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The T.M. Field

The T.M. has its source and vector potential four-vectors lie strictly in the 2-d Lorentz plane:
\begin{displaymath}
( S_{\tau}, S_{\xi}, S_{r}, S_{\theta})=\left(
\xi\frac{\par...
...tial \tau},0,0 \right)\quad
\textrm{in }I\textrm{ or in }II~,
\end{displaymath} (31)


\begin{displaymath}
( A_{\tau}, A_{\xi}, A_{r}, A_{\theta})=\left(
\xi\frac{\par...
...rtial \tau},0,0\right)\quad
\textrm{in }I\textrm{ or in }II~,
\end{displaymath} (32)

and
\begin{displaymath}
( A_{\xi}, A_{\tau}, A_{r}, A_{\theta})=\left(
\frac{1}{\xi}...
...\partial \xi},0,0\right)\quad
\textrm{in }F\textrm{ or in }P~
\end{displaymath} (33)

The components of the T.M. Maxwell field are
  In $I$ or in $II$ In $F$ or in $P$
$\hat E_{long.}:$ $ \displaystyle \frac{1}{\xi}F_{\xi\tau}=
\left(
\frac{1}{\xi}\frac{\partial}{\...
...}{\partial \xi}
-\frac{1}{\xi^2}\frac{\partial^2}{\partial\tau^2} \right) \psi$ $\displaystyle \frac{1}{\xi}F_{\tau\xi}=
-\left(
\frac{1}{\xi}\frac{\partial}{\...
...{\partial \xi}
-\frac{1}{\xi^2}\frac{\partial^2}{\partial\tau^2} \right) \psi $
$\hat E_r:$ $ \displaystyle\frac{1}{\xi}F_{r\tau}=
\frac{\partial}{\partial r}\left(
\frac{\partial \psi}{\partial \xi} \right)$ $ \displaystyle F_{r\xi}=
\frac{\partial}{\partial r}\left(
\frac{1}{\xi}\frac{\partial \psi}{\partial \tau} \right) $
$\hat E_\theta:$ $ \displaystyle\frac{1}{\xi r}F_{\theta\tau}=
\frac{1}{r} \frac{\partial}{\partial \theta}
\left( \frac{\partial \psi}{\partial \xi}
\right)$ $\displaystyle \frac{1}{r}F_{\theta\xi}=
\frac{1}{r} \frac{\partial}{\partial \theta}\left
( \frac{1}{\xi}\frac{\partial \psi}{\partial \tau} \right)$
$\hat B_{long.}:$ $ \displaystyle \frac{1}{r}F_{r\theta}=0$ $ \displaystyle \frac{1}{r}F_{r\theta}=0$
$\hat B_r:$ $\displaystyle \frac{1}{r}F_{\theta\xi}=
\frac{1}{r} \frac{\partial}{\partial \theta}\left
( \frac{1}{\xi}\frac{\partial \psi}{\partial \tau} \right)$ $ \displaystyle\frac{1}{\xi r}F_{\theta\tau}=
\frac{1}{r} \frac{\partial}{\partial \theta}
\left( \frac{\partial \psi}{\partial \xi}
\right)$
$\hat B_\theta:$ $ \displaystyle F_{\xi r}=
-\frac{\partial}{\partial r} \left( \frac{1}{\xi}\frac{\partial \psi}{\partial \tau}
\right)$ $\displaystyle \frac{1}{\xi}F_{\tau r}=
-\frac{\partial}{\partial r} \left(
\frac{\partial \psi}{\partial \xi} \right)
$



Ulrich Gerlach 2001-10-09