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A question regarding Y=B+S by a nuclear physics toddler

by rpndixit5
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rpndixit5
#1
May7-14, 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.
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kjhskj75
#2
May7-14, 06:27 AM
P: 11
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 anti-K 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 anti-baryons 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.
ChrisVer
#3
May7-14, 12:38 PM
P: 891
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 Weinberg-Salam-Glashow 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

Bill_K
#4
May7-14, 01:14 PM
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A question regarding Y=B+S by a nuclear physics toddler

Quote Quote by ChrisVer View Post
In the context of the Weinberg-Salam-Glashow 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
ChrisVer, you need to be careful to distinguish between isospin and hypercharge (which is what the OP is about) from weak isospin and weak hypercharge (which is what you're describing!)

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 = I3 + Y/2. For example the strange quark has Q = -1/3, I3 = 0 and hypercharge -2/3.

Weak isospin T arises from the symmetry of left-handed lepton pairs under weak interactions. Right-handed leptons form a singlet under weak isospin. Weak hypercharge comes from Q = T3 + Y/2. (Unfortunately the same symbol Y is often used for both) For the same example the left-handed strange quark has Q = - 1/3, T3 = -1/2 and weak hypercharge +1/3.
ChrisVer
#5
May7-14, 01:25 PM
P: 891
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)...
Bill_K
#6
May7-14, 02:18 PM
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Quote Quote by ChrisVer View Post
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...
No, strong isospin and weak isospin are mathematically quite different. Strong isospin is an "accidental" degeneracy that results from pretending the first two quark flavors are identical. Similarly one can also define U-spin (s and d identical) and V-spin (s and u identical). None of them is related to the underlying symmetry group of the strong interactions, color SU(3).
andrien
#7
May7-14, 02:18 PM
P: 1,020
by finding the subalgebras with dynkin diagrams
Are you going to find the subalgebra of su(2) using a Dynkin diagram? You know the Dynkin diagram here is a point, or you have something else in mind.
ChrisVer
#8
May7-14, 02:28 PM
P: 891
SU(2) is just a single case... removing the 1 root will give you the U(1) alone...
ChrisVer
#9
May7-14, 02:35 PM
P: 891
Quote Quote by Bill_K View Post
No, strong isospin and weak isospin are mathematically quite different. Strong isospin is an "accidental" degeneracy that results from pretending the first two quark flavors are identical. Similarly one can also define U-spin (s and d identical) and V-spin (s and u identical). None of them is related to the underlying symmetry group of the strong interactions, color SU(3).
I still cannot understand how this defies what I'm saying...
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)...
andrien
#10
May7-14, 03:24 PM
P: 1,020
Quote Quote by ChrisVer View Post
SU(2) is just a single case... removing the 1 root will give you the U(1) alone...
You need to use an embedding in general, the other part also matters since SU(2)≠ U(1) U(1)


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