3
$\begingroup$

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 $21 eV$?
$\endgroup$
1
$\begingroup$

The frequency shift (ratio) between two locations $r_1$ and $r_2$: $$\frac{v_2}{v_1} \simeq 1 + \left(\frac{1}{r_2}-\frac{1}{r_1}\right)\frac{G M}{c^2}$$

ref., e.g.: https://arxiv.org/abs/gr-qc/0403082

$\endgroup$
5
  • $\begingroup$ correction: instead v1/v2 should be v2/v1 $\endgroup$ May 9 '19 at 11:38
  • 1
    $\begingroup$ Hi! You can edit your answer too instead of mentioning the correction in the comments. $\endgroup$
    – Eagle
    May 9 '19 at 11:40
  • 1
    $\begingroup$ For future reference, please link to the abstract page on Arxiv, as some users may be on mobile devices $\endgroup$
    – Kyle Kanos
    May 9 '19 at 13:00
  • $\begingroup$ Equation $(1)$ is using the Schwarzschild metric - which reduces to equation $(2)$ in a weak gravitational fields, e.g., the Sun. The calculated value is approximately $\Delta \nu/\nu=2.1\times 10^{-6}$ - or roughly $0.01\;A$ for $5,000 \;A$ - or green light. This is undetectable due to the Doppler broading of the spectral line. $\endgroup$ Oct 22 '19 at 5:16
  • $\begingroup$ And we derived equation $(2)$ using Newtonian gravity and $m=h\nu/c^{2}$ in my modern physics course with just brief introduction to special relativity. $\endgroup$ Oct 22 '19 at 5:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy