cosmic microwave background - CMB anisotropy, temperature and doppler effect

How would you derive the following equation:

$$T' = T_{CMB} \left( \frac{\sqrt{1-v^2}}{1+v\cos{\theta}} \right)$$

which describes how the temperature of the CMB varies due to the speed $v$ of the Earth. The angle $\theta$ is the direction of observation with respect to the Earth's velocity.

Answer

The relativistic doppler shift of frequency is given by

$$f_o = \frac{f_ s}{\gamma (1 + (v/c) \cos \theta)},$$ where $f_o$ is the observed frequency, $f_s$ is the emitted (source) frequency, $\gamma = (1 - v^2/c^2)^{-1/2}$ and $\theta$ is as you define. This is a standard result - e.g. http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Transverse_Doppler_effect

For blackbody radiation, the temperature is obtained from the peak frequency $h f_{\rm max} = 2.82kT$. This latter is an expression of Wien's law for a blackbody spectrum in terms of frequency. So you can replace the frequencies with temperatures and there you have it.

$$T_o = T_s \frac{(1 - v^2/c^2)^{1/2}}{1 + (v/c)\cos\theta}$$

And this is your equation, where I presume your $v$ is in units where $c=1$.