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Isospin breaking, charge symmetry and charge indepedence 
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#1
May1714, 04:43 PM

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Hi, I'm just getting a little confused with all the definitions here and I need some confirmation on what I say is correct or not;
Isospin symmetry: The property that an interaction is independent of the [itex] T_3 [/itex] value? Isospin breaking: The property that it is dependent on [itex] T_3 [/itex]? Charge symmetry: The property that the interaction is unchanged upon swapping protons with neutrons and neutrons with protons. Charge independence: The interaction is the same for pp, pn and nn. I've honestly looked but I haven't managed to get a full confirmation on what isospin symmetry actually means. Are these correct? I swear I also have two sources telling me that the nuclear force is and isn't charge symmetric. The latter case because; 1) The proton and neutron have different masses 2) They have different EM contributions. But the nuclear force is also isospin breaking right? Especially for one pion exchange. Because the one pion exchanges for pp/nn are different for np (the former only have [itex] \pi^0 [/itex]). So is the nuclear force not charge symmetric, not charge independent, and isospin breaking? 


#2
May1714, 07:29 PM

Sci Advisor
P: 908

Sam 


#3
May1814, 06:01 AM

Sci Advisor
Thanks
P: 4,160




#4
May1814, 02:27 PM

Sci Advisor
P: 908

Isospin breaking, charge symmetry and charge indepedence
[tex]T^{a}= \int d^{3} x N^{ \dagger } \frac{ \sigma^{ a } }{ 2 }N + \epsilon^{ a b c } \phi^{ b } \dot{ \phi }^{ c } \ \ [/tex] The relation [itex]Q_{e} = B / 2 + T^{ 3 }[/itex] is correct in the lowest order, i.e. when you ignore the em interaction and it follows from the [itex]SU(2)[/itex] symmetric Lagrangian [tex] \mathcal{ L } = i \bar{ N } \gamma^{ \mu } \partial_{ \mu } N + (1 / 2 ) ( \partial_{ \mu } \phi^{ a } )^{ 2 } + i g \bar{N} \gamma^{5} \sigma . \phi N / 2 [/tex] which is also [itex]U_{B}(1)[/itex] invariant with conserved Bcharge given by [tex]B = \int d^{3} x N^{\dagger} N[/tex] 


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