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.