Cosmology · Matter–antimatter asymmetry and hypothetical relics

Connecting proposals to testable consequences

Trace a conditional source scaling through a supplied transport and spectrum, then compare exact band responses without turning a template match into proof of baryogenesis.

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

Keep production, propagation and a declared comparison rule separate.
Calculate a source-parameter degeneracy and finite logarithmic-band areas.
State exactly what a selected template passing or failing can establish.

A proposed mechanism needs an observable consequence

A baryogenesis or relic scenario becomes testable through its predicted connections: particle reactions, annihilation radiation, a defect signal or another specified observable. The relevant prediction depends on the actual mechanism.

A gravitational-wave spectrum from a proposed first-order transition also needs a source model, evolution and propagation. A resemblance between a spectrum and data would not by itself determine how baryon number was generated.

Field-and-fluid calculations can motivate conditional scaling relations. Their source lifetime, velocity, length and approximation regime must remain visible when such a relation is used.

The supplied synthetic comparison below treats one conditional scaling as an exact teaching definition, then adds its own transport and triangular spectrum. No measured background, instrument sensitivity or current detection is asserted.

Dimensionless source combinations expose a degeneracy

Let H have units s⁻¹, source duration τ have units s, characteristic length R have units m and U = v_rms/c be a small dimensionless fluid speed. Supply w = 1/3 and a dimensionless correlator coefficient equal to 1.

Use Ω_* = [3/(4π²)](Hτ)(HR/c)(1+w)²U⁴ = [4/(3π²)](Hτ)(HR/c)U⁴ as the stipulated production energy fraction. The factor HR/c is dimensionless; HR alone has speed units.

Scenario A has Hτ = 1/16, HR/c = 1/100 and U = 1/10. Scenario B keeps HR/c fixed, increases Hτ to 1/4 and reduces U to 1/(10√2).

Because U⁴ falls by a factor of four while the lifetime factor rises by four, A and B have the same Ω_* = 1/(12,000,000π²). Scenario C keeps A’s lifetime and velocity but doubles R, giving twice that production fraction.

Supplied dimensionless source parameters and imposed peak frequencies
Scenario	Hτ	HR/c	U	Ω_/Ω_A	f_peak
A	1/16	1/100	1/10	1	1 Hz
B	1/4	1/100	1/(10√2)	1	1 Hz
C	1/16	1/50	1/10	2	1/2 Hz

Production and the present spectrum are different quantities

For the same supplied production epoch and background, impose a dimensionless transport factor S = 1/1000 and define Ω₀ = SΩ_*. Ω₀ is normalized to today’s critical energy density; S includes background normalization as well as redshifting and is not just a⁻⁴.

Let x = ln(f/f_ref), with f_ref = 1 Hz, and supply the triangular shape h(x;x_p) = max(1−|x−x_p|,0), where x_p = ln(f_peak/f_ref). Its area over x is exactly 1.

Define P₀(f) = dΩ₀/d ln f = Ω₀h(x;x_p). The integral over logarithmic frequency is Ω₀; P₀ is not strain amplitude or energy per unit frequency.

Prescribe the same 1 Hz peak for A and B. For C, impose the inverse-size frequency rule so doubling R moves the peak to 1/2 Hz. The triangle, S and frequency rule are original diagnostic inputs, not a solved physical spectral template.

Finite bands can miss part of a real template

Supply ideal bands x∈[−1,0] and x∈[0,1]. A and B each give area Ω₀A/2 in each band, with no support outside their combined interval.

For C let L = ln 2, so its peak is at x = −L and its full support is [−1−L,1−L]. Its normalization is 2Ω₀A.

Piecewise integration gives left-band area Ω₀A(1+2L−2L²), right-band area Ω₀A(1−L)² and unobserved low-frequency area Ω₀A L². All three areas sum to the full 2Ω₀A.

