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Next: Flow of Radiant T.E. Up: Axially Symmetric Source and Previous: Magnetic Dipole and its

Electric Dipole and its Radiation Field

The most important electric dipole radiators are two linear antennas each of length $a$ aligned parallel to the $z$-axis, located at $\xi'=\xi'_I$ and $\xi'=\xi'_{II}$ located in Rindler sectors $I$ and $II$, and hence accelerated into opposite directions. Suppose each antenna has electric dipole moment

\begin{displaymath}
q_{I,II}(\tau')\,a\equiv\textbf{d}_{I,II}(\tau')\quad\quad\q...
...harge)}\times \textrm{(length)}=\textrm{dipole moment}~\right]
\end{displaymath}

Its charge-flux four-vector obtained from Eq.(31) is
$\displaystyle \left( \hat S_{\tau'} ,\hat S_{\xi'} ,\hat S_{r'} ,\hat S_{\theta'} \right)$ $\textstyle =$ $\displaystyle \left(
\xi\frac{\partial S}{\partial \xi'}, \frac{1}{\xi'}\frac{\partial S}{\partial \tau'},
0,0, \right)$  
  $\textstyle =$ $\displaystyle \left( q_{I,II}(\tau')a\frac{d}{d\xi'}\delta(\xi'-\xi'_{I,II}),
\...
...\theta'_0)}{r'}
~,\quad \left[\frac{\textrm{charge}}{\textrm{length}^3} \right]$ (62)

and the corresponding scalar source function $S$ for Eq.(27) is
\begin{displaymath}
S:\quad S_{I,II}(\tau',\xi',r',\theta')=
q_{I,II}(\tau')a~\...
...uad
\left[\frac{\textrm{charge}}{\textrm{length}^2} \right]~.
\end{displaymath} (63)

This source is symmetric around the $z$-axis because it is non-zero only at $r'=0$. The electric dipole moment is the proper volume integral of this source,

\begin{displaymath}
\int_0^\infty \int_0^\infty \int _0^{2\pi}
S_{I,II}(\tau',\x...
...uad
\left[~\textrm{(charge)} \times \textrm{ (length)}~\right]
\end{displaymath}

The axial symmetry of the source implies that its radiation is independent of the polar angle $\theta$. Consequently, except some for a source-dependent factor, the scalar field $\psi_F$ in $F$ is the same as Eq.(61). One finds
$\displaystyle \psi_F(\xi,\tau,r,\theta)=2a \!\!\!\!\!\!\!\!$   $\displaystyle \left[
\frac{\xi'_i}{\sqrt{(\xi^2-\xi^{'2}_I -r^2)^2+(2\xi \xi'_I)^2}}
~q_I(\tau+\sinh^{-1}u_I) \right.$  
  $\textstyle -$ $\displaystyle \left. \frac{\xi'_i}{\sqrt{(\xi^2-\xi^{'2}_{II} -r^2)^2+(2\xi \xi'_{II})^2}}
~q_{II}(\tau-\sinh^{-1}u_{II})
\right]~.\quad\quad [\textrm{charge}]$ (64)

This is the T.M. scalar field due to a pair of localized linear antennas, both situated on the $z$-axis, each one with its own time-dependent dipole moment $q(\tau')\,a$. By setting one of them to zero one obtains the radiation field due to the other.


next up previous
Next: Flow of Radiant T.E. Up: Axially Symmetric Source and Previous: Magnetic Dipole and its
Ulrich Gerlach 2001-10-09