The first Friedmann equation can be rewritten as Ω_total - 1 = kc²/(a²H²), where Ω_total = ρ/ρ_crit. What is the critical density ρ_crit, and what does it determine?
Aρ_crit = 3H²/(8πG) — it determines whether the universe is spatially flat (k=0), positively curved (k=+1), or negatively curved (k=-1)
Bρ_crit = c²/(8πG) — it determines the age of the universe
Cρ_crit = H²/(4πG) — it determines whether the universe will expand forever
Dρ_crit = 3H/(8πG) — it determines the deceleration parameter
The critical density ρ_crit = 3H²/(8πG) is the density at which the universe is spatially flat (k = 0). If ρ > ρ_crit, the universe is positively curved (k = +1, closed); if ρ < ρ_crit, it is negatively curved (k = -1, open). The density parameter Ω = ρ/ρ_crit is the central observable in cosmology. Current observations give Ω_total ≈ 1.000 ± 0.002, indicating the universe is very close to spatially flat. The present critical density is about 9.5 × 10⁻²⁷ kg/m³ — roughly 6 hydrogen atoms per cubic meter.
Question 2 True / False
In a universe containing only matter (p = 0, Λ = 0, k = 0), the scale factor grows as a(t) ∝ t^{2/3}.
TTrue
FFalse
Answer: True
For a flat, matter-dominated universe with Λ = 0, the first Friedmann equation gives (ȧ/a)² = (8πG/3)ρ. Since matter density dilutes as ρ ∝ a⁻³ (volume expansion), this becomes ȧ² ∝ a⁻¹, which integrates to a(t) ∝ t^{2/3}. The expansion decelerates (ä < 0) because gravity slows it down. For comparison, a radiation-dominated universe (p = ρc²/3) gives a(t) ∝ t^{1/2}, and a cosmological-constant-dominated universe gives a(t) ∝ exp(Ht), exponential expansion.
Question 3 Short Answer
Explain why the second Friedmann equation shows that ordinary matter and radiation always decelerate the expansion, while a cosmological constant accelerates it.
Think about your answer, then reveal below.
Model answer: The acceleration equation ä/a = -(4πG/3)(ρ + 3p/c²) + Λ/3 shows that the contribution of matter/radiation to ä is proportional to -(ρ + 3p/c²). For ordinary matter (p ≥ 0), ρ + 3p/c² > 0, so the matter contribution is negative — the expansion decelerates. For radiation (p = ρc²/3), the deceleration is even stronger. The cosmological constant Λ contributes +Λ/3, which is positive (for Λ > 0) and drives acceleration. Equivalently, a cosmological constant acts like a fluid with equation of state p = -ρc², giving ρ + 3p/c² = ρ - 3ρ = -2ρ < 0, which means its 'gravitational' effect is repulsive. The observed acceleration of cosmic expansion (discovered 1998) requires Λ > 0 or an equivalent dark energy component.
The key insight is that in GR, pressure gravitates — and negative pressure (tension) produces gravitational repulsion. A cosmological constant has maximally negative pressure (p = -ρc²), making it the most efficient driver of accelerating expansion. This is why the universe's expansion is accelerating despite the decelerating effect of matter.
Question 4 Short Answer
Derive how the energy density of radiation scales with the scale factor a(t), starting from the continuity equation dρ/dt + 3(ȧ/a)(ρ + p/c²) = 0.
Think about your answer, then reveal below.
Model answer: For radiation, p = ρc²/3. Substituting into the continuity equation: dρ/dt + 3(ȧ/a)(ρ + ρ/3) = dρ/dt + 4(ȧ/a)ρ = 0. This gives dρ/ρ = -4 da/a, which integrates to ρ ∝ a⁻⁴. The a⁻³ factor comes from the dilution of photon number density as the volume expands, and the additional a⁻¹ factor comes from the cosmological redshift — each photon loses energy as its wavelength stretches with the expansion: E = hν ∝ a⁻¹. For comparison, matter (p = 0) gives ρ ∝ a⁻³ (volume dilution only), and a cosmological constant (p = -ρc²) gives ρ = const (energy density of vacuum does not dilute).
The different scaling laws — ρ_matter ∝ a⁻³, ρ_radiation ∝ a⁻⁴, ρ_Λ = const — explain why the universe passes through radiation-dominated, matter-dominated, and dark-energy-dominated eras as it expands. Each era has a different expansion law a(t) determined by the Friedmann equations.