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  How a uniform gravitational field affects the wavelength of a light beam?

+ 0 like - 0 dislike

I was reading Einstein's 1911 paper named "On the Influence of Gravitation on the Propagation of Light" when is stated a formula for frequencies measured by observers at different fixed positions (heights) on Earth surface. One observer is at the origin of some coordinate system and measures a frequency $\nu_0$of a light beam. The second observer is at some position $x$ in the same chart and measures a frequency $\nu$ of the same light beam. Einstein obtais that

$$ \nu = \nu_0 \left( 1 + \frac{\phi}{c^2}\right) \tag1$$

where $\phi \le 0$ such that $\nu \le \nu_0$, the famous gravitational redshift.

However, at the end of section 3 of this paper, Einstein came to other result about speed of light measured by these observers:

$$ c= c_0 \left( 1 + \frac{\phi}{c^2} \right) \tag2$$

and so $c\le c_0$. Here $c$ is measured by the "$\nu$"  observer an its called "coordinate speed of light" and $c_0$ is measured by the "$\nu_0$" observer, and it is the usual vaccuum speed of light.

My question is: How the wavelength should transform from one observer to the other? I mean, if I take eq. (1), I could use

$$\frac{c}{\lambda} = \frac{c_0}{\lambda_0}\left( 1+ \frac{\phi}{c^2} \right) $$

Then by eq.(2) we obtain $\lambda =\lambda_0$,   which I think is non-sense because since the observers have two different frequencies of the same light beam, they certainly should disagree about its wavelength. Where is my mistake?

asked Jul 19, 2021 in Theoretical Physics by Lil' Gravity (0 points) [ no revision ]
recategorized Jul 20, 2021 by Lil' Gravity

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