Editing Earnshaw's theorem (section)

===Fixed-orientation magnetic dipole===
It will be proven that at all points in free space
<math display="block"> \nabla \cdot (\nabla U) = \nabla^2 U = {\partial^2 U \over {\partial x}^2} + {\partial^2 U \over {\partial y}^2} + {\partial^2 U \over {\partial z}^2} = 0.</math>

The energy ''U'' of the magnetic dipole '''M''' in the external magnetic field '''B''' is given by
<math display="block">U = -\mathbf{M}\cdot\mathbf{B} = -M_x B_x - M_y B_y - M_z B_z.</math>

The Laplacian will be
<math display="block">\nabla^2 U = -\frac{\partial^2}{{\partial x}^2} \left(M_x B_x + M_y B_y + M_z B_z\right) - \frac{\partial^2}{{\partial y}^2} \left(M_x B_x + M_y B_y + M_z B_z\right) - \frac{\partial^2}{{\partial z}^2} \left(M_x B_x + M_y B_y + M_z B_z\right)</math>

Expanding and rearranging the terms (and noting that the dipole '''M''' is constant) we have
<math display="block">\begin{align}
  \nabla^2 U &= -M_x\left({\partial^2 B_x \over {\partial x}^2} + {\partial^2 B_x \over {\partial y}^2} + {\partial^2 B_x \over {\partial z}^2}\right) - M_y\left({\partial^2 B_y \over {\partial x}^2} + {\partial^2 B_y \over {\partial y}^2} + {\partial^2 B_y \over {\partial z}^2}\right) - M_z\left({\partial^2 B_z \over {\partial x}^2} +{\partial^2 B_z \over {\partial y}^2} +{\partial^2 B_z \over {\partial z}^2}\right)\\[3pt]
             &= -M_x \nabla^2 B_x - M_y \nabla^2 B_y - M_z \nabla^2 B_z
\end{align}</math>

but the Laplacians of the individual components of a magnetic field are zero in free space (not counting electromagnetic radiation) so
<math display="block">\nabla^2 U = -M_x 0 - M_y 0 - M_z 0 = 0,</math>

which completes the proof.