PASA 2024

# Under an Iron Sky:
On the Entropy at the Start of the Universe

Luke A. Barnes1 and Geraint F. Lewis2<EMAIL_ADDRESS>1School of
Science, Western Sydney University, Locked Bag 1797, Penrith South, NSW 2751,
Australia  2Sydney Institute for Astronomy, School of Physics A28, The
University of Sydney, NSW, 2006, Australia

###### Abstract

Curiously, our Universe was born in a low entropy state, with abundant free
energy to power stars and life. The form that this free energy takes is
usually thought to be gravitational: the Universe is almost perfectly smooth,
and so can produce sources of energy as matter collapses under gravity. It has
recently been argued that a more important source of low-entropy energy is
nuclear: the Universe expands too fast to remain in nuclear statistical
equilibrium (NSE), effectively shutting off nucleosynthesis in the first few
minutes, providing leftover hydrogen as fuel for stars. Here, we fill in the
astrophysical details of this scenario, and seek the conditions under which a
Universe will emerge from early nucleosynthesis as almost-purely iron. In so
doing, we identify a hitherto-overlooked character in the story of the origin
of the second law: matter-antimatter asymmetry.

###### doi:

10.1017/pas.2024.xxx

###### keywords:

Initial conditions of the universe – Big Bang nucleosynthesis – Baryogenesis

##  1 Introduction

The early Universe was relatively simple. A few minutes after the Big Bang,
the Universe was composed of electromagnetic radiation, dark matter,
electrons, and baryons, mainly in the form of hydrogen and helium (Mathews et
al., 2017). Dark energy was present, lurking in the background at a level that
would be irrelevant for several billion years (Velten et al., 2014). The
Universe’s various forms of energy were (almost) uniformly distributed through
space.

Slight inhomogeneities, perhaps seeded by inflation, were growing under the
attractive pull of gravity, with dark matter and gas pooling into the sites
that would become present day galaxies (White & Rees, 1978; Springel et al.,
2006). As the density of gas increased, so did its ability to cool, fragment
and collapse. Within this compressed gas, the first proto-stars formed as
nuclear reactions ignited inside their cores. These stars ended their lives in
supernovae explosions, polluting the interstellar medium with heavier
elements. This enriched medium formed into future generations of stars
(Aguirre, 2017), which soon reionized the Universe (Gunn & Peterson, 1965).

The formation and redistribution of heavier elements was essential for the
emergence of life in the Universe. Stars form the raw materials for planets
and creatures (Armitage, 2018). Stars also bathe planets with low entropy
radiation, the ultimate power source for life (Schrödinger, 1944). Stars in
our Universe emit photons whose energy is roughly the same as the typical
energy of chemical bonds, making processes such as photo-synthesis possible.
This fact traces back to a remarkable coincidence between fundamental
constants: $m_{p}^{3}\approx\alpha^{6}~{}m_{e}^{2}~{}m_{Pl}$, where $m_{p}$,
$m_{e}$ and $m_{Pl}$ are the proton, electron and Planck masses respectively,
and $\alpha$ is the fine-structure constant (Carter, 1974).

This stelliferous period of the life in the Universe, though it may last for
trillions of years, is fleeting compared to the indefinite amount of time
ahead, where energy will become sparse and the cosmos will tend to a state of
maximum entropy (Adams & Laughlin, 1999). As with all isolated physical
systems, the Universe seems headed for a state in which all its energy is
trapped in useless forms, such as inside black holes or evenly dispersed in
low-energy radiation.

Statistical mechanics explains this tendency — the second law of
thermodynamics – roughly as follows. A complete description of every detail of
a physical system is called a _microstate_. As well as the actual microstate
of a system at a particular time, we also consider the space of all possible
microstates of the system (_phase space_). The laws of nature describe which
future microstate a given microstate will evolve into after a certain amount
of time. However, for many systems of interest, this will involve an
infeasibly large amount of detail. More practically, we use a statistical
approach: armed with a method — known as _coarse graining_ — that calculates
macroscopic quantities (e.g. temperature) from microscopic ones (e.g. the
position and momentum of every particle) and a probability distribution over
microstates, we can seek to derive the assured results of classical
thermodynamics. We find that the space of microstates is typically dominated
by equilibrium states of maximum entropy, that is, maximum entropy states are
the most probable states. Thus, a system that has had sufficient time to
explore its possible states is very likely to be found in a maximum entropy
state. (Consult your local statistical mechanic for a more nuanced account).

This raises a puzzle: we have explained why low entropy states tend to evolve
toward high entropy states, but why are there _any_ low entropy states? If low
entropy states are improbable, why are any observed at all? This puzzle
becomes all the more acute as we rewind the second law back to the beginning
of the Universe: why wasn’t the Universe born in a maximum entropy state?
There are, after all, plenty of them! The need for a low-entropy boundary
condition in the past to explain the second law of thermodynamics is known as
the Past Hypothesis. We will not resolve these questions here; the interested
reader is directed to Penrose (1979); Price (1996); Albert (2000); Carroll &
Chen (2004); Wald (2005); Earman (2006); Frigg (2008); Wallace (2011);
Winsberg (2012); Goldstein et al. (2016) and beyond.

Those who subscribe to a low-entropy beginning have sought to identify the
form of free energy that is available in the early Universe, that is, the
aspect of the arrangement of the early Universe that supplies its low entropy.
This presents another puzzle: the early Universe was a homogeneous plasma with
uniform temperature, and to our thermodynamic intuition, trained on boxes of
classical ideal gas, this may appear to be a state of maximum entropy.

The standard solution to this puzzle was presented in Penrose (1979, 1989),
and popularised in his book The Emperor’s New Mind (Penrose, 1990). Penrose
identifies the crucial role of gravity: the highest-entropy arrangement of a
box of matter is that in which all the matter has collapsed into a black hole.
A high-entropy big bang would be a expanding Universe that is born full of
black holes. By contrast, in our Universe, while the initial homogeneity will
not allow for thermal energy to be extracted by heat flowing from hot to cold
regions, gravity will cause the matter can collapse into clumps, transforming
potential energy into kinetic energy. By comparing our observable Universe
with a black hole of the same mass, Penrose calculates the entropy of our
Universe relative to its maximum, which in turn implies that the fraction of
phase space that is at least as low entropy as our Universe is extremely
small, one part in $10^{10^{123}}$.

However, Wallace (2010) and Rovelli (2019) have argued that most important
source of low entropy is not gravitational but nuclear. Our Universe at early
times is in Nuclear Statistical Equilibrium (NSE). At high temperatures, NSE
favours small nuclear species: protons and neutrons. As the temperature and
density of the Universe decrease via expansion, this equilibrium is maintained
as long as the proton $\leftrightarrow$ neutron reaction rate is fast compared
to the expansion of the Universe. However, at about 1 second ($T\sim 0.8MeV$)
the reaction rate has slowed relative to the expansion such that the proton to
neutron ratio freezes out. After this, nucleosynthesis begins in earnest about
at $\sim 100$ seconds. By $\sim 1000$ seconds, the Universe has expanded and
cooled to the point that nucleosynthesis is over, leaving the baryonic
component in the form of hydrogen and helium, with a trace of heavier
elements.

Our thermodynamic, second law arrow of time is driven largely by low-entropy
(hotter than the CMB) radiation from the Sun. While the Sun was _ignited_ by
gravitational collapse over its first $\sim$ million years as a protostar, its
energy output for the last 4.5 billion years has been powered by the fusion of
hydrogen to helium. However, as noted by Rovelli (2019) and as we will show in
later section, if the Universe had maintained NSE through just one more decade
in temperature (down to $10^{9}~{}$K, instead of $10^{10}~{}$K as in our
Universe), with the nuclear reaction rates remaining fast compared to cosmic
expansion, all protons and neutrons would have been bound into heavy nuclei.
The Universe would be a homogeneous plasma of [56]Fe. “Stars” formed from such
strongly-bound nuclei would be unable to undergo fusion (unless their initial
collapse was so violent that [56]Fe was disintegrated into smaller elements).
Furthermore, with a periodic table consisting of a single element, the
chemical complexity required for life would be unavailable. Rovelli (2019)
notes that our Universe, by failing to maintain NSE, exits nucleosynthesis in
a metastable state, stranded in a low entropy state until gravitational
collapse greatly accelerates the nuclear reaction chain toward the NSE state
of a low temperature plasma: iron.

