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Energy radiated by the dipole

Recall our old buddy the complex Poynting vector for harmonic fields (J6.132):

$\displaystyle \vS = \frac{1}{2} {\rm Re}\left\{ \vE \times \vH^\ast \right\}.$ (13.53)

The factor of $ 1/2$ comes from time averaging the fields. This is the energy per unit area per unit time that passes a point in space. To find the time average power per solid angle, we must relate the normal area through which the energy flux passes to the solid angle:

$\displaystyle d A_n = r^2 d \Omega$ (13.54)

and project out the appropriate piece of S, i. e. -- $ {\bf n} \cdot
{\bf S}$ . We get (with $ \mu = 1$ )

$\displaystyle \frac{d P}{d \Omega} = \frac{1}{2}$   Re$\displaystyle [r^2 {\bf n} \cdot ({\bf E} \times {\bf H}^\ast)].$ (13.55)

where we must plug in E and H from the expressions above for the far field.

After a bunch of algebra that I'm sure you will enjoy doing, you will obtain:

$\displaystyle \frac{d P}{d \Omega} = \frac{c^2}{32\pi^2} \sqrt{\frac{\mu_0}{\epsilon_0}} k^4 \mid ({\bf n} \times \vp) \times {\bf n} \mid^2.$ (13.56)

The polarization of the radiation is determined by the vector inside the absolute value signs. By this one means that one can project out each component of $ \vp$ (and hence the radiation) before evaluating the square independently, if so desired. Note that the different components of $ \vp$ need not have the same phase (elliptical polarization, etc.).

If all the components of $ \vp$ (in some coordinate system) have the same phase, then $ \vp$ necessarily lies along a line and the typical angular distribution is that of (linearly polarized) dipole radiation:

$\displaystyle \frac{d P}{d \Omega} = \frac{c^2}{32\pi^2} \sqrt{\frac{\mu_0}{\epsilon_0}} k^4 \mid \vp \mid^2 \sin^2 \theta$ (13.57)

where $ \theta$ is measured between p and n. When you integrate over the entire solid angle (as part of your assignment) you obtain the total power radiated:

$\displaystyle P = \frac{c^2 k^4}{12\pi} \sqrt{\frac{\mu_0}{\epsilon_0}} \mid \vp \mid^2$ (13.58)

The most important feature of this is the $ k^4$ dependence which is, after all, why the sky is blue (as we shall see, never fear).

Example: A centerfed, linear antenna

In this antenna, $ d << \lambda$ and

$\displaystyle I(z,t) = I_0 \left( 1 - \frac{2 \mid z \mid }{d} \right) e^{-i \omega t}.$ (13.59)

From the continuity equation (and a little subtle geometry),

$\displaystyle \deldot \vJ = \frac{dI}{dz} = - \frac{\partial \rho'(z) e^{-i \omega t}}{\partial t} = i \omega \rho'(z)$ (13.60)

and we find that the linear charge density (participating in the oscillation, with a presumed neutral background) is independent of $ z$ :

$\displaystyle \rho'(z) = \pm \frac{2 i I_0}{\omega d}$ (13.61)

where the $ +/-$ sign indicates the upper/lower branch of the antenna and the $ '$ means that we are really treating $ \rho/(dxdy)$ (which cancels the related terms in the volume integral below). We can then evaluate the dipole moment of the entire antenna for this frequency:

$\displaystyle p_z = \int_{-d/2}^{d/2} z \rho'(z) dz = \frac{i I_0 d}{2 \omega}.$ (13.62)

The electric and magnetic fields for $ r > d$ in the electric dipole approximation are now given by the previously derived expressions. The angular distribution of radiated power is

$\displaystyle \frac{d P}{d \Omega} = \frac{I_0^2}{128\pi^2} \sqrt{\frac{\mu_0}{\epsilon_0}} (kd)^2 \sin^2 \theta$ (13.63)

and the total radiated power is

$\displaystyle P = \frac{I_0^2 (kd)^2}{48\pi}\sqrt{\frac{\mu_0}{\epsilon_0}} .$ (13.64)

Remarks. For fixed current the power radiated increases as the square of the frequency (at least when $ kd << 1$ , i. e. - long wavelengths relative to the size of the antenna). The total power radiated by the antenna appears as a ``loss'' in ``Ohm's Law'' for the antenna. Factoring out $ I_0^2/2$ , the remainder must have the units of resistance and is called the radiation resistance of the antenna:

$\displaystyle R_{\rm rad} = \frac{2P}{I_0^2} = \frac{(kd)^2}{24\pi}\sqrt{\frac{\mu_0}{\epsilon_0}} \approx 5 (kd)^2$     ohms$\displaystyle )$ (13.65)

where we do the latter multiplication to convert the resulting units to ohms. Note that this resistance is there for harmonic currents even if the conductivity of the metal is perfect. Note further that by hypothesis this expression will only be valid for small values of $ R_{\rm rad}$ .

Good golly, this is wonderful. We hopefully really understand electric dipole radiation at this point. It would be truly sublime if all radiators were dipole radiators. Physics would be so easy. But (alas) sometimes the current distribution has no $ \ell = 0$ moment and there is therefore no dipole term! In that case we must look at the next term or so in the multipolar expansions

Lest you think that this is a wholly unlikely occurrance, please note that a humble loop carrying a current that varies harmonically is one such system. So let us proceed to:

next up previous contents
Next: Magnetic Dipole and Electric Up: Dipole Radiation Previous: Asymptotic properties in the   Contents
Robert G. Brown 2017-07-11