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\date{21 January 2006}
\author{Ulrich~\textsc{Bruchholz}\footnote{Dipl.-Ing.
\textsl{Ulrich Bruchholz},~http://www.bruchholz-acoustics.de}}
\title{Why~Material~Quantities~Must~be Integration~Constants}
% Actually trivial but refused to believe

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Let us take force and energy in electromagnetic fields.
This is derived for example from several charges, e.g.
{\large \begin{equation}
K = -\frac{Q_{1} Q_{2}}{4 \pi \varepsilon |\mathbf{r}_{2} - \mathbf{r}_{1}|^2}
\quad , \qquad
W = \frac{Q_{1} Q_{2}}{4 \pi \varepsilon |\mathbf{r}_{2} - \mathbf{r}_{1}|}
\quad .
\end{equation}\par}
In which, one charge does not act to itself.

The transition to very many and very small charges and currents
leads, with tensor notation, to
{\large \begin{equation}
K^\mu = F^\mu{}_{\nu} S^{\nu} \quad , \qquad
T_{\mu\nu} = F_{\mu\alpha}F_\nu{}^{\alpha} - \frac{1}{4} g_{\mu\nu}
F_{\alpha\beta}F^{\alpha\beta} \quad , 
\end{equation}\par}
with
{\large \begin{displaymath}
K^\mu = T^{\mu\nu}{}_{;\nu} \quad ,
\end{displaymath}\par}
that means, the force density is the divergence of the energy-momentum
tensor.

This transition is fairly difficult, because it does not more distinguish
what acts to itself and what not. But one can see that the sources,
i.e. distributed charges and currents, do not appear in the energy
and momentum components. Moreover, this energy-momentum tensor meets
the \textsc{Einstein} equation only, if the sources vanish. That
identically means, the energy law is met only under this condition.

Each try introducing additional terms for the energy-momentum
tensor, in order to save the sources, has failed up to now.
But there is an alternative: \emph{The integration constants of
the source-free \textsc{Maxwell} equations.} Only with them, the
unification of gravitation with electromagnetism becomes consistent.

Actually, the discrete charges in equ.(1) are integration constants
anyway. Otherwise, above mentioned transition were impossible.

Taking the energy of the field according to equ.(2), one does
usually not know how to integrate it.\footnote{An exception is
to find in [3].} One could doubt if these components are real
energies and momenta. For that reason, energy must be an
integration constant. With the equivalence of mass and energy,
mass is an integration constant too.

Taking the energy-momentum tensor of the distributed mass
and the resulting force density
{\large \begin{equation}
T^{\mu\nu} = \rho~\frac{\mathsf{d}x^\mu}{\mathsf{d}s}
~\frac{\mathsf{d}x^\nu}{\mathsf{d}s} \quad , \qquad
K^\mu = \rho~k^\mu \quad ,
\end{equation}\par}
the clean solution consists in~$ \rho = 0 $~.
The alternative $ k^\mu = 0 $ may have heuristic meaning, because
it provides equations of motion.\footnote{Geometrically, it describes
geodesics.} But the needed motion is not given in each case.

Under these aspects, only the geometric theory of fields
provides the way out of the inconsistencies from any sources.
\\ \\ \\
\textsl{References~:}\\
{}[1] \textsc{Wunsch}, G.: Theoretische Elektrotechnik. Lectures
at Technische Universit\"at Dresden, 1966-1968.\\
{}[2] \textsc{Bruchholz}, U.: http://bruchholz.psf.net/article2.pdf~,
with further references.\\
{}[3] \textsc{Bruchholz}, U.: http://bruchholz.psf.net/h-article.pdf~.\\
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{\footnotesize This document has been composed with \LaTeX{}~.}
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