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At hyperphysics the formula is:

$$ \frac{\nu_\infty}{\nu_0} = \sqrt{1 - \frac{2GM}{c^2r}} $$

I found this article in which the authors

have studied the wavelengths of a large number of solar iron lines observed near the center of the disk. They show that the observed shifts (solar wavelength minus laboratory wavelength) for strong lines agree with the predicted values given by

$$\Delta\lambda/\lambda = \frac{GM}{rc^2}= 2.12^{-6}$$

Can you please say if this formula is correct and where it comes from?

Also, if it is, I would imagine that the formula for the frequancy at infinity (or else) would be

$$\nu = \nu_{r_0}\left( 1-\frac{GM}{r_0 c^2}\right)$$

Can you tell me where I am wrong or why the official formula is different? Is that formula expressly mentioned in SR or is it derived in some way?

John Rennie
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user157860
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  • I would imagine That's not very useful to respond to. You need to explain why you expect that form or we can't help you with any conceptual issue that may be involved. – StephenG - Help Ukraine Aug 08 '18 at 07:14

1 Answers1

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The first formula is correct, it comes from the Schwarzschild metric and assumes that the Sun is a spherically symmetric mass. It compares the proper time of a stationary observer at infinity with the proper time of an observer at $r_0$.

Your third equation contains a transcription error, since you have missed the factor of two and the square root from the first equation?

The second equation is an approximation and arises if $r \gg 2GM/c^2$ and you do a binomial expansion of the first expression and then express the frequencies in terms of wavelength.

i.e. $$ \frac{\nu_{\infty}}{\nu_0} =\frac{\lambda_0}{\lambda_{\infty}} \simeq 1 - \left(\frac{1}{2}\right)\left(\frac{2GM}{c^2r_0}\right)+ ...$$

and then $$\frac{\Delta \lambda}{\lambda_{\infty}} = 1-\frac{\lambda_0 }{\lambda_{\infty}} \simeq \frac{GM}{c^2r_0} $$

ProfRob
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  • It's not a transcription error, if the formula were not approximated, if GM/rc^2 is the exact value of the ratio between the decrease of frequency and the original photon, then the frequency of the photon at infinity $f_\infty$ is given by the original frequency f minus the ratio times the frequency = f - f * ratio = $f_{r_0}\times (1-\frac{GM}{r_0c^2})$ – user157860 Aug 08 '18 at 07:44
  • I really don't follow. $\nu$ is the frequency in your first formula. You literally multiply both sides by $\nu_0$ to get what should be your third formula. – ProfRob Aug 08 '18 at 08:04
  • I was just using f instead of nu. I didn't know that the second formula was an approximation. – user157860 Aug 08 '18 at 08:06