Timeline for Transformations of electroweak gauge field $W_\mu$ under $U(1)_{e.m.}$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 12, 2015 at 18:03 | vote | accept | Riccardo Buscicchio | ||
| Jun 12, 2015 at 17:49 | answer | added | Riccardo Buscicchio | timeline score: 0 | |
| Jun 11, 2015 at 17:06 | comment | added | Riccardo Buscicchio | So the $U(1)_{e.m.}$ mixes $W_\mu^+$ and $W_\mu^-$ fields as any other doublet? I wasn't very sure if the $U(1)e.m.$ as a residual of $SU(2)xU(1)$ break would behave so simple. | |
| Jun 11, 2015 at 12:01 | comment | added | ACuriousMind♦ | The $\mathrm{U}(1)_{e.m.}$ trafo is simply specified by the electric charge, $\mathrm{U}(1)$ representations are boring 1D reps which are phase multiplications, after all. | |
| Jun 11, 2015 at 11:57 | history | edited | Qmechanic♦ |
edited tags
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| Jun 11, 2015 at 11:55 | comment | added | Riccardo Buscicchio | Well,the gauge Group is broken, but not completely.there is a residual part,that is the elecromagnetic $U(1)$.then my question is: How does the fields defined in theory transform under this subgroup? Photon and leptons are quite straightforward,but what about $W\mu$? | |
| Jun 11, 2015 at 11:50 | history | edited | Riccardo Buscicchio | CC BY-SA 3.0 |
SU(1) into SU(2)
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| Jun 10, 2015 at 19:36 | comment | added | ACuriousMind♦ | What do you mean? Having the gauge symmetry broken means that there are no real gauge transformations left. | |
| Jun 10, 2015 at 16:51 | history | asked | Riccardo Buscicchio | CC BY-SA 3.0 |