Cosmic microwave background: Blog 34, Worksheet 11.1, Problem 2

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 (the particle content of the electromagnetic radiation is called photon) satisfies (1). At about the redshift z « 1100 when the universe had the temperature T = 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 its environment, as we will see, the spectrum still maintains an identical form to the Planck spectrum, albeit characterized by a different temperature.

Source

Part A: If the photons was emitted at redshift z with frequency ν, what is its frequency ν' today?

We know from two weeks ago that \[ z = \frac{v}{c} = \frac{ \lambda_0 - \lambda_e }{ \lambda_e } \]  Combining this with the fact that \[ \lambda = \frac{c}{\nu} \]  \[ z = \frac{ \frac{c}{\nu'}  -  \frac{c}{\nu}  }{  \frac{c}{\nu}  }= \frac{ \nu }{\nu'} \] \[ \nu' = \frac{\nu}{z + 1} \]

Part B:  If a photon at redshift z had the energy density uνdν, what is its energy density uν'dν' today? (Hint, consider two effects: 1) the number density of photons is diluted; 2) each photon is redshifted so its energy, E = hν, is also redshifted.)

We know energy density is a function of energy and volume \(u_ν'dν' = \frac{E}{V} \) so we just need to figure out how energy and volume have changed from the past to today.

Once again, we know \[ E = h \nu \] so \[ E' = \frac{h \nu}{z + 1} \] based on the conversion found above. We also know distances scale to volume as \( R^3 ~ V \) so we have the relationship: \[V_{today} \propto a_0^3 ( 1 + z )^3 V_{tomorrow} \] This means \[ n = n_0 a_0^{-3} ( 1 + z )^{-3} \] Because \[u_νdν = E n = h \nu n \] \[u_ν'dν' = \frac{h \nu}{z + 1}n_0 a_0^{-3} ( 1 + z )^{-3} \] So the energy density today would be \[u_ν'dν' = a_0^{-3} ( 1 + z )^{-4} u_νdν \]

Part C: Plug in the relation between ν and ν' into the Planck spectrum: \[u_ν dν  = \frac{ 8 \pi h \nu^3 }{c^3} \frac{1}{e^{\frac{h \nu}{kT}} - 1 } d \nu \] and also multiply it with the overall energy density dilution factor that you have just figured out above to get the energy density today. Write the final expression as the form uν'dν' . What is uν'? This is the spectrum we observe today. Show that it is exactly the same Planck spectrum, except that the temperature is now T' =  T(1 + z)^-1 .

Plugging in our scaling relation and multiplying by our dilution factor give the rather complicated expression: \[u_ν' dν' = \frac{ 8 \pi h ( \nu'(z+1))^3}{c^3} \frac{1}{e^{\frac{h( \nu'(z+1))}{kT}} - 1 } d \nu' ( 1 + z) \cdot a_0^{-3} ( 1 + z )^{-4} \] This simplifies to \[u_ν' dν' = \frac{ 8 \pi h  \nu'^3}{c^3} \frac{1}{e^{\frac{h( \nu'(z+1))}{kT}} - 1 } d \nu' \] This is the same result that we would have obtainted if we replaced T with \(\frac{T}{1 + z} \).

Part D: As you have just derived, according to 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 this temperature T' today? (This was indeed observed first in 1964 by American radio astronomers Arno Penzias and Robert Wilson, who were awarded the 1978 Nobel Prize.)

This problem is now just a matter of plugging in things we know: \[ z = 1100 \] \[ T = 3000 \: K \] \[ T' = \frac{T}{1 + z} \] So our final temperature of the CMB today is \[ T' = \frac{3000}{1101} = 2.725 \: K \]  This is pretty close to what Penzias and Wilson found.