Rovelli (2019) characterises the situation by saying that,

> “the dominant source of low-entropy that feeds the observed irreversible
> behaviour of the universe is a single degree of freedom, the scale factor,
> which is (in a precise technical sense specified below) out of equilibrium.”

This description is correct, but it doesn’t answer the question: why is the
scale factor of our universe out of equilibrium? The normalisation of the
scale factor is arbitrary; only the evolution of its relative value has
physical meaning. The evolution of the scale factor depends on the fundamental
constants of cosmology. Which of these, in concert with the physics of nuclear
reactions, determines that our Universe begins out of nuclear equilibrium?

In this paper, we fill in some of the astrophysical details of the argument of
Wallace (2010) and Rovelli (2019), and in so doing identify the crucial role
played by an as-yet overlooked character: matter-antimatter asymmetry. The
scale factor of our universe is out of equilibrium (in the sense that Rovelli
explains) in part because the matter-antimatter asymmetry in our universe is
small. Universes with larger matter-antimatter asymmetry burn more of their
nucleons to heavy elements in their early stages.

The layout of this paper is as follows. In Section 2, we revisit
nucleosynthesis in our Universe, reviewing the impact of reaction rates and
expansion histories. In Section 3, we consider nuclear statistical equilibrium
and its perturbation by an expanding Universe. In Section 4, we explain how to
make an almost-iron Universe, and the implications for the fundamental
constants of cosmology. Finally, in Section 6 we discuss the implications for
the initial entropy of the Universe and the second law.

##  2 Big Bang Nucleosynthesis

In this section, we will review the relationship between cosmic expansion and
nuclear reaction rates in the early Universe. Numerical codes such as those of
Arbey et al. (2018) and Lewis et al. (2016) that follow primordial
nucleosythesis consider only the lightest few elements in the periodic table,
up to oxygen. Hence, they cannot follow the equilibrium state of the Universe
all the way to iron. Thus, we will focus in this section on the equilibrium
between protons and neutrons for a family of cosmic expansion histories,
expanding to consider NSE in more detail in later sections.

We follow the account of nucleosynthesis in Mukhanov (2004). The abundance of
free neutrons is,

$X_{n}=\frac{n_{n}}{n_{n}+n_{p}}~{},$ (1)

where $n_{n}$ $(n_{p})$ is the number density of free neutrons (protons). In
NSE, the ratio of free neutrons to free protons is given by,

$\frac{n_{n}}{n_{p}}=e^{-Q_{np}/T}$ (2)

where $T$ is the temperature and $Q_{np}=1.293$ MeV is the mass difference
between the neutron and proton. As shown by Mukhanov (2004), the relative
number density of neutrons is given by an asymptotic series of the form,

$X_{n}=X_{n}^{eq}\left(1-\frac{1}{\lambda_{np}}\left(1+e^{-\frac{Q_{np}}{T}}\right)^{-1}\frac{{\dot{X}}_{n}^{eq}}{X_{n}^{eq}}+...\right)~{}.$
(3)

Hereafter, we will ignore the small corrections represented by the ellipses.
In this equation, $\lambda_{np}$ is the neutron to proton interaction rate,
given by,

$\lambda_{np}=\frac{255}{\tau_{n}x^{5}}\left(x^{2}+6x+12\right)$ (4)

where $\tau_{n}\approx 886$ seconds is the neutron lifetime and
$x=\frac{Q_{np}}{T}$ (Bernstein et al., 1989). If the rate of reaction
$\lambda_{np}$ is high compared to the rate at which the expansion of the
Universe perturbs the abundance away from equilibrium
($\sim\dot{X}_{n}^{eq}/X_{n}^{eq}$), then the system remains close to its
equilibrium state, $X_{n}\approx X_{n}^{eq}$.

Figure 1:  Graphical representation of the inequality presented in Equation 7.
In this and all subsequent plots, temperature decreases to the right. The
left-hand side depends only on the temperature at one second, represented as
red horizontal lines in the above. The right-hand side of the inequality is
shown as coloured lines, for a set of Universes with expansion index $n$.
Nuclear statistical equilibrium for the proton to neutron ratio is maintained
when a coloured line is _above_ a given pink dashed line. Our Universe is
shown by the thick red dotted and thick purple solid lines.

In the following, we will parameterise the relationship between temperature
and cosmic time as,

$T=T_{o}t^{-n}$ (5)

where $T_{o}$ is the temperature at a cosmic age of 1 second. This equation is
accurate if the Universe’s energy content is dominated by a single fluid. The
standard cosmological model is radiation dominated through nucleosynthesis,
which implies $n=1/2$. Then, Equation 3 can be written as,

$X_{n}=X_{n}^{eq}\left(1+\frac{n}{\lambda_{np}\left(1+e^{-x}\right)^{2}}\frac{x_{o}^{1/n}}{x^{1/n-1}}\right)$
(6)

where $x_{o}=\frac{Q}{T_{o}}$. Statistical equilibrium will be maintained so
long as the second term in the parentheses is much smaller than one. We can
illustrate this condition by separating the inequality as follows,

$x_{o}\ll\left[\left(\frac{255}{\tau_{n}}\right)\frac{x^{1/n-6}}{n}(x^{2}+6x+12)\left(1+e^{-x}\right)^{2}\right]^{n}$
(7)

The left-hand side depends on only the temperature of the Universe at 1
second, while the right-hand side is a function of temperature that depends on
the expansion index $n$. In Figure 1, the horizontal red dashed lines
represent the left-hand side, and the solid coloured lines represent the
right-hand side of the inequality. Statistical equilibrium holds when a
coloured line is above a given red dashed line.

For the standard cosmological model, dominated by radiation through
nucleosynthesis,

$T=\frac{0.8595}{\sqrt{t}}~{}\textrm{MeV}$ (8)

where $t$ is measured in seconds. This expansion is presented as a thick
purple line in Figure 1. This line crosses the $T_{o}=1~{}$MeV pink dashed
line at around $T=1~{}$MeV, so, as expected, neutrons and protons go out of
equilibrium at a cosmic age of about $t\sim 1~{}$second.

To consider an alternative Universe, we have previously explored
nucleosynthesis in the $R_{h}=ct$ Universe (Lewis et al., 2016). This model
has linear expansion, and the relationship between the temperature and time is
given by,

$T=\frac{1.036\times 10^{8}}{t}~{}\textrm{MeV}$ (9)

where the temperature is constrained by the Cosmic Microwave Background
temperature today. In this Universe, that protons and neutrons go out of
equilibrium at about $T\sim 5\times 10^{-3}~{}$MeV, which implies a time scale
of hundreds of years. As shown in Lewis et al. (2016), at these comparatively
low temperatures, ($\sim 6\times 10^{7}~{}$K), the NSE abundances of heavier
elements are non-negligible.

This section has illustrated the conditions for departure from NSE in an
expanding Universe. The expansion of the Universe perturbs the abundances of
the nuclear species away from their equilibrium values at a particular
temperature and total density. Nuclear reactions drive the system back to
equilibrium. Each of these processes can be characterised by a timescale, and
the fastest process wins: either the rapid expansion of the Universe freezes
the nuclear abundances in place (except for radioactive decay), or nuclear
reactions maintain NSE.

