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mitchell porter
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In the standard model, electric charge isn't fundamental, it's a combination of "hypercharge" and what could be called "W0 boson charge".
In quantum field theory, a charge is (from the particle perspective) the quantum probability for emission/absorption of a force particle; or (from the field perspective) a coefficient in a differential equation, specifying how strongly a field is affected by a force field. Electric charge is related to the photon field, so the electric charge of the electron is (1) amplitude for electron to emit/absorb a photon (2) strength of the coupling between electron field and photon field.
And in the standard model, the photon field is actually a component of the combination of two fundamental fields, the "hypercharge boson" field B and the "W0 boson" field (the W0 being one out of three weak bosons, W0, W+, W-). The Higgs field interacts with all of (W+,W-,W0,B), and the positive ground-state energy of the Higgs field gives mass to all those fields, except in the special combination of W0,B that we call photon...
So the "photon charge" is actually a combination of "B charge" (hypercharge) and "W0 charge". Specifically, it's half the B charge, plus all the W0 charge. At the fundamental level, e.g. the electron is interacting with hypercharge field and W0 field with certain strengths, but we combine those in this way, to get the overall strength of its interaction with the "electromagnetic field". And that strength is its electric charge.
Now if you look at the generalized charges of particles in the standard model, you find that although e.g. left-handed electrons and right-handed electrons have the same electric charge, -1, they differ in these ingredients. See this table.
http://math.ucr.edu/~huerta/guts/node11.html
Y is hypercharge, I3 is W0 charge. The left-handed electron has hypercharge -1 and W0 charge -1/2. The right-handed electron has hypercharge -2 and W0 charge 0 (it doesn't interact with weak bosons at all). But applying the recipe "half the hypercharge plus all the W0 charge", in both cases you come out with an electric charge of -1.
So it seems strange because the similarity of behavior between left- and right-handed electrons, when they interact with photons, masks an underlying dissimilarity in how they interact with the hypercharge field and the W0 field, a dissimilarity which isn't visible when those interactions are aggregated appropriately.
There are two things I want to understand. First, is there some reason of symmetry or group theory that explains this "coincidence"? Second, what is the relationship between the formula that electric charge is "half the hypercharge plus all the W0 charge", and the formula that the photon field is "(cos theta_W times the B0 field) plus (sin theta_W times the W0 field)", where theta_W is an empirical parameter?
These are probably elementary questions, but I don't have as much time as I used to have, for figuring things out for myself. And a public discussion might be educational.
(Acknowledgement: This post started life as an email to M.G. and S.S.)
In quantum field theory, a charge is (from the particle perspective) the quantum probability for emission/absorption of a force particle; or (from the field perspective) a coefficient in a differential equation, specifying how strongly a field is affected by a force field. Electric charge is related to the photon field, so the electric charge of the electron is (1) amplitude for electron to emit/absorb a photon (2) strength of the coupling between electron field and photon field.
And in the standard model, the photon field is actually a component of the combination of two fundamental fields, the "hypercharge boson" field B and the "W0 boson" field (the W0 being one out of three weak bosons, W0, W+, W-). The Higgs field interacts with all of (W+,W-,W0,B), and the positive ground-state energy of the Higgs field gives mass to all those fields, except in the special combination of W0,B that we call photon...
So the "photon charge" is actually a combination of "B charge" (hypercharge) and "W0 charge". Specifically, it's half the B charge, plus all the W0 charge. At the fundamental level, e.g. the electron is interacting with hypercharge field and W0 field with certain strengths, but we combine those in this way, to get the overall strength of its interaction with the "electromagnetic field". And that strength is its electric charge.
Now if you look at the generalized charges of particles in the standard model, you find that although e.g. left-handed electrons and right-handed electrons have the same electric charge, -1, they differ in these ingredients. See this table.
http://math.ucr.edu/~huerta/guts/node11.html
Y is hypercharge, I3 is W0 charge. The left-handed electron has hypercharge -1 and W0 charge -1/2. The right-handed electron has hypercharge -2 and W0 charge 0 (it doesn't interact with weak bosons at all). But applying the recipe "half the hypercharge plus all the W0 charge", in both cases you come out with an electric charge of -1.
So it seems strange because the similarity of behavior between left- and right-handed electrons, when they interact with photons, masks an underlying dissimilarity in how they interact with the hypercharge field and the W0 field, a dissimilarity which isn't visible when those interactions are aggregated appropriately.
There are two things I want to understand. First, is there some reason of symmetry or group theory that explains this "coincidence"? Second, what is the relationship between the formula that electric charge is "half the hypercharge plus all the W0 charge", and the formula that the photon field is "(cos theta_W times the B0 field) plus (sin theta_W times the W0 field)", where theta_W is an empirical parameter?
These are probably elementary questions, but I don't have as much time as I used to have, for figuring things out for myself. And a public discussion might be educational.
(Acknowledgement: This post started life as an email to M.G. and S.S.)