Cosmology · The cosmological constant and other dark-energy models

Parameter constraints and model degeneracy

Solve two supplied response equations, track a shared calibration shift and separate local cosmological sensitivity from a finite probability experiment.

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

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After reading, you can

Identify which parameter combination one response constrains.
Propagate shared and individual errors through a two-response calculation.
State the model and statistical assumptions behind a parameter constraint.

Begin with two unknowns and one relation

Let p and q be two dimensionless parameter offsets. Supply a normalized noiseless response d₁ = 2p+q. If d₁ = 0, all points on q = −2p give the same response.

This is a degeneracy: one response does not distinguish changes that compensate in this way. It is a statement about the specified map from parameters to response, rather than evidence that the parameters have no physical meaning.

Now supply a second response d₂ = 5p+q. Its zero-response line q = −5p crosses the first at p = q = 0. Two differently directed sensitivities can resolve the degeneracy in this algebraic model.

Solving the two equations gives p̂ = (d₂−d₁)/3 and q̂ = (5d₁−2d₂)/3. The hats mark values inferred using the supplied response model; their reliability depends on what enters d₁ and d₂.

A physical sensitivity motivates the response

For a spatially flat model with separately conserved pressureless matter and a constant-w component, write F(a) = [H(a)/H_ref]² = Ω_m a⁻³+(1−Ω_m)a⁻³⁽¹⁺ʷ⁾. The reference rate H_ref is supplied and known in this example.

Choose the reference parameters Ω_m = 1/2 and w = −2/3, with a = 1 as the reference epoch. Define p = δΩ_m and q = (ln 2)δw. These are small offsets about that reference, rather than fitted values from observations.

At a = 1/2 the reference F is 5 and its first-order change is δF ≈ 6p+3q. At a = 1/4 the reference F is 34 and δF ≈ 60p+12q. Dividing the two changes by 3 and 12 motivates the two different response directions above.

The physical FLRW expression is nonlinear for finite parameter changes. The separate finite experiment below defines its response equations as exactly linear; it is not an exact finite substitution into F(a). Changing curvature, H_ref, the component set or evolving w would change the inference problem.

Give the calibration and noise a probability model

Define the finite experiment exactly by d₁ = 2p+q+s+e₁ and d₂ = 5p+q+s+e₂. Let the shared calibration shift s be equally likely to be −0.02 or +0.02, and each individual error e₁,e₂ be equally likely to be −0.01 or +0.01.

The three signs are independent. Therefore all eight sign combinations have probability 1/8. Set the true offsets to p = q = 0 for the displayed outcome table; every number is a supplied synthetic outcome, not a measured H value.

A shared s moves both responses equally. It cancels in their difference and hence in p̂, but it remains in q̂. Two correlated summaries do not automatically eliminate a common calibration error.

The response covariance is Cov(d) = [[0.0005, 0.0004], [0.0004, 0.0005]]. The off-diagonal entry comes from the shared shift and must be retained when propagating uncertainty.

All eight equiprobable outcomes of the exact finite linear experiment, at true p = q = 0
s	e₁	e₂	d₁	d₂	p̂	q̂
−0.02	−0.01	−0.01	−0.03	−0.03	0	−0.030000
−0.02	−0.01	0.01	−0.03	−0.01	0.006667	−0.043333
−0.02	0.01	−0.01	−0.01	−0.03	−0.006667	0.003333
−0.02	0.01	0.01	−0.01	−0.01	0	−0.010000
0.02	−0.01	−0.01	0.01	0.01	0	0.010000
0.02	−0.01	0.01	0.01	0.03	0.006667	−0.003333
0.02	0.01	−0.01	0.03	0.01	−0.006667	0.043333
0.02	0.01	0.01	0.03	0.03	0	0.030000

Different combinations inherit different uncertainty

Averaging all eight states gives zero mean for both inferred offsets. Their variances are Var(p̂) = 2/90000 and Var(q̂) = 65/90000, with Cov(p̂,q̂) = −7/90000.

The standard deviations are √2/300 ≈ 0.004714 and √65/300 ≈ 0.026874. For q = (ln 2)δw, the corresponding standard deviation of the inferred δw is √65/(300 ln 2), within this finite experiment.

Holding s fixed at zero would reduce Var(q̂) to 29/90000. That would be a different assumed experiment, not evidence that the original shared error had disappeared. Var(p̂) is unchanged because the shared term cancels algebraically.

Covariance records second moments; it does not turn the eight states into a Gaussian distribution. An interval one standard deviation wide on each side of the true value contains only half of this experiment’s outcomes for either individual inferred parameter, rather than an automatic 68 percent.

