A Linear Center-Fed Half-Wave Antenna

Suppose we are given a center-fed dipole antenna with length $\lambda/2$ (half-wave antenna). We will assume further that the antenna is aligned with the z axis and centered on the origin, with a current given by:

$$I = I_0 \cos(\omega t) \cos \left(\frac{2 \pi z}{\lambda} \right)$$ (15.24)

Note that in ``real life'' it is not easy to arrange for a given current because the current instantaneously depends on the ``resistance'' which is a function of the radiation field itself. The current itself thus comes out of the solution of an extremely complicated boundary value problem. For atomic or nuclear radiation, however, the ``currents'' are generally matrix elements associated with transitions and hence are known.

In any event, the current density corresponding to this current is

$$\vJ = \hz I_0 \cos\left(\frac{2 \pi r}{\lambda} \right) \frac{\delta(1 - \abs{\cos \theta})}{2 \pi r^2 \sin \theta}$$ (15.25)

for $r \le \lambda/4$ and

$$\vJ = 0$$ (15.26)

for $r > \lambda/4$ .

When we use the Hansen multipoles, there is little incentive to convert this into a form where we integrate against the charge density in the antenna. Instead we can easily and directly calculate the multipole moments. The magnetic moment is m_L & = &&int#int;·_L^0&ast#ast; d^3r
& = & I_02&pi#pi; &int#int;_0^2&pi#pi; &int#int;_0^&lambda#lambda;/4 (kr) j_&ell#ell;(kr) {·&ell#ell; &ell#ell;m&ast#ast;(0,&phis#phi;) + ·&ell#ell;&ell#ell;m&ast#ast;(&pi#pi;,&phis#phi;) } d&phis#phi; dr (where we have done the integral over $\theta$ ). Now,

$$\hat{z} \cdot \vsh{\ell \ell}{m} = \frac{1}{\sqrt{\ell(\ell+1)}} m Y_L$$ (15.27)

(Why? Consider $(\hz \cdot \vL) Y_L)$ ...) and yet

$$Y_L(0,\phi) = \delta_{m0} \left[ \frac{2\ell+1}{4\pi} \right]^{1/2}$$ (15.28)

$$Y_L(\pi,\phi) = (-1)^\ell \delta_{m0} \left[ \frac{2\ell+1}{4\pi} \right]^{1/2} .$$ (15.29)

Consequently, we can conclude ( $m \delta_{m0} = 0$ ) that

$$m_L = 0.$$ (15.30)

All magnetic multipole moments of this linear dipole vanish. Since the magnetic multipoles should be connected to the rotational part of the current density (which is zero for linear flow) this should not surprise you.

The electric moments are n_L & = & &int#int;·_L^0 &ast#ast; d^3r
& = & I_02 &pi#pi; &int#int;_0^2 &pi#pi; &int#int;_0^&lambda#lambda;/4 (kr) { &ell#ell;+12&ell#ell;+1 j_&ell#ell;-1(kr) [ ·&ell#ell;,&ell#ell;-1m &ast#ast;(0,&phis#phi;) + ·&ell#ell;,&ell#ell;+1m &ast#ast;(&pi#pi;,&phis#phi;) ] d&phis#phi; dr .
& & - . &ell#ell;2&ell#ell;+ 1 j_&ell#ell;+ 1(kr) [ ·&ell#ell;, &ell#ell;+1m &ast#ast;(0,&phis#phi;) + ·&ell#ell;,&ell#ell;-1m &ast#ast;(&pi#pi;, &phis#phi;) ] }. If we look up the definition of the v.s.h.'s on the handout table, the z components are given by:

$$\hat{z} \cdot \vsh{\ell, \ell-1}{m \ast}(0,\phi) = \delta_{m0} \sqrt{\frac{\ell}{4 \pi}}$$ (15.31)

$$\hat{z} \cdot \vsh{\ell, \ell-1}{m \ast}(\pi,\phi) = (-1)^{\ell - 1} \delta_{m0} \sqrt{\frac{\ell}{4 \pi}}$$ (15.32)

$$\hat{z} \cdot \vsh{\ell, \ell+1}{m \ast}(0,\phi) = - \delta_{m0} \sqrt{\frac{\ell+1}{4 \pi}}$$ (15.33)

$$\hat{z} \cdot \vsh{\ell, \ell+1}{m \ast}(\pi,\phi) = - (-1)^{\ell - 1} \delta_{m0} \sqrt{\frac{\ell+1}{4 \pi}}$$ (15.34)

so the electric multipole moments vanish for $m \ne 0$ , and

$$n_{\ell,0} = I_0 \delta_{m0} \sqrt{\frac{\ell(\ell+1)}{4\pi(2 \el... ...) \int_0^{\lambda/4} \cos(kr) \Big[j_{\ell - 1}(kr) + j_{\ell+1}(kr) \Big] dr .$$ (15.35)

