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The T.E. Field

For the T.E. degrees of freedom the charge-current $S_\mu dx^\mu$ and the vector potential $A_\mu dx^\mu$ have the form given by
\begin{displaymath}
( S_{\tau}, S_{\xi}, S_{r}, S_{\theta})=\left(0,0,
\frac{1}{...
... S}{\partial \theta},
-r\frac{\partial S}{\partial r}\right)~,
\end{displaymath} (29)

and
\begin{displaymath}
( A_{\tau}, A_{\xi}, A_{r}, A_{\theta})=\left(0,0,
\frac{1}{...
...{\partial \theta},
-r\frac{\partial \psi}{\partial r}\right)~.
\end{displaymath} (30)

The electromagnetic field,

\begin{eqnarray*}
\frac{1}{2}F_{\mu\nu}dx^\mu \wedge dx^\nu &\equiv&
\hat E_{lon...
...d\theta
\qquad\qquad \qquad\quad\textrm{in}~P~\textrm{and}~F~,
\end{eqnarray*}



has the following components:
  In $I$ or in $II$ In $F$ or in $P$
$\hat E_{long.}:$ $ \displaystyle{\frac{1}{\xi}F_{\xi\tau}=0}$ $\displaystyle \frac{1}{\xi}F_{\tau\xi}=0$
$\hat E_r:$ $ \displaystyle\frac{1}{\xi}F_{r\tau}=
-\frac{1}{r} \frac{\partial}{\partial\theta}\left( \frac{1}{\xi}
\frac{\partial \psi}{\partial \tau} \right)$ $ \displaystyle F_{r\xi}=
-\frac{1}{r} \frac{\partial}{\partial\theta}\left(
\frac{\partial \psi}{\partial \xi} \right) $
$\hat E_\theta:$ $ \displaystyle\frac{1}{\xi r}F_{\theta\tau}=
\frac{\partial}{\partial r}\left( \frac{1}{\xi}\frac{\partial \psi}{\partial \tau}
\right)$ $\displaystyle \frac{1}{r}F_{\theta\xi}=
\frac{\partial}{\partial r}\left( \frac{\partial \psi}{\partial \xi} \right)$
$\hat B_{long.}:$ $ \displaystyle \frac{1}{r}F_{r\theta}=
-\left(
\frac{1}{r}\frac{\partial}{\par...
...l}{\partial r}
+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2} \right) \psi
$ $ \displaystyle \frac{1}{r}F_{r\theta}=
-\left(
\frac{1}{r}\frac{\partial}{\par...
...l}{\partial r}
+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2} \right) \psi
$
$\hat B_r:$ $\displaystyle \frac{1}{r}F_{\theta\xi}=
\frac{\partial}{\partial r}\left( \frac{\partial \psi}{\partial \xi} \right)$ $ \displaystyle\frac{1}{\xi r}F_{\theta\tau}=
\frac{\partial}{\partial r}\left( \frac{1}{\xi}\frac{\partial \psi}{\partial \tau}
\right)$
$\hat B_\theta:$ $ \displaystyle F_{\xi r}=
\frac{1}{r}\frac{\partial}{\partial\theta} \left(\frac{\partial \psi}{\partial \xi}
\right)$ $\displaystyle \frac{1}{\xi}F_{\tau r}=
\frac{1}{r}\frac{\partial}{\partial\theta} \left(
\frac{1}{\xi}\frac{\partial \psi}{\partial \tau} \right)
$
The carets in the first column serve as a reminder that these components are relative to the orthonormal basis of the metric, Eqs.(25) and (26).



Ulrich Gerlach 2001-10-09