[Click here for a PDF of this post with nicer formatting]
Q:
Show that without the displacement current \( \PDi{t}{\BD} \), Maxwell’s equations will not satisfy conservation relations.
A:
Without the displacement current, Maxwell’s equations are
\begin{equation}\label{eqn:continuityDisplacement:20}
\begin{aligned}
\spacegrad \cross \BE( \Br, t ) &= – \PD{t}{\BB}(\Br, t) \\
\spacegrad \cross \BH( \Br, t ) &= \BJ \\
\spacegrad \cdot \BD(\Br, t) &= \rho_{\mathrm{v}}(\Br, t) \\
\spacegrad \cdot \BB(\Br, t) &= 0.
\end{aligned}
\end{equation}
Assuming that the continuity equation must hold, we have
\begin{equation}\label{eqn:continuityDisplacement:40}
\begin{aligned}
0
&= \spacegrad \cdot \BJ + \PD{t}{\rho_\mathrm{v}} \\
&= \spacegrad \cdot \lr{ \spacegrad \cross \BH } + \PD{t}{} (\spacegrad \cdot \BD) \\
&= \PD{t}{} (\spacegrad \cdot \BD) \\
&\ne 0.
\end{aligned}
\end{equation}
This shows that the current in Ampere’s law must be transformed to
\begin{equation}\label{eqn:continuityDisplacement:60}
\BJ \rightarrow \BJ + \PD{t}{\BD},
\end{equation}
should we wish the continuity equation to be satisfied. With such an addition we have
\begin{equation}\label{eqn:continuityDisplacement:80}
\begin{aligned}
0
&= \spacegrad \cdot \BJ + \PD{t}{\rho_\mathrm{v}} \\
&= \spacegrad \cdot \lr{ \spacegrad \cross \BH – \PD{t}{\BD} } + \PD{t}{} (\spacegrad \cdot \BD) \\
&= \spacegrad \cdot \lr{ \spacegrad \cross \BH } – \spacegrad \cdot \PD{t}{\BD} + \PD{t}{} (\spacegrad \cdot \BD).
\end{aligned}
\end{equation}
The first term is zero (assuming sufficient continity of \(\BH\)) and the second two terms cancel when the space and time derivatives of one are commuted.