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A question regarding Y=B+S by a nuclear physics toddler 
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#1
May714, 01:51 AM

P: 6

If Q=e[I+0.5Y] and Y=B+S. What is the Q/e and S value for ρ and k mesons, Ω and Δ baryons?
I means third component of isospin and Y,B,S,Q,e have usual meanings? This is the question. I don't even know what these symbol means. Can someone please explain the symbols and solve this problem. 


#2
May714, 06:27 AM

P: 10

There are 3 ρ mesons, with Q = +1, 0, 1. For all three B=0 and S=0
There are 2 K mesons with Q=+1, 0. For both B=0 and S=1 There are also antiK mesons with Q=0, 1, B=0 & S=1 There are 4 Δ baryons, with Q=+2,+1,0,1, for all B=1 and S=0 There is one Ω baryon, with Q=1, B=1 and S=3 Q is the electric charge, normally measured in units of "e", the electron charge, or more correctly the positron charge since electrons are negative. B is the baryon number, +1 for baryons, 1 for antibaryons and 0 for everything else. S is the strangeness, and is equal to the number of strange antiquarks minus the number of strange quarks. (So the strange quark has S=1, this is historical) Y is called the hypercharge, someone else can explain the importance of that. 


#3
May714, 12:38 PM

P: 754

For the Y (hypercharge), you can just see it as the generator for local abelian U(1) transformations of your fields (it's something like their charges). It's not exactly the electric charge of the particle, but in the end it contributes to it... For example, imposing local symmetry for the Dirac spinor under a phase change:
[itex] \psi'= e^{iYa(x)} \psi[/itex] you'll eventually get something like a photon (and Electromagnetism)... In the context of the WeinbergSalamGlashow model, where the Standard model contains the SU(2)xU(1) symmetry, Y is the charge corresponding to U(1)... After the SU(2)xU(1) is broken by Higgs mechanism into U(1) the new U(1)'s charge (the electric charge Q) gets some contribution both from Y and I (of SU(2) ). So you have some formulas like Q=I+0.5Y 


#4
May714, 01:14 PM

Sci Advisor
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P: 4,160

A question regarding Y=B+S by a nuclear physics toddler
Isospin I pertains to hadrons only, and arises from the symmetry of the up and down quark under strong interactions. The strange quark forms a singlet under isospin. Hypercharge comes from Q = I_{3} + Y/2. For example the strange quark has Q = 1/3, I_{3} = 0 and hypercharge 2/3. Weak isospin T arises from the symmetry of lefthanded lepton pairs under weak interactions. Righthanded leptons form a singlet under weak isospin. Weak hypercharge comes from Q = T_{3} + Y/2. (Unfortunately the same symbol Y is often used for both) For the same example the lefthanded strange quark has Q =  1/3, T_{3} = 1/2 and weak hypercharge +1/3. 


#5
May714, 01:25 PM

P: 754

comment/wondering:
hypercharge is just the charge of a U(1) group no need to distinguish between isospin or anything for SU(2), whether you work with the flavor or isospin or left SU(2) the mathematics are the same... In a complete theory for example, you can add up B,L,S,... to each definition, why? because it takes all the contributions of each U(1) that exists... When the symmetry of SU(2)xU(1) breaks, the resulting U(1)_Q will get the contribution from U(1)_Y and also a contribution from the SU(2) (by finding the subalgebras with dynkin diagrams)... 


#6
May714, 02:18 PM

Sci Advisor
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P: 4,160




#7
May714, 02:18 PM

P: 1,020




#8
May714, 02:28 PM

P: 754

SU(2) is just a single case... removing the 1 root will give you the U(1) alone...



#9
May714, 02:35 PM

P: 754

For example, whether you are working with spin or isospin or any quantity, the generators of SU(2) can be always chosen to be the pauli matrices (so the algebra won't change because of the physics for example I can treat isospin and spin in exactly the same way without making any distinction between them ).... The same for the hypercharge, being the generator/charge of a U(1)... 


#10
May714, 03:24 PM

P: 1,020




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