Hertz Potentials | M.E. Irizarry-Gelpí

Wed 30 January 2019
Physics
#hertz

In classical electrodynamics, you have three important objects: the \(A\) potential, the \(F\) field, and the \(J\) current. The potential and the field are related via

\begin{equation*} F^{ab} = \partial^{a} A^{b} - \partial^{b} A^{a} \end{equation*}

The field and the source density are (dynamically) related via

\begin{equation*} J^{b} = \partial_{a} F^{ab} \end{equation*}

These are a subset of the Maxwell equations. Since \(F^{ab}\) is antisymmetric, you have the (dynamic) continuity equation:

\begin{equation*} \partial_{a} J^{a} = 0 \end{equation*}

From the potential, differentiation gets you to the field. From the field, differentiation gets you to the source density.

The Lorenz equation for the \(A\) potential is

\begin{equation*} \partial_{a} A^{a} = 0 \end{equation*}

When the Lorenz equation for \(A\) holds, you have

\begin{equation*} \partial_{a} F^{ab} = \partial^{2} A^{b} = J^{b} \end{equation*}

Thus, \(J\) acts as the source term in the inhomogeneous wave equation for \(A\).

The Lorenz equation for \(A\) looks just like the continuity equation, which suggests writing the potential as

\begin{equation*} A^{b} = \partial_{a} B^{ab} \end{equation*}

Here \(B\) is the Hertz potential. The Hertz potential is antisymmetric, since it plays the role of the \(F\) field. In terms of the Hertz potential, you have

\begin{equation*} F^{ab} = \partial^{a} \partial_{c} B^{cb} - \partial^{b} \partial_{c} B^{ca} \end{equation*}

Suppose that the \(B\) potential satisfies a Bianchi equation:

\begin{equation*} \partial^{c} B^{ab} + \partial^{b} B^{ca} + \partial^{a} B^{bc} = 0 \end{equation*}

Then it follows that

\begin{equation*} F^{ab} = \partial^{2} B^{ab} \end{equation*}

That is, \(F\) acts as the source term in the inhomogeneous wave equation for \(B\).

Now suppose that \(B\) can be further written as

\begin{equation*} B^{ab} = \partial^{a} C^{b} - \partial^{b} C^{a} \end{equation*}

Here \(C\) is a secondary Hertz potential. If you impose a Hertz equation on \(C\),

\begin{equation*} \partial_{a} C^{a} = 0 \end{equation*}

then it follows that

\begin{equation*} A^{a} = \partial^{2} C^{a} \end{equation*}

That is, \(A\) acts as the source term in the inhomogeneous wave equation for \(C\).

At this stage, after all these assumptions, you have \(A\) playing the role of \(J\), \(B\) playing the role of \(F\), and \(C\) playing the role of \(A\). Since \(J\) can contribute singular terms from monopoles (1-poles), and \(F\) can contribute singular terms from dipoles (2-poles), I wonder if it is also true that \(A\) can contribute singular terms from quadrupoles (4-poles), that \(B\) can contribute singular terms from octupoles (8-poles), and that \(C\) can contribute singular terms from hexadecapoles (16-poles).

More generally, you have the \(H\) field related to the current:

\begin{equation*} \partial_{a} H^{ab} = J^{b} \end{equation*}

The \(H\) and \(F\) fields are related via

\begin{equation*} H^{ab} = F^{ab} - P^{ab} \end{equation*}

Here, \(P\) describes the polarization and magnetization. Both \(H\) and \(F\) are antisymmetric.