Examining this equation, we see that all the even $\ell$ terms vanish! However, all the odd $\ell$ , $m = 0$ terms do not vanish, so we can't quit yet. We use the following relations:

$$j_{\ell-1} + j_{\ell+1} = \frac{2\ell+1}{kr}j_\ell$$ (15.36)

(the fundamental recursion relation),

$$n_0(kr) = -\frac{\cos(kr)}{kr}$$ (15.37)

(true fact) and

$$\int dz \: f_\ell(z) g_{\ell'}(z) = \frac{z^2}{[\ell'(\ell'+1) - \ell(\ell+1)]} \Big(f'_\ell g_{\ell'} - f_\ell g'_{\ell'} \Big)$$ (15.38)

for any two spherical bessel type functions (a valuable thing to know that follows from integration by parts and the recursion relation). From these we get

$$n_{\ell,0} = \frac{\pi I_0}{2 k} \delta_{m0} \sqrt{\frac{2 \ell + 1}{4\pi \ell(\ell + 1)}} \Big( 1 + (-1)^{\ell+1} \Big) j_\ell(\pi/2).$$ (15.39)

Naturally, there is a wee tad of algebra involved here that I have skipped. You shouldn't. Now, let's figure out the power radiated from this source. Recall from above that: P & = & k^22&mu#mu;_0&epsi#epsilon;_0 &sum#sum;_L { &mid#mid;m_L &mid#mid;^2 + &mid#mid;n_L &mid#mid;^2 }
& = & k^22&mu#mu;_0&epsi#epsilon;_0 &sum#sum;_&ell#ell; odd n_&ell#ell;,0^2
& = & &pi#pi;I_0^28 &mu#mu;_0&epsi#epsilon;_0 &sum#sum;_&ell#ell; odd (2&ell#ell;+1&ell#ell;(&ell#ell;+1) ) [ j_&ell#ell;(&pi#pi;/2) ]^2

Now this also equals (recall) $\frac{1}{2}I_0^2 R_{\rm rad}$ , from which we can find the radiation resistance of the half wave antenna:

$$R_{\rm rad} = \frac{\pi}{4} \sqrt{\frac{\mu_0}{\epsilon_0}} \sum_... ...d}} \left(\frac{2\ell+1}{\ell(\ell+1)} \right) \left[ j_\ell(\pi/2) \right]^2 .$$ (15.40)

We are blessed by this having manifest units of resistance, as we recognize our old friend $Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} \approx 377\Omega$ (the impedance of free space) and a bunch of dimensionless numbers! In terms of this:

$$R_{\rm rad} = Z_0 \left(\frac{\pi}{4} \sum_{\ell {\rm ~odd}} \left(\frac{2\ell+1}{\ell(\ell+1)} \right) \left[ j_\ell(\pi/2) \right]^2 \right).$$ (15.41)

We can obtain a good estimate of the magnitude by evaluating the first few terms. Noting that

$$j_1(\pi/2) = \left( \frac{2}{\pi} \right)^2$$ (15.42)

$$j_3(\pi/2) = \left( \frac{2}{\pi} \right)^2 \left[\frac{60}{\pi^2} - 6 \right]$$ (15.43)

and doing some arithmetic, you should be able to show that $R_{\rm rad} = 73.1\Omega$ .

Note that the ratio of the first (dipole) term to the third (octupole) term is |n_3n_1|^2 & = & 712 23 [60&pi#pi;^2 - 6 ]^2
& = & 718 [60&pi#pi;^2 - 6 ]^2 &ap#approx;0.00244 That means that this is likely to be a good approximation (the answer is very nearly unchanged by the inclusion of the extra term). Even if the length of the antenna is on the order of $\lambda$ , the multipole expansion is an extremely accurate and rapidly converging approximation. That is, after all, why we use it so much in all kinds of localized source wave theory.

However, if we plug in the ``long wavelength'' approximation we previously obtained for a short dipole antenna (with $d = \lambda/2$ ) we get: R_rad = (kd)^224&pi#pi;&mu#mu;_0&epsi#epsilon;_0 &ap#approx;48 &Omega#Omega; which is off by close to a factor of 50%. This is not such a good result. Using this formula with a long wavelength approximation for the dipole moment (only) of n_1,0 &ap#approx;I_0k 23&pi#pi; yields $R_{\rm rad} \approx 80\Omega$ , still off by 11%.