This original example first supplies two normalized noiseless linear responses d₁ = 2p+q and d₂ = 5p+q, with dimensionless offsets p and q. Panel A plots their zero-response lines q = −2p and q = −5p on linear axes p from −0.02 to 0.02 and q from −0.05 to 0.05. The steeper line is clipped geometrically to the actual displayed q domain: its endpoints are p = −0.01, q = 0.05 and p = 0.01, q = −0.05. The shallower line extends from (−0.02, 0.04) to (0.02, −0.04). Solid blue and dashed orange distinguish the lines, which cross at the directly marked origin. The box is a supplied drawing domain, not a confidence region. The physical motivation is a flat separately conserved matter plus constant-w expansion with known supplied H_ref. F(a) = [H(a)/H_ref]² = Ω_m a⁻³+(1−Ω_m)a^(−3(1+w)), evaluated at reference Ω_m = 1/2, w = −2/3. With p = δΩ_m and q = (ln 2)δw, first-order δF at a = 1/2 is 6p+3q, while at a = 1/4 it is 60p+12q; reference F values are 5 and 34. Dividing by 3 and 12 motivates only the local linear directions. Finite parameter changes in that physical FLRW expression are nonlinear; the following finite experiment is defined independently as exactly linear. It has d₁ = 2p+q+s+e₁ and d₂ = 5p+q+s+e₂. Three independent equally likely signs give s = ±0.02 and e₁,e₂ = ±0.01, hence eight states each with probability 1/8. True p = q = 0 is stipulated for Panel B. State 1: (s, e₁, e₂) = (-0.02, -0.01, -0.01); (d₁, d₂) = (-0.03, -0.03); (p̂, q̂) = (0, -0.03). State 2: (s, e₁, e₂) = (-0.02, -0.01, 0.01); (d₁, d₂) = (-0.03, -0.01); (p̂, q̂) = (0.006667, -0.043333). State 3: (s, e₁, e₂) = (-0.02, 0.01, -0.01); (d₁, d₂) = (-0.01, -0.03); (p̂, q̂) = (-0.006667, 0.003333). State 4: (s, e₁, e₂) = (-0.02, 0.01, 0.01); (d₁, d₂) = (-0.01, -0.01); (p̂, q̂) = (0, -0.01). State 5: (s, e₁, e₂) = (0.02, -0.01, -0.01); (d₁, d₂) = (0.01, 0.01); (p̂, q̂) = (0, 0.01). State 6: (s, e₁, e₂) = (0.02, -0.01, 0.01); (d₁, d₂) = (0.01, 0.03); (p̂, q̂) = (0.006667, -0.003333). State 7: (s, e₁, e₂) = (0.02, 0.01, -0.01); (d₁, d₂) = (0.03, 0.01); (p̂, q̂) = (-0.006667, 0.043333). State 8: (s, e₁, e₂) = (0.02, 0.01, 0.01); (d₁, d₂) = (0.03, 0.03); (p̂, q̂) = (0, 0.03). The rounded coordinates in this text use six decimals; actual calculation uses exact fractions and saved points retain six-pixel-decimal precision. Panel B uses the same linear p,q drawing ranges for inferred p̂ and q̂. Blue circles identify negative s and orange squares positive s; direct state numbers distinguish individual outcomes without color. The inverse response is p̂ = (d₂−d₁)/3 and q̂ = (5d₁−2d₂)/3. A common s cancels in p̂ and remains as s in q̂. Cov(d) has diagonal 0.0005 and off-diagonal 0.0004. Parameter variances are 2/90000 and 65/90000, with covariance −7/90000. Their STDs are √2/300 and √65/300. For either individual inferred parameter, the interval from true value minus its STD to true value plus STD contains exactly four of the eight equally likely outcomes. This is 1/2 repeated-experiment marginal coverage, not an automatic 68 percent, confidence construction or posterior probability. No Gaussian contour, credible band, real observation, parameter constraint, preferred mechanism or candidate discovery is displayed. Changing flatness, H_ref, component set, evolving w or nuisance assumptions changes the physical inference. Original Ark teaching model.

A parameter interval is conditional evidence

In a real analysis, calibrated observables and a stated probability model replace this supplied table. The response may be nonlinear, the nuisance parameters uncertain, and additional observables correlated.

A parameter constraint therefore needs its model family, assumed data response, calibration treatment, priors or repeated-experiment construction, and covariance. Removing those conditions changes what the reported interval means.

The local FLRW derivatives here explain why different epochs can constrain different combinations. They do not supply observed expansion data, a posterior, a confidence region, a model-selection result or a microscopic identification of dark energy.

Use complementary probes to test different physical responses and common systematics explicitly. A narrower conditional parameter spread is useful, while deciding between complete physical mechanisms requires their additional predictions and evidence.

Worked reasoning

Add the same calibration shift to both responses

What does adding s to d₁ and d₂ do to the inferred offsets?

The difference d₂−d₁ stays unchanged, so p̂ does not change.
The numerator 5d₁−2d₂ increases by 3s.
Therefore q̂ increases by s and retains the shared variance 0.02².

The common shift cancels from p̂ but survives fully in q̂.

Try the reasoning

Does d₁ = 0 alone require p = q = 0 in the noiseless map?
Show the answer and reasoning
No: every point q = −2p gives d₁ = 0.
A single relation leaves a degeneracy line in the two-parameter plane.
For d₁ = −0.03 and d₂ = −0.01, find the inferred offsets.
Show the answer and reasoning
p̂ = 1/150 ≈ 0.006667 and q̂ = −13/300 ≈ −0.043333.
Use the two stated inverse response equations; the rounded table follows those exact fractions.
Does a covariance matrix alone justify Gaussian 68 percent intervals?
Show the answer and reasoning
No.
The full distribution here has eight discrete states, and its actual one-standard-deviation marginal coverage is one half.

What to carry forward

Different response directions can resolve a conditional parameter degeneracy.
A shared calibration can cancel in one inferred combination and remain in another.
Local physical sensitivity, finite response assumptions and statistical coverage must be stated separately.

References

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