In our Universe, once freeze-out occurs at $T\approx 0.84~{}$MeV $=9.7\times
10^{9}~{}$K (Mukhanov, 2004), the abundances of nuclear species can differ
significantly from NSE. Big Bang Nucleosynthesis is far from a quasi-static,
near-equilibrium process in our Universe. Figure 2 shows the abundances of the
light elements in our Universe as a function of temperature and time,
calculated by AlterBBN (Arbey et al., 2018). AlterBBN numerically integrates
production and destruction rates over time for a complete nuclear reaction
network that extends from protons and neutrons to [16]O. As we will see in
later sections, under conditions of NSE the abundance of deuterium (for
example) peaks at $10^{-11}$; in our Universe, deuterium reaches abundances
that are $10^{8}$ times higher than this, thanks to out-of-equilibrium nuclear
reactions.

Figure 2:  Abundances of the light elements in our Universe as a function of
temperature (bottom axis, decreases to the right) and time (top). As we will
see in later sections, under conditions of NSE the abundance of deuterium (for
example) peaks at $10^{-11}$; in our Universe, deuterium reaches abundances
that are $10^{8}$ times higher than this. These nucelosynthesis pathways were
integrated using AlterBBN (Arbey et al., 2018).

##  3 Nuclear Statistical Equilibrium and Expansion

###  3.1 Nuclear Abundances in NSE

This section will consider the effect of the expansion of the universe on
Nuclear Statistical Equilibrium. We begin by briefly reviewing the principles
of NSE, which have been expounded at length elsewhere (Clayton, 1968; Kolb &
Turner, 1990; Mukhanov, 2005; Iliadis, 2015).

For non-relativistic matter in _thermal_ equilibrium, integrating the Maxwell-
Boltzmann distribution over momentum gives the number density ($n_{i}$) of a
particle species $i$ with particle mass $m_{i}$, degeneracy $g_{i}$ (also
called the internal nuclear partition function) and chemical potential111As
explained in Lippuner & Roberts (2017), the chemical potential can be
redefined as $\hat{\mu}=\mu-m_{i}-B_{i}$. This is particularly useful for
numerical calculations of the NSE abundances. $\mu_{i}$ at a temperature $T$,

$n_{i}=g_{i}\left(\frac{m_{i}kT}{2\pi\hbar^{2}}\right)^{\frac{3}{2}}\exp\left(\frac{\mu_{i}-m_{i}c^{2}}{kT}\right)$
(10)

The nuclear species $i$ has $Z_{i}$ protons and $N_{i}$ neutrons, giving
atomic mass number $A_{i}=Z_{i}+N_{i}$. Hereafter, we will set the usual
physical constants to unity: $c=\hbar=k=1$.

If the reaction chain from neutrons and protons to species $i$ is in _secular_
equilibrium — that is, if the rates of _all_ forward and reverse reactions are
equal — then the chemical potentials are related as:

$\mu_{i}=Z_{i}\mu_{p}+N_{i}\mu_{n}~{}.$ (11)

We define $\theta=(m_{u}T/2\pi)^{3/2}$, where $m_{u}=1.6605$ g is the atomic
mass constant, and the binding energy of species $i$ is
$B_{i}=Z_{i}m_{p}+N_{i}m_{n}-m_{i}$. It is useful to express these number
densities in terms of mass abundances relative to the total number density
$n_{b}$ of protons and neutrons (inside and outside of nuclei). Then, given
that $g=2$ for the proton and neutron, we can combine the previous equations
to give the mass abundances $X_{i}$,

$\displaystyle X_{i}$ $\displaystyle\equiv\frac{A_{i}~{}n_{i}}{n_{b}}$ (12)
$\displaystyle=\frac{g_{i}~{}A_{i}^{\frac{3}{2}}}{2^{A_{i}}}\left(\frac{n_{b}}{\theta}\right)^{A_{i}-1}X_{p}^{Z_{i}}X_{n}^{N_{i}}\exp\left(\frac{B_{i}}{T}\right)$
(13) $\displaystyle n_{b}$ $\displaystyle=\sum\limits_{i}A_{i}n_{i}$ (14)

The relative abundances sum to one, and charge conservation ensures that that
the number density of protons is equal to the net number density of electrons
(that is, electrons minus positrons),

$\sum\limits_{i}X_{i}=1\qquad\sum\limits_{i}\frac{Z_{i}}{A_{i}}X_{i}=Y_{e}~{}.$
(15)

These two conditions close the set of equations, given $T$, $Y_{e}$ and
$n_{b}$ (or equivalently, the mass density $\rho_{b}=m_{u}n_{b}$). The
parameter $Y_{e}$ is related to the the ratio of total neutrons (inside and
outside nuclei) to protons: $f_{N}=(1-Y_{e})/Y_{e}$. We integrate these
equations using a modified version of the code of Seitenzahl et al.
(2008)222http://cococubed.asu.edu/code_pages/nse.shtml.

In the context of big bang cosmology, the temperature and baryon number
density $n_{b}$ are linked by the baryon-to-photon ratio,

$\displaystyle\eta$ $\displaystyle\equiv\frac{n_{b}}{n_{\gamma}}$ (16)
$\displaystyle n_{\gamma}$ $\displaystyle=\frac{2\zeta(3)}{\pi^{2}}~{}T^{3}$
(17) $\displaystyle=3.37\times 10^{4}\textrm{
mol/cm}^{3}~{}\left(\frac{T}{10^{9}\textrm{K}}\right)^{3}~{},$ (18)

where $\zeta(3)\approx 1.202$ is the Riemann zeta function evaluated at 3.

In general, NSE will favour protons and neutrons at high temperatures, while
at low temperatures, strongly-bound nuclei will be favoured, subject to the
condition of fixed $Y_{e}$. It is not immediately obvious from Equation (15)
which nuclei will be most abundant, but is crucial to what follows and so we
will briefly elaborate.

Suppose that at a given temperature, two species $i$ and $j$ dominate the
relative abundances,

$\displaystyle 1$ $\displaystyle\approx X_{i}+X_{j}\quad$ $\displaystyle
Y_{e}\approx\frac{Z_{i}}{A_{i}}X_{i}+\frac{Z_{j}}{A_{j}}X_{j}$
$\displaystyle\Rightarrow X_{i}$
$\displaystyle=\frac{Y_{e}-Z_{j}/A_{j}}{Z_{i}/A_{i}-Z_{j}/A_{j}}\quad$
$\displaystyle X_{j}=\frac{Z_{i}/A_{i}-Y_{e}}{Z_{i}/A_{i}-Z_{j}/A_{j}}$

Note that, if species $i$ has the same proton/nucleon ratio as that required
by $Y_{e}$, then $X_{i}=1$; in other words, species $i$ can incorporate all
the available protons and neutrons without any leftovers. More generally, if
$Z_{i}/A_{i}<Y_{e}$, then $Z_{j}/A_{k}>Y_{e}$ (or vice versa). That is, if one
species has a higher proton/nucleon ratio than required by $Y_{e}$, then the
other must have a lower ratio.

Given $X_{i}$ and $X_{j}$, we can solve for the mass abundances of the proton
$X_{p}$ and neutron $X_{n}$ in Equation (13), and thus derive the abundance of
any third species $k$. To be consistent with our assumption that two species
dominate, it must be the case that $X_{k}\approx 0$.

  * •

At high temperatures, the exponential term in Equation (13) will tend to
unity, and the dependence on temperature comes from the $\theta$ term:
$X_{k}\propto T^{-\frac{3}{2}(A_{k}-1)}$. Thus, all species with $A_{k}>1$
will have vanishing abundance at large $T$, leaving only the proton and
neutron.

