Timeline for The vacuum as trigger
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 14, 2011 at 21:26 | vote | accept | Andersi2 | ||
| May 5, 2011 at 7:10 | comment | added | jdm | @Anders I: I don't think it makes sense in this case to talk about positions or spatial proximity. The virtual particle is not a billiard ball that knocks out something from the nucleus. In the case of beta decay, if you look at the Feynman diagram (en.wikipedia.org/wiki/…), a neutron turns into a proton, while "emitting" a virtual W boson (that turns into an antineutrino and an electron). The decay probability depends on the probability of the neutron to emit the W, among other factors. | |
| May 4, 2011 at 23:15 | comment | added | Andersi2 | Thanks, but does the hypothetical "homerun" depend on the spatial proximity to a given proton or neutron of the manifestation of the virtual particle? | |
| May 4, 2011 at 18:26 | comment | added | anna v | continued: that are allowed energetically. No, in general they cannot knock off a proton or neutron. | |
| May 4, 2011 at 18:24 | comment | added | anna v | @Anders I Nuclear interactions and the energies available for them are of the order of MeVs and keVs . The nuclear "force" can be well approximated by a potential well, as in the shell model which describes the behavior of nuclei quite well. This force is a manifestation of a spill over from the fundamental interactions with quarks and gluons that hold the protons and neutrons together, a bit like the Van der Waals forces for electromagetism in atomic physics. The virtual particles are way off their mass shell because of the small energies available and can play role only in decays | |
| May 4, 2011 at 7:53 | comment | added | Andersi2 | A virtual particle then can't "knock" a neutron or proton out of the nucleus? | |
| May 4, 2011 at 6:10 | history | answered | Luboš Motl | CC BY-SA 3.0 |
