At hyperphysics they give this general formula for a Schwarzschild background:
$$\frac{v_{\infty}}{v_0}= \left( 1-\frac{2GM}{r_0 c^2} \right)^{\frac{1}{2}}, \tag{1}$$
and a formula (2) "to express the frequency shift between two locations":
$$\frac{v}{v_0}\approx 1 - \frac{GM}{r_0 c^2}. \tag{2}$$
Can you explain how to use these formulae to find the change in frequency/wavelength of a photon? For example one of $2 eV$ energy (green-cyan) = $4.836 \times 10^{14} h\nu$ emitted on the surface of the Sun and observed on Earth? I suppose we can ignore the blueshift produced by the Earth.
A Google search gave one specific article, which explains changes in terms of velocity. Do you know of actual experimental data, precise figures concerning the redshift of light coming from the Sun? How do they find the redshift, do they compare specific frequencies with the ones emitted in the lab?
I found a recurring value:
- $2.1 \times 10^{-6}$, how and when we can use it?
If we substitute values in the formula (2) we get the ratio $$\left(1- \frac{2950}{7\times 10^8}\right) =.9999958$$ and the frequency of the shifted photon is $4.8359 7962$ :
- $2\times 10^{-5}$.
Is this approximated value correct? Can you tell me how to find the exact value predicted by the theory, since we have nowhere so far considered the distance travelled by the photon? Do you know of a particular frequency of solar photon tested and what is the difference with the predicted value?
- If the photon has $1 MeV$ does it lose $2$ or $21 eV$?