we wish to show that:
z = (lambda_o - lambda_e ) / lambda_e

This is the formula for the redshift, denoted by z, is related to the expansion of the universe

• z is the redshift
• lambda_o is the wavelength of the observed light
• lambda_e is the wavelength of the emitted light
Light travelles along a null geodesic. This comes from general relativity, and is in fact one of the inputs into that theory. What this means is that the distance travelled by a piece of light in spacetime = 0 (honestly :) The distance is measured by the metric. The metric distance for a photon is given by the following formula:
ds^2 = c^2dt^2 - a(t)^2dr^2
• ds^2 is the square of the proper distance
• c^2dt^2 is the speed of light squared times the proper time interval squared
• a^2 dr^2 is the scale factor of the universe (how big it is kinda) times the distance interval squared
No angular dependance as light travells radially. Now cos the geodesic is null then ds^2 = 0. Solve the equation for r to find in integral form

r = int_(obs)^(emit) (c dt)/(a(t))

• obs is the time of observation of the photon
• emit is the time of emission of the photon
r is a comoving quantity and is invariant (I know this does not make much sense, my apologies, at some point I will elucidate furthur) In order to remain invariant if you tweak the limits on the integral the integrand has to remain the same. That is the crux of the argument. For this to hold the following equation must hold

dt(emit) / dt (obs) = a(obs) / a(emit)
Time is inversly related to the frequency and hence related to the wavelength. So we see that for the photon we have related changes in its wavelength over time to changes in the size of the universe over time. This is the cosmological redshift (as opposed to gravitational or doppler. The really important thing to note is that it is a property of the metric. For furthur reading try "Cosmological Physics" by John Peacock, who is a very nice man and a cosmologist. Its a bit heavy but dosen't fuck around.