Basically, matter is the rarest thing in the universe (or at least it used to be)

We are bounded in a nutshell of Infinite space: Blog Post #35, Worksheet # 11.1, Problem #3: Basically, matter is the rarest thing in the universe (or at least it used to be)

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 the Plank Spectrum Equation, 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 \[\int_0^\infty \frac{x^2}{e^x - 1} dx \approx 2.4.\]

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

(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 \(9.2 \times 10^{-30} g~cm^{-3}\) . The baryon density is about 4% of it. The masses of proton and neutron are very similar \((\approx 1.7 \times  10^{-24} G )\).

What is the number density of baryons?

(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).

(a) Knowing the energy of a single photon, a simple scale value defines all photon’s energy: \[E = h\nu \] \[E_n = n\cdot h\nu \] \[n_\nu = \frac{E_{n_\nu}}{h\nu},\] which can be defined with the equations for energy density as: \[ n_\nu = \frac{u_\nu}{h\nu},\] so \[u_\nu d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,\]becomes: \[\frac{u_\nu}{h\nu} d\nu = \frac{1}{h\nu} \cdot \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,\] \[n_\nu d\nu =  \frac{8\pi \nu^2}{c^3} \frac{1}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,\] and integrating it for all frequencies: \[\int_0 ^\infty n_\nu d\nu = \int_0 ^\infty \frac{8\pi \nu^2}{c^3} \frac{1}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,\] \[n_{all \nu} =  \frac{8\pi}{c^3} \left(\frac{k_B T}{h_P}\right)^3 \int_0^\infty \frac{\nu^2}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,\] and knowing a particular equality established by the problem: \[\int_0^\infty \frac{x^2}{e^x - 1} dx \approx 2.4 ,\] we have the final description of the number of photons as: \[n_{all ~\nu} =  \frac{8\pi}{c^3} \left(\frac{k_B T}{h_P}\right)^3 (2.4)\]

(b) And now plugging in the values given in the problem, we have: \[n_{all~ \nu} =  \frac{8\pi}{c^3} \left(\frac{k_B T}{h_P}\right)^3 (2.4)\] \[k_B = 1.38 \times 10^{-16} erg~ K^{-1}, \] \[c = 3.00 \times 10^{10} cm ~s^{-1},\] \[h = 6.62 \times 10^{-27} erg~s, \] so:  

\[n_{all~ \nu} =  \frac{8\pi}{3.00 \times (10^{10} cm ~s^{-1})^3} \left(\frac{1.38 \times 10^{-16} erg~ K^{-1} \cdot 2.7K}{ 6.62 \times 10^{-27} erg~s }\right)^3 (2.4),\] and thus the density of photons is: \[n_{all ~\nu} = 4 \times 10^2 ~photons/cm^3\]

(c)  Being given the density of mass in the universe, we can now solve for baryon density in particular: \[\rho_{universe} = 9.2 \times 10^{-30} g~cm^{-3}\] \[\rho_{baryons} = 0.04 \times \rho_{universe} ,\] and knowing the mass of a baryon, we can solve for the number density of baryons: \[\rho_{baryons} = 3.68*10^{-31} g~cm^{-3}\] \[n_{baryons} = \rho_{baryons} / m_{baryon}\] \[ n_{baryons} =3.68*10^{-31} g~cm^{-3} / 1.7 * 10^{-24} \frac{g}{baryon}\] \[ n_{baryon} = 2.16*10^{-7} \frac{baryons}{cm^3}\]

(d) Simply doing what the problem establishes, we have: \[n_{all \nu} = 4 \times 10^2 ~photons/cm^3\] \[ n_{baryon} = 2.16471*10^{-7} \frac{baryons}{cm^3}\] \[n_{all ~\nu}/ n_{baryon} = 1.85*10^9 \frac{photons}{baryons},\] and indeed, there are billions of photons for every baryon.