Analysis: 120 Orders of Magnitude Increase in Cosmological Constant

1. Understanding the Question

The cosmological constant Λ in Einstein's equations relates to dark energy density ρ_Λ by:

ρ_Λ = Λc²/8πG

In natural units (c = ℏ = 1), Λ has units of [length]⁻². The measured value is approximately:

Λ_obs ≈ 1.1 × 10⁻⁵² m⁻²

The Planck length is ℓ_P ≈ 1.6 × 10⁻³⁵ m, and the Planck epoch is t_P ≈ 5.4 × 10⁻⁴⁴ s.

The question considers Λ increased by 10¹²⁰ times its observed value, and asks whether the resulting expansion rate corresponds to timescales shorter than the Planck time.

2. Relating Λ to Expansion Rate

In a de Sitter universe dominated by Λ, the Hubble parameter is:

H = c√(Λ/3)

For the observed Λ:

H_obs ≈ 1.82 × 10⁻¹⁸ s⁻¹

Hubble time T_obs = 1/H_obs ≈ 5.5 × 10¹⁷ s (∼17 billion years)

3. Scaling Λ by 10¹²⁰

New cosmological constant:

Λ_new = 10¹²⁰ × Λ_obs ≈ 1.1 × 10⁶⁸ m⁻²

New Hubble parameter:

H_new ≈ 1.82 × 10⁴² s⁻¹

4. Resulting Timescale

New Hubble time:

T_new = 1/H_new ≈ 5.5 × 10⁻⁴³ s

Planck time t_P ≈ 5.4 × 10⁻⁴⁴ s

T_new is about 10 times longer than t_P

5. Planck Units Interpretation

In Planck units (ℓ_P = t_P = 1):

Λ_obs/ℓ_P⁻² ≈ 2.8 × 10⁻¹²²

Multiplying by 10¹²⁰ gives Λ ∼ ℓ_P⁻², so H ∼ t_P⁻¹

This means T_H ∼ t_P exactly

6. Conclusion

With Λ increased by 10¹²⁰ in Planck units, the Hubble time equals the Planck time. The expansion rate reaches the Planck scale, entering the quantum gravity regime.

Answer: Yes