Editing Displacement current (section)

===Wave propagation===
The added displacement current also leads to wave propagation by taking the curl of the equation for magnetic field.<ref name=Slater2>{{cite book |title=Electromagnetism |page=91 |author=JC Slater and NH Frank |edition=op. cit. |isbn=978-0-486-62263-7 |url=https://books.google.com/books?id=GYsphnFwUuUC&pg=PA91|year=1969 |publisher=Courier Corporation }}</ref>

<math display=block>\mathbf{J}_\mathrm{D} = \epsilon_0\frac{\partial \mathbf{E}}{\partial t}\,.</math>

Substituting this form for {{math|'''J'''}} into [[Ampère's law]], and assuming there is no bound or free current density contributing to {{math|'''J'''}}:

<math display=block>\nabla \times \mathbf{B} = \mu_0 \mathbf{J}_\mathrm{D}\,,</math>

with the result:

<math display=block>\nabla \times \left(\nabla \times \mathbf{B} \right) = \mu_0 \epsilon_0 \frac{\partial}{\partial t} \nabla \times \mathbf{E}\,.</math>

However,
<math display=block>\nabla \times \mathbf{E} = -\frac{\partial}{\partial t} \mathbf{B}\,,</math>

leading to the [[wave equation]]:<ref name=King>{{cite book |page=182 |title=Wave Motion |author=J Billingham, A C King |isbn=978-0-521-63450-2 |publisher=Cambridge University Press |url=https://books.google.com/books?id=bNePaHM20LQC&pg=PA182|year=2006}}</ref>
<math display=block>-\nabla \times \left( \nabla \times \mathbf{B} \right) = \nabla^2 \mathbf{B} =\mu_0 \epsilon_0 \frac {\partial^2}{\partial t^2} \mathbf{B} = \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \mathbf{B}\,,</math>

where use is made of the vector identity that holds for any vector field {{math|'''V'''('''r''', ''t'')}}:

<math display=block>\nabla \times \left(\nabla \times \mathbf{V}\right) = \nabla \left(\nabla \cdot \mathbf{V}\right) - \nabla^2 \mathbf{V}\,,</math>

and the fact that the divergence of the magnetic field is zero. An identical wave equation can be found for the electric field by taking the ''curl'':

<math display=block>\nabla \times \left(\nabla \times \mathbf{E} \right) = -\frac {\partial}{\partial t}\nabla \times \mathbf{B} = -\mu_0 \frac {\partial}{\partial t} \left(\mathbf{J} + \epsilon_0\frac {\partial}{\partial t} \mathbf{E} \right)\,.</math>

If {{math|'''J'''}}, {{math|'''P'''}}, and {{mvar|ρ}} are zero, the result is:

<math display=block>\nabla^2 \mathbf{E} = \mu_0 \epsilon_0 \frac{\partial^2}{\partial t^2} \mathbf{E} = \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \mathbf{E}\,.</math>

The electric field can be expressed in the general form:

<math display=block>\mathbf{E} = - \nabla \varphi - \frac{\partial \mathbf{A}}{\partial t}\,,</math>

where {{mvar|φ}} is the [[electric potential]] (which can be chosen to satisfy [[Poisson's equation]]) and {{math|'''A'''}} is a [[vector potential]] (i.e. [[magnetic vector potential]], not to be confused with surface area, as {{math|'''A'''}} is denoted elsewhere). The {{math|∇''φ''}} component on the right hand side is the Gauss's law component, and this is the component that is relevant to the conservation of charge argument above. The second term on the right-hand side is the one relevant to the electromagnetic wave equation, because it is the term that contributes to the curl of {{math|'''E'''}}. Because of the vector identity that says the curl of a gradient is zero, {{math|∇''φ''}} does not contribute to {{math|∇×'''E'''}}.