- #1

- 6,002

- 2,625

$$\nabla\cdot\vec{B}=0\Rightarrow \nabla\cdot(\nabla\chi+\nabla\times\vec{A})=0\Rightarrow\nabla^2\chi=0$$

Maxwell's Ampere's law seems to give us no additional information about this potential since it will be :

$$\nabla\times\vec{B}=\nabla\times (\nabla\chi+\nabla\times\vec{A})=\nabla\times(\nabla\chi)+\nabla\times(\nabla\times\vec{A})=0+\nabla\times(\nabla\times\vec{A})$$

So can we prove somehow that in most cases we have ##\chi=0## . Are there any specific systems known that ##\chi\neq 0##?