Blog Post 35, Worksheet 11.1, Problem 3

3. Baryon-to-photon ratio of our universe. 

(a) Despite the fact that the CMB has a very low temperature (that you have calculated above), the number of photons is enormous. Let us estimate what that number is. Each photon has energy \(h\nu\). From equation (1), figure out the number density, \(n_\nu\), of the photon per frequency interval \(d\nu\). Integrate over \(d\nu\) to get an expression for total number density of photon given temperature T. Now you need to keep all factors, and use the fact that \[\int_0^\infty \frac{x^2}{e^x-1}dx \approx 2.4\]

In problem 1, we are given the following equation for the energy density per frequency interval \(d\nu\) of the black body radiation: \[u_{\nu}d\nu = \frac{8πh_p\nu^3}{c^3}\frac{1}{e^{\frac{h_p\nu}{e_BT}}-1}d\nu\] To get the number density, \(n_\nu\), from the above equation, we need to divide by the energy of a photon, \(h\nu\):
\[n_\nu = \frac{u_\nu d\nu}{h_p \nu} = \frac{1}{h_p \nu} \frac{8πh_p\nu^3}{c^3}\frac{1}{e^{\frac{h_p\nu}{e_BT}}-1}d\nu\] This gives us the following formula for \(n_\nu\): \[n_\nu = \frac{8π}{c^3}\frac{\nu^2}{e^{\frac{h_p\nu}{e_BT}}-1} d\nu\] Now we can use u substitution to simplify the integral, with \(u = \frac{h\nu}{kT}\) and \(du = \frac{h\:d\nu}{kT}\). We substitute for \(\nu^3\) and \(d\nu\) to get: \[n_\nu = \frac{8πk^3T^3}{c^3h^3} \frac{u^2}{e^u-1}du\] Now we integrate over du: \[n_\nu = \frac{8πk^3T^3}{c^3h^3} \int_0^\infty \frac{u^2}{e^u-1}du\] Using the given value for the integral, we get that \[n_\nu = \frac{19.2 πk^3T^3}{c^3h^3}\]

(b) Use the following values for the constants: \(k_B = 1.38 • 10^{-16} \text{erg K}^{-1}\), \(c = 3.00 • 10^{10} \text{ cm s}^{-1}\), \(h = 6.62 • 10^{27} \text{ erg s}\), and use the temperature of CMB today that you have computed from 2(d) to calculate the number density of photon today in our universe today (i.e. how many photons per cubic centimeter). 

This is just a matter of plugging in the given constants and our value for the temperature, 2.7 K: \[n_\nu = \frac{19.2π(1.38 • 10^{-16})^3(2.7)^3}{(3 • 10^{10})^3(6.62 • 10^{-27})^3} \approx 400 \text{ photons cm}^{-3}\]

(c) Let us calculate the average baryon number density today. In general, baryons refer to protons or neutrons. The present-day density (matter + radiation + dark energy) of our Universe is \(9.2 • 10^{-30} \text{g cm}^{-3}\). The baryon density is about 4% of it. The masses of proton and neutron are very similar \((\approx 1.7 • 10^{-25} \text{ g}\)). 
What is the number density of baryons?

The baryon density is about 4% of the present-day density, or\[ 0.4(9.2 • 10^{-30} \text{g cm}^{-3}) = 3.68 • 10^{-31} \text{g cm}^{-3}\] To find the number density of baryons, we divide this baryon density by the masses of each particle (proton or neutron): \[\frac{3.68 • 10^{-31} \text{g cm}^{-3}}{1.7 • 10^{-25} \text{ g}} = 2.16 • 10^{-7} \frac{\text{baryons}}{\text{cm}^3}\]

(d) Divide the above two numbers, you get the baryon-to-photon ratio. As you can see, our universe contains much more photons than baryons (proton and neutron):

Our ratio is obtained by dividing our answer from part (b) by our answer from part (c), which gives a ratio of approximately \(2.9 • 10^9 \frac{\text{photons}}{\text{baryon}}\).