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.

Answer 1

For fixed $r$ and $\phi$, all but the time differentials vanish, so

$$ds^2 = c^2 d\tau^2 = \left(1-\cfrac{a}{r}\right)c^2dt^2$$

so that

$$ d\tau^2 =\left(1-\cfrac{a}{r}\right)dt^2$$

$$ \frac{d\tau^2}{dt^2}= \left(1-\cfrac{a}{r}\right)$$

$d\tau$ 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

$$ \frac{d \tau}{dt} = \frac{ c\lambda_s} { c \lambda}= \sqrt{ 1- \frac{GM}{r c^2}}$$

where the subscript $s$ stands for source and $\lambda$ is the shifted wavelength. Finally we can write

$$ \frac{\lambda} {\lambda_s} = ( 1- \frac{GM}{r c^2})^{-\frac{1}{2}}$$