Timeline for Non-relativistic limit of complex scalar field Lagrangian
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:39 | history | edited | CommunityBot | replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/ | |
| Jul 1, 2015 at 5:07 | vote | accept | Physics_Plasma | ||
| Jun 27, 2015 at 19:05 | history | edited | Qmechanic♦ | CC BY-SA 3.0 | Added explanation |
| Jun 27, 2015 at 18:45 | comment | added | Physics_Plasma | Oh, I might see it now. If the energy E that is remaining is not including the rest energy, then $E = \frac{p^2}{2m}$. Since we are investigating the case where $E\ll m$, then given E is just kinetic energy, then we get $E \ll \frac{p^2}{2E}$, or $E \ll p$. | |
| Jun 27, 2015 at 18:10 | comment | added | Qmechanic♦ | Only indirectly. | |
| Jun 27, 2015 at 17:48 | comment | added | Physics_Plasma | Ok, so if I plug the plane wave expansion into the Lagrangian and keep the terms in the same order as they are in my second equation from the bottom, then I get $\mathcal{L} = E\Psi^{\dagger}\Psi + \frac{E^2}{2m}\Psi^{\dagger}\Psi - \frac{p^2}{2m}\Psi^{\dagger}\Psi -\frac{\lambda}{4m^2}(\Psi^{\dagger}\Psi)^2. $ I can clearly see how the second term is negligible compared to the first. Is there a reason why the second term is negligible compared to the third term, with the square of the momentum? | |
| Jun 27, 2015 at 9:02 | history | edited | Qmechanic♦ | CC BY-SA 3.0 | added 19 characters in body |
| Jun 27, 2015 at 8:49 | history | answered | Qmechanic♦ | CC BY-SA 3.0 |