The discussion revolves around the uniqueness of the magnetic vector potential in relation to the magnetic field. Participants explore whether the magnetic vector potential can be considered unique given that the magnetic field has a unique solution.
Participants generally agree that the magnetic vector potential is not uniquely determined due to the existence of infinitely many equivalent forms. However, there is ongoing uncertainty regarding the implications of this for the uniqueness of the magnetic field itself.
The discussion highlights the dependence on the choice of scalar fields in determining the magnetic vector potential and the implications for the uniqueness of the magnetic field. Unresolved assumptions regarding the definitions and conditions under which these potentials are considered may affect the conclusions drawn.
Shyan said:For any magnetic field ## \vec B ##, there are an infinite number of equivalent magnetic vector potentials##\vec A##(##\vec B=\vec \nabla\times\vec A ##), related by ## \vec {A'}=\vec A+\vec \nabla \phi ## for some scalar field ## \phi ##. So the magnetic vector potential of a magnetic field is not uniquely determined.
As I said, all those infinite number of vector potentials are equivalent, which means they all give the same magnetic field. That's because ## \vec \nabla \times \vec \nabla \phi=0 ## for any scalar field ## \phi ##. So ## \vec{A'}=\vec A+\vec\nabla \phi \Rightarrow \vec\nabla\times\vec {A'}=\vec\nabla\times\vec A ##.fricke said:thank you very much for the reply!
As what you have explained, magnetic vector potential is not uniquely determined since there are an infinite number of equivalent vector potential A, for magnetic field B. But does it mean B also not uniquely determined since we could choose any scalar field of A?
Shyan said:As I said, all those infinite number of vector potentials are equivalent, which means they all give the same magnetic field. That's because ## \vec \nabla \times \vec \nabla \phi=0 ## for any scalar field ## \phi ##. So ## \vec{A'}=\vec A+\vec\nabla \phi \Rightarrow \vec\nabla\times\vec {A'}=\vec\nabla\times\vec A ##.