Cosmology Fine Tuning Problem

[TRB8985]

Homework Statement

Suppose (incorrectly) that H scales as temperature squared all the way back until the time when the temperature of the universe was 10¹⁹ GeV/k_B (i.e., suppose the universe was radiation dominated all the way back to the Planck time).

Also suppose that today the dark energy is in the form of a cosmological constant Λ, such that ρ_Λ today is equal to 0.7*ρ_critical and ρ_Λ remains constant throughout the history of the universe. What was ρ_Λ / (3H²/8πG) back then?

(From Modern Cosmology by Dodelson, pg. 25)

Homework Equations

ρ_critical = (3H₀²/8πG)

T = 10¹⁹ GeV/k_B = 1.16045* 10³² K

T₀ = 2.725 K

For a radiation-dominated universe, a ∝ t^1/2.

The Attempt at a Solution

I understand a part of the solution wherein ρ_Λ / ρ_critical = 0.7, but I'm supposed to multiply this value by something.

In the answer key, Dodelson multiplies 0.7 by the ratio of (H₀ / H)². The text states:

"By assumption, the universe is forever radiation dominated (clearly not true today, but a good approximation early on), so H / H₀ = a^-2."

Given this, the inverse of H / H₀ would result in a², and since H scales as temperature squared, then (a²)² = a⁴ which can then be applied to the ratio of the temperature. That latter part makes sense. However, I'm not quite understanding where Dodelson pulled the ratio of H₀ / H from to get things started.

Could anyone provide any insight on this? Thank you very much for your help.

(This question is being attempted via an independent study and not a homework question. Additionally, there are no cosmology specialists at my university who could provide any useful feedback on how to attack this situation.)

 

[George Jones]

Homework Statement

Suppose (incorrectly) that H scales as temperature squared ...

Multiply the expression in the exercise by one in the form $1 = \rho_{cr} / \rho_{cr}$, then use equation (1.3) to substitute for $\rho_{cr}$ in the numerator (but not in the denominator).

 

[TRB8985]

George, your input was incredibly helpful and brought the entire picture together. Thank you so much for your help! I appreciate that.