- #1
RedX
- 970
- 3
A free proton cannot turn into a neutron (via beta decay) because the neutron is heavier and that would violate energy conservation laws.
However, in the nucleus, a proton can turn into a neutron (and eject a positron in the process). Why doesn't this happen all the time? Reducing the charge on your nucleus seems to lower the electromagnetic energy, and the nucleon-nucleon strong force is the same for protons and neutrons so that doesn't change. So I'm seeing nothing but good things from this process.
Also I noticed that in the chiral Lagrangian the weak interaction is almost left chiral, e.g.:
[tex]\mathcal L=\bar{n}\gamma^\mu(1-g_A\gamma_5)p W^{-}_\mu[/tex]
where [tex]g_A[/tex] is the numeric value 1.27 and p is the proton Dirac field and n is the neutron Dirac field and W is the W-boson.
So 1.27 is almost close to 1, in which case it looks like the weak interaction for leptons. What is the significance of this? Does this mean for protons traveling at high speeds, such as the LHC, if the proton is right-handed, it will not undergo a weak interaction?
However, in the nucleus, a proton can turn into a neutron (and eject a positron in the process). Why doesn't this happen all the time? Reducing the charge on your nucleus seems to lower the electromagnetic energy, and the nucleon-nucleon strong force is the same for protons and neutrons so that doesn't change. So I'm seeing nothing but good things from this process.
Also I noticed that in the chiral Lagrangian the weak interaction is almost left chiral, e.g.:
[tex]\mathcal L=\bar{n}\gamma^\mu(1-g_A\gamma_5)p W^{-}_\mu[/tex]
where [tex]g_A[/tex] is the numeric value 1.27 and p is the proton Dirac field and n is the neutron Dirac field and W is the W-boson.
So 1.27 is almost close to 1, in which case it looks like the weak interaction for leptons. What is the significance of this? Does this mean for protons traveling at high speeds, such as the LHC, if the proton is right-handed, it will not undergo a weak interaction?