Timeline for Gauge fixing for a semi-relativistic model involving electromagnetism
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 5, 2021 at 11:55 | comment | added | Jakob Möller | Got it, thanks for the clarification! | |
| Feb 5, 2021 at 11:53 | comment | added | Carlo Beenakker | $\nabla\cdot J=0\Rightarrow \nabla\cdot (\Delta A)=0\Rightarrow \Delta(\nabla\cdot A)=0$ and hence either $\nabla\cdot A=0$ or the vector potential is given by $A=A_0+\nabla\lambda$ with $\nabla\cdot A_0=0$ and $\Delta\nabla\lambda=0$; this remaining gauge freedom to choose $\lambda$ will force $\lambda\equiv 0$ if you add the condition that $\lambda\rightarrow 0$ for $r\rightarrow\infty$. | |
| Feb 5, 2021 at 11:11 | comment | added | Jakob Möller | Sorry I'm still confused, that was not my question: How does taking the divergence of the equation $-\Delta A = J$ imply $\nabla \cdot A =0$? | |
| Feb 4, 2021 at 16:32 | comment | added | Carlo Beenakker | let's see what freedom is left: $\nabla\cdot A=0$ satisfies current conservation, if you try $A'=A+\nabla \lambda$ then you need $\Delta\nabla\lambda=0$; that leaves some room, but if you wish $\lambda$ to become constant at infinity only $\lambda\equiv 0$ remains. | |
| Feb 4, 2021 at 13:08 | comment | added | Jakob Möller | Just to be sure, when taking the divergence you get $\Delta \nabla \cdot A = 0$ because $\nabla \cdot J=0$ but how does that imply $\nabla \cdot A = 0$? | |
| Feb 4, 2021 at 12:51 | history | undeleted | Carlo Beenakker | ||
| Feb 4, 2021 at 12:50 | history | deleted | Carlo Beenakker | via Vote | |
| Feb 4, 2021 at 12:49 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |