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In the previous section, we examined the radiation emitted by a short electric dipole of oscillating dipole moment
![\begin{displaymath}
{\bf p}(t) = p_0 \sin(\omega t) \hat{\bf z},
\end{displaymath}](img2263.png) |
(1116) |
where
. We found that, in the far field, the mean
electromagnetic energy flux takes the form [see Eq. (1094)]
![\begin{displaymath}
\langle {\bf u}\rangle = \frac{\omega^4 p_0^{ 2}}{32\pi^2 \epsilon_0 c^3}\frac{\sin^2\theta}{r^2} \hat{\bf r},
\end{displaymath}](img2265.png) |
(1117) |
assuming that the dipole is centered on the origin of our spherical polar coordinate system.
The mean power radiated into the element of solid angle
, centered on the angular coordinates (
,
),
is
![\begin{displaymath}
dP = \langle {\bf u}(r, \theta, \varphi)\rangle\!\cdot\!{\hat{\bf r}} r^2 d{\mit\Omega}.
\end{displaymath}](img2267.png) |
(1118) |
Hence, the differential power radiated into this element of solid angle is
simply
![\begin{displaymath}
\frac{dP}{d{\mit\Omega}} = \frac{\omega^4 p_0^{ 2}}{32\pi^2 \epsilon_0 c^3} \sin^2\theta.
\end{displaymath}](img2268.png) |
(1119) |
This formula completely specifies the radiation pattern of an oscillating electric dipole (provided that the dipole is much shorter in length than the
wave-length of the emitted radiation). Of course, the power radiated into
a given element of solid angle is independent of
, otherwise energy would not be conserved. Finally, the total radiated power is the integral of
over all solid angles.
Next: Thompson scattering
Up: Electromagnetic radiation
Previous: The Hertzian dipole
Richard Fitzpatrick
2006-02-02