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Buzz Bloom
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- I assume the temperature of the CMB and H atoms both start at about the same temperature of 2491 K at the time decoupling completed when the universe was 487,000 years old. The RMS value of the velocity of hydrogen atoms was 7852 m/s.
Q1: Was there any significant subsequent interactions of the hydrogen atoms with the CMB photons?
Q2: How did the H atom temperature vary with time due to the expansion of the universe?
I begin with the values of certain variables obtained from the references cited.
I am guessing there was no significant interaction between H atoms and the CMB after decoupling completed, but I have no confidence that this is correct. If someone knows of a reference discussing this question, I would much appreciate seeing a link.
My math in the remainder of this post (if it has no errors) shows that the temperature of H atoms does not vary with the expansion of the universe in the same way that CMB temperature varies. If this is correct, then any batch of hydrogen which did not end up in galaxies might have a current temperature approximately the same as that predicted by the effect of the universe expansion on the H temperature.
The calculated temperature results for today (a=1) are as follows.
The remainder of this post is the calculations for TH(a=1).
It is assumed that the temperature of both CMB and H atoms were the same, at the time that decoupling completed.
Let x be the coordinate in which an H atom moves. V is the speed which an atom is moving:
The value for C is calculated from Eq 10:
Ωm = 0.315
H0 = 70.0 (km/s)Mpc = 1/(4.408 × 1017) s
H0 was calculated from [2] as a weighted average of 30 values dated between 2001 and 2020. I used the inverse square of the error range as weights.tdc ~= 487,000 yr = 1.537 × 1013 s
". . . decoupling took place over roughly 115,000 years, and when it was complete, the universe was roughly 487,000 years old."
Note: 1 yr = 3.1556952 × 107 s.zdc = 913
This was calculated using [4] with values H = 70.0 varying z until the age was 487,000 years.adc = 0.001094
This was calculated using a = 1/(1+z).TCMB(a=1) = 2.725 K
I am guessing there was no significant interaction between H atoms and the CMB after decoupling completed, but I have no confidence that this is correct. If someone knows of a reference discussing this question, I would much appreciate seeing a link.
My math in the remainder of this post (if it has no errors) shows that the temperature of H atoms does not vary with the expansion of the universe in the same way that CMB temperature varies. If this is correct, then any batch of hydrogen which did not end up in galaxies might have a current temperature approximately the same as that predicted by the effect of the universe expansion on the H temperature.
The calculated temperature results for today (a=1) are as follows.
TCMB(a=1) = 2.725 K
TH(a=1) = 0.002959 K
The remainder of this post is the calculations for TH(a=1).
It is assumed that the temperature of both CMB and H atoms were the same, at the time that decoupling completed.
(Eq 1) TH(a=0.001094) = TCMB(a=0.001094)
= TCMB(a=1)/0.001094 K = 2.725/0.001094 = 2491 K
The H atom RMS velocity is calculated using the equation:(Eq 2) VRMS = (3kT/M)1/2.
Using Eq 1, the RMS velocity of H atoms at decoupling completion is calculated as follows.(Eq 3) VRMSdc = (3kTdc/M)1/2 = 7883 m/s
Here,k = Boltzmann constant
= 1.380649 ×10−23 J/K
= 1.380649 ×10−23 kg m2 s-2/K,
M = molecular mass m of H atom = 1.008 dalton, where
1 dalton = 1.66053906660(50) ×10−27 kg.
Let x be the coordinate in which an H atom moves. V is the speed which an atom is moving:
(Eq 4) V = dx/dt.
As the particle moves from x to x+dx, there is a reduction in velocity due to the expansion of the universe:(Eq 5) dV = - H dx.
Combining Eq 4 and Eq 5 yields:(Eq 6) dV/V = - H dt
The definition of H is:(Eq 7) H = (da/dt)/a.
This yields:(Eq 8) dt = (1/H) (da)/a.
Combining Eq 4 and Eq 7 yields:(Eq 9) dV/V = - da/a.
Integrating Eq 9 yields:(Eq 10) V = C/a,
where C derives from a constant of integration.The value for C is calculated from Eq 10:
(Eq 11) C = adc VRMSdc
= 0.001094 x 7852 m/s = 8.59 m/s.
I now calculate the average H atom velocity V(a=1) using Eq 10 and Eq 11:(Eq 12) V(a=1) = 8.59 m/s.
I now use Eq 1 and Eq 2 to calculate TH(a=1).(Eq 13) TH(a=1)/ TH(a=0.001094) = V2(a=1)/V2(a=0.001094)
(Eq 14 TH(a=1) = 2491 K × (8.59 m/s)2 / (7883 m/s)2
= 0.002959 K
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