Timeline for Massless charged particles
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 11, 2011 at 14:33 | comment | added | user4552 | You don't have to appeal to collider experiments. If there were massless charged particles, there would be obvious effects that we would notice in everyday life. We'd see pair production when visible-light photons interacted with matter. | |
| Apr 2, 2011 at 20:26 | comment | added | Gerard | What does it mean when you say: "the fine-structure constant is running" - and then "above the mass of the electron"? | |
| Apr 2, 2011 at 17:37 | comment | added | Luboš Motl | BTW preemptively, you could also ask about the running of the $SU(3)$ QCD coupling. Indeed, this $SU(3)$ is not broken - but rather confined - which means that it's running to arbitrarily long distances, not only the Compton wavelength of the lightest quarks. In some sense, the potential energy goes like $|r|$ for long distances (although it gets converted to quark-antiquark pairs if $|r|$ is too large). | |
| Apr 2, 2011 at 17:34 | comment | added | Luboš Motl | Dear John, the $SU(2)\times U(1)$ symmetry is broken at the electroweak scale, 246 GeV or so, which means that the corresponding potential isn't just $g^2/r$, the Coulomb Ansatz, but $g^2 \exp(-vr)/r$: it is exponentially decreasing at distances longer than the W-boson Compton wavelength. This "classical" exponential decrease is far more important than some logarithmic corrections from the running. Your question effectively assumes that the potential is $g(r)^2/r$ even at long distances which is surely wrong. But yes, neutrino loops of course make some impact on all processes for $E>m_\mu$. | |
| Apr 2, 2011 at 10:21 | comment | added | Marek | @John: for weak interactions you need to consider weak isospin (because the group is SU(2) and acts on doublets). Because it has to be conserved you always need to include some massive particles in your weak diagrams too. In your particular case, you'd have e.g. $W^- \to e^- \bar{\nu}$. | |
| Apr 2, 2011 at 8:21 | comment | added | John | Using that logic, does that mean the weak force constant should run all the way to the neutrino mass, since the neutrinos have weak charge? | |
| Apr 2, 2011 at 5:43 | history | answered | Luboš Motl | CC BY-SA 2.5 |