  * •

At low temperature, the exponential term will dominate. If we create a 3D
vector $\textbf{a}_{i}=(B_{i},Z_{i},N_{i})$ for each nuclear species, then
(after some algebra) the dependence of $X_{k}$ on the exponential term is,

$X_{k}\propto\exp\left(\frac{1}{T}~{}\frac{\textbf{a}_{k}\cdot(\textbf{a}_{i}\times\textbf{a}_{j})}{\left(Z_{i}N_{j}-N_{i}Z_{j}\right)}\right)~{},$
(19)

which uses the usual vector cross and dot products. To be consistent with our
assumptions, the term in parentheses must be negative (and the denominator not
equal to zero). Thus, for a given $Y_{e}$, we search for the pair of nuclei
$i$ and $j$ such that a) $Z_{i}/A_{i}\leq Y_{e}$ and $Z_{j}/A_{j}\geq Y_{e}$,
and b) for _all_ other species $k$, the term in parentheses above is
negative.333Writing $f_{i}=N_{i}/Z_{i}$ and $b_{i}=B_{i}/Z_{i}$, we can
characterise the available species as a cloud of points in $(f_{i},b_{i})$
space. We can write $Y_{e}$ equivalently as a neutron/proton ratio:
$f_{e}=(1-Y_{e})/Y_{e}$. Then, equivalently and after even more algebra, we
search for the pair of nuclei $i$ and $j$ such that a) $f_{i}\geq f_{e}\geq
f_{j}$, and b) the line that passes through $(f_{i},b_{i})$ and
$(f_{j},b_{j})$ passes above (larger values of $B/Z$) than the rest of the
cloud. All things being equal, this will preference species with larger $B/Z$.
We have performed this search numerically for a set of nuclei from the
Evaluated Nuclear Structure Data File (ENSDF)444Accessed via https://www-
nds.iaea.org/relnsd/NdsEnsdf/QueryForm.html.; the results are shown in Figure
3. The top panel shows all available nuclei, while the bottom panel shows only
nuclei that are stable to all forms of radioactive decay. A set of highly-
bound (large $B_{i}/A_{i}$) nuclei dominates the abundances at the value of
$Y_{e}$ that matches their own proton/nucleon ratio: $Y_{e}=Z_{i}/A_{i}$. In
between these values, the abundance is shared between neighbouring species. In
particular, at small values of $Y_{e}$, free neutrons accompany the nucleus
with the largest value of $B_{i}/Z_{i}$; at large values of $Y_{e}$, free
protons accompany the nucleus with the largest value of $B_{i}/N_{i}$.

  * •

If we assume that the free neutron-proton ratio is maintained by the weak
force in equilibrium, rather than the fixed ratio assumed above, then
$X_{n}=X_{p}\exp(-Q_{np}/T)$, as in Equation (2). In this case, some (simpler)
algebra shows that at low temperature, the most abundant nucleus has the
largest value of $B_{i}/A_{i}-Q_{np}N_{i}/A_{i}$, which is [56]Fe. (It is not,
perhaps surprisingly, the most bound species per nucleon with the largest
value of $B_{i}/A_{i}$, which is 62Ni.)

Figure 3: _Top_ : The low-temperature NSE mass abundances of _all_ nuclei, as
a function of the overall proton/nucleon ratio $Y_{e}$. This plot uses every
known nuclear species in the NSDD database, including unstable ones. Each
species (80Ni, 78Ni, 80Zn, etc.) has a coloured line that matches its coloured
text label. Each species dominates the abundance at the value of $Y_{e}$ that
matches its own proton/nucleon ratio: $Y_{e}=Z_{i}/A_{i}$. In between these
values, the abundance is shared between neighbouring species. _Bottom_ : As
above, but using only nuclear species that are stable to all forms of
radioactive decay.

A potentially confusing aspect of NSE is that it takes no account of the
stability of a nuclear species, beyond its binding energy. As we will see in
the next section (Figure 4), small but non-zero abundances of famously highly-
unstable elements such as [5]Li (half-life: $4\times 10^{-22}$ s) and [8]Be
(half-life: $8\times 10^{-17}$ s) are present in NSE. In our Universe, the
instability of these nuclei is one reason for BBN essentially ending at [4]He,
and in stars it means that the nuclear path from protons to heavier elements
requires a three-body interaction: the triple-alpha process. While [9]Be is
stable to all forms of decay, [8]Be is more abundant in NSE because its
binding energy per nucleon is larger.555More precisely, its larger value of
$B_{i}/A_{i}-N_{i}Q_{np}/A_{i}$.

How do such short-lived elements remain in equilibrium? An unstable element
will be in equilibrium if its rate of decay is equal to its rate of formation.
For example, [5]Li and [8]Be require:

$\isotope[4]{He}+p\rightleftharpoons\isotope[5]{Li}\qquad\isotope[4]{He}+\isotope[4]{He}\rightleftharpoons\isotope[8]{Be}~{}.$
(20)

Equating the rates of production and decay, we find the conditions for secular
equilibrium:

$\begin{split}~{}\langle\sigma
v\rangle_{5}~{}n_{\isotope[4]{He}}n_{p}=\lambda_{\isotope[5]{Li}}n_{\isotope[5]{Li}}\\\
\frac{1}{2}~{}\langle\sigma
v\rangle_{8}~{}n^{2}_{\isotope[4]{He}}=\lambda_{\isotope[8]{Be}}n_{\isotope[8]{Be}}\end{split}$
(21)

where $\lambda_{\isotope[5]{Li}}$ $~{}(\lambda_{\isotope[8]{Be}})$ is the
decay rate (in s-1) for [5]Li ([8]Be), $~{}\langle\sigma v\rangle_{5}$ and
$~{}\langle\sigma v\rangle_{8}$ are the thermally-averaged reaction rates per
particle pair for the reactions in Equation 20, and $n_{i}$ is the number
density of species $i$.

Both NSE (10) and secular equilibrium (21) can be rearranged to constrain, for
example, the combination $n_{\isotope[4]{He}}n_{p}/n_{\isotope[5]{Li}}$, given
that equilibrium sets the chemical potentials:
$\mu_{\isotope[5]{Li}}=\mu_{\isotope[4]{He}}+\mu_{p}$. This might seem to
overdetermine the abundances; how do the decays and reactions conspire to
maintain both NSE and secular equilibrium? The answer is microreversibility
(Messiah, 1968, pg. 868), otherwise known as the _reciprocity theorem_
(Iliadis, 2015, pg. 75). Since nuclear interactions are time-reversal
symmetric666The detection of $T$ violation in particle physics — specifically,
kaon interactions mediated by the weak force — has lead to searches for
detectable $T$-violations in low-energy nuclear physics. These tests,
including experiments specifically aimed at reciprocity or detailed balance
(Blanke et al., 1983), support the conclusion that nuclear reactions are
invariant under time-reversal (Iliadis, 2015, pg. 75). Any effect of $T$
violation from the weak force is negligible., there is a simple relationship
between the forward and reverse decay / reaction rates per particle pair,
which depends on the phase space available in the entrance and exit channels.
The interaction doesn’t care which state we call “initial” and which we call
“final”. For example, in the case of [8]Be, the decay rate is related to the
[4]He + [4]He reaction rate per particle pair as
($g_{\isotope[8]{Be}}=g_{\isotope[4]{He}}=1$),

$\frac{\frac{1}{2}~{}\langle\sigma
v\rangle_{8}}{\lambda_{\isotope[8]{Be}}}=\left(\frac{m_{u}T}{\pi}\right)^{\\!\\!-\frac{3}{2}}\exp\left(\frac{-m_{\isotope[8]{Be}}+2m_{\isotope[4]{He}}}{T}\right)~{}.$
(22)

While the NSE relationship between the abundances holds only in equilibrium,
this relationship between the decay / reaction rates per particle pair holds
at all times. In equilibrium, both the right and left hand sides are equal to
$n_{\isotope[8]{Be}}/n^{2}_{\isotope[4]{He}}$. Perhaps counter intuitively,
this implies that — _ceteris paribus_ — the rate of formation for an unstable
nucleus increases with its decay rate.

###  3.2 Reaction Rates Perturbed by Expansion

The decay and reaction rates determine how quickly NSE is established from a
given set of initial abundances. Given a reaction network that connects
protons and neutrons to whichever nuclear species are most abundant, we can
calculate the rate at which equilibrium will be established if the system is
perturbed.