The unobserved area is not zero emission. A comparison restricted to two bands must retain the supplied spectrum and its support instead of silently equating the measured-band sum with the entire signal.

Exact template areas in units Ω₀A, with numerical values rounded for reading
Scenario	Left band	Right band	Outside the two bands	Full area
A	1/2	1/2	0	1
B	1/2	1/2	0	1
C	1+2L−2L² ≈ 1.425388	(1−L)² ≈ 0.094159	L² ≈ 0.480453	2

A supplied synthetic production scaling is proportional to (H tau)(H R/c)U^4. A and B have different lifetime and speed but the same production fraction, the same imposed transport S = 1/1000 and the same 1 Hz peak; their complete spectra coincide and are drawn as one line. C doubles the source size, so its supplied normalization doubles and its imposed peak shifts to 1/2 Hz. The exact triangular templates are drawn against x = ln(f/1 Hz), with vertical P0/ Omega0A, where P0 = dOmega0/d ln f. They are not detector strain or energy per Hz. Vertical guides delimit the declared bands [−1,0] and [0,1]. C extends to lower x outside both bands, with area (ln 2)^2 in the common units. The complete area table retains both observed bands and the unobserved part. Template comparison uses a declared deterministic tolerance of 0.1 per band; it supplies no likelihood, confidence interval, detection or baryon yield. Supplied source parameters and imposed frequency rule; U is a small speed divided by c. Columns: Case; Hτ; HR/c; U; Ω*/Ω*A; Peak Hz. Rows: A; 1/16; 1/100; 1/10; 1; 1 / B; 1/4; 1/100; 1/(10√2); 1; 1 / C; 1/16; 1/50; 1/10; 2; 1/2. All logarithmic-band areas in units Ω₀A; outside plus both bands equals the full area. Columns: Case; Left area; Right area; Outside; Full area. Rows: A; 0.500000; 0.500000; 0.000000; 1 / B; 0.500000; 0.500000; 0.000000; 1 / C; 1.425388; 0.094159; 0.480453; 2.

A template test leaves both scope and degeneracy

Supply synthetic reference readouts (1/2,1/2) in units Ω₀A. Predeclare that a template passes only if the absolute residual in each band is at most 1/10 in those units.

A and B both pass exactly and remain indistinguishable in these two channels. C fails both bands: its left residual is about +0.925388 and its right residual about −0.405841 relative to the references.

This is a deterministic tolerance rule, not a confidence interval, likelihood or discovery threshold. Changing a template parameter or a background assumption requires its own new forward comparison.

C failing does not disprove all first-order transitions or baryogenesis. A and B matching does not identify their microscopic origin or baryon yield; additional linked predictions and independently assessed uncertainties are still needed.

Worked reasoning

Compare A and B before looking at their spectra

Why do A and B have the same source fraction?

At fixed H and R, B’s Hτ is four times A’s.
Its U is smaller by √2, so U⁴ is smaller by a factor of four.
The two changes cancel in Ω_*; with the same supplied S and peak they also give the same complete triangular spectrum.

They are exactly degenerate under the supplied source scaling and spectral response.

Try the reasoning

At fixed other inputs, what factor follows from doubling U within the assumed regime?
Show the answer and reasoning
16.
The stipulated amplitude scales as U⁴; a larger U must still satisfy the regime assumptions before applying that scaling.
Why is HR/c needed when H is in s⁻¹ and R in metres?
Show the answer and reasoning
It is dimensionless.
HR has units m/s; dividing by c removes those units.
Does C failing this rule disprove baryogenesis or first-order phase transitions?
Show the answer and reasoning
No.
It rejects only this supplied template under this deterministic rule, transport and background; other mechanisms need their own predicted tests.

What to carry forward

A test follows the full chain from source dynamics to an observable response.
Different source parameters can produce the same specified spectrum.
A selected template comparison has a narrower meaning than confirming or excluding a whole class of mechanisms.

References

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