Gravitational redshift in terms of wavelength

Question

I know that Einstein’s theory of general relativity predicts that the wavelength of electromagnetic radiation will lengthen as it climbs out of a gravitational well. Photons must expend energy to escape, but at the same time must always travel at the speed of light, so this energy must be lost through a change of frequency rather than a change in speed. If the energy of the photon decreases, the frequency also decreases. This corresponds to an increase in the wavelength of the photon, or a shift to the red end of the electromagnetic spectrum – hence the name: gravitational redshift.
How can I get the gravitational redshift in terms of the wavelength?
I would really appreciate your help.

joseph h · Accepted Answer

For fixed $r$ and $phi$, all but the time differentials vanish, so
$$ds^2 =  c^2 dtau^2  = left(1-cfrac{a}{r}right)c^2dt^2$$
so that
$$  dtau^2  =left(1-cfrac{a}{r}right)dt^2$$
$$ frac{dtau^2}{dt^2}= left(1-cfrac{a}{r}right)$$
$dtau$ is the clock time of an observer at distance $r$ from the source, $dt$ is the
time measured by a distant observer, and $a$ is the Schwarzshild radius given by
$$a = frac{GM}{c^2}$$
Now
$$ frac{d tau}{dt} = frac{ clambda_s} { c lambda}=  sqrt{ 1- frac{GM}{r c^2}}$$
where the subscript $s$ stands for source and $lambda$ is the shifted wavelength. Finally we can write
$$ frac{lambda} {lambda_s} = ( 1- frac{GM}{r c^2})^{-frac{1}{2}}$$

Gravitational redshift in terms of wavelength

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