In the Big Bang model, the universe started with a hot radiation-dominated soup in thermal equi- librium. 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.
(a) If the photons was emitted at redshift z with frequency ν, what is its frequency ν' today?
(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.)
(c) Plug in the relation between ν and ν' into the Planck spectrum:
\[u_{\nu}d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P\nu}{k_B T}}-1}d\nu\]
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 uν1dν1. What is uν 1 ? This is the spectrum we observe today. Show that it is exactly the same Planck spectrum, except that the temperature is now \(T'=T(1+z)^{-1}\).
(d) As you have just derived, according to Big Bang model, we should observe a black body radia- tion 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.)
(a) If the photons was emitted at redshift z with frequency ν, what is its frequency ν' today?
Let's use the Doppler equation.
\[z=\frac{\lambda'-\lambda}{\lambda}\]
Converting wavelength to frequency, we find that:
\[z=\frac{\nu-\nu'}{\nu'}\]
Solving for \(\nu'\), we get:
\[\boxed{\nu'=\frac{\nu}{z+1}}\]
(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.)
For this problem we will use our diagram above. Firstly, energy density decreases by a factor of 3 because the universe is expanding in 3 spatial dimensions. Then, the energy of each photon decease by an additional factor because the expansion increases the photon's wavelength. This sums to a factor of 4. Mathematically, it looks like this, where \(a\) is the expansion factor, and \(a'\) is the expansion factor today.
\[\frac{u_{\nu'}d\nu'}{u_{\nu}d\nu}=\left(\frac{a}{a'}\right)^4=(z+1)^{-4}\]
(c) Plug in the relation between ν and ν' into the Planck spectrum:
\[u_{\nu}d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P\nu}{k_B T}}-1}d\nu\]
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 uν1dν1. What is uν 1 ? This is the spectrum we observe today. Show that it is exactly the same Planck spectrum, except that the temperature is now \(T'=T(1+z)^{-1}\).
\[u_{\nu}d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P\nu}{k_B T}}-1}d\nu\]
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 uν1dν1. What is uν 1 ? This is the spectrum we observe today. Show that it is exactly the same Planck spectrum, except that the temperature is now \(T'=T(1+z)^{-1}\).
It's time for more algebra.
Firstly we recall that \(\nu=\nu'(z+1)\). Taking the derivative, we find that \(d\nu=d\nu'(z+1)\).
Pluggin in we get:
\[u_{\nu'}d\nu' =\frac{1}{(z+1)^4} \frac{8\pi h_P (\nu'(z+1))^3}{c^3} \frac{1}{e^{\frac{h_P(\nu'(z+1))}{k_B T}}-1}d\nu'(z+1)\]
Simplifying, we get:
\[u_{\nu'} =\frac{8\pi h_P \nu'}{c^3} \frac{1}{e^{\frac{h_P(\nu'(z+1))}{k_B T}}-1}\]
This is identical to the original expression except for temperature, which is now scaled by a factor of 1/(z+1).
Thus:
\[\boxed{T'=T(z+1)^{-1}}\]
(d) As you have just derived, according to Big Bang model, we should observe a black body radia- tion 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.)
Let's plug and chug.
\[T'=T(z+1)^{-1}\]
\[T'=3000K(1100+1)^{-1}\]
\[\boxed{T'=2.72K}\]
I worked with B. Brzycki, G. Grell, and N. James on this problem.
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