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multiple "colors" of Weak Force charge ? gluons have R/G/B, so do W-bosons have O/Y ?? Rate Topic: -----

#1 Widdekind 


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When quarks interact, via the Strong Force, the exchanged gluon changes the "colors" of both quarks. Seemingly, an initially "red" quark, "exudes" its "red-ness", into a gluon, which the other quark absorbs, so becoming "red".

Now, when particles interact, via the Weak Force, the exchanged W boson changes the "charges" of both particles. So, an initially "negative" electron, "extrudes" its "negative-ness", into a W-boson, which the other neutrino absorbs, so becoming "negative". So, is there an additional "Weak Color" that also changes?

Note, the Weak Force, like the Strong Force, is a short-ranged force, modeled mathematically as a "non-Abelian gauge field" (Nambu. Quark). Does that imply, that the Weak Force grows with distance; or evidences asymptotic freedom & infra-red-slavery; or that W/Z bosons can generate other W/Z bosons, i.e. is "EM glue" ?

This post has been edited by Widdekind: 8 January 2012 - 12:51 PM

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#2 mathematic 


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The weak force as a function of distance is nothing like the strong force. Look at chart in the following:

http://en.wikipedia....tal_interaction
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#3 Widdekind 


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If the 'Weak Hyper-Charges', of the fundamental quanta, are Y_W \equiv 2 ( Q - T_3 ), then apparently, if only for left-handed fermions:

Y_{W,u} = 1/3
Y_{W,d} = 1/3

Y_{W,\nu} = -1
Y_{W,e} = -1

So, apparently, all (left-handed) quarks interact Weakly, with a "Weak Charge", of 1/3. And, all leptons interact Weakly, with a "Weak Charge" of -1. For example, neutrinos carry the same Weak Charge as electrons, but carry no EM Charge.

Now, if, in the EM & Strong interactions, "like repels like" whilst "opposites attract"; then does that imply, that quarks repel themselves Weakly, and leptons repel themselves Weakly, but quarks attract leptons Weakly ??



For right-handed fermions, T = 0. So, the 'Weak Hyper-Charges', of those fundamental quanta, would be Y_W \equiv 2 ( Q - T_3 ) = 2 Q:

Y_{W,u} = 4/3
Y_{W,d} = -2/3

Y_{W,\nu} = 0
Y_{W,e} = -2

But, right-handed fermions, having no "weak isospin", "do not undergo weak interactions", implying that their effective Y_W = 0. Since Y_W is conserved in all reactions, i.e. is a "good quantum number", then when left-handed fermions enter into Weak interactions, they must emerge from those interactions left-handed, e.g. neutrinos which approach an interaction left-handed, i.e. spin directed anti-parallel to momentum, they must not "spin flip" upon absorbing or emitting Weak bosons; or, at least if they do "spin flip", to account for the S=1 of the W-bosons, then their momenta must also change direction too. Does this mean, that when a neutrino approaches a down quark, that the emission of a W- boson from the latter, to the former, reverses both of their spins (to account for the emission / absorption, of an S=1 quanta), and so reverses both of their momenta as well (to preserve the left-handed, anti-parallel, relation, between spin & momenta) ????????????????

\left( \downarrow \nu_e \right) \rightarrow \leftarrow \left( d \uparrow \right)

\left( \downarrow  \nu_e \right) \rightarrow \leftarrow W^{-} \; \left( \downarrow u \right) \rightarrow

\leftarrow \left( e^{-} \uparrow \right) \; \left( \downarrow u \right) \rightarrow

I guess the exchange, of a super-massive 80 GeV Weak boson represents a large momentum transfer, so that both previously impinging particles "bounce back" from the interaction, having had both their angular momenta, and linear momenta, reversed ??



Note, too, that, for all of the EW bosons, T_3 = Q, so that Y_W = 2(Q - T_3) = 0, i.e. explicitly:

Y_{W,W^{+}} = 0
Y_{W,W^{-}} = 0

Y_{W,Z^0} = 0
Y_{W,\gamma^{\;}} = 0

Does that imply, that, unlike gluons, W-bosons carry no Weak charge, and so cannot emit other W-bosons ??
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