Why don't the nuclear fusion processes inside the sun produce electron antineutrinos $(\bar\nu_e)$?

Question

Why don't the nuclear fusion processes inside the sun produce $\bar\nu_e$ despite having the same mass as $\nu_e$? Is the reason as simple as "there is no production channel for $\bar\nu_e$s." ?

Answer 1

The sun starts with electrons and protons and fuses the protons into heavier nuclei. Almost all stable nuclei contain neutrons, which means that in this process, protons have to be converted to neutrons. This requires the destruction of electrons by charge conservation, which in turn requires the creation of electron neutrinos by lepton number conservation. So we expect the sun to produce almost all electron neutrinos.

Answer 2

Yes, the reason is as simple as "there is no production channel for antineutrinos".

The primary stellar fusion reaction families, the proton-proton chain and the CNO cycle (significant in stars with mass > $1.3 M_\odot$) both convert some protons to neutrons, and that conversion releases a positron & an electron neutrino.

There are no other significant nuclear reactions happening in the Sun. Of course, the Sun contains all the naturally occurring elements, which it inherited from the primordial gas & dust cloud that formed the Solar System. And that includes elements like uranium and thorium, whose decay chains include isotopes that undergo beta- decay, which does release antineutrinos.

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Sufficiently massive stars (> $8 M_\odot$) create heavier nuclides via the alpha ladder, but those reactions (generally) do not affect the neutron balance, and therefore don't emit neutrinos or antineutrinos.

In general, the nuclides produced by the major stellar fusion processes have equal numbers of protons & neutrons. When we get to the heavier nuclides of the alpha ladder, those nuclei are unstable: they need more neutrons than protons to be stable, so they tend to decay by emitting a positron & a neutrino.

As we progress up the alpha ladder, the temperature required to overcome the Coulomb barrier becomes enormous. At such high temperatures, the stellar core's thermal spectrum contains significant numbers of very high energy photons. (Astrophysicists call them gamma rays, even though they are not emitted from a nucleus). These gamma photons have enough energy to disrupt nuclear structure, creating free alpha particles, a process known as photodisintegration. (For most of the alpha ladder, the alpha particles are actually created by photodisintegration, rather than being primordial helium or helium produced by the p-p chain or the CNO cycle).

Gamma photons with sufficiently high energy can also interact with nuclei to induce pair production, i.e., an electron + positron pair. Usually such pairs quickly annihilate, but about 1 in $10^{19}$ pairs decay to a neutrino + antineutrino instead. Wikipedia gives some details in its article on carbon burning. That article also mentions that as well as the main carbon + helium reaction, there are also some carbon + carbon fusion reactions, one of which releases a free neutron. There are other side reactions for heavier elements which also release a neutron, eg during neon burning.

A free neutron decays (with a mean lifetime of around 14.7 minutes) into a proton, electron, and antineutrino, unless it's consumed in some other reaction. These free neutrons can be harnessed by the slow s-process reactions that occur with heavy seed nuclei outside the stellar core. (These seed nuclei are inherited from the primordial cloud that formed the star). The s-process is responsible for the creation of many nuclides heavier than iron. It mainly operates in asymptotic giant branch stars. These s-process nuclides may have an excess of neutrons and thus can undergo beta- decay, releasing antineutrinos.

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For really significant stellar antineutrino production, we have to go to a large star that's beginning to undergo core collapse. When that happens, most of the protons & electrons in the core get converted into neutrons and neutrinos.

From Wikipedia's article on Type II Supernova:

The core collapse phase is so dense and energetic that only neutrinos are able to escape. As the protons and electrons combine to form neutrons by means of electron capture, an electron neutrino is produced. In a typical Type II supernova, the newly formed neutron core has an initial temperature of about 100 billion kelvins, $10^4$ times the temperature of the Sun's core. Much of this thermal energy must be shed for a stable neutron star to form, otherwise the neutrons would "boil away". This is accomplished by a further release of neutrinos. These 'thermal' neutrinos form as neutrino-antineutrino pairs of all flavors, and total several times the number of electron-capture neutrinos. The two neutrino production mechanisms convert the gravitational potential energy of the collapse into a ten-second neutrino burst, releasing about $10^{46}$ joules (100 foe).

These thermal neutrino-antineutrino pairs are produced by the photoneutrino process: a high energy photon interacts with an electron (or positron): $$e^\pm + \gamma \to e^\pm + \nu + \bar\nu$$ This is similar to the pair production that occurs when a high energy photon interacts with a nucleus. This process is discussed in The photo-neutrino process in astrophysical systems by Sharada Iyer Dutta, Sasa Ratkovic, and Madappa Prakash.

Answer 3

You must convert proton to neutron somehow (pp chain or CNO chain) and this means by charge conservation that a positron must be emitted, and by lepton number conservation to balance lepton number that anti-electron is accompanied by neutrino and not anti-nu.