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Authors
Affiliations
Université Paris-Saclay
IJCLab
Université Paris-Saclay
IJCLab
Université Paris-Saclay
IJCLab

1Neutron freeze-out

Γn≈H\Gamma_n \approx H
Γn=σwnνc\Gamma_n = \sigma_w n_\nu c
nν(Tγ)=34411nγ(Tγ)(onlyνe)n_\nu(T_\gamma) = \frac{3}{4}\frac{4}{11}n_\gamma(T_\gamma)\quad (only \nu_e)
nγ=2ζ(3)(kBT)3π2ℏ3c3n_\gamma = \frac{2 \zeta(3) (k_B T)^3}{\pi^2 \hbar^3 c^3}
σw=(GF2E2)/(ℏc)2=GF2(kBT/(ℏc))2=5×10−48(kBT1 MeV)2 m2\sigma_w = (G_F^2 E^2)/(\hbar c)^2 = G_F^2 (k_B T/(\hbar c))^2 = 5\times 10^{-48} \left(\frac{k_B T}{1\,\mathrm{MeV}}\right)^2\,\mathrm{m}^2
GF=1.166×10−5 GeV−2G_F = 1.166 \times 10^{-5}\,\text{GeV}^{-2}
gA=1.26(Kolb Turner p.91)g_A = 1.26 \quad\text{(Kolb Turner p.91)}
H=8π3G90ℏ3c5g⋆1/2(T)(kBT)2H = \sqrt{\frac{8 \pi^3 G}{90 \hbar^3 c^5}} g_\star^{1/2}(T) (k_B T)^2
g⋆(100 MeV>T>511 keV)=10.75g_\star(100\,\mathrm{MeV} > T > 511\,\mathrm{keV}) = 10.75
g⋆(T<511 keV)=3.36g_\star(T < 511\,\mathrm{keV}) = 3.36
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(<Quantity 3.15287605e+09 K>, <Quantity 10. s>, <Quantity 9.97026951e+09 K>)

2Neutrino temperature and time to temperature accurate conversion

See Weinberg p.540 p.529

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3Orders of magnitude

From equilibrium equations and rate with GFG_F and gAg_A :

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Freeze out temperature

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4Exact computation

The freeze-out temperature obtained with orders of magnitude is too high to explain the helium aboundance (by a factor of 3, but here numerical factors do count). We now use exact (Weinberg integrals) and semi-exact (Dodelson approximation) methods.

Using Weinberg and Dodelson equations

Check equality of constant gVg_V in Weinberg eq 15.7.9 p.547 with with Fermi constant GFG_F

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In Weinberg, neutron life time is 1013s but current value if 878s. We rescale its amplitude constant:

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Computation of interaction rates:

(<Quantity 0.65549597 1 / s>, <Quantity 0.13069837 1 / s>, <Quantity 0.64641477 1 / s>, <Quantity 0.4335013 1 / s>)

Check that λpn∼λnpe−Qn/T\lambda_{pn} \sim \lambda_{np} e^{-Qn/T} :

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/Users/jneveu/miniforge3/envs/m2-cosmo/lib/python3.11/site-packages/astropy/units/quantity.py:671: RuntimeWarning: overflow encountered in exp
  result = super().__array_ufunc__(function, method, *arrays, **kwargs)
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Solve

(np.float64(0.2546931106785803), np.float64(0.2554279084910855))
(np.float64(-0.059484795266252305), np.float64(0.23618253503009692), <Quantity 0.50298636 1 / s>, <Quantity 0.50298636 1 / s>, <Quantity 0.99406274 s>)
(np.float64(0.0590577858660713), np.float64(0.23588394768970708), <Quantity 0.5 1 / s>, <Quantity 0.5 1 / s>, <Quantity 0.85917135 MeV>)

Differential equation 3.27 from Dodelson p.67

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array([-3.83979787e-11])
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(<Quantity 1.77604491 1 / s>, <Quantity 1.16045181e+10 K>)
(<Quantity 0.67481545 MeV>, <Quantity 7.83090811e+09 K>)
(<Quantity 0.64293845 MeV>, <Quantity 7.46099084e+09 K>)
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np.float64(0.25901803917730154)

5Deutérium ratio

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(<Quantity 132.03654981 s>, <Quantity 127.90464621 s>)
(np.float64(0.25901803917730154), np.float64(0.14734740090432036), np.float64(0.12842439943489092))

Maximum helium 4 abondance (if all neutrons end in helium nuclei, no D nor Li nor...):

(np.float64(0.25684879886978185), np.float64(0.25684879886978185))

6Plots

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Taux de désintégration du neutron (plateau de la courbe)`

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