Editing Earnshaw's theorem (section)

===Laplacian of individual components of a magnetic field===
It is proven here that the Laplacian of each individual component of a magnetic field is zero. This shows the need to invoke the properties of magnetic fields that the [[divergence]] of a magnetic field is always zero and the [[curl (mathematics)|curl]] of a magnetic field is zero in free space. (That is, in the absence of current or a changing electric field.) See [[Maxwell's equations]] for a more detailed discussion of these properties of magnetic fields.

Consider the Laplacian of the x component of the magnetic field
<math display="block">\begin{align}
\nabla^2 B_x &= \frac{\partial^2 B_x}{\partial x^2} + \frac{\partial^2 B_x}{\partial y^2} + \frac{\partial^2 B_x}{\partial z^2} \\
&= \frac{\partial}{\partial x} \frac{\partial B_x}{\partial x}  + \frac{\partial}{\partial y} \frac{\partial B_x}{\partial y} + \frac{\partial}{\partial z} \frac{\partial B_x}{\partial z}
\end{align}</math>

Because the curl of '''B''' is zero,
<math display="block">\frac{\partial B_x}{\partial y} = \frac{\partial B_y}{\partial x},</math>
and
<math display="block">\frac{\partial B_x}{\partial z} = \frac{\partial B_z}{\partial x},</math>
so we have
<math display="block">\nabla^2 B_x = \frac{\partial}{\partial x} \frac{\partial B_x}{\partial x} + \frac{\partial}{\partial y}\frac{\partial B_y}{\partial x} + \frac{\partial}{\partial z}\frac{\partial B_z}{\partial x}.</math>

But since ''B<sub>x</sub>'' is continuous, the order of differentiation doesn't matter giving
<math display="block">\nabla^2 B_x = {\partial \over \partial x}\left({\partial B_x \over \partial x} +{\partial B_y \over \partial y} +{\partial B_z \over \partial z} \right) = {\partial \over \partial x}(\nabla \cdot \mathbf{B}).</math>

The divergence of '''B''' is zero,
<math display="block">\nabla \cdot \mathbf{B} = 0,</math>
so
<math display="block">\nabla^2 B_x = {\partial \over \partial x}(\nabla \cdot \mathbf{B}) = 0.</math>

The Laplacian of the ''y'' component of the magnetic field ''B<sub>y</sub>'' field and the Laplacian of the ''z'' component of the magnetic field ''B<sub>z</sub>'' can be calculated analogously.  Alternatively, one can use the [[vector calculus identities|identity]]
<math display="block">\nabla^{2}\mathbf{B} = \nabla \left(\nabla \cdot \mathbf{B}\right) - \nabla \times \left( \nabla \times \mathbf{B} \right),</math>
where both terms in the parentheses vanish.