Free Fractionally Charged Particles

1. Sep 4, 2014

stevendaryl

Staff Emeritus
It is intriguing (to me) that while fractionally charged particles exist in the standard model, they are always bound into composite particles of integer charge. The standard model explains this by QCD: fractionally charged particles all have nonzero color charge, and so can't be free. But I'm wondering whether it would be possible to introduce a fractionally charged free particle, or whether the theory becomes inconsistent in that case. Is there some deep reason that colorless particles must have integral charge, other than the fact that the standard model carefully assigns color and electric charges to make this happen?

2. Sep 4, 2014

Staff Emeritus
You can't do this by itself. The sum of the charges under all forces for all particles (not antiparticles) must be zero. So if you create a Q=4/3 "fatlectron", there needs to be a Q=-4/3 partner, *plus* their antiparticles.

3. Sep 4, 2014

ChrisVer

Vanadium, if it's not antiparticles, than what's the partner of the electron?
But that is actually weird for me...since you can have free quarks [fractional charged particles] at high energies[temperatures] as for example in the Quark-Gluon Plasma... So it's not a matter of theory I guess, but a matter of observation tthat we get only integer-quantized charges.
The most intriguing particle quantization is their masses.... :) so chaotic/irrational (eg from some MeV for the u,d up to 4 GeV for b and then a really high jump to 170GeV for t). And so on... especially if those particles are [maybe] one single thing...

4. Sep 4, 2014

Staff Emeritus
The total charge for *all* the particles has to be zero. So you have 3 (colors) +2/3 quarks, and 3 -1/3quarks, and an electron -1, and that sums to zero. (x3 generations, and it still sums to zero). So if I add a new particle with charge q, I must at a minimum add a new particle of charge -q.

On top of that, there are antiparticles.

5. Sep 4, 2014

Orodruin

Staff Emeritus
You forgot: Electron 511 keV, neutrino sub-eV ... The last jump down is at least 5 orders of magnitude.

Also, Vanadium is not saying it right out, but he seems to be referring to (one of) the requirements for anomaly cancellation.

6. Sep 4, 2014

ChrisVer

aha so I guessed, but I had misunderstood the requirement -for eg U[1]- $\sum_{i} Y_{i}^{3}=0$ for anomaly cancelation **, considering $i$ running over everything [particles & antiparticles (or Left/Right reprs)- so in that case electron would cancel positron and so on].

**: which once summed over all particles gives the 1.3 condition here:
http://cds.cern.ch/record/439081/files/0005015.pdf

Vana's post thouggh referred to the next two...1,4 and 1.5..so I just didn't *remember* [or know] those conditions...

Last edited: Sep 4, 2014