## Saturday, November 28, 2015

### 34 | Worksheet 11.1, Problem 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.

 CMB (image: cosmology.berkeley.edu)

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 (the particle content of the electromagnetic radiation is called a photon) satisfies the Planck spectrum. At about the redshift $$z \approx 1100$$ when the universe had the temperature $$T \approx 3000 \text{ K}$$, 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 a redshift $$z$$ with frequency $$\nu$$, what is the its frequency $$\nu'$$ today?

We have shown before that the frequency observed today $$\nu'$$ is equal to the emitted frequency $$\nu$$ multiplied by the scale factor $$a(t)$$ at the time the light was emitted. This gives:
$\nu' = a(t)\nu$
In a previous post, we've also shown that the scale factor is related to the redshift (\z\) corresponding to light emitted at time $$t$$:
$a(t) = \frac{1}{1+z}$
Plugging this into our frequency expression, we get:
$\boxed{\nu' = \frac{\nu}{1+z}}$

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

To answer this question, we need to consider two components of the energy density term. In a most basic sense, energy density is defined as $$\frac{\text{energy}}{\text{volume}}$$. As the universe expands, both of these values change. Let's start with the energy.

As the photon travels, it is redshifted. Since it's frequency changes, its energy is also "redshifted." The energy of a photon is given as $$E = h\nu$$. In (a), we showed that $$\nu$$ changes by a factor of $$(1+z)^{-1}$$. Since frequency and energy are directly proportional, we know that the change in energy must also be proportional to $$(1+z)^{-1}$$.

As the universe expands, the volume gets larger. This essentially "dilutes" photons in the universe. We know that volume is proportional to $$d^3$$, where $$d$$ represents some distance. In a previous worksheet, we showed that the proper distance $$d(t) = a(t)d$$. In terms of volume, this gives $$V(t) = a^3(t)V$$. In terms of redshift, we know have $$V(t) = (1+z)^3 V$$. This changes our energy density term by a factor of $$(1+z)^{-3}$$.

Between the change in energy and change in volume, we know that the energy density must overall change by a factor of $$(1+z)^{-4}$$. So,
$\boxed{u_{\nu'} \mathrm{d}\nu' = (1+z)^{-4} u_{\nu}\mathrm{d}\nu}$

(c) Plug the relation between $$\nu$$ and $$\nu'$$ into the Planck spectrum:
$u_{\nu} \mathrm{d}\nu = \frac{8\pi h \nu^3}{c^3} \frac{1}{\mathrm{e}^{\frac{h\nu}{kT}}-1} \mathrm{d}\nu$
and also multiply it with the overall energy dilution factor that you have just figured out to get the energy density today. Write a final expression as the form $$u_{\nu'}\mathrm{d}\nu$$. Show that it is exactly the same Plank spectrum except that the temperature is now $$T' = T(1+z)^{-1}$$.

If we rewrite the Planck spectrum in terms of the energy density today, we get:
$u_{\nu'} \mathrm{d}\nu' (1+z)^{-4} = \frac{8\pi h (\nu' (1+z)^{-1})^3}{c^3} \frac{1}{\mathrm{e}^{\frac{h\nu' (1+z)^{-1}}{kT}}-1} \mathrm{d}\nu'(1+z)^{-1}$
When we solve for the energy density today $$u_{\nu'} \mathrm{d}\nu'$$, we get:
$u_{\nu'} \mathrm{d}\nu' = \frac{8\pi h \nu'^3}{c^3} \frac{1}{\mathrm{e}^{\frac{h\nu' (1+z)^{-1}}{kT}}-1} \mathrm{d}\nu'$
Note that this equation is the same as above, except for the exponential term which is now:
$\mathrm{e}^{\frac{h\nu' (1+z)^{-1}}{kT}}$
Note that we can simply rewrite this as:
$\mathrm{e}^{\frac{h\nu'}{kT'}}$
where $$T' = T(1+z)^{-1}$$.

(d) As you have just derived, according to the Big Bang model we should observe a black body radiation with temperature $$T'$$ throughout the entire universe. This is the Cosmic Microwave Background. Using the information given at the beginning of this problem, what is this temperature $$T'$$ today?

Above, we found:
$T' = T (1+z)^{-1}$
At the beginning of the problem, we were told that the CMB was formed at a redshift of about $$z \approx 1100$$ when the universe had a temperature $$T = 3000\text{ K}$$. Plugging these values in, we can get an approximate temperature for the CMB today:
$T' = (3000 \text{ K}) (1 + 1100)^{-1} = \boxed{2.72 \text{ K}}$