In particular, the expansion of the Universe (with scale factor $a$) will
perturb both the number density of all species $n_{i}\propto a^{-3}$ and the
reaction rates via the temperature $T\propto a^{-1}$. This section models a
Universe that expands quasi-statically, that is, in very small increments with
a sufficiently long pause at each step for the Universe to reestablish
equilibrium. We ignore the relationship between expansion rate and energy
density, given by the Friedmann equations, instead treating the expansion rate
as an independent variable. This quasi-static scenario is very different to
BBN in our universe.

Suppose that when the Universe has scale factor $a$, temperature $T$ and
baryon number density $n_{b}$, a reaction $A+B\rightleftharpoons C+D$ of 4
non-relativistic species is in secular and thermal equilibrium. By
microreversibility and Equation (10), the forward and reverse reaction rates
are related at all times by:

$\displaystyle\frac{~{}\langle\sigma v\rangle_{CD}}{~{}\langle\sigma
v\rangle_{AB}}=f_{r}(T)\equiv\frac{g_{A}g_{B}}{g_{C}g_{D}}\left(\frac{m_{A}m_{B}}{m_{C}m_{D}}\right)^{\\!\frac{3}{2}}\exp\left(-\frac{Q}{T}\right)$

where $Q=m_{A}+m_{B}-m_{C}-m_{D}$, and $f_{r}(T)$ depends on the reaction $r$,
but is a reminder that temperature is the only dynamic variable on the right
hand side. In secular equilibrium, the rate of change of the number density of
species $A$ is zero,

$\displaystyle\dot{n}_{C}$ $\displaystyle=~{}\langle\sigma
v\rangle_{AB}~{}n_{A}n_{B}-~{}\langle\sigma v\rangle_{CD}~{}n_{C}n_{D}$ (23)
$\displaystyle=~{}\langle\sigma
v\rangle_{AB}\left[n_{A}n_{B}-f_{r}(T)n_{C}n_{D}\right]$ (24)
$\displaystyle=0$ (25)

Now suppose that the Universe instantaneously expands by a small amount
$\textrm{d}a$, altering the temperature and number density. This will induce a
small reaction rate $\textrm{d}\dot{n}_{C}$. The perturbed reaction rate can
be shown to obey,

$\frac{\textrm{d}\dot{n}_{C}}{\textrm{d}\log a}=~{}\langle\sigma
v\rangle_{AB}~{}n_{A}n_{B}~{}\frac{Q}{T}~{}.$ (26)

If $Q$ is positive, energy must be supplied for the reverse reaction to occur.
As the Universe expands ($\textrm{d}a>0$) and cools, the reaction products are
energetically favoured. Thus, the net production rate of the reactant $C$ is
positive.

If some of the reactants or products are relativistic, then instead of
Equation (10), we use Equation (18), or its equivalent for fermions. At the
relevant temperatures ($>10^{9}$ K), neutrinos and electrons/positrons are
relativistic (Mukhanov, 2005, pg. 91). For simplicity, we will assume that
they are effectively massless and have zero chemical potential; see the
extended discussion of this point in Mukhanov (2005, pg. 89ff). Following the
same approach, the perturbed reaction rate is shown in Table 1 for a range of
reaction forms.

Reaction | Relativistic | $\textrm{d}\dot{n}_{C}/\textrm{d}\log a$
---|---|---
species
$A+B\rightleftharpoons C+D$ | none (or, $B$ and $D$) | $~{}\langle\sigma v\rangle_{AB}~{}n_{A}n_{B}~{}\frac{Q}{T}$
$A+B\rightleftharpoons C+D$ |  $D$ (e.g. photodisintegration) | $~{}\langle\sigma v\rangle_{AB}~{}n_{A}n_{B}\left(\frac{Q}{T}-\frac{3}{2}\right)$
$A+B\rightleftharpoons C$ | none | $~{}\langle\sigma v\rangle_{AB}~{}n_{A}n_{B}\left(\frac{Q}{T}-\frac{3}{2}\right)$
$A\rightleftharpoons B+C$ | none (e.g. $\alpha$ decay) | $\lambda_{A}~{}n_{A}\left(\frac{Q}{T}+\frac{3}{2}\right)$
$A\rightleftharpoons B+C+D$ |  $C$ and $D$ (e.g. $\beta$ decay) | $\lambda_{A}~{}n_{A}~{}\frac{Q}{T}$
$A+B+C\rightleftharpoons D$ | none (e.g. triple $\alpha$) | $~{}\langle\sigma v\rangle_{ABC}~{}n_{A}~{}n_{B}~{}n_{C}~{}\left(\frac{Q}{T}-3\right)$
Table 1: Perturbed reaction rate for a variety of reaction forms, and with the
relativistic species shown in the middle column. The quantity $Q$ is the total
mass of reactants minus the total mass of products.

After the small expansion, the number density of species $i$ changes due to
expansion at a rate $\textrm{d}n_{i}/\textrm{d}a$. We can also solve the NSE
equations at the new temperature and density of the Universe, to find the new
NSE abundances that have changed by
$\textrm{d}n^{\textrm{eq}}_{i}/\textrm{d}a$. Then,

$t_{eq,i}=\frac{\textrm{d}n^{\textrm{eq}}_{i}/\textrm{d}a-\textrm{d}n_{i}/\textrm{d}a}{\textrm{d}\dot{n}_{i}/\textrm{d}a}~{},$
(27)

gives the timescale for species $i$ to return to equilibrium. Given that
$T\propto 1/a$, we replace the derivative with respect to $\log a$ with the
derivative with respect to $-\log T$. We consider only reactions that produce
the larger nuclei when they are under-abundant; thus $t_{eq,i}$ is always
positive.

We require a set of reactions that connect the relevant particles — protons,
neutrons, electrons/positrons, neutrinos — with the largest nucleus that
dominates elemental abundances in NSE, [56]Fe. Ideally, we would integrate all
possible reactions until the system returned to equilibrium. However, we can
simplify the calculation by considering the fastest reaction pathways.
Further, given that our quasi-static calculation is only ever infinitesimally
far from equilibrium — and we thus do not specify a particular cosmological
expansion history or temperature-density-time relation — we use Equation (27),
rather than an explicit time integrator like AlterBBN. We add the
equilibration times (27) for the production of each element in a given pathway
in series, and add times for alternative pathways in parallel.

Ideally, we would integrate all possible reactions until the system returned
to equilibrium. However, we can simplify the calculation by considering the
fastest reaction pathway, and adding the equilibration times (27) for the
production of each element in the pathway.

First, protons and neutrons interact via the weak force,

$p+e^{-}\rightleftharpoons n+\nu\qquad p+\bar{\nu}\rightleftharpoons n+e^{+}$
(28)

The reaction rates are calculated from the integrals in Mukhanov (2005, pg.
100-1).

For the production of nuclei, we use the “21 isotope network” of
Cococubed777http://cococubed.asu.edu/code_pages/burn_helium.shtml, described
as the “default workhorse network of MESA” (Paxton et al., 2011), which is a
widely-used stellar evolution code. The temperature of BBN is comparable to
the temperature of the cores of the most massive stars. Further, the baryon
density of the universe is lower, so that no additional high-density (e.g.
three-body) interactions are expected. Thus, we expect the MESA reaction chain
to provide the appropriate reactions for our purposes. Reaction rates were
drawn from the online database
NETGEN888http://www.astro.ulb.ac.be/Netgen/form.html (Aikawa et al., 2005; Xu
et al., 2013, 2011). [4]He is produced by the following chain:

$\isotope[1]H(n,\gamma)\isotope[2]H(p,\gamma)\isotope[3]{He}(\isotope[3]{He},2p)\isotope[4]{He}~{}.$
(29)

[4]He is connected to [56]Fe by an alpha-chain to [52]Fe, followed by free-
neutron capture,

$\begin{split}&\isotope[4]{He}(2\alpha,\gamma)\isotope[12]{C}(\alpha,\gamma)\isotope[16]{O}(\alpha,\gamma)\isotope[20]{Ne}(\alpha,\gamma)\isotope[24]{Mg}\ldots\\\
&\isotope[24]{Mg}(\alpha,\gamma)\isotope[28]{Si}(\alpha,\gamma)\isotope[32]{S}(\alpha,\gamma)\isotope[36]{Ar}(\alpha,\gamma)\isotope[40]{Ca}\ldots\\\
&\isotope[40]{Ca}(\alpha,\gamma)\isotope[44]{Ti}(\alpha,\gamma)\isotope[48]{Cr}(\alpha,\gamma)\isotope[52]{Fe}(n,\gamma)\isotope[53]{Fe}\ldots\\\
&\isotope[53]{Fe}(n,\gamma)\isotope[54]{Fe}(n,\gamma)\isotope[55]{Fe}(n,\gamma)\isotope[56]{Fe}~{}.\end{split}$
(30)

In addition, we include the effect of the following parallel pathways.

