# How were atoms able to form after the big bang?

How where atoms able to form after the big bang? From my understanding all the materials to form atoms were floating around and by random chance they collided and created atoms. If the universe was quickly expanding after the big bang, would the materials to form atoms be spread to thin. I am assuming that the materials would be spread very thin, since the universe was in existence for a long time before the first atoms were created.

• you will find answers to all your questions in the excellent book by Stephen Weinberg, "The First Three Minutes". – niels nielsen Mar 8 '18 at 18:47
• Another, technical review is arxiv.org/pdf/0809.0631.pdf – Anders Sandberg Mar 8 '18 at 18:54

There is an important difference between the first atomic nuclei and the first atoms forming. Nucleosynthesis began when the temperature cooled enough that nuclei were not immediately disrupted by free-flying photons and nucleons, and ended when the temperature and density became too low to allow fusion processes (a few minutes after big bang).

At this point the temperature was still too high for atoms to form: any electron that ended up bound to a nucleus would nearly immediately be bumped away by a high-energy photon. The result was that the universe remained a plasma of free nuclei and electrons. It took until the era of recombination about 378,000 later for the temperature to become less than 3,000 Kelvin, when atoms could properly form. Since nuclei and electrons are oppositely charged they could easily find each other even at the recombination era densities.

If we assume everything is just hydrogen, and use $x$ to denote the free electron fraction (also equal to the fraction unbound protons) the Saha equation for the relative abundance equlibrium reads $$\frac{x^2}{1-x}=\frac{1}{n_H + n_p}\left ( \frac{m_e k_B T}{2\pi \hbar^2}\right) ^{3/2} e^{-\frac{E_I}{k_B T}}$$ where $m_e$ is the electron mass and $E_I=13.6$ eV the ionization energy of hydrogen. The temperature behaves as $T=2.728(1+z)$ K as a function of redshift and the density also conveniently behaves as $n_p+n_H = 1.6(1+z)^3$ per cubic meter. If you solve it for $x=1/2$ you get the time for recombination, $z\approx 1500$.

For the purpose of the question the interesting thing is that it doesn't matter how fast the redshift $z$ changes. In a universe that expanded slowly (a long period going from $z=2000$ to $z=1000$) the recombination would have proceeded just like a rapidly expanding universe - as long as the expansion was not so fast that there was no time to reach equilibrium.

How fast will we reach equilibrium? The mean free path of a particle will be $l\approx 1/\sqrt{2} n \sigma$ where $\sigma$ is the cross section. The mean speed in the Maxwell-Boltzmann distribution is $v=\sqrt{8k_BT/\pi m}$ so the mean time for a collision will be $\tau=l/v=\sqrt{\pi m/16kTn^2\sigma^2}$ The time Equilibrium will happen after every particle have had the time to collide a few times. So if we plug in $z=1500$ we get a density $n_p+n_H = 5\times 10^9$ per cubic meter and $T=4094$ K. $m\approx m_p = 1.67262\times 10^{-27}$ kg and $\sigma\approx 10^{-20}$ square meter. This gives $\tau \approx 1.524\times 10^6$ seconds (assuming my numbers and math are correct). About 17 days - way faster than the slow transition from plasma to atomic gas.

During nucleosynthesis this equilibration did not happen completely, partially because of the deuterium bottleneck that delayed everything. Had the process had time to continue long enough most hydrogen would have been burned into helium. Fortunately the expansion was fast enough back then to only produce some primordial helium.