun and Earth. – Mikael Jensen Feb 07 '22 at 15:05