Uniqueness of magnetic vector potential

[fricke]

I able to prove magnetic field is uniquely determined but I am confused how to prove that magnetic vector potential is also unique.

Can I say that magnetic vector potential is uniquely determined since magnetic field has unique solution?

Thanks.

 

[ShayanJ]

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.

 

[fricke]

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.

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?

 

[ShayanJ]

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?

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]

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 $\vec{A'}=\vec A+\vec\nabla \phi \Rightarrow \vec\nabla\times\vec {A'}=\vec\nabla\times\vec A$.

thank you very much! I understand now! thank you.