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The title pretty much explains the question, but I've always thought that it'd be a neutron because of its 0 charge.

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    $\begingroup$ Because there is no other lighter baryon that proton can decay into. And the baryon number is conserved in the standard model. $\endgroup$ – Andrei Geanta Nov 28 '17 at 21:24
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    $\begingroup$ Baryonic number is not conserved by electroweak interactions in general due to quantum chiral anomaly. And proton decay (into pions and positrons, say) was actively discussed and searched for experimentally, it is not ruled out even today despite the lack of experimental confirmation. So there can be no a priori reason why proton should be stable. $\endgroup$ – Conifold Nov 28 '17 at 21:41
  • $\begingroup$ Are you asking why the neutron (in isolation) decays to a proton (plus electron and anti-neutrino) but not the other way around? $\endgroup$ – Alfred Centauri Nov 28 '17 at 22:03
  • $\begingroup$ @Conifold Proton decay is theoretically impossible in the Standard Model and the "Standard Model" tag on the question implies that this is not a request for conceivable beyond the Standard Model theories that would differ from this theoretical prohibition. Sphaleron interactions can't give rise to proton decay. $\endgroup$ – ohwilleke Aug 2 at 18:29
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Protons, neutrons and in general hadrons are in the quantum mechanical regime and are not elementary. Elementary are the quarks with their positive and negative charges. Here is the main quantum mechanical image of a proton in rough lines

proton

and here is a neutron

neutron

Again in rough lines, they are quantum mechanically bound by the potential with the additions of two fundamental forces, the strong QCD force, and the electromagnetic force. In quantum mechanics, as stated also in the comments, systems are stable when they are in the lowest energy state, and this state happens to be the proton.

The title pretty much explains the question, but I've always thought that it'd be a neutron because of its 0 charge.

The neutron "bag" contains two down quarks , and ends up heavier than the proton bag, which has two up, lighter than down , quarks, so the neutron is in a higher energy level than the proton. The picture is more complicated because of QCD , and the hadronic bags contain a large number of virtual gluon and quark antiquark pairs, as illustrated here for the proton, but qualitatively the argument of lowest energy level holds.

myproton

A lot of theoretical effort is expended in trying to calculate the mass of the proton in lattice QCD, where all the contributing potentials are taken into account,

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  • $\begingroup$ Another key part of the puzzle is conservation of baryon number. The aggregate number of quarks in a system minus the aggregate number of anti-quarks in a system is conserved. So, a proton, which has three quarks more than the number of anti-quarks it has (on shell and virtual combined), can't decay into an end state that has, for example, four, two, one, or zero on shell quarks in it. The proton is not the lowest mass-energy hadron (that would be the pion at about 1/7th the rest mass of a proton), but it is the lowest mass-energy hadron with three quarks (which is a baryon number of 1). $\endgroup$ – ohwilleke Aug 2 at 18:25
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the usual answer is that there is no lighter hadron state available into which it could possibly decay. according to this picture, the proton is absolutely stable i.e., if left alone, it lasts forever.

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