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Cosmology Fine Tuning Problem
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[QUOTE="TRB8985, post: 6052863, member: 355135"] [h2]Homework Statement [/h2] Suppose (incorrectly) that H scales as temperature squared all the way back until the time when the temperature of the universe was 10[SUP]19[/SUP] GeV/k[SUB]B[/SUB] (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[SUB]Λ[/SUB] today is equal to 0.7*ρ[SUB]critical[/SUB] and ρ[SUB]Λ[/SUB] remains constant throughout the history of the universe. What was ρ[SUB]Λ[/SUB] / (3H[SUP]2[/SUP]/8πG) back then? (From [I]Modern Cosmology[/I] by Dodelson, pg. 25) [h2]Homework Equations[/h2] ρ_critical = (3H[SUB]0[/SUB][SUP]2[/SUP]/8πG) T = 10[SUP]19[/SUP] GeV/k[SUB]B[/SUB] = 1.16045* 10[SUP]32[/SUP] K T[SUB]0[/SUB] = 2.725 K For a radiation-dominated universe, a ∝ t[SUP]1/2[/SUP]. [h2]The Attempt at a Solution[/h2] I understand a part of the solution wherein ρ[SUB]Λ[/SUB] / ρ[SUB]critical[/SUB] = 0.7, but I'm supposed to multiply this value by something. In the answer key, Dodelson multiplies 0.7 by the ratio of (H[SUB]0[/SUB] / H)[SUP]2[/SUP]. The text states: "[I]By assumption, the universe is forever radiation dominated (clearly not true today, but a good approximation early on), so H / H[SUB]0[/SUB] = a[SUP]-2[/SUP]."[/I] Given this, the inverse of H / H[SUB]0[/SUB] would result in a[SUP]2[/SUP], and since H scales as temperature squared, then (a[SUP]2[/SUP])[SUP]2[/SUP] = a[SUP]4[/SUP] 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 H[SUB]0[/SUB] / 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.) [/QUOTE]
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