## Derivation of the Magnetic Vector Potential for Stationary Electric and Magnetic Fields

For stationary electric and magnetic fields, Maxwells' equations give
$\mathbf{\nabla} \cdot \mathbf{H} =\mathbf{0}$
and
$\mathbf{\nabla} \times \mathbf{H} = \mathbf{J}$

Since
$\mathbf{\nabla} \cdot \mathbf{H} =\mathbf{0}$
we can write
$\mathbf{H} = \mathbf{\nabla} \times \mathbf{A}$
for a suitable vector field
$\mathbf{A}$
,  called the magnetic vector potential.
The choice of
$\mathbf{H}$
is not uniqu. In fact we can find
$\mathbf{A, \;A'}$
such that
$\mathbf{H} = \mathbf{\nabla} \times \mathbf{A} = \mathbf{\nabla} \times \mathbf{A'}$

In this case
$\mathbf{\nabla} \times (\mathbf{A} - \mathbf{A'})= \mathbf{0}$
.
It then follows that
$\mathbf{A}- \mathbf{A'} = \mathbf{\nabla} \phi \rightarrow \mathbf{A} = \mathbf{A'} + \mathbf{\nabla} \phi$

Where
$\phi$
is some scalar function.
Taking the divergence of the second equation above gives
$\mathbf{\nabla} \cdot \mathbf{A} =\mathbf{\nabla} \cdot \mathbf{A'} + \nabla^2 \phi$

We can choose
$\phi$
so that
$\nabla \cdot \mathbf{A} = 0$

Hence, in magnetostatics, the magnetic vector potential is chosen so that
$\mathbf{H} =\mathbf{\nabla} \times \mathbf{A}$
and
$\mathbf{\nabla} \cdot \mathbf{A} = \mathbf{0}$

Substitute
$\mathbf{H} = \mathbf{\nabla} \times \mathbf{A}$
into
$\mathbf{\nabla} \times \mathbf{H} = \mathbf{J}$
to obtain
$\mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{A}) = \mathbf{J}$

Now use the identity
$\mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{A}) = \mathbf{\nabla} (\mathbf{\nabla} \cdot \mathbf{A})- \nabla^2 \mathbf{A}$

Since
$\mathbf{\nabla} (\mathbf{\nabla} \cdot \mathbf{A})=0$
, we have

$\nabla^2 \mathbf{A}= - \mathbf{J}$