Baryon-to-photon ratio of our universe: Blog 35, Worksheet 11.1, Problem 3

Part 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 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 $$\int_0^{\infty} \frac{x^2}{e^x - 1} \approx 4.2$$ We already know $$E = h \nu$$ and $$E n d \nu = \frac{ 8 \pi h \nu }{c^3} \frac{1}{e^{\frac{h \nu}{kT}} - 1 } d \nu$$ The integral of this is equivalent to $$\int E n d \nu = \int \frac{ 8 \pi h \nu }{c^3} \frac{1}{e^{\frac{h \nu}{kT}} - 1 } d \nu = E n$$ Adding this with our definition of energy and rearranging the integral gives us: $$n = \frac{ 8 \pi K^3 T^3 }{c^3 h^3} \int_0^{\infty} \frac{x^2}{e^x - 1}$$ Where $$x = \frac{ h\nu}{KT}$$ and $$d \nu = \frac{ KT dx}{h}$$ With the approximation given above we get a cleaner answer for n $$n_{\nu} = \frac{ 8 \pi K^3 T^3 }{c^3 h^3}(2.4)$$

Part B: Use the following values for the constants: kB =1.38 x 10^-16 erg/K, c = 3.00 x 10^10 cm/s, h =  6.62 x 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?)

In problem 2b, we found the temperature of CMB today to be about 2.725 Kelvin. Plugging this in with the constants specified above should give us: $$n_{\nu} = \frac{ 8 \pi K^3 T^3 }{c^3 h^3}(2.4) = \frac{ 8 \pi (1.38 \times 10^{-16})^3 2.725^3 }{( 3 \times 10^{10})^3 (6.62 \times 10^{-27})^3}(2.4) \approx 170.62 \frac{photons}{cm^3}$$

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

This means that the density of our universe that is baryons is $$9.2 \times 10^{-30} \times 0.04 = 3.68 \times 10^{-31}$$ Dividing this number by the mass of a typical baryon gives us the density: $$n_b = \frac{\rho}{m} = \frac{ 3.68 \times 10^{-31}}{1.7 \times 10^{-24} } \approx 2.16 \times 10^{-7}$$

Part 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).
$$\frac{photons}{baryons} = \frac{2.16 \times 10^{-7} }{170.62} = 1.26 \times 10^{-9}$$