Cosmology · Spacetime, curvature and cosmological models
Critical density, curvature and model parameters
Use a common critical-density denominator to compare three hypothetical reference-epoch accounts, distinguish energy shares from density parameters and keep the curvature sign convention explicit.
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
- Define critical energy density and a component density parameter with consistent units.
- Apply the stated GR/FLRW reference-epoch closure relation including vacuum once.
- Distinguish a ratio to critical density from a fraction of total physical energy density.
Critical density belongs to a rate and a model
A critical density is a useful reference quantity in a specified cosmological model. Here use general relativity with an FLRW background, a reference scale factor a_ref = 1 and a chosen positive reference expansion rate H_ref. No measured present-day values are supplied.
Write critical energy density as ε_crit = 3c²H_ref²/(8πG), in J/m³. It is the total energy density corresponding to zero spatial curvature at that rate under the stated Friedmann equation and component convention.
A mass-density version is ρ_crit = 3H_ref²/(8πG), in kg/m³. Multiplying that mass-density quantity by c² gives ε_crit. The two formulas use different density units; placing kilograms per cubic metre into the energy-density formula without conversion would be an error.
The reference quantity changes if the chosen epoch and its H change. It is not a fixed universal material density that is identical throughout every possible history. A quoted parameter must therefore identify its reference epoch and model.
Density parameters need not sum to one by themselves
Define each physical component parameter as Ω_i = ε_i/ε_crit at this reference epoch. All components use the same critical-density denominator. Let Ω_total be their sum, including vacuum energy if it is present in this convention.
Treat the cosmological constant as an equivalent vacuum component already included in ε_total. Do not also add a separate explicit Λ term. With dimensionless a and K = k/R_ref² in m⁻², the reference Friedmann equation becomes 1 = Ω_total − Kc²/H_ref².
Define Ω_k = −Kc²/H_ref² at a_ref = 1. Then Ω_total + Ω_k = 1. Positive Ω_k corresponds to negative spatial curvature, and negative Ω_k corresponds to positive spatial curvature. The minus sign is part of the definition, not a statement that curvature is a negative amount of matter.
Ω_k is an effective geometric term in this algebraic account. It is not an identified physical energy species whose particles must be added to the physical component total. The relation uses all included physical components, not only matter.
Keep the reference rate fixed and vary the contents
Compare three separate hypothetical reference accounts at the same positive H_ref. Set Ω_m = 2/5 in each, and choose vacuum parameters Ω_vac = 2/5, 3/5 or 4/5. Radiation and other components are omitted from these chosen examples, not declared absent from our actual Universe.
The total physical critical-density ratios are respectively 4/5, 1 and 6/5. The closure relation gives Ω_k = +1/5, 0 and −1/5. Under the stated conventions their spatial curvatures are negative, zero and positive.
Matter alone is below the critical benchmark in all three examples. Comparing Ω_m with 1 therefore cannot distinguish their curvature signs. The other physical components contribute to the density appearing in the equation.
These are choices illustrating the equation, not observational fits or probabilities assigned to curvature. Their common denominator makes a bar comparison meaningful. Renormalizing every physical total to one before plotting would conceal the very difference being examined.
| Ω_m | Ω_vac | Ω_total | Ω_k = 1 − Ω_total | Spatial curvature |
|---|---|---|---|---|
| 2/5 | 2/5 | 4/5 | +1/5 | Negative |
| 2/5 | 3/5 | 1 | 0 | Zero |
| 2/5 | 4/5 | 6/5 | −1/5 | Positive |
Geometry still needs its own scope
At a general epoch the corresponding geometric parameter is Ω_k(t) = −Kc²/[a(t)²H(t)²]. It can change with time even when K is fixed. The reference values in the table should not be treated as constants at every epoch without evolving the model.
An optional length check uses the instantaneous nonflat curvature-radius magnitude R_curv = a/√|K|. At the chosen reference epoch this is (c/H_ref)/√|Ω_k|. For either nonflat example, |Ω_k| = 1/5, so R_curv = √5 times c/H_ref. The flat example has no finite curvature radius from this expression.
That curvature-radius calculation does not supply global topology or total spatial volume. The previous flat periodic construction already shows why local curvature alone cannot establish infinite extent.
Nor does a curvature sign alone settle future fate when contents and pressure behaviour are not fully specified. Use the expansion dynamics and conditions for that question. Critical-density closure is a precisely defined relation within the stated model, rather than a replacement for the whole history.
Worked reasoning
Calculate the geometric term and the actual matter share
At the stated reference epoch, Ω_m = 2/5 and Ω_vac = 4/5. What are Ω_k and matter’s share of the two-component physical total?
- Add the physical critical-density ratios: Ω_total = 2/5 + 4/5 = 6/5.
- Closure gives Ω_k = 1 − 6/5 = −1/5, corresponding to positive spatial curvature in this convention.
- Normalize by the physical total: Ω_m/Ω_total = (2/5)/(6/5) = 1/3.
Ω_k = −1/5; matter is 1/3 of the specified physical total, although Ω_m is 2/5.
Try the reasoning
May Ω_m alone determine the curvature sign when vacuum or other components contribute?
Show the answer and reasoning
No.
The stated Friedmann closure relation uses the total included physical energy density.
If vacuum is included in Ω_total, should its explicit Λ contribution be added again?
Show the answer and reasoning
No.
That would count the same contribution twice.
Do all physical Ω_i necessarily sum to one in a nonflat model?
Show the answer and reasoning
No.
They use critical density as the denominator; Ω_total + Ω_k = 1 in the stated convention. Physical shares instead divide by ε_total.
What to carry forward
- Critical density uses an epoch’s expansion rate and a stated gravitational/component convention.
- Density parameters and physical energy shares have different denominators.
- The geometric curvature term needs its sign convention and does not alone fix topology or future fate.
References
Sources supporting the explanations, physical distinctions and examples in this lesson.