Blog #34: WS11.1, #2

2. Cosmic microwave background. One of the successful predictions of the Big Bang model is the cosmic microwave background (CMB) existing today. In this exercise let us figure out the spectrum and temperature of the CMB today. 
In the Big Bang model, the universe started with a hot radiation-dominated soup in thermal equilibrium. In particular, the spectrum of the electromagnetic radiation satisfies the Planck spectrum $$u_{\nu}d\nu=\frac{8\pi h_P\nu^3}{c^3}\frac{1}{e^{\frac{h_P\nu}{k_BT}}-1}d\nu$$ At about the redshift $z\approx 1100$ when the universe had the temperature $T\approx 3000K$, almost all the electrons and protons in our universe are combined and the universe becomes electromagnetically neutral. So the electromagnetic waves (photons) no longer get absorbed or scattered by the rest of the contents of the universe. They started to propagate freely in the universe until reaching our detectors. Interestingly, even though the photons are no longer in equilibrium with their environment, as we will see, the spectrum still maintains an identical form to the Planck spectrum, albeit characterized by a different temperature. 

a) If the photon was emitted at redshift z with frequency $\nu$, what is its frequency $\nu'$ today? 

Based on the redshift equation, we can find that $z=\frac{\nu-\nu'}{\nu'}=\frac{\nu}{\nu'}-1$
Solving for $\nu'$, we get $\nu'=\frac{\nu}{z+1}$.

b) If a photon at redshift z had the energy density $u_{\nu}d\nu$, what is its energy density $u_{\nu'}d\nu'$ today? 

The energy of the photon decreases as its wavelength gets stretched with the expanding of space. This is quantified by the changing frequency calculated above: the energy decreases by a factor of $\frac{1}{z+1}$. The number density of photons also decreases as space expands. If we imagine a region of space expanding:
$v\propto r^3\propto a^3r^3\propto(\frac{1}{z+1})^3r^3$, since the volume depends on the scale factor a, which is itself also proportional to $\frac{1}{z+1}$.
The combined effect means that $u_{\nu'}d\nu'=u_{\nu}d\nu(\frac{1}{z+1})^4$.

c) Plug in the relation between $\nu$ and $\nu'$ into the Planck spectrum and multiply it with the overall energy density dilution factor that you have just figured out to get the energy density today. Write the final expression as the form \(u_{\nu'}d\nu'\. What is $u_{\nu'}$? This is the spectrum we observe today. Show that it is exactly the same as the Planck spectrum, except that the temperature is now $T'=T(1+z)^{-1}$. 

We know the following:
$T=T'(z+1)$
$u_{\nu}d\nu=u_{\nu'}d\nu'(z+1)^4$
$\nu=\nu'(z+1)$, which means that $d\nu=d\nu'(z+1)$
We can then plug these into the Planck equation.
$u_{\nu'}d\nu'(z+1)^4=\frac{8\pi h_P[\nu'(z+1)]^3}{c^3}\frac{1}{e^{\frac{h_P\nu'(z+1)}{k_BT'(z+1)}}}d\nu'(z+1)$
$u_{\nu'}d\nu'=\frac{8\pi h_P\nu'^3}{c^3}\frac{1}{e^{\frac{h_P\nu'}{k_BT'}}}d\nu'$
This is the same form as the original Planck spectrum.
$u_{\nu'}=\frac{8\pi h_P\nu'^3}{c^3}\frac{1}{e^{\frac{h_P\nu'}{k_BT'}}}$

d) According to the Big Bang model, we should observe a black body radiation with temperature T' filled in the entire universe. This is the CMB. Using the information given at the beginning of this problem, what is the temperature T' today? 

$T'=T(z+1)^{-1}$
$T'=\frac{3000K}{1100+1}=2.72K$