GLOBAL KNOWLEDGE ARK

Cosmology · Matter–antimatter asymmetry and hypothetical relics

Evidence for a matter–antimatter imbalance

Separate evidence about matter sign, abundance inference and a conserved baryon excess, then calculate what pair annihilation can actually change.

Preparation: ratios, powers of ten and ordinary arithmetic. Physical quantities and notation are introduced as they are needed.

Back to the chapter

After reading, you can

  • Distinguish matter-sign evidence from a model-dependent abundance estimate.
  • Track an entropy-normalized excess through pair annihilation.
  • Separate survival of an imposed imbalance from its physical generation.
01

Define the imbalance before interpreting evidence

The baryon-asymmetry question concerns a difference between baryons and antibaryons. In the simplest number ledger a baryon carries baryon number +1 and an antibaryon −1; this is different from net electric charge.

Ordinary matter dominance is tested through the composition of accessible material and through possible consequences of matter–antimatter contact. Ordinary atomic spectra alone do not identify the baryon sign of a distant otherwise similar object.

Annihilation radiation can test specified mixtures or contacting domains, but the prediction depends on geometry, transport and cosmological assumptions. A historical domain calculation is not a universal proof about every possible antimatter distribution.

Abundance inference asks another question. CMB baryon-density inference and conversion of deuterium abundance through nucleosynthesis use specified models and nuclear inputs; they are not direct sign-sensitive counts of all cosmic baryons and antibaryons.

02

Separate net yield from fractional imbalance

Supply a homogeneous dilute species model with baryon and antibaryon number densities n_B and n_anti in m⁻³. Let a prescribed bath have s_b equal to entropy density divided by k_B, also in m⁻³, with s_ba³ constant.

Define dimensionless yields Y_B = n_B/s_b and Y_anti = n_anti/s_b. Their difference is the net baryon yield; their fractional imbalance is (Y_B−Y_anti)/(Y_B+Y_anti).

Choose the invented reference Y_* = 10⁻¹² and initial b = Y_B/Y_* = 101, ā = Y_anti/Y_* = 99. The initial net yield is therefore 2Y_* while the fractional imbalance is only 1/100.

Only pair annihilation acts after these initial conditions. Inverse pair creation and baryon-number violation are omitted, and the dilute species do not alter the prescribed bath entropy or background. These are continuum mean yields, not literal finite integer counts.

03

Annihilation removes equal numbers of both species

Let k = ⟨σv⟩ have units m³/s and define nonnegative dimensionless exposure θ = Y_*∫ks_bdt. In the stated bath the yield equations become db/dθ = dā/dθ = −bā.

Subtracting them gives d(b−ā)/dθ = 0, so b−ā = 2 throughout. A process that removes one baryon and one antibaryon at a time preserves their number difference.

The exact ratio is r = ā/b = (99/101)exp(−2θ). Together with the fixed difference this gives ā = 2r/(1−r) and b = 2/(1−r).

Their sum decreases, so the fractional imbalance can grow even though the net yield stays fixed. This is the survival of an already supplied excess, rather than a mechanism that creates it.

04

The excess survives while pairs disappear

At θ = (ln 99)/2, the ratio r is 1/101. The yields are b = 101/50 and ā = 1/50, still differing by 2.

The fractional imbalance is then 50/51, about 0.980392. Its large value reflects the removal of most pairs, not an increase in the conserved baryon excess.

As θ tends to infinity, b tends to 2 and ā tends to zero. At every finite θ in this continuous model, ā remains positive; the limiting zero is not a finite annihilation endpoint.

For a symmetric control with initial b = ā = 100, both equal 100/(1+100θ). Their net yield remains zero, so the same annihilation process does not generate an excess from symmetric initial conditions.

