## Wednesday, February 12, 2014

### Eddington's number

can you estimate Eddington's number, the total number of protons in the observable Universe?

1. first variant: you can estimate the total baryonic mass $M_b$ by multiplying the baryonic density in terms of the critical density $\rho_\mathrm{crit}=3H_0^2/(8\pi G)$ with the spherical Hubble volume, $$M_b = \frac{4\pi}{3}\chi_H^3\rho_\mathrm{crit}\Omega_b$$ with the Hubble distance $\chi_H=c/H_0$, yielding $$M_b=\Omega_b\frac{c/H_0}{G/c^2}$$ resulting in a number $n_p=M_b/m_p\simeq 10^{80}$ by dividing with the proton mass $m_p$. This requires a measurement of $H_0$ and $\Omega_b$.

1. effectively, $n_p$ is given by the ratio between the Hubble distance and the Schwarzschild radius of a proton.

2. second variant: you can predict the number density of photons from the CMB-temperature by using the Planck-formula, multiply with the Hubble volume and use the baryon-to-photon-ratio known from BBN-models for scaling down the number of photons in the Hubble sphere to the number of protons.

1. specifically, the number density of photons follows from integration of the Planck-formula, and is $\zeta(3)/\pi^2 (k_BT/\hbar/c)^3$, multiplied with the Hubble volume $4\pi/3(c/H_0)^3$ gives $$n = \frac{64}{3\pi}\zeta(3)\left(\frac{k_BT}{\hbar H_0}\right)^3$$ which has a numerical value of $\simeq10^{86}$, which the photon to baryon ratio of $\eta=10^{-9}$ brings down to a value close to the previous one. This calculation requires a measurement of $T$, $H_0$ and $\eta$ (from BBN).