Cosmology · Consistency, tensions and claims of new physics
Comparing structure-growth estimates
Separate bias, growth response and matter-power amplitude in a supplied linear redshift-space model before comparing inferred growth quantities.
Preparation: ratios, powers of ten and ordinary arithmetic. Physical quantities and notation are introduced as they are needed.
Back to the chapterAfter reading, you can
- Distinguish a power amplitude from a growth response and tracer bias.
- Derive an exact degeneracy in a supplied angular power spectrum.
- State the assumptions needed to compare structure-growth estimates.
A growth estimate refers to a defined response
A clustering measurement does not directly label every contribution as matter amplitude, tracer bias or growth rate. Inferring these quantities requires a relation between matter, velocities and the measured tracer.
In a linear single-growth-factor description, write the positive growth factor as D(a) and its logarithmic derivative as f = d ln D/d ln a. The value of f at one epoch does not specify the full history D(a).
A comparison must also match scale, epoch, calibration and the quantity being estimated. An amplitude at one wavenumber, a smoothed variance and a growth-rate product are different summaries.
The following supplied ensemble-power model illustrates an exact degeneracy. Its numbers are teaching inputs and carry no observed growth discrepancy, uncertainty distribution or empirical likelihood.
Follow density and velocity into redshift space
At one epoch and one fixed positive comoving wavenumber k, assume linear perturbations, the plane-parallel limit, irrotational velocities, deterministic tracer bias δ_g = bδ_m and no velocity bias. Take b>0 and f≥0.
Use the linear continuity relation θ_v = −fδ_m, with the dimensionless velocity-divergence normalization absorbed into θ_v. For direction cosine μ to the line of sight, the redshift-space response is δ_s = (b+fμ²)δ_m.
For a homogeneous isotropic matter ensemble, define δ(k) = ∫d³x δ(x)e^(−ik·x) and ⟨δ(k)δ(k′)⟩ = (2π)³δ_D(k+k′)P_m(k). The Fourier coefficient has units of comoving volume and P_m has units of comoving m³.
It follows that P_s(k,μ) = (b+fμ²)²P_m(k). This ensemble-power relation is conditional on the stated linear response; no distribution of individual Fourier coefficients has been supplied.
The ideal angular response fixes two products
Choose a positive reference power P_ref(k) in the same units and define S = √[P_m(k)/P_ref(k)]. This S is a dimensionless amplitude at the selected k.
Supply the positive transverse amplitude T = √[P_s(k,0)/P_ref(k)] = 0.10 and radial amplitude R = √[P_s(k,1)/P_ref(k)] = 0.14. Then bS = 0.10 and fS = R−T = 0.04.
Consequently β = f/b = 0.4. These products determine the complete ideal angular power P_s/P_ref = (0.10+0.04μ²)² without separately determining b, f and S.
Different positive S values therefore give the same full angular curve when b and f change consistently. Treating the three cases below as three independently measured angular spectra would misrepresent their exact equality.
| S | P_m/P_ref | b = 0.10/S | f = 0.04/S | β |
|---|---|---|---|---|
| 1/25 = 0.04 | 1/625 | 5/2 | 1 | 2/5 |
| 1/20 = 0.05 | 1/400 | 2 | 4/5 | 2/5 |
| 2/25 = 0.08 | 4/625 | 5/4 | 1/2 | 2/5 |
Square the amplitude to obtain the power
At μ = 0,1/2,1 the amplitude √(P_s/P_ref) is respectively 0.10,0.11,0.14. The corresponding power ratios are 0.0100,0.0121,0.0196.
At μ = 1/2, μ² = 1/4. The response is therefore 0.10+0.04/4 = 0.11 before squaring; replacing μ² with μ would give a different model.
P_m, P_s and P_ref all have comoving-volume units, so their ratios and S are dimensionless. S is neither a Fourier coefficient nor the real-space density contrast.
A small S does not by itself establish linearity because its magnitude depends on the chosen P_ref. Linearity and the validity of the velocity response are assumptions of this example, not consequences of its normalized numbers.
| μ | μ² | √(P_s/P_ref) | P_s/P_ref |
|---|---|---|---|
| 0 | 0 | 0.10 | 0.0100 |
| 1/2 | 1/4 | 0.11 | 0.0121 |
| 1 | 1 | 0.14 | 0.0196 |
Additional information must break a defined degeneracy
An external constraint on S would select b and f within this ideal response. Without it, the measured products permit a family of factorizations. An additional observable helps only through a stated and tested response to the unknowns.
The selected-k amplitude S is not σ₈, which involves smoothing an entire matter power spectrum over a specified radius. It is also not S₈ or an empirically measured fσ₈.
Real comparisons must account for survey selection, geometry, nonlinear evolution, velocity effects, scale-dependent bias and shared calibration where relevant. Those additions require their own models and uncertainty treatment.
Changing the inferred f alone does not establish a gravity theory or a viable alternative growth history. A physical proposal must connect the expansion, perturbation and observational responses under coherent assumptions.
Worked reasoning
Recover one factorization and its angular power
For supplied S = 1/20, find b, f and P_s/P_ref at μ = 1/2.
- Use bS = 0.10 to obtain b = 2, and fS = 0.04 to obtain f = 4/5.
- At μ = 1/2, the normalized amplitude is S[b+f/4] = (1/20)(2+1/5) = 0.11.
- Square the amplitude: P_s/P_ref = 0.11² = 0.0121.
b = 2, f = 4/5, and the power ratio is 0.0121.
Try the reasoning
What b and f give the same curve when S = 2/25?
Show the answer and reasoning
b = 5/4 and f = 1/2.
The products remain bS = 1/10 and fS = 1/25.
Does this one-k S equal a measured σ₈?
Show the answer and reasoning
No.
σ₈ requires a specified smoothing integral over the matter power spectrum; the example supplies a ratio at one k.
Does choosing another factorization establish a new theory of gravity?
Show the answer and reasoning
No.
The factorization supplies no complete dynamics, growth history or additional observable predictions.
What to carry forward
- Compare quantities defined at the same epoch and scale under stated responses.
- An angular clustering response can determine products while leaving factors degenerate.
- A selected-scale power amplitude is different from a smoothed variance or a physical explanation.
References
Sources supporting the explanations, physical distinctions and examples in this lesson.