  * •

[4]He can also be produced via neutron capture
$\isotope[3]{He}(n,\gamma)\isotope[4]{He}$, and via tritium
$\isotope[2]H(n,\gamma)\isotope[3]H(p,\gamma)\isotope[4]{He}$; tritium has a
lifetime of 12.32 years, long enough to be measured in underground water
sources (Barnes, 1971, 1973).

  * •

For each reaction in the “alpha-chain” between [24]Mg and [52]Fe, there is an
alternative two-step reaction that involves a proton catalyst, e.g.
$\isotope[24]{Mg}(\alpha,p)\isotope[27]{Al}(p,\gamma)\isotope[28]{Si}$.

  * •

The reactions
$\isotope[12]C+\isotope[12]C\rightleftharpoons\isotope[20]{Ne}+\alpha$,
$\isotope[12]C+\isotope[16]O\rightleftharpoons\isotope[24]{Mg}+\alpha$ and
$\isotope[16]O+\isotope[16]O\rightleftharpoons\isotope[28]{Si}+\alpha$ are
included.

  * •

We include an alternative pathway in the Cococubed network between [54]Fe and
[56]Fe via [57]Co.

  * •

The alpha-chain becomes slower as we approach [52]Fe, as it can create only
nuclei with $Z=N$ and these elements become less bound: [44]Ti, [48]Cr, and
[52]Fe are radioactive. The slowest of the alpha-chain reactions over much of
the relevant temperature range is
$\isotope[44]{Ti}(\alpha,\gamma)\isotope[48]{Cr}$. We consider an alternative
pathway that involves beta-decay and electron capture:
$\isotope[44]{Ti}(e^{-},\nu)$$\isotope[44]{Sc}(,e^{+}\nu)$$\isotope[44]{Ca}(\alpha,\gamma)$$\isotope[48]{Ti}(\alpha,\gamma)$
$\isotope[52]{Cr}(\alpha,\gamma)$$\isotope[56]{Fe}$.

###  3.3 Maintaining NSE in an Expanding Universe

Figure 4:  _Top Panels:_ NSE mass abundances as a function of temperature
(decreasing to the right) for a Universe with baryon-to-photon ratio $\eta$.
The top left panel is for the value of $\eta=6\times 10^{-10}$, as in our
Universe, and the right panel shows the effect of increasing the baryon-to-
photon ratio to $\eta=6\times 10^{-6}$. Note that the line styles are not the
same for the two panels. To reduce clutter, not all nuclei are shown: the
plotted nuclei are those that, at some temperature, are in the top five
abundances. Both panels show a similar pattern: protons and neutrons at high
temperature, [4]He coming to dominate at intermediate temperatures, and highly
bound nuclei (particularly [56]Fe) dominating at low temperatures. _Bottom
Panels:_ The equilibration timescales from Equation (27) for the Universe with
$\eta=6\times 10^{-10}$ (left) and $\eta=6\times 10^{-6}$ (right). As shown in
the legend, the solid lines are for the equilibrium between protons and
neutrons, from neutrons to [4]He, and from [4]He to [56]Fe. (Neutrons are
always overabundant, and thus their production is not a rate-limiting step.
The neutron equilibration time is shown for comparison with freeze-out in our
Universe.) The horizontal dashed line shows $t_{eq}=1$ second. The dotted
green line shows $1/H$, the Hubble time, from Equation (32). The dashed
vertical lines show the temperature at which the [4]He (left, red) and [56]Fe
(right, blue) have NSE abundances greater than 0.9. The thin grey lines show
the many reactions in the chain from [4]He to [56]Fe — because they span many
orders of magnitude, the overall $t_{eq}$ is approximately equal to the
timescale of the slowest reaction. In both panels, the production of [4]He is
the rate-limiting step.

Figure 4 (top) shows the NSE mass abundances as a function of temperature
(decreasing to the right) for a Universe with a fixed baryon-to-photon ratio
($\eta$), and with $X_{n}=X_{p}\exp(-Q/T)$. The left panel is for the value of
$\eta=6\times 10^{-10}$, as in our Universe. Beginning at high temperatures,
the Universe is dominated by protons and neutrons, with a small number of
deuterons. As the temperature decreases, [4]He comes to dominate the
abundances, due to its anomalously-large binding energy per nucleon amongst
small elements. Small amounts of other light elements are also present,
including unstable species. At about $2.7\times 10^{9}$ K, the abundances of
highly-bound elements (chromium, iron, nickel) rises rapidly. As expected, at
small temperatures, [56]Fe is the most abundant nucleus.

The top right panel of Figure 4 shows the effect of increasing the baryon-to-
photon ratio to $\eta=6\times 10^{-6}$. The qualitative features are very
similar. The lower ratio of photons reduces the photo-disintegration rate,
allowing nuclei species to exist at greater abundances at larger temperatures.
The transition to larger, bound nuclei happens at higher temperatures, with
[56]Fe again becoming the most abundant nucleus at low temperatures. If we
decrease $\eta$, the pattern makes a similar shift to the lower temperatures.

The lower panels of Figure 4 show the equilibration time (Equation 27) for a
given value of $\eta$, for the reaction network outlined above. As shown in
the legend, the solid lines are for the equilibrium between protons and
neutrons, from neutrons to [4]He, and from [4]He to [56]Fe. The horizontal
line shows $t_{eq}=1$ second. The dashed vertical lines show the temperature
at which the [4]He (left, red) and [56]Fe (right, blue) have NSE abundances
greater than 0.9. The dotted green line shows the Hubble time, relevant to the
next section,

$\displaystyle t_{H}\equiv H^{-1}$ $\displaystyle=\left(\frac{8\pi
G}{3}\kappa\right)^{\\!\\!-\frac{1}{2}}T^{-2}$ (31)
$\displaystyle=198.9\left(\frac{T}{10^{9}\textrm{ K}}\right)^{-2}\textrm{ s}$
(32) $\displaystyle\textrm{where}\quad\kappa$
$\displaystyle=\frac{\pi^{2}}{30}~{}\left(g_{b}+\frac{7}{8}g_{f}\right)~{},$
(33)

where the degeneracy factor for bosons is $g_{b}=2$ (2 photon polarisation
states), and for fermions is $g_{f}=10$ (3 neutrino species + 3 antineutrinos
+ 2 electron spin states + 2 positron spin states) at the relevant
temperatures (Mukhanov, 2005, pg. 94). The thin grey lines show the many
reactions in the chain — because they span many orders of magnitude, the
overall $t_{eq}$ is approximately equal to the timescale of the slowest
reaction. The lines stop at low temperature because the NSE abundance begins
to decline; the species is over-abundant, and so the equilibration timescale
no longer depends on the rate of production of the species. _Note well_ that,
in this quasi-static case, neutrons are always overabundant, and thus their
production is not a rate-limiting step. The neutron equilibration time is
shown for comparison with freeze-out in our Universe.

