# M.E. Irizarry-Gelpí

Physics impostor. Mathematics interloper. Husband. Father.

# Hertz Potentials

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.