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This is Neutron decay: $$n^o \to p^+ + e^- + \overline {\nu_e}.$$ and this is proton one: $$p^+ \to n^o + e^+ + \nu_e$$

so when the $e^+ =e^-$ and $\nu_e=\overline {\nu_e}$ why $n \not= p$?

my idea is that:

$$\nu_e \not= \overline {\nu_e}$$

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closed as unclear what you're asking by user1504, Manishearth Jun 26 '13 at 22:54

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Note that proton decay does not proceed through the indicated channel because the energy of the RHS is greater than that of the LHS (in every frame of reference). Further, there is no reason to expect an exact match between the nucleon masses unless isospin is a good symmetry and it is known to be broken. Further still, the neutrino mass differences squared extracted from neutrino beams and solar neutrinos agree with those for the antineutrinos. –  dmckee Aug 15 '12 at 18:22
    
@dmckee why not put that into an answer? –  Rody Oldenhuis Aug 15 '12 at 18:29

4 Answers 4

The neutron is made of two down quarks and an up quark; the proton of two up quarks and a down quark. This leads to two effects that differentiate their masses. One is that the up and down quark themselves have different masses. The other is that the proton is charged, and so quantum corrections involving virtual photons affect its mass. The details are extraordinarily difficult to calculate, but lattice QCD simulations allow the effects to be disentangled, and the claimed result is that the down quark mass is about 4.8 MeV versus an up quark mass of about 2.0 MeV. So, it's mostly the fact that the down quark is heavier than the up quark that makes the neutron heavier than the proton.

As others said, the "proton decay" process you wrote down is forbidden, precisely because the neutron is heavier than the proton.

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As you may already know, nucleons are made of quarks; protons (uud) and neutrons (udd) were the masses are 938.3 MeV and 939.6 MeV, respectively. The key point is that the majority of the nucleon mass comes from quark interactions.

To see this, consider the following points:

  1. if you were to mass a “free” up or down quark, they would have a mass of only a few MeV.

  2. if you restricted these quarks to within the nucleus (about 10^(-15) m), they would have masses on the order of several hundred MeV. Memory tells me somewhere on the order of a few hundred MeV (300-400 MeV). This is derived from the zero point energy for 10^(-15) m well.

  3. So to account for the missing 600-700 MeV, one has to look at the quark interactions within the nucleons. And the current ability to calculate this (via QCD) is currently beyond our theoretical abilities.

So to say that the nucleons have different masses because of the differences in the masses of the neutrinos, is not correct. The majority of the nucleon masses comes from quark interactions!

If we take a native model to try to explain the differences in mass, the down quark is a few MeV more massive than the up quark (no one knows why). Because of the zero point energy of being trapped within a distance of 10^(-15) m, each nucleon should each have a mass of around 1 GeV. So the only difference between the proton and neutron will be on the order of MeV, and this difference can be thought of as a result of

• the differences between the proton (udu) and neutron (udd) is that the has neutron’s second down quark is heavier than the proton’s second up quark. So the greater mass of this down quark gives the neutron a greater mass than the proton.

• the electrostatic forces among two ups and a down (proton) will differ from those between those two downs and an up (neutron).

So the instability of the neutron (beta decay) is due to it having a slightly greater mass than the proton. And as the saying goes, nature always seeks the lowest state of energy (mass).

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The positron is the electron's antiparticle. The antineutrino is the neutrino's antiparticle. By the CPT theorem, we expect the masses of these particles to be equal to those of the antiparticle.

The neutron is not the proton's antiparticle (the fact that the neutron's charge is not -1 should be enough for this to be clear). We therefore don't have a constraint on their masses from QFT. It turns out that the proton is less massive than the neutron, so the proton decay process you describe above requires an energy input (or thermal temperatures so high that the energy difference is irrelevant).

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There is no proton decay - proton is a stable particle.

$p^+ \to n + e^+ + \nu_e$ is not decay, it's reaction. It should be write as:

$\gamma+p^+ \to n + e^+ + \nu_e$, or
$p^+ + p^+\to p^+ +n + e^+ + \nu_e$

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Well, you're right that we haven't observed it, but it would be very hard to make sense of the asymmetry of matter and anti-matter if protons don't decay (since that would imply baryon number is conserved). –  Mark M Aug 15 '12 at 21:08
    
@Mark M, In your case proton decay may be $p^+ \to \mu^+ \pi$, for example, not $p^+ \to n e^+ \nu_e$ –  voix Aug 15 '12 at 21:17
    
@Fabian: That doesn't work with inflation. The proton is already unstable in the standard model, by anomalies, as shown by t'Hooft, but the rate at ordinary energies is vanishing. –  Ron Maimon Aug 16 '12 at 3:56

protected by dmckee Jun 25 '13 at 3:28

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