advanced20 min studyLesson 856 in Physics

Dark matter freezeout and freezein models

Integrate simple relic-yield equations while exposing their particle and cosmological assumptions.

Before you begin

On this page

The question for today

What you’ll understand

Write a number-density equation with explicit particle-counting conventions.
Integrate a post-freezeout annihilation tail.
Distinguish freezein production from departure out of chemical equilibrium.

The gravitational evidence for dark matter does not identify how its abundance was established. Thermal freezeout and feeble freezein are two conditional production mechanisms, each built from a particle model and an expansion history. Freezeout begins with chemical equilibrium and retains a relic after annihilations become inefficient. Freezein starts with a negligible population that accumulates through rare reactions without attaining chemical equilibrium. The same observed density can arise from different microscopic rates, masses, and histories, so a successful abundance calculation is a consistency test rather than unique identification of a dark-matter particle.

Remove expansion dilution before following reactions

Take a stable self-conjugate species with number density $n$ and pair annihilation into an equilibrium bath. Use a thermally averaged annihilation cross section with the convention in which loss of two particles cancels the one-half identical-pair event-counting factor. The resulting density equation includes a dilution term from expansion and a difference between forward annihilation and inverse production. For distinct particles and antiparticles, define their separate or combined densities before importing any numerical factor.

Introduce the yield as number density divided by bath entropy density. During adiabatic expansion, this removes dilution and isolates changes from reactions. The useful inverse-temperature variable $x=m/T$ increases as the bath cools. Its simplest evolution equation assumes constant entropy count, radiation domination for the usual prefactor, kinetic distributions suitable for thermal averaging, and no additional source. Varying entropy counts, coannihilations, asymmetries, or a separately heated dark sector require modified equations.

\dot n+3Hn=-\langle\sigma v\rangle(n^2-n_{\rm eq}^2),\qquad Y=\frac ns,\qquad x=\frac mT,\qquad \frac{dY}{dx}=-\frac{s\langle\sigma v\rangle}{Hx}(Y^2-Y_{\rm eq}^2)

The yield equation uses conserved comoving bath entropy and constant

g_{*s}

m

is the particle mass,

n_{\rm eq}

its equilibrium density, and

s

the entropy density used consistently in both yields.↔ Scroll wide equations horizontally.

Freezeout has a residual annihilation tail

A massive equilibrium abundance becomes exponentially small during cooling. Eventually annihilations cannot keep the actual population tracking that rapidly changing target. A rate comparison estimates the departure, but residual annihilations continue afterward and can materially lower the asymptotic yield. Chemical freezeout also need not coincide with kinetic decoupling: elastic scattering can maintain a thermal momentum shape after number-changing processes have become ineffective.

For a velocity-independent annihilation coefficient in a fixed-count radiation era, the post-departure yield equation reduces to an inverse-square coefficient in inverse temperature. Integrating its reciprocal gives a transparent approximation for the late abundance. The familiar inverse relation between relic density and annihilation cross section holds within such assumptions. Extra entropy release, resonant annihilation, coannihilation, or an unusual expansion rate can change the relation; it is not a general law for all dark matter.

\frac{dY}{dx}\simeq-\frac{\lambda}{x^2}Y^2\quad\Longrightarrow\quad \frac1{Y_\infty}=\frac1{Y_f}+\frac{\lambda}{x_f},\qquad Y_{\rm FI,\infty}=Y_{\rm initial}+\int S(x)\,dx

The first result neglects inverse production after a chosen

x_f

, with constant positive dimensionless

\lambda

. The second is a schematic feeble-production equation with source

S=dY/dx

when backreaction is negligible.↔ Scroll wide equations horizontally.

Freezein remembers a source rather than an equilibrium departure

In freezein, bath particles can be thermal while the dark species is not. A sufficiently small coupling suppresses inverse reactions and annihilation of the accumulated population, so production is approximately linear in the collision source. Increasing that coupling then increases the relic yield within the feeble regime. Renormalizable examples can be dominated near a parent-particle mass, whereas higher-dimensional interactions can weight the highest available temperatures and introduce sensitivity to reheating. Initial abundance must also be specified.

As a dated observational benchmark, the Planck 2018 parameter analysis, in its 2021 arXiv revision, reports $\Omega_c h^2=0.120\pm0.001$ at 68 percent confidence for base flat six-parameter $\Lambda$ CDM. Here $\Omega_c$ is the cold-dark-matter density fraction and $h$ normalizes the present Hubble constant to $100\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$ . That fitted density neither selects freezeout nor establishes a coupling. Translating a calculated yield into this density additionally requires the particle mass, present entropy normalization, stability, and intervening cosmic history.

Work it through

Continue annihilation after a chosen departure point

A toy post-freezeout model has $x_f=20$ , $Y_f=10^{-9}$ , and $dY/dx=-\lambda Y^2/x^2$ for all later times, with $\lambda=2\times10^{10}$ . Neglect inverse reactions and later entropy production. Compute the asymptotic yield and determine how much annihilation remains after the chosen starting point.

Differentiate the reciprocal to obtain $d(1/Y)/dx=\lambda/x^2$ . This removes the nonlinear power of the yield without changing its positive initial condition. The integral from the starting point to infinity is $\lambda/x_f$ .
Evaluate the two reciprocal contributions: $1/Y_f=10^9$ and $\lambda/x_f=10^9$ . Their sum is $2\times10^9$ , so the reciprocal yield doubles after the chosen departure point.
Invert the sum to find $Y_\infty=5\times10^{-10}$ . The late yield is half the starting yield, demonstrating that setting the abundance equal to its value at a rate crossing can miss significant residual annihilation. The starting values were stipulated rather than derived from a full equilibrium solution.

The result

The final yield is $5\times10^{-10}$ : half the starting comoving population is removed by the modeled annihilation tail. A particle mass is still needed to turn this dimensionless yield into a present mass density.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

A source-amplitude analogy for feeble production

Reset and set both controls to coefficient $a=1$ , position $x=1$ ; the signed area is $1/3$ . Reset and set both controls to $a=2$ , $x=1$ ; the area is $2/3$ . This illustrates linear dependence of an integrated source on its amplitude. The polynomial is not the dark-matter collision source, and the lab includes neither inverse reactions nor freezeout, expansion, or a coupling-to-cross-section conversion.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA source integral scales linearly with its amplitude only while the evolution remains in the assumed source-dominated regime.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Freezein accumulates rare production from a bath. Its increasing-coupling trend is conditional on remaining out of chemical equilibrium and on fixed background and initial conditions.

Your turn

Try it yourself

For a separate toy freezein calculation, use $dY/dx=Ae^{-x}$ for $x\ge0$ , with $Y(0)=0$ and $A=3\times10^{-12}$ . Find the final yield. If the source amplitude is proportional to a portal coupling squared, what happens when that coupling doubles, assuming all freezein approximations still hold?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Integrate the normalized exponential source and distinguish source amplitude from the coupling itself.

Compare with the solution

Integration gives $Y(x)=A(1-e^{-x})$ , hence $Y_\infty=A=3\times10^{-12}$ . Doubling the coupling multiplies the source amplitude by four, yielding $1.2\times10^{-11}$ . This is a deliberately simplified source integral over a dimensionless variable, not a claimed physical collision kernel. The scaling fails if the larger coupling establishes equilibrium or makes destruction of the produced population appreciable.

Keep the idea

Take it with you

Relic-density calculations require both a particle model and a cosmic history.
Post-departure annihilations can change a freezeout abundance substantially.
Freezein yield grows with a feeble source, with possible sensitivity to reheating and initial conditions.

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