I am looking into neutrino decoupling in the early universe, which is said to occur when the reactions: $$e^++n \leftrightarrow p+\bar{\nu}$$ $$\nu+n \leftrightarrow p+e^-$$ $$e^-+e^+\rightarrow \bar{\nu}+\nu$$ are prevented from occuring. However after this time the decay if neutrons still occurs: $$n\rightarrow p+e^-+\bar{\nu}$$ I am therefore wondering why neutrino decoupling is not taken to be after this latter reaction stops occurring (some 3 minutes latter) rather then the former three.
$\begingroup$
$\endgroup$
asked May 4, 2017 at 5:57
Quantum spaghettificationQuantum spaghettification
7,75513 gold badges71 silver badges170 bronze badges
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
It is based on the word "decoupling" . It means that the probability of neutrinos to find a target and interact with it is very very small.
It is the same meaning given to photon decoupling. It is not that the photons will not interact if they meet a target and get deflected. It is because the probability of finding and hitting a target particle or nucleus is very small. Thus we get a snapshot of the universe at the time of the decoupling with the cosmic microwave background radiation. The small inhomogeneities come from large targets like those that will develop as clusters of galaxies, whose density is such that photons still at that time would be absorbed.
The neutrino decoupling and generation of the nuclei is seen in this timeline.. It is called neutrino ( and photon) transparency.
If or when we develop accurate neutrino detectors we will be able to see at the time of neutrino decoupling a snapshot of the universe.
Neutrons will always decay when free. After neutrino decoupling those neutrinos will join the rest, but their numbers will be small, and at the time (energy density) of creation of nuclei there are no longer free neutrons.
