Etudes des proprietes des neutrinos dans les contextes ...

tel-00450051, version 1 - 25 Jan 2010
¯νe + p ⇋ n + e + , (8.10)
n ⇋ p + e − + ¯νe. (8.11)
We denote the forward and reverse rates of the first process as λνen and λ e − p,
respectively. Likewise, the forward and reverse rates of the second process are
λ¯νep and λe + n, respectively, while those of the third process are λn decay and λpe¯νe,
respectively. At high enough temperature (T ≫ 1 MeV), where these rates are
very fast, the isospin of any nucleon will flip from neutron to proton and back at a
rate which is rapid compared to the expansion rate of the universe, establishing a
steady state equilibrium. As the universe expands and the temperature drops the
rates of the lepton capture processes will drop off quickly. Eventually the lepton
capture rates will fall below the expansion rate and n/p will be frozen in (except
for free neutron decay). However, there is no sharp freeze-out, the neutron-toproton
ratio n/p is modified by the lepton capture reactions down to temperatures
of several hundred keV and by neutron decay through the epoch of alpha particle
formation Tα. The evolution of the electron fraction Ye = 1/(1+n/p) throughout
the expansion is governed by
dYe
dt = Λn − Ye (Λn + λ¯νep + λe − p + λpe¯νe) (8.12)
where we took into account the rates of the neutron destroying processes Λn =
λνen + λe + n + λndecay and the weak isospin changing rates. In the limit where the
isospin flip rate is fast compared to the expansion rate H, the neutron-to-proton
ratio has a steady state equilibrium value (dYe/dt = 0) given by [99]:
n
p =
λ¯νep + λe−p + λpe¯νe
, (8.13)
λνen + λe + n + λn decay
≈ λ¯νep + λe−p λνen + λe + .
n
If the electron neutrinos and antineutrinos and the electrons and positrons all
have Fermi-Dirac energy spectra, then Eq.(8.13) can be reduced to 2 [37]
n
p ≈ (λe−p/λe + n) + e−ξe+ηe−δmnp/T (λe−p/λe + n) eξe−ηe+δmnp/T , (8.14)
+ 1
where ηe = µe/T is the electron degeneracy parameter and (mn − mp)/T ≡
δmnp/T ≈ 1.293 MeV/T is the neutron-proton mass difference divided by temperature.
If chemical equilibrium is achieved, then we have µe − µνe = µn − µp,
where µn and µp are the neutron and proton total chemical potentials, respectively,
and in this case Eq.(8.14) reduces to
n
p ≈ e(µe−µνe −δmnp)/T . (8.15)
2 We assume here identical neutrino and plasma temperatures and neglect the neutron
decay/three-body capture processes of Eq.(8.11) as done in [1].
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