For $\eta=6\times 10^{-10}$ (as in our Universe, bottom left panel), the
timescale to [4]He increases rapidly as the temperature drops and its
reactants (tritium and [3]He) have very small abundances. To make a Universe
that is at least 90% helium by mass at some temperature, the Universe must
expand on a timescale of $\sim 2\times 10^{9}$ years at $\sim 3\times
10^{9}~{}$K.

Once [4]He has been produced, the alpha-chain reactions can produce [56]Fe on
a timescale of $2\times 10^{5}$ years. This time is set by the “bump” in the
equilibration timescale at $2.4\times 10^{9}$ K, which is caused by a
transition between the neutron pathway to [56]Fe and the alternative pathway
via beta-decay and electron capture.

The bottom right panel of Figure 4 shows the the equilibration time for the
larger value of $\eta=6\times 10^{-6}$. The qualitative features are similar.
Because the abundance of [4]He and [56]Fe dominates at larger temperatures,
where the reaction rates are larger, the respective timescales are shorter.
[4]He reaches 90% abundance by mass after $\sim 0.12$ years (45 days), at
$4.3\times 10^{9}~{}$K. At lower temperatures, [56]Fe reaches 90% abundance by
mass after a further $\sim 23$ hours. Smaller values of $\eta$ show the
opposite trend — lower temperatures, slower reactions, longer equilibration
timescales.

We can illustrate the trend with $\eta$ using the AlterBBN nucleosynthesis
code (Arbey et al., 2018). Though the reaction chain extends only to
$\isotope[16]O$, the production of [4]He illustrates the trend above. Figure 5
shows the late-time elemental abundances for deuterium, [3]He and [4]He. As
$\eta$ increases, there are fewer photo-disintegrating photons, so species can
survive at higher temperatures. NSE abundances peak at higher temperatures,
where the rates of NSE-establishing rates are faster. As a result, the
abundance of $\isotope[4]He$ increases with $\eta$, as expected999AlterBBN
seems to encounter numerical issues above $\eta=10^{-2}$. from Figure 4.

Figure 5: Late-time elemental abundances for [4]He, deuterium and [3]He as a
function of the baryon to photon ratio $\eta$, as calculated by the
nucleosynthesis code AlterBBN (Arbey et al., 2018). As $\eta$ increases, there
are fewer photo-disintegrating photons, so species can survive at higher
temperatures. NSE abundances peak at higher temperatures, and the rates of
NSE-establishing rates are faster. As a result, the abundance of [4]He
increases with the baryon-to-photon ratio.

Summarising, for a Universe with $\eta=6\times 10^{-10}$ (like ours) to have
fused a substantial fraction of its baryons into [56]Fe , it would have had to
cool from $\sim 10^{10}~{}$K to $\sim 2\times 10^{9}~{}$K over a period of
about a billion years. This is — comfortably — longer than the minute or so
that this cooling period lasted in our Universe. The rate limiting step turns
out to be the production of [4]He, due to the very small NSE abundances of its
reactants, tritium and [3]He.

##  4 How to Make an Iron Universe

NSE can be maintained only when cosmological expansion is slow compared to the
reaction rates that maintain equilibrium. Can we make a Universe that burns
all the way to [56]Fe?

We have little scope for making an iron Universe by slowing the expansion of
the Universe. Consider an expanding Universe that includes radiation (with
density $\rho_{r}$), matter ($\rho_{m}$), a cosmological constant
($\rho_{\Lambda}$), a curvature term ($-kc^{2}/R^{2}$) and an extra form of
energy ($\rho_{X}$) with equation of state $w=P_{X}/\rho_{X}$. For each of
these components in the Friedmann equation for the expansion rate of the
Universe, we can define characteristic times: $t_{r}=(8\pi
G\rho_{r}/3)^{-1/2}$ etc., and $t_{c}=R/c$. Then,

$H^{2}=\frac{1}{t_{r}^{2}}+\frac{1}{t_{m}^{2}}+\frac{1}{t_{X}^{2}}+\frac{\textrm{sign}(\rho_{\Lambda})}{t_{\Lambda}^{2}}-\frac{k}{t_{c}^{2}}~{}.$
(34)

If we want the Universe to expand quasi-statically between two temperatures on
a timescale $t_{eq}$ — recalling that $1/H$ is a characteristic timescale for
expansion — we need $H<1/t_{eq}$ between the two temperatures. However, at a
temperature of $3\times 10^{9}$ K, the energy density of radiation
(relativistic species) implies that $t_{r}\approx$ a few minutes. Additional
matter and “X” terms do not help: adding energy density can only make $H$
larger. We can cancel the positive contributions in Equation (34) with a
negative cosmological constant, but this is a version of Einstein’s static
universe: it cannot be made to subsequently expand and cool. If it departs
from $H=0$, it can only recollapse.

The best we can do is a “stalling” Universe: if we include positive curvature
and a positive cosmological constant, we can arrange for the Universe to
stall, with $H\approx 0$ as the Universe almost recollapses under the
attractive pull of matter and radiation, only to be saved by the repulsion of
the cosmological constant. However, because each of the terms in Equation (34)
varies differently with the scale factor $a$, the Universe must stall at a
particular value of $a$, that is, a particular temperature. This is not quasi-
static expansion, but would still allow reactions to proceed. This scenario is
extremely fine-tuned — the terms in Equation (34), each of order (a few
minutes)-2, would have to fortuitously cancel to many decimal places to stall
the Universe for the required nuclear timescales.

However, as shown in the previous section, we can effectively speed up the
nuclear reactions by increasing the baryon-to-photon ratio $\eta$. This shifts
the NSE abundances to higher temperatures, where the reaction rates are
substantially larger.

Figure 6: The same as Figure (4), but for a Universe with $\eta=6\times
10^{-3}$. The dotted green line in the right panel shows $1/H$, the Hubble
time, from Equation (32). The [4]He and [56]Fe equilibration times in the
right panel remain below the Hubble time for temperatures above $2.8\times
10^{9}~{}$K, meaning that these reactions remain in equilibrium. At $2.8\times
10^{9}~{}$K, the NSE abundance of [56]Fe is $X=0.87$.

Figure 6 shows the NSE abundances (left) and equilibration timescale (right)
for a Universe with $\eta=6\times 10^{-3}$. The dotted green line in the right
panel shows $1/H$, the Hubble time, from Equation (32). The [4]He and [56]Fe
equilibration times remain below the Hubble time for temperatures above
$2.8\times 10^{9}~{}$K, meaning that these reactions remain in equilibrium. At
$2.8\times 10^{9}~{}$K, the NSE abundance of [56]Fe is $X=0.87$. A Universe
with sufficiently large baryon-to-photon ratio will burn a significant
fraction of its baryonic content into [56]Fe.

##  5 Discussion

In this section, we discuss the approximations made in our calculation. Our
calculation considered a limited number of pathways between protons/neutrons
and highly-bound nuclei such as iron. We assume that Cococubed and MESA have
identified the fastest reaction path, or at least, there is no alternative
path that is significantly faster.

We have not integrated the reaction network to calculate abundances, but
instead have relied on timescales calculated from equilibrium reaction rates
perturbed by the expansion of the universe. This approach is appropriate to
the scenario we consider, in which the abundances stay close to their
equilibrium value until the large elements are produced. Our conclusions are
drawn in light of this order-of-magnitude calculation.

Our assumption of a Maxwell-Boltzmann distribution for non-relativistic
species (Equation 10) and $n\propto T^{3}$ for relativistic species breaks
down if any of the relevant species becomes degenerate. This depends on the
chemical potentials of the relevant species. As shown in detail in Mukhanov
(2005, pg. 91ff.), in the Lepton era — when electrons are relativistic but
protons are not — the ingredients of the universe are photons, leptons ($e$,
$\mu$, $\tau$), neutrinos ($\nu_{e}$, $\nu_{\mu}$, $\nu_{\tau}$), light
baryons ($p$, $n$, $\Lambda$), and mesons ($\pi^{0}$, $\pi^{\pm}$). Taking
into account relations between the chemical potentials due to reactions such
as $\mu^{-}\rightarrow e^{-}+\bar{\nu}_{e}+\nu_{\mu}$, only five are
independent, which can be taken to be $\mu_{e^{-}}$, $\mu_{n}$,
$\mu_{\nu_{e}}$, $\mu_{\nu_{\mu}}$, and $\mu_{\nu_{\tau}}$.

