neutron decay

[RedX]

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.:

\mathcal L=\bar{n}\gamma^\mu(1-g_A\gamma_5)p W^{-}_\mu

where g_A 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?

[Bob S]

A free proton at rest cannot decay into a neutron. A 7 TeV proton, at rest in its own restframe, cannot decay into a neutron either.
Bob S

[RedX]

A free proton at rest cannot decay into a neutron. A 7 TeV proton, at rest in its own restframe, cannot decay into a neutron either.
Bob S

Is the proton really free or can it interact with the collider-ring material?

Also, at energies greater than .2 GeV (the QCD scale), how is the hadronization of quarks treated? With lattice theory? Because for energies below .2 GeV, a simple effective Lagrangian can be built (the chiral Lagrangian) that describes hadrons in a very simple manner. But 7 TeV is much higher, so I don't even know how to begin doing calculations on protons at that energy.

[Vanadium 50]

Is the proton really free or can it interact with the collider-ring material?

It's free. If it interacts with the collider ring material it stops going around.

[JustinLevy]

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.
Neutrons are fermions and thus there is pauli exclusion to consider as well. For the simplest example, deuterium (proton + neutron) is a bound state, but neutron + neutron is not a bound state.

[RedX]

Neutrons are fermions and thus there is pauli exclusion to consider as well. For the simplest example, deuterium (proton + neutron) is a bound state, but neutron + neutron is not a bound state.

Can't one neutron have up spin, and the other down spin?

A free neutron is heavy so it breaks up into a proton which is lighter. In atoms, protons can turn into neutrons, and the only driving force seems to be to relieve the electromagnetic forces.

[Vanadium 50]

Can't one neutron have up spin, and the other down spin?

Yes, but the dineutron still isn't bound.

[clem]

It turns out that the n-n force in the triplet, spin 1 case is a bit too weak to bind two neutrons. The spin 1 n-p force is a bit stronger and has a bound state, the deuteron, which is weakly bound.