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
 
\[ \nabla^2 \mathbf{A}= - \mathbf{J} \]