# B Does every elementary particle eventually decay into quarks?

1. Mar 27, 2017

### zinq

Everything in the universe that has a beginning seems to also have an end. Stars, galaxies, molecules, atoms. But what about elementary particles? In case of heat death would it become a uniform quark soup?

2. Mar 27, 2017

### Staff: Mentor

Not just quarks, since those (more precisely, up and down quarks) are not the only particles that, according to the Standard Model, cannot decay into other particles. Electrons and electron neutrinos also have that property among the fermions.

3. Mar 27, 2017

### mathman

Why should stable nuclei decay at all?

4. Mar 27, 2017

### Chalnoth

Eventually proton decay should get rid of stable nuclei.

5. Mar 28, 2017

### 256bits

Is that a given?
I thought some experiments suggested the proton decay was virtually non-existent.

6. Mar 28, 2017

### Chalnoth

Most likely, yes. Proton decay is virtually guaranteed by simple symmetry arguments: some process created the protons, and the time reverse of that process would necessarily cause them to decay (note: it wouldn't quite be that simple, but that's the basic idea). The properties of that process may suppress it to an extremely low level (e.g. requiring a very massive intermediate particle, akin to how the masses of the weak force bosons suppresses the weak nuclear force), but simple arguments seem to indicate it's impossible to eliminate it entirely.

7. Mar 28, 2017

### 256bits

That makes tons of sense. I knew about the symmetry argument, but had never gone farther in depth to entertain and explore that the creation implies a destroy time reverse process.
Should have known though. Thanks for the explanation.

8. Mar 28, 2017

### Comeback City

Just to go along with the stable nuclei thing, is it not possible at all that certain "stable" nuclei have half-lives that are simply too long for us to determine?

9. Mar 28, 2017

Staff Emeritus
We know exactly which nuclei are stable and which are merely long lived. If there is an allowed transition with a summed mass of the daughters below that of the parent, it will occur at some rate. Possibly trillions of years, but at some rate. There are exactly 164 nuclei that will decay at some rate but their decays have not been observed, and exactly 90 stable nuclei.

10. Mar 28, 2017

### snorkack

What arguments are used to claim that any process ever created protons?

Simple symmetry arguments would suggest that time reversal of big bang would be big crunch, yet big crunch seems not to be expected now. Time reversal of a black hole is a white hole, yet white holes are also not expected.

11. Mar 28, 2017

### Chalnoth

They exist, and aren't balanced by anything in our universe with a negative baryon number. Look up baryogenesis if you want to see some of the current science surrounding this topic.

The difference is that proton decay is a microscopic reaction, while black holes and a "big crunch" are macroscopic.

The relevant difference between the macroscopic realm and the microscopic realm is that if something is a valid solution to the equations in the microscopic realm, it is guaranteed to happen with some frequency. So if you can go from a collection of particles that have no baryon number, and get out a result that has a net baryon number (as must have happened in the early universe), then there is some magnitude for that process to happen in reverse.

12. Mar 28, 2017

### Khashishi

Seems reasonable, but until we actually detect a baryon number non-conservation, it's a conjecture. It's possible that the universe always had a positive baryon number.

13. Mar 29, 2017

### Chronos

The question of proton decay remains an open question in particle physics. One of the fundamental rules in particle physics is the total number of quarks minus the total number of antiquarks must not change in a decay. Given that the proton is the lightest particle that has more quarks than antiquarks, it appears to have no viable decay mode. Some theorists question whether this rule of quark conservation is necessarily inviolable, but, no violations are known.

14. Apr 11, 2017

### nikkkom

There is no fundamental requirement for this in SM. SM is a QFT and as such, _local_ gauge symmetries are its fundamental building blocks.
Baryon conservation is one of accidental _global_ symmetries of the SM Lagrangian. Not local.
SM has sphalerons, which do not conserve baryon number: they convert baryons to antileptons.