These five chemical potentials are in turn specified by five conserved
quantities: the baryon-to-entropy ratio $B$, net electric charge-to-entropy
ratio $Q$, and three lepton number-to-entropy ratios $L_{e}$, $L_{\mu}$, and
$L_{\tau}$. The entropy decreases with the scale factor as $a^{-3}$, and is
dominated by relativistic species: $s\sim n_{\gamma}\sim T^{3}$. It follows
that $B\sim\eta$.

For non-relativistic species, the Maxwell-Boltzmann distribution is an
accurate approximation at the relevant temperatures if $m\gg T$ and
$(m-\mu)\gg T$. For the proton, the mass (938 MeV) is significantly larger
than the temperature of BBN ($\sim$ MeV). Further, Mukhanov (2005, pg. 93)
shows that:

$\frac{m_{p}-\mu_{p}}{T}\approx\ln\left(\frac{1}{B}\left(\frac{m_{p}}{T}\right)^{3/2}\right)\approx\ln\left(\frac{1}{B}\right)+10~{}.$
(35)

Hence, even for $B\sim\eta\sim 1$, the Maxwell-Boltzmann distribution is
sufficiently accurate.

For relativistic species, the relation $n\propto T^{3}$ is accurate if
$|\mu|\ll T$. For photons, $\mu_{\gamma}=0$. The most important relativistic
species during BBN is the electron. As shown in Mukhanov (2005, pg. 93), the
electric neutrality of the universe implies that:

$\frac{|\mu_{e}|}{T}\approx B.$ (36)

Thus, our approximation for electrons is accurate so long as $B\sim\eta\ll 1$.

That leaves the electron neutrino, which could have a significant chemical
potential if there is a large asymmetry between neutrinos and anti-neutrinos.
In our universe, observations of the CMB and BBN constrain $\mu_{\nu_{e}}/T$
(which is conserved) to the range $[-0.01,0.22]$ (Hansen et al., 2002), which
is consistent with zero but not particularly tightly constrained. Here, in
common with most BBN calculations, we assume that $\mu_{\nu_{e}}/T$ is
negligible.

In summary, our calculation gives the correct order-or-magnitude timescales
for $\eta\ll 1$ (assuming no significant neutrino asymmetry). A substantial
abundance of heavy elements are be produced for $\eta=6\times 10^{-3}$, where
our assumptions should still be valid.

A confirming set of calculations have been made by Aguirre (1999, 2001), who
considered nucleosynthesis in a cold big bang Universe. They integrate the
reaction rates, accounting for chemical potentials and degeneracy, for a
reaction chain that terminates at oxygen. They note that “by varying [the
baryon-photon ratio] and [the lepton-baryon ratio], almost any desired yield
of primordial helium and metals can be obtained.” Our calculations confirm
that the reactions can proceed to highly-bound elements such as iron.

A more general discussion of the connection between matter-antimatter
asymmetry, BBN, fine-tuning and life can be found in Steigman & Scherrer
(2020), who speculate that high values of $\eta$ may produce highly-bound
nuclei such as iron.

##  6 Conclusions: Implications for Initial Entropy

The baryon-to-photon ratio in our Universe is set by the asymmetry between
matter and antimatter in the early Universe. At very early times, there is
roughly a billion-and-one particles for every billion antiparticles. As the
Universe cools below their rest-mass energy, annihilation turns a billion
particle-antiparticle pairs into two billion photons, leaving a baryon-photon
ratio of $\sim$ 1 / two billion.

Thus, an unexpected character in the story of the low entropy of our Universe
emerges — matter-antimatter asymmetry. It is well known that an exactly
matter-antimatter symmetric Universe would contain essentially no baryons at
late times, after annihilation. The Universe would be pure photons, unable to
form structures more complex than slight overdensities of photons. If, as some
have argued, black hole evaporation into photons via Hawking radiation
indicates that (contra-Penrose) a pure photon gas is a higher entropy state
than a black hole, then a matter-antimatter symmetric Universe would very
quickly reach a state of maximum entropy, with neither nuclear nor
gravitational entropy available.

However, as we have shown here, in a universe with a sufficiently high degree
of matter-antimatter _asymmetry_ — which results, after annihilation, in a
large value of the baryon-to-photon ratio ($\eta$) — primordial nuclear
reactions bind a substantial fraction of baryons into [56]Fe. Following the
discussion of Rovelli (2019), in such a Universe the metastable state into
which the Universe evolved would differ from its true nuclear equilibrium
state at late times only with regard to a small residue of non-iron nuclei.
The nuclear energy that could have been released in stars to power life has
instead become a near-negligible addition to the cosmic microwave background,
and is uselessly redshifting away with the expansion, unable to be put to good
use.

In such a Universe, the gravitational aspect of low entropy identified by
Penrose would still be available. There would still be a thermodynamic arrow
of time. Matter would be able to collapse into bound structures and release
energy. However, unless this collapse was able to unbind iron nuclei into its
constituents, it would not ignite long-lived nuclear reaction-powered objects
such as stars.

Could collapse unbind iron? If such a process were possible, there are
circumstances in our Universe where it might occur. The most extreme cases of
gravitational collapse in our Universe include the accretion disks around
black holes, but these show evidence of iron in their innermost regions, and
no evidence of its disintegration. Type II supernovae result from the collapse
of a massive star onto an iron core, but this results in nuclear reactions
beyond iron (up to uranium) in the ejecta, and either a black hole or a
neutron star as the relic. Again, no large scale disintegration of iron
occurs. The entropic value of a neutron star seems minimal — the nuclear
energy of the neutrons does not seem to be _available_.

In the absence of nuclear fuel, astrophysical sources of low-entropy radiation
would have to be powered by gravitational energy. Such systems in our Universe
— protostars, accretion disks, hot stellar relics — are shorter-lived and/or
less-stable in their radiative output than stars as we know them. Note well,
of course, that these scenarios may not exhaust what is possible in a pure
iron Universe.

The central point is that the thermodynamic arrow of time in our Universe —
pointing from the low-entropy beginning of the Universe, through the ignition
of stars by gravitational collapse, to nuclear reaction-powered low-entropy
photons from the Sun, which are the source of almost all of the second-law
processes around us, including ourselves — depends crucially on matter-
antimatter asymmetry in the early Universe. At the moment, we do not know what
caused this asymmetry (that is, the physics of baryogenesis). We know some of
the necessary conditions, thanks to Sakharov (1967). The standard model of
particle physics has the required asymmetries, but not to a degree sufficient
to explain our Universe. The as-yet unknown physics that explains matter-
antimatter asymmetry is a major character in the story of the arrow of time.

###### Acknowledgements.

This project was conceived at the meeting “The Chimera of Entropy II” held at
the John Bell Center for the Foundations of Physics on the beautiful island of
Hvar, Croatia in July 2019. Extreme thanks to the very useful data and code
collection at cococubed.asu.edu by F.X. Timmes. This research made use of the
primordial nucleosynthesis code AlterBBN (https://alterbbn.hepforge.org Arbey
et al. (2018)), the SciPy ecosystem Virtanen et al. (2020), NumPy Harris et
al. (2020) and Matplotlib Hunter (2007). LAB acknowledges that this
publication was made possible through the support of a grant from the John
Templeton Foundation. The opinions expressed in this publication are those of
the author and do not necessarily reflect the views of the John Templeton
Foundation. We thank the anonymous referee for useful comments.

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