The Galois Group of Quarks: How is a Group Assigned to a Particle?

[Ray]

Can anyone give an answer (or give a web reference) to the following question: How is a group assigned to a particle? I've seen groups assigned to shapes, polynomials, permutations, rotations and transformations. But how is a group assigned to a point particle?

 

[RGevo]

Is your question why particles are chosen to be in certain representations of a group?

Or something else I'm not aware of?

 

[Ray]

No, it's just that whenever I've seen groups in the past there has been a clear mapping between the objects in question and the group elements and operation. E.g., the vertices of an equilateral triangle and S3.

 

[Ray]

PS the reference to Galois was meant to be ironic...

 

[ChrisVer]

Why do we assign groups to particles?
We don't assign groups to particles.
We assign groups in nature. Those groups reflect some symmetry there is, or some symmetry that there is not. Plus those groups are not spacetime groups, but gauge groups.
The SU(3)xSU(2)xU(1) group for example, explains the symmetries that the particles we know obey. For example the U(1) corresponds to some phase change of your field, and we know that if that's a global ohase (so you can't distinguish between points) nothing changes, while if the phase is given locally to each different spacetime point you get the existence of massless field similar to the EM field in order to keep it.
And so on...
I hope I was clear and correct.

 

[RGevo]

The galois reference confused me ;)

I agree with ChrisVer.
1) I either observe some particles, such as left handed neutrinos and massive charged leptons. I can then try and embed these in my theory. Which requires some understanding of the representation of different groups.

For example, the measurements of the ratio of hadrons to muons in e+e- suggest there are 3 colours. So, the choice of placing the quarks in a 3 of SU(3) in colour space followed.

There are lots of other things you can do, for example the approximate symmetries in QCD. The model of placing the u,d and s quarks in a 3 of SU(3) flavour space. as proposed by gell-man etc.

Sometimes knowing that you want to place particle in for example, a doublet of SU2 when you have only so far seen one of these particles (such as the strange in the strange/charm doublet) allows the prediction of undiscovered particles.

 

[ChrisVer]

Well, the reason of the strange-charm quark lies in GIM mechanism (which explained the lack of observation of flavor changing neutral currents), which in fact predicted the existence of the extra quark that was discovered later putting s and c in the same flavor SU2 doublet.
The other thing with SU2 doublets was the left-right movers which we introduced to explain the parity violation of the weak interactions. Someone before that would expect that both left/right handed movers would interact in the same way in every case. SU2 allowed us to distinguish between them two, putting the left movers in SU2 doublets and the right ones in SU2 singlets, thus under SU(2) they would be seen differently.

Why SU(3)xSU(2)xU(1)? Well this in fact, I don't know to answer. I don't think that people know... since we already know that it's just a broken subgroup in lower energies of higher symmetries (eg leptogenesis, baryogenesis, dark matter, massive neutrinos etc)

 

[Ray]

Many thanks for the replies, still don't get it. Suppose we have a family of related particles which share in several properties, then anyone particle can be thought of as a state of a generic particle. Using column vectors of ones and zeros we can describe an interaction between two particular particles by giving their 'before' and 'after' reaction vectors. Interactions can then be modeled by a collection of linear transformations which would have group properties reflecting the internal symmetries. I suppose that what I'm asking for is a matrix group version of Feynman diagrams.

 

[Vanadium 50]

Feynman diagrams are computational shorthand.

When you do a calculation in QCD, at each vertex you place a T-matrix, which are the generators of SU(3) color.

When you do a calculation in QED, at each vertex you place a number, and a single number is the U(1) equivalent of the T-matrix.

So they are already there.

 

[Ray]

Many thanks for all the replies.
Found a really useful web reference that does it for me: http://web.mit.edu/molly/Public/8.06/final.pdf
Thanks again, over and out.