Invented finite exposure snapshots; b and ā are yields divided by Y_*
θbāb−āFractional imbalance
01019921/100
(ln 3)/2101/3433/34234/67
(ln 9)/2101/4511/45245/56
(ln 99)/2101/501/50250/51
Removing pairs preserves an imposed excessA supplied homogeneous dilute continuum model begins with normalized baryon yield b = 101 and antibaryon yield a_bar = 99. Pair annihilation alone gives b_prime = a_bar_prime = −b a_bar in dimensionless exposure theta. The upper logarithmic plot shows both positive yields and their constant difference 2, only over 0 <= theta <= ln(99)/2. The lower panel shows the growing fractional imbalance (b−a_bar)/(b+a_bar); it rises from 1/100 to 50/51 while the net stays fixed. The mean yields are not literal integer particles. At the final finite point a_bar = 1/50 remains positive; zero is only an infinite-exposure limit. Y_star = 10^(−12) is an invented normalization and the bath entropy is supplied without backreaction, inverse pair creation or baryon-number violation. The model preserves an imposed imbalance and does not explain its origin.Removing pairs preserves an imposed excessInvented continuum yields: b = Y_B/Y*, ā = Y_anti/Y*.Y* = 10⁻¹²; pair annihilation only; prescribed dilute bath.Finite exposure interval: 0 ≤ θ ≤ (ln 99)/2.Baryon yield bAntibaryon yield āConserved difference b − ā = 2Vertical normalized yield · logarithmic axis0.010.11101000(ln3)/2(ln9)/2(ln99)/2Horizontal dimensionless annihilation exposure θFractional imbalance (b−ā)/(b+ā)00.250.50.7510(ln3)/2(ln9)/2(ln99)/2Horizontal θ · the fraction changes, the difference does notθbāb−āFraction01019921/100(ln3)/2101/3433/34234/67(ln9)/2101/4511/45245/56(ln99)/2101/501/50250/51θ = Y* ∫ k s_b dt; k is in m³/s and s_b in m⁻³.s_b is entropy density divided by k_B, not entropy in J/(m³ K).Equal reaction losses preserve b − ā = 2 at every exposure.At the last finite point ā = 1/50 > 0; no finite zero is drawn.Symmetric initial yields would keep zero net excess.No inverse creation, entropy injection or origin mechanism is solved.These are supplied mean yields, not observed cosmic abundances.
A supplied homogeneous dilute continuum model begins with normalized baryon yield b = 101 and antibaryon yield a_bar = 99. Pair annihilation alone gives b_prime = a_bar_prime = −b a_bar in dimensionless exposure theta. The upper logarithmic plot shows both positive yields and their constant difference 2, only over 0 <= theta <= ln(99)/2. The lower panel shows the growing fractional imbalance (b−a_bar)/(b+a_bar); it rises from 1/100 to 50/51 while the net stays fixed. The mean yields are not literal integer particles. At the final finite point a_bar = 1/50 remains positive; zero is only an infinite-exposure limit. Y_star = 10^(−12) is an invented normalization and the bath entropy is supplied without backreaction, inverse pair creation or baryon-number violation. The model preserves an imposed imbalance and does not explain its origin. Exact supplied finite snapshots; all yield columns use the same Y* normalization. Columns: θ; b; ā; b−ā; Fraction. Rows: 0; 101; 99; 2; 1/100 / (ln3)/2; 101/34; 33/34; 2; 34/67 / (ln9)/2; 101/45; 11/45; 2; 45/56 / (ln99)/2; 101/50; 1/50; 2; 50/51.
05

A surviving excess still needs an origin explanation

For a separate conversion illustration, stipulate n_γ/s_b = 1/5 throughout this toy interval. Then η = (n_B−n_anti)/n_γ = 5(Y_B−Y_anti) = 10⁻¹¹, constant here.

This artificial reservoir ratio and η are not measured cosmological values. Photon or entropy injection would require revising the assumptions before using the same conversion.

A real abundance comparison must retain the observational response, selected systems or datasets, nuclear calculations and theoretical uncertainties. Agreement between methods does not make all of those assumptions disappear.

Baryogenesis asks how a nonzero net yield can arise dynamically and survive subsequent processing. Evidence for an imbalance, a fitted abundance and an explanation of its origin are connected questions with distinct requirements.

Worked reasoning

Recover the yields from their ratio and difference

Find b,ā and the imbalance fraction when θ = (ln 99)/2.

  1. The exact ratio becomes r = (99/101)/99 = 1/101.
  2. Solve ā = rb and b−ā = 2 to obtain b = 101/50 and ā = 1/50.
  3. Divide their difference by their sum: 2/(102/50) = 50/51.

The yields are 101/50 and 1/50 in units Y_*; the net stays 2Y_* while the fractional imbalance becomes 50/51.

Try the reasoning

  1. What are b and ā at θ = (ln 9)/2?

    Show the answer and reasoning

    b = 101/45 and ā = 11/45.

    Their ratio is 11/101 and their difference remains 2.

  2. Can the fractional imbalance rise while the net yield remains fixed?

    Show the answer and reasoning

    Yes.

    Pair annihilation reduces the sum of the yields while preserving their difference.

  3. Does any finite θ make ā exactly zero in this continuous model?

    Show the answer and reasoning

    No.

    The exponential ratio is positive at finite exposure; zero is only a limiting value, and a finite stochastic-particle model would be a different experiment.

What to carry forward

  • Matter-sign evidence, abundance inference and an origin mechanism answer different questions.
  • Pair annihilation preserves a supplied baryon excess.
  • A larger fractional imbalance does not imply that more net baryon number was generated.

References

Sources supporting the explanations, physical distinctions and examples in this lesson.

  1. The origin of the matter–antimatter asymmetryMichael Dine; Alexander KusenkoTeaching resourcearxiv.orgRead source
  2. A Matter–Antimatter Universe?A. G. Cohen; A. De Rújula; S. L. GlashowResearch paperarxiv.orgRead source
  3. Planck 2018 results. VI. Cosmological parametersPlanck Collaboration (N. Aghanim et al.)Research paperarxiv.orgRead source
  4. One percent determination of the primordial deuterium abundanceRyan Cooke; Max Pettini; Charles C. SteidelResearch paperarxiv.orgRead source
  5. Observation of the 1S–2S transition in trapped antihydrogenM. Ahmadi et al. (ALPHA Collaboration; full byline retained in the primary preview)Research paperdoi.orgRead source