- #1
Labboi
- 5
- 0
- Homework Statement
- Find H(T)
- Relevant Equations
- H^2 = 8πG/3 *ρR = 4π^3G/ 45 g∗*T^4
Here's the problem:
It is more common to define the “effective number of relativistic degrees of freedom” by writing the total radiation energy as ρR = ργ + ρν + ρe± = (π^2/30) g∗*T^4 , where g∗ = 2 + 7/8(6 + 4) = 43/4 . (1.52) With this, the expansion rate during the radiation era is given by H^2 = 8πG/3 *ρR = 4π^3G/ 45 g∗*T^4. Therefore, when we calculate the expansion rate during the radiation era, we must be careful about how many relativistic degrees of freedom we have in the universe at a given time. For g∗ = 43/4, we obtain 1/H(T) = 1.48 1 (MeV/T^2) sec. (1.54) As the age of the universe during the radiation era is t = 1/(2H), we also have t = 1 2H(T) = 0.74 1 MeV T 2 sec
I don't see how he get's the 1.48. I'm trying to solve H(T) at 1 MeV. But I can't seem to get the right number. When I plug in the numbers I get 5.44.
It is more common to define the “effective number of relativistic degrees of freedom” by writing the total radiation energy as ρR = ργ + ρν + ρe± = (π^2/30) g∗*T^4 , where g∗ = 2 + 7/8(6 + 4) = 43/4 . (1.52) With this, the expansion rate during the radiation era is given by H^2 = 8πG/3 *ρR = 4π^3G/ 45 g∗*T^4. Therefore, when we calculate the expansion rate during the radiation era, we must be careful about how many relativistic degrees of freedom we have in the universe at a given time. For g∗ = 43/4, we obtain 1/H(T) = 1.48 1 (MeV/T^2) sec. (1.54) As the age of the universe during the radiation era is t = 1/(2H), we also have t = 1 2H(T) = 0.74 1 MeV T 2 sec
I don't see how he get's the 1.48. I'm trying to solve H(T) at 1 MeV. But I can't seem to get the right number. When I plug in the numbers I get 5.44.
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