Blog #35, Worksheet #11.1, Problem #3, Baryons and Photons

3. a) Despite the fact that the CMB has a very low temperature (that you have calculated above), the number of photon is enormous. Let us estimate what that number is. Each photon has energy hν. From equation (1), figure out the number density, nν, of the photon per frequency interval dν. Integrate over dν 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

We start with the given equation 1) of \[\frac{8 \pi}{c^3} \cdot \frac{v^2}{e^{\frac{hv}{kT}} - 1} dv = n_v dv\]To integrate dv, we will use a u substitution and call \[u = \frac{hv}{kT}\] Which gives us \[du = \frac{h}{kT}dv \qquad and \qquad v^2 = \frac{k^2 T^2 u^2}{h^2}\] Plugging these relationships into our original equation 1) we are left with \[\frac{8 \pi k^3 T^3}{c^3 h^3} \cdot \frac{u^2}{e^u - 1} du = n_v dv \]We then integrate the left side from 0 to \(\infty\) and using the relationship that \[\int^{\infty}_0 \frac{x^2}{e^x - 1} dx \thickapprox 2.4 \]We end up with \[n_v dv = \frac{19.2 \pi k^3 T^3}{c^3 h^3}\]

b) (b) Use the following values for the constants: kB = 1.38x10^-16 erg K^-1 , c = 3.00x10^10cm s^-1 , h = 6.62x10^27 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?)

For this part, we simply use our relationship from part a to solve for the number density of photons in our universe today. We solve \[\frac{19.2 \pi (1.38x10^{-16})^{3} (2.7)^{3}}{(3x10^10)^3 (6.62x10^{-27})^3} \thickapprox 400 \frac{photon}{cm^3}\]

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 9.2x10^30g cm^3 . The baryon density is about 4% of it. The masses of proton and neutron are very similar (1.7x 10^24g). What is the number density of baryons?

We are told that the total density of the universe is equal to \[\rho_{total} = 9.2x10^{-30} \frac{g}{cm^3}\]and we are told that baryons make up 4% of the total density meaning \[\rho_{baryon} = 3.68x10^{-31} \frac{g}{cm^3} \]We also know that the average baryon weighs \(1.7x10^{-24} g \) so the number density of baryons is \[n_{baryon} = \frac{\rho_{baryon}}{1.7x10^{-24} g } = 2.16x10^{-7} \frac{baryon}{cm^3}\]

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).

Finally, we compare the amount of baryons to photons and get \[\frac{400 \frac{photons}{cm^3}}{2.16x10^{-7} \frac{baryons}{cm^3}} \thickapprox 2x10^9 \frac {photon}{baryon}\] Showing that there are far far more photons in our universe than there are baryons, or there is a lot more light than matter!