# Stability of quarks

1. Jan 15, 2014

### Hluf

I'm a new comer to study the hadron physics.Why light quarks are more stable than heavy quarks and which one is easy to study? why? Thank you

2. Jan 15, 2014

### RGevo

Hi hluf, a heavy quark or baryon has a probability to decay to something with less mass. This decay process cannot occur to something heavier (in isolation) without getting energy from nothing.

The most stable particles are the lightest. The quark decays tend to be through the weak force.

The lightest particles you may wish to look at as examples are the pions and the kaons (being made up of some the lightest 3 quarks)

3. Jan 15, 2014

### clem

A quark (or any other particle) can only decay into a lighter quark.
Only the lightest u quark is stable.

4. Jan 15, 2014

### Staff: Mentor

The light quarks do not exist as isolated particles, they are always bound in hadrons*. For light quarks, you have to consider the mass of the hadron - and the proton (with two up-quarks and one down-quark) is stable**. Neutrons (with two down-quarks and one up-quark) can be stable as part of nuclei.

The other 4 quarks form unstable hadrons that decay after a while.

*with a few known exceptions: the very early universe, for very short times in particle accelerators and maybe in the core of neutron stars
**or extremely long-living. Some theories predict a decay, but none was observed so far.

Depends on the property you want to study.

5. Jan 16, 2014

### RocketSci5KN

Are there any good theories why neutrons in *most* nucleii are somehow stabilized against beta decay, whereas free neutrons have a known half-life? I'd like something better than 'it's all due to the weak force'...

Also, would the light quarks theoretically be stable in their free form?

6. Jan 16, 2014

### Staff: Mentor

If the sum of the masses of the post-decay nucleus and the emitted electron is less than the mass of the pre-decay nucleus, then beta decay is possible. The "extra" pre-decay mass becomes kinetic energy of the outgoing electron and antineutrino, and of the recoiling post-decay nucleus.

If the sum of the masses of the post-decay nucleus and the emitted electron is greater than the mass of the pre-decay nucleus, then beta decay is not possible. Extra energy would have to be supplied from outside in order to "make" the "extra" post-decay mass.

The difference in mass one way or the other is basically due to different amounts of binding energy for different nuclei.

7. Jan 16, 2014

### Staff: Mentor

It is mainly due to the strong force and energy conservation, see jtbell's post.
"Most" nuclei are unstable. Most of our everyday world is made out of stable or extremely long-living atoms because the short-living ones decay so quickly (or do not get produced at all).

There is no free form.

8. Jan 18, 2014

### ChrisVer

Well the mass differences that were mentioned are the right way to think of neutron's (in)stability.

What helped me in that was that image- I don't know whether it's correct or not but it makes sense-.
Suppose you have a free neutron, it will remain a neutron forever (not interacting) until it decomposes due to beta decay. Beta decay is a weak interaction process, so it's characteristic time is generally larger than the strong's interaction.
Now suppose that the neutron is in the nuclei. What happens then? it interacts with the protons, via puons (Yukawa mesons). If you draw the procedure of that, you will see that the proton at point A emits a puon, becoming a neutron, and the neutron at point B receives the puon and becomes a proton. And this goes on and on. So in fact you never have one neutron waiting to decay. The neutrons change with protons over and over again in times of order of strong interaction characteristic time which is mass lesser than the weak's.
So by that image, the neutron will be stable in the nuclei because strong interactions don't allow it to decay.

9. Jan 19, 2014

### Staff: Mentor

- the particles are called pions, not puons
- a free neutron does not have to "wait" for a decay, so changing the identity of nucleons frequently would not prevent a decay.
- that model cannot explain unstable nuclei, or their difference to stable ones.