\documentclass[12pt,a4paper]{article}
\usepackage{latexsym}
\linespread{1.4}
\date{16. August 2005}
\author{Ulrich~\textsc{Bruchholz}\footnote{Dipl.-Ing.
\textsl{Ulrich Bruchholz},~http://www.bruchholz-acoustics.de}}
\title{Derivation of \textsc{Planck}'s~Constant
from~\textsc{Maxwell}'s~Electrodynamics}
\begin{document}
\sloppypar
\maketitle
Let us go from \textsc{Maxwell}'s equations of the vacuum that culminate
in wave equations for the electric potential
\begin{equation} \Box \varphi = 0 \end{equation}
and
\begin{equation} \Box \mathcal{A} = 0 \end{equation}
for the magnetic vector potential.
Take the wave solution from equ. (2), in which the vector
potential consist of a single component vertical to the propagation
direction
\begin{equation}
A_y = A_y \Bigl( \omega {} \cdot (t-x) \Bigr) \qquad ,
\end{equation}
in which is set $ c = 1 $ (normalization). \\
$ \omega $ is a constant, and is identical with the circular frequency
at waves. \\
\textit{x} means the direction of the propagation. \\
$ A_y $ is an \emph{arbitrary} real function from
$ \omega \cdot (t-x) $~, and independent on $ y , z $.
The field strengthes respectively flow densities (which are the same
in the vacuum) become
\begin{equation}
E_y = \frac{\partial A_y}{\partial t} =
\omega A_y ' \Bigl( \omega \cdot (t-x) \Bigr) \qquad ,
\end{equation}
and
\begin{equation}
B_z = - \frac{\partial A_y}{\partial x} =
\omega A_y ' \Bigl( \omega \cdot (t-x) \Bigr) \qquad .
\end{equation}
$ A_y ' $ means the total derivative. \\
The energy density of the field results in
\begin{equation}
\eta = \frac{\varepsilon_{\circ}}{2} \cdot ( E_y ^2 + B_z ^2 ) =
\omega ^2 \varepsilon_{\circ} A_y ' ^2 \Bigl( \omega \cdot (t-x) \Bigr) \qquad .
\end{equation}
The geometric theory of fields allows geometric boundaries from the
non-linearities in the equations of this theory. [1]
If one assumes such boundary, like those in stationary solutions
of the non-linear equations, the included energy becomes
the volume integral within this boundary
\begin{equation}
\int\!\!\!\int\!\!\!\int \eta~\mathsf{d} (t-x)~\mathsf{d} y~\mathsf{d} z
\quad = \quad
\omega \quad \varepsilon_{\circ} \int\!\!\!\int\!\!\!\int A_y ' ^2
\Bigl( \omega \cdot (t-x) \Bigr)~\mathsf{d} \Bigl( \omega \cdot (t-x) \Bigr)
~\mathsf{d} y~\mathsf{d} z \quad .
\end{equation}
This volume integral were not possible without the boundary, because
the linear solution alone is not physical for the infinite extension.
We can write the last equation as
\begin{equation} E = \omega~\hbar \qquad , \end{equation}
or
\begin{equation} E = h~\nu \qquad , \end{equation}
because the latter volume integral has a constant value.
The fact that this value is always the same also means, that
exactly one solution exists with $ \omega $~as parameter. \\
With it, the fundamental relation of quantum mechanics follows
from classical fields.
Summarizingly, the derivation involves two predictions: \\
1) The photon has a geometric boundary. That may be the reason
of the particle behaviour. \\
2) There is exactly one wave solution. \\
This could be supported by numerical simulations, when the value
of above volume integral became ~$ \hbar / \varepsilon_{\circ} $~. \\
\\
\\
\\
{}\textsl{Reference~:}{} \\
{}[1]~\textsc{Bruchholz}, U.~: http://bruchholz.psf.net/article2.txt~, \\
with more references. \\
See also http://bruchholz.psf.net/Geometry.pdf~. \\
\\
\\
\\
\\
{}\footnotesize This document has been composed with \LaTeX{} .
\end{document}