Energy that was there 13.68 billion years ago

We are bounded in a nutshell of Infinite space: Blog Post #34, Worksheet # 11.1, Problem #2: Energy that was there 13.68 billion years ago

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 (the particle content of the electromagnetic radiation is called photon) satisfies The Plank Spectrum. 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 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 \(\nu\), what is its frequency  \(\nu^{\prime}\) today?

(b) If a photon at redshift z had the energy density \(u_\nu d\nu\), what is its energy density \(u_{\nu^\prime}d\nu^\prime\) today? (Hint, consider two effects: 1) the number density of photons is diluted; 2) each photon is redshifted so its energy, \(E = hv\), is also redshifted.)

(c) Plug in the relation between \(\nu\) and \(\nu^\prime\) 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 \(u_{\nu^\prime}d\nu^\prime\). What is \(u_{\nu^\prime}\)? This is the spectrum we observe today. Show that it is exactly the same Planck spectrum, except that the temperature is now \(T^\prime = T(1+z)^{-1}\)  .

(d) As you have just derived, according to Big Bang model, we should observe a black body radiation with temperature \(T^\prime \) filled in the entire universe. This is the CMB. Using the information given at the beginning of this problem, what is this temperature \(T^\prime\) 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) Starting off with the normal redshift equation, it can be turned into a redshift equation that describes the same value with photon frequencies: \[z = \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} \] \[\lambda = \frac{c}{\nu}\] \[z = \frac{\frac{c}{\nu_{observed}} - \frac{c}{\nu_{emitted}}}{\frac{c}{\nu_{emitted}}} ,\] and simplyfing the equation a bit to find a sole value for the \(\nu^\prime\): \[z = \frac{\nu_{emitted}} {\nu_{observed}} - 1 \] \[z + 1= \frac{\nu_{emitted}} {\nu^\prime} \]\[ \nu^\prime= \frac{\nu_{emitted}}{ z + 1},\] so we now see how the frequency seen today is always larger than the original frequency with which the photon was emitted.

(b)If we know a basic tenant of cosmology, the way the scale factor changes with redshift: \[a =\frac{1}{1+z},\] we can use it to describe the rest of the proportions which describe how the universe changes over periods of time. Knowing that \[ \lambda \propto a(t) \] from \[z = \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} \] \[z + 1= \frac{\lambda_{observed}} {\lambda_{emitted}} \] and focusing in on \(\lambda_{observed}\), we know how  \[ \lambda \propto a(t) \] and how\(\lambda = \frac{c}{\nu}\) so the relationship can be described as: \[ \frac{1}{\nu} \propto a(t) \] \[ \nu \propto \frac{1}{a(t)} \] which can be rewritten as:\[ \nu^\prime \propto \frac{1}{a^\prime (t)} .\] Another aspect that the problem describes is how volume is related to the scale factor, which is simply: \[V \sim a(t)^3,\] and so \[V^\prime \sim a(t)^{\prime~ 3}.\] With these, we can start doing some squiggle math to correctly describe the relationships of the variables: \[E\sim \nu\] \[E^\prime\sim \nu^prime,\] and the energy density of the universe is described as \[\rho_E = \frac{E}{V},\] so it can also be: \[\rho_E \sim \frac{\nu}{V}\] \[\rho_E \sim \frac{\frac{1}{a}}{\frac{1}{a^3}}\] \[\rho_E = \frac{1}{a^4},\] which is just scaled as anything else has for this problem: \[\rho_E^\prime = \frac{1}{a^{\prime 4}},\] and since the scale factor in the present is one: \[\rho_E^\prime = 1 ,\] and thus:  \[\frac{\rho_E}{\rho_E^\prime} \sim \frac{\frac{1}{a^4}}{1}\] \[\frac{\rho_E}{\rho_E^\prime} \sim \frac{1}{a^4},\] and applying the definition of the scale factor from redshift: \[\frac{\rho_E}{\rho_E^\prime} \sim (1+z)^4,\] so the energy density is what the Plank Spectrum calculates: \[\frac{u_\nu d\nu}{u_{\nu^\prime} d\nu^\prime} \sim (1+z)^4\] \[u_\nu d\nu \sim u_{\nu^\prime} d\nu^\prime (1+z)^4 \] \[ u_{\nu^\prime} d\nu^\prime \sim \frac{u_\nu d\nu}{(1+z)^4},\] which is how energy density scales with redshift (i.e. time).

(c) Knowing the Plank Spectrum equation: \[u_\nu d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,\] and knowing how it scales with time (redshift): \[z + 1= \frac{\nu_{emitted}} {\nu^\prime} \] \[ \nu_{emitted} =\nu^\prime (z + 1)\]  \[ d\nu_{emitted} =d\nu^\prime (z + 1),\] we can solve for how it changes:\[u_\nu d\nu = \frac{8\pi h_P (\nu^\prime (z + 1))^3}{c^3} \frac{1}{e^{\frac{h_P (\nu^\prime (z + 1))}{k_B T}} - 1} (d\nu^\prime (z + 1))\] \[ u_{\nu^\prime} d\nu^\prime \sim \frac{u_\nu d\nu}{(1+z)^4}\] \[\frac{u_\nu d\nu}{(1+z)^4} = \frac{\frac{8\pi h_P (\nu^\prime (z + 1))^3}{c^3} \frac{1}{e^{\frac{h_P (\nu^\prime (z + 1))}{k_B T}} - 1} (d\nu^\prime (z + 1))}{ (1+z)^4},\] \[ u_{\nu^\prime} d\nu^\prime = \frac{8\pi h_P \nu^\prime3}{(z+1) c^3} \frac{1}{e^{\frac{h_P (\nu^\prime (z + 1))}{k_B T}} - 1} (d\nu^\prime (z + 1))\]  \[u_{\nu^\prime} d\nu^\prime = \frac{8\pi h_P \nu^\prime3}{c^3} \frac{1}{e^{\frac{h_P (\nu^\prime (z + 1)}{k_B T}} - 1} (d\nu^\prime ),\] and after all that algebra, we can solve for the final (z +1) by incorporating the change in temperature equation: \[T^\prime = T(1+z)^{-1},\] so the final equation becomes:  \[u_{\nu^\prime} d\nu^\prime = \frac{8\pi h_P \nu^\prime3}{c^3} \frac{1}{e^{\frac{h_P (\nu^\prime}{k_B T^\prime}} - 1} (d\nu^\prime)\]

(d) For the last part, we simply have to incorporate the definitions given in the beginning of the problem and the temperature change equation: \[T^\prime = T(1+z)^{-1}\] \[T = 3000 K\] \[ z= 1100\] \[ T^\prime = 3000K \cdot (1+1100)^{-1}\] \[T^\prime = 2.7 K,\] which is the temperature of the Cosmic Microwave background in the present.