Vector Potential and Zero Divergence

[hellsingfan]

I'm trying to understand when a vector field is equal to the curl of a vector potential. Why is it possible that there is always a vector potential with zero divergence?

relevant Equation:

B=∇χA

I'm trying to understand the proof that the above vector potential A can be one with zero divergence.

 

[WannabeNewton]

We know that if we perform a gauge transformation $A = A' + \triangledown \xi$, where $\xi$ is an arbitrary scalar field, then both $A'$ and $A$ result in the same observed magnetic field i.e. $B = \triangledown \times A' = \triangledown \times A$ (and of course, as usual, we have to perform the associated gauge transformation of the scalar potential to keep the observed electric field the same).

Say we are given a vector potential $A'$. We can find a $\xi$ that solves $\triangledown ^{2}\xi = -\triangledown \cdot A'$. Performing the gauge transformation $A = A' + \triangledown \xi$ we see that $\triangledown \cdot A = \triangledown \cdot A' + \triangledown ^{2}\xi = 0$ hence we can fix this gauge (again, after performing the associated gauge transformation of the scalar potential) so that we have $B = \triangledown \times A, \triangledown \cdot A = 0$. This is called the Coulomb gauge.

 

[hellsingfan]

Thank You!

 

[WannabeNewton]

No problem!