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According to Grand Unification Theory, protons can decay into electron (even at low energy; just the probability is very low). It doesn't mean you can replace proton with electron.

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Both single beta-decay and double beta-decay may occur with $e^+$ as well as $e^-$. However, in both cases, the emission of $e^-$ is predicted to appear (and in the single beta case, is also observer to appear) in a larger number of processes essentially because it's energetically easier for neutrons to decay to protons plus electrons; than it is for ...

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It depends who you ask. A string theorist would answer that string theory is the idea that the point-like particles of elementary particle physics can also be modeled as one-dimensional objects called strings. According to string theory, strings can oscillate in many ways. On distance scales larger than the string radius, each oscillation mode gives rise to ...

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No. The atoms are protons, electrons and neutrons. The fact that neutrons beta decay into a proton + electron + electron antineutrino does not mean that neutrons are made of a proton and electron and a neutrino.

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No. The decay products of a certain particle are not equivalent to its constituents. This is evident especially in the context of fundamental particles: quarks can decay into other particles, but that does not mean that a quark is not elementary (see my answer to this question). Nuclei are made of neutrons and protons, which in turn consist of quarks and ...

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The point is that there is no mass difference between left-handed and right-handed neutrinos. The mass is only defined for the complete field $\nu=\nu_L+\nu_R$. For this reason, I think it's not correct to use the oscillation probability formula to compute oscillations between LH and RH neutrinos.

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I think I now have an answer. My problem was that I assumed how much neutrinos oscillated depended solely on their level of mixing. With that intuition it seems that neutrinos should oscillate significantly into their right handed counterparts. However, there is more to the story. Oscillations are also dependent on the difference of masses between the mass ...

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Neutrinos interact in the Standard Model only through their left-handed component, via electroweak interactions. However, the propagating neutrinos, which are mass eigenstates, are described by a field that is a Dirac spinor, i.e. with both chiralities $$\nu=\nu_L+\nu_R.$$ Therefore, when neutrinos are created or measured, the Dirac spinor is projected ...

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This is exactly what the Mu2e experiment at Fermilab is designed to investigate. As the name suggests this will look for evidence of muons changing either directly to electrons or to an electron plus a photon. If you have a lot of spare time the 562 page project description is available on the Arxiv. It'll be a few years before they have any results. I'm ...

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Another take: Neutrinos are a member of the elementary particle set on which the Standard Model of particle physics is founded. Included in the definition of elementary particles is the concept of their being point particles, i.e. no extension in three dimensional space. The Standard Model and its calculational tools have been validated by the existing ...

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The temptation is to think of a particle as a little ball whizzing around space, and it therefore makes sense to ask about the size and density of the ball. The trouble is that this is a fundamental misunderstanding of what a particle is. Our current best description of particles is using quantum field theory. This describes particles as excitations in ...

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You do understand that short-range neutrino measurements show the expected flavor mix, right? That is if we set up a detector a few meters from a nuclear core we observe neutrino interactions involving electrons. If we set one up just downstream of a neutrino beam-line we see mostly interactions involving muons (with the expected admixture of those ...

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Let's just focus on the top vertex. First, draw an arrow pointing down. Then, you can determine from charge conservation (or looking for the relevant term in the lagrangian) that the particle must be $W^+$. Because $(W^+)^\ast=W^-$, this vertex has precisely the same value if you reverse that arrow and replace the $W^+$ with a $W^-$. Here I am just using ...

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