Timeline for Proton scattering off electrons
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| S Nov 21, 2016 at 10:26 | history | suggested | Mass | CC BY-SA 3.0 |
fixed math typos
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| Nov 21, 2016 at 10:10 | review | Suggested edits | |||
| S Nov 21, 2016 at 10:26 | |||||
| Nov 21, 2016 at 7:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 22, 2016 at 1:08 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 20, 2016 at 5:35 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 17, 2016 at 20:57 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added 5 characters in body; edited tags
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| Aug 17, 2016 at 19:31 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 13, 2016 at 18:53 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 11, 2016 at 10:56 | comment | added | Marty Green | What about scattered radiation? At some point, if the speeds are high enough, isn't radiation generated by the collision and consequent acceleration of the two particles? Then the energy of the outgoing particles must be less than the incoming energy. | |
| Jun 11, 2016 at 10:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 3, 2015 at 16:46 | answer | added | Neoh | timeline score: 1 | |
| Apr 17, 2013 at 21:19 | comment | added | dmckee --- ex-moderator kitten | Yes, but in DIS you don't get a proton out. You get a haronic spray. You can also scatter off a proton and create a $\Delta^+$, but again, you don't have a proton in the final state. Or produce pions or kaons, but then you have more than two particles in the final state. If you have an electron and proton in and exactly an electron and a proton out there is no freedom to be inelastic. | |
| Apr 17, 2013 at 20:45 | comment | added | Dmist | @dmckee sorry I don't quite follow your second sentence, doesn't deep inelastic scattering contradict what you're saying (I may well be confusing myself here) | |
| Apr 17, 2013 at 12:29 | comment | added | dmckee --- ex-moderator kitten | "I'm not sure whether this is inelastic scattering or elastic scattering" Electrons are structureless, point particles at attainable energies and protons get a new name if they are excited. There is no place for the energy to go in a inelastic case which wouldn't be reflected in the description of the interaction. | |
| Apr 17, 2013 at 10:13 | history | asked | Dmist | CC BY-SA 3.0 |