Changing particle and antiparticle populations

The total population and its net excess can change differently

Pair creation and annihilation can change the number of particles plus antiparticles while preserving their difference for a conserved charge. A declining symmetric pair population therefore need not erase a previously supplied net excess.

This is different from explaining why the Universe has an observed matter excess. A calculation that inserts a positive initial or conserved difference has not generated that difference. Its origin requires additional physics and evidence.

To isolate the account, consider one hypothetical equal-mass particle–antiparticle pair with rapid reactions maintaining chemical equilibrium with a bath. Assume negligible backreaction and conserved comoving bath entropy throughout the stated interval.

The example uses a leading classical nonrelativistic approximation, rather than a precision relativistic abundance calculation. No actual particle identity, measured baryon asymmetry or final relic yield is assigned.

Normalize populations without hiding expansion dilution

Let s_b denote the bath entropy density divided by k_B, so it has number-density units m⁻³. Define dimensionless yields Y_+ = n_+/s_b and Y_− = n_−/s_b. Conserved comoving bath entropy means that a conserved net particle charge gives a constant yield difference.

Choose x = mc²/(k_B T) from 20 to 30; cooling increases x. The supplied zero-chemical-potential yield is b(x) = 2 × 10⁻⁹ (x/20)^(3/2) exp(20 − x). It is positive and becomes smaller as x rises over this interval.

The form follows the leading nonrelativistic density proportional to T^(3/2) exp(−mc²/(k_B T)), divided by a bath entropy density proportional to T³ with constant entropy degrees of freedom. Its normalization is a chosen input.

Impose the conserved difference Δ = Y_+ − Y_− = 2 × 10⁻¹⁰. The reaction bath has zero chemical potential for the pair-annihilation products, so equilibrium sets opposite pair chemical potentials and gives Y_+Y_− = b².

The small pair component falls while the excess remains

Solving the positive-yield branch gives Y_+ = [√(Δ² + 4b²) + Δ]/2 and Y_− = [√(Δ² + 4b²) − Δ]/2. Their difference is exactly Δ in this approximation; their product is exactly b².

For numerical work when b is very small, the equivalent expression Y_− = 2b²/[√(Δ² + 4b²) + Δ] avoids subtracting nearly equal numbers. It preserves the small positive antiparticle yield rather than rounding it to zero prematurely.

At x = 20, both yields substantially exceed the imposed difference. By x = 30, Y_+ is very close to Δ, while Y_− is still positive at about 1.39128 × 10⁻¹⁶. The remaining net excess has not been created by the pair calculation.

The table displays yields in units of 10⁻¹⁰. A value of 2 in that display means the dimensionless yield 2 × 10⁻¹⁰, not two particles per cubic metre. Physical densities require the bath entropy density as well.

A nonzero chemical potential keeps track of the imposed charge

In this approximation, write Y_+ = b exp(μ/(k_B T)) and Y_− = b exp(−μ/(k_B T)). Multiplication removes μ and gives the product b², while subtraction ties μ to the imposed Δ.

For the chosen rows, μ/(k_B T) rises from about 0.049979 at x = 20 to 2.371451 at x = 25 and 7.089218 at x = 30. A growing chemical-potential ratio here accounts for the conserved asymmetry within equilibrium.

The occupation remains dilute over this constructed interval: the approximate zero-momentum particle factor exp[−x + μ/(k_B T)] is at most about 2.17 × 10⁻⁹ at the tabulated stages. This check supports the classical treatment of the supplied example, without quantifying the error of its leading nonrelativistic expansion.

The displayed decimal precision is arithmetic within that chosen approximation. It is not a measured precision or a claim that relativistic corrections and real reaction dynamics are negligible in every physical application.

An equilibrium account is conditional on reactions keeping up

The product law assumes sufficiently rapid pair creation and annihilation to maintain the declared chemical equilibrium. If the relevant reactions fail to keep up with expansion, these algebraic equilibrium yields need not describe the subsequent population.

A real freeze-out calculation evolves the populations with the appropriate collision terms, expansion and initial conditions. It can leave a different surviving symmetric component from the equilibrium continuation.

The example nevertheless establishes a useful distinction: reducing a pair population, retaining a conserved excess and generating an asymmetry are three different questions. They should not be treated as interchangeable explanations.

When reading an abundance claim, identify the entropy normalization, conserved charges, equilibrium assumptions, particle masses, reaction rates and approximation regime. Only then can its output be compared with an observed abundance or a mechanism for its origin.

Worked reasoning

Try the reasoning

1. Does the chosen positive Δ explain why the actual Universe has a matter excess?

   Show the answer and reasoning

   No.

   Δ was inserted as a conserved input, so the calculation does not provide an asymmetry-generation mechanism.

2. Is the antiparticle yield exactly zero at x = 30 in the supplied model?

   Show the answer and reasoning

   No; it is about 1.39128 × 10⁻¹⁶.

   A small positive equilibrium yield should not be replaced by a fabricated exact zero.

3. May these equilibrium formulas be continued through an actual chemical freeze-out without a kinetic check?

   Show the answer and reasoning

   No.

   Their premise of maintained chemical equilibrium would then fail.

What to carry forward

• Pair suppression can reduce the symmetric population while preserving an imposed net charge.
• The yield normalization and approximation regime are part of the result.
• An equilibrium pair account neither generates the supplied asymmetry nor replaces a freeze-out calculation.