In the context of describing the departures from equilibrium in the early universe (i.e. Big Bang Nucleosynthesis, recombination), one uses Boltzmann equations. Let's suppose we have a binary interaction of the form $1+ 2 \leftrightarrow 3+ 4$, where $1$, $2$, $3$ and $4$ represent different particles. Under some assumptions (particles involved are non-relativistic, they are in kinetic equilibrium etc) one derives the following differential equation for the evolution of the number density $n_1$ $$ a^{-3} \frac{d(n_1 a^3)}{dt} = n_1^{(0)} n_2^{(0)} \langle \sigma v \rangle \left( \frac{n_3 n_4}{n_3^{(0)}n_4^{(0)}} - \frac{n_1 n_2}{n_1^{(0)}n_2^{(0)}}, \right) $$ where $a(t)$ is the scale factor and $\langle \sigma v \rangle$ is the thermally-averaged interaction cross-section. The quantities $n_i^{(0)}$ are given by $n_i^{(0)} = \frac{g_i}{(2\pi)^3} \int d^3 p \ e^{-E_i/T}$, and are related to the number densities as follows: $n_i = e^{\mu_i /T} n_i^{(0)}$, where $\mu_i$ denotes the chemical potential of species $i$. When interactions are very strong (i.e. interaction rate $\Gamma (t) = n_2(t)\langle \sigma v \rangle$ is big compared to Hubble rate $H(t)$), then the term in parenthesis in the previous equation vanishes, meaning that particles are in chemical equilibrium. This condition can be written in two equivalent ways: $$ \frac{n_1 n_2}{n_3 n_4} = \frac{n_1^{(0)} n_2^{(0)}}{n_3^{(0)}n_4^{(0)}}, \ \ \ \ \rm{or} \ \ \ \ \mu_1 + \mu_2 = \mu_3 + \mu_4. $$ It is then typically said that forward and backward rates become equal. And here is where it comes the problem, since I don't see why this has to be always the case. Let's take the example of Deuterium production $n + p \leftrightarrow D + \gamma$. Since photons have $n_\gamma^{(0)} = n_\gamma$, one can write the chemical equilibrium condition in the following way $$ \frac{n_D}{n_n n_p} = \frac{3}{4} \left( \frac{4\pi}{m_pT} \right)^{3/2} e^{B_D/T}, $$ where $B_D = m_n + m_p -m_D$ is the binding energy of deuterium. We see from previous equation that the density of Deuterium will start to increase with respect the one of protons and neutrons when the temperature decreases below the binding energy, $T < B_D$. That is, the reaction of Deuterium production, $n + p \rightarrow D + \gamma$, is more favorable than the inverse process, $D + \gamma \rightarrow n+ p$.

How can this be compatible with the idea that in chemical equilibrium forward and backward rates should become equal?
Many thanks in advance, this is driving me crazy


2 Answers 2


The Universe is never exactly in equilibrium, but it is approximately in equilibrium on timescales short compared to the Hubble time and large compared to the interaction rates.

On timescales short compared to the Hubble time, the temperature is approximately constant. Your equation relating $n_D$, $n_N$, and $n_P$ therefore says that these three species will approximately be in equilibirum (since their ratio will be approximately constant in time, since the temperature is approximately constant in time). The reaction rates balance. (I'm assuming you have in mind BBN, so we can ignore neutron decay).

On longer timescales, the temperature does change because the Universe expands. Of course, in equilibrium the temperature wouldn't change, but the Universe isn't exactly in equilibrium. Since the temperature is changing slowly, an intuitive picture to have in mind is that the Universe passes through a series of states that are each approximately in equilibirum at temperature $T$. Within each "slice" (for times small enough that the temperature doesn't change), reaction rates are balanced. However, the reaction rates of two different slices will in general be different, because the temperatures are different. Additionally, the relative amounts of different species can also change "between" slices (please understand I'm trying to use language that's intuitive, but in more detail what is happening is that there is a small but continuous change in the relative amounts of different species because of the Universe's expansion making the Universe not an equilibrium state; these small changes build up to become large changes over long times, tracking the approximate thermal equilibrium which is constantly adjusting as the temperature changes).

  • $\begingroup$ For all practical purposes, I'd certainly agree that the local universe is never in equilibrium, but, since Penrose has proposed a cosmological model (that he described throughout his 2020 Nobel Prize speech, at youtube.com/watch?v=DpPFn0qzYT0 , which requires that the thermodynamic equilibrium of one iteration of a non-multiversal universe should become the Big Bang of the next, I have, in view of the OP's "cosmology" tag, added a simpler answer that's intended to complement your own. $\endgroup$
    – Edouard
    Commented Mar 7, 2022 at 21:25
  • $\begingroup$ @Edouard We have to disagree about whether your answer is "simpler" :-) My answer is based on the thermal history of the Universe, which is very well understood theoretically and observationally, and is largely independent of the early history of the Universe (whether that be inflation or some alternative), about which our knowledge is much more speculative. Any model of the early Universe, including Penrose's, must "look like" the thermal history I am describing in my answer, from at least BBN to the present day. $\endgroup$
    – Andrew
    Commented Mar 7, 2022 at 21:35
  • 2
    $\begingroup$ @Andrew many thanks for your detailed answer! it was truly enlightening. I was forgetting the fact that the chemical equilibrium condition is actually valid at a given time/temperature. $\endgroup$ Commented Mar 7, 2022 at 22:32
  • $\begingroup$ @Andrew -staff or a robot are preventing my upvote of your comment (-we both know that comments don't affect "reputation", so that doesn't matter). Penrose feels that the approach to equilibrium, rather than the "next Big Bang" aftermath of it, is what would most resemble inflation in his model. Aside from that (another "directional" issue), and the fact that Penrose's claim that no natural or artificial clock would work between aeons leaves some unconventionally blank spaces in physical history, I can't see anything wrong with your answer. $\endgroup$
    – Edouard
    Commented Mar 8, 2022 at 4:23
  • $\begingroup$ Uh, except maybe a challenge to an aspect of Big Bang nucleosynthesis that was posed by Planck/WMAP satellite data, as described in the Wiki on BBN. $\endgroup$
    – Edouard
    Commented Mar 8, 2022 at 4:41

The simplest explanation might be 2002's "Steady-state eternal inflation", formulated by Aguirre & Gratton (in a preprint found by its name on Arxiv), which has dual universes where time passes in opposite directions on opposite sides of a Cauchy surface: It was accepted as compatible with the Borde-Guth-Vilenkin Theorem (which is often misconstrued as prohibiting any cosmological models that are both past- and future-eternal) in the last footnote to the last (2003's) revision of that theorem by its own authors. (Aguirre has, much more recently, been working with Deutsch on a similar model, "State-to-state cosmology", whose preprint is at https://arxiv.org/abs/2106.15692 .)

  • $\begingroup$ The "measurements and status of theory" section of Wikipedia's article "Big Bang nucleosynthesis" is the part that mentions some difficulties with the center of the mainstream theory. (AG's cosmological model, with its tacit endorsement by BGV, is certainly mainstream.) $\endgroup$
    – Edouard
    Commented Apr 1, 2022 at 13:12

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