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11.1 Dipole Radiation

11.1.1 Electric Dipole Radiation

Consider a dipole which is "oscillating" at some particular frequency. This looks like two metal spheres separated by distance d d and connected by a small wire; at time t t the charge on the upper sphere is q(t) q(t) and the charge on the lower sphere is q(t) -q(t) . We drive the charge back and forth through the wire from one sphere to the other at an angular frequency ω \omega

q(t)=q0cos(ωt) q(t) = q_0 \cos (\omega t)

The dipole moment is

p(t)=p0cos(ωt)z^p0q0d \vec p(t) = p_0 \cos (\omega t) \vu z \qquad p_0 \equiv q_0 d

Of course, we consider the harmonically oscillating dipole because we can build any other oscillations out of this basis.

Figure 11.2

The retarded potential of this oscillating dipole is the superposition of the two point charges:

V(r,t)=14πϵ0[q0cos[ω(tγ+/c)]γ+q0cos[ω(tγ/c)]γ] V(\vec r, t) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_0 \cos [\omega(t - \gr_+ / c)]}{\gr_+} - \frac{q_0 \cos[\omega(t - \gr_- / c)]}{\gr_-} \right]

where

γ±=r2rdcosθ+(d/2)2 \gr_{\pm} = \sqrt{ r^2 \mp r d \cos \theta + (d/2)^2}

To take this physical dipole towards a perfect dipole, we want

dr while q0d=const. d \ll r \quad \text{ while } \quad q_0 d = \text{const.}

This approximation gives us

γ±r(1d2rcosθ) \gr_{\pm} \approx r \left( 1 \mp \frac{d}{2r} \cos \theta \right)

so that

1γ±1r(1±d2rcosθ) \frac{1}{\gr_{\pm}} \approx \frac{1}{r} \left( 1 \pm \frac{d}{2r} \cos \theta \right)

and

cos[ω(tγ±/c)]cos[ω(tr/c)±ωd2ccosθ]=cos[ω(tr/c)]cos(ωd2ccosθ)±sin[ω(tr/c)]sin(ωd2ccosθ) \begin{aligned} \cos[\omega(t - \gr_{\pm} / c)] & \approx \cos \left[ \omega(t - r/c) \pm \frac{\omega d}{2c} \cos \theta \right] \\ & = \cos [ \omega(t - r/c)] \cos \left( \frac{\omega d}{2c} \cos \theta \right) \\ & \quad \pm \sin[\omega(t - r/c)] \sin \left( \frac{\omega d}{2c} \cos \theta \right) \end{aligned}

In the limit of a perfect dipole, we take the dipole approximation

dcω d \ll \frac{c}{\omega}

Under that condition,

cos[ω(tγ±/c)]cos[ω(tr/c)]ωd2ccosθsin[ω(tr/c)] \cos [ \omega(t - \gr_{\pm}/c)] \approx \cos [\omega(t - r/c)] \mp \frac{\omega d}{2c} \cos \theta \sin [\omega(t - r/c)]

Finally, we do not really care what happens near the origin. Rather, we are looking for the far-field behavior of the radiation, so we must consider fields that survive at large distances from the source:

rcω r \gg \frac{c}{\omega}

In this region, the retarded potential reduces to

V(r,θ,t)=p0ω4πϵ0c(cosθr)sin[ω(tr/c)] V(r, \theta, t) = - \frac{p_0 \omega}{4 \pi \epsilon_0 c} \left( \frac{\cos \theta}{r} \right) \sin [ \omega(t - r/ c)]

What about the vector potential? In our model of spheres connected by a wire, it is determined by the current flowing in the wire:

I(t)=dqdtz^=q0ωsin(ωt)z^ \vec I(t) = \dv{q}{t} \vu z = - q_0 \omega \sin(\omega t) \vu z

so

A=μ04πJ(r,tr)γdτ=μ04πd/2d/2q0ωsin[ω(tγ/c)]z^γdz \begin{aligned} \vec A & = & \frac{\mu_0}{4 \pi} \int \frac{ \vec J(\vec r', t_r)}{\gr} \dd \tau' \\ & = & \frac{\mu_0}{4 \pi} \int_{-d/2} ^{d/2} \frac{- q_0 \omega \sin [ \omega(t - \gr / c)]\vu z}{\gr} \dd z \end{aligned}

Given our previous approximations, since the integration itself happens over the assumed short distance d d , we can replace the integrand by its value at the center and introduce a factor of d d

A(r,θ,t)=μ0p0ω4πrsin[ω(tr/c)]z^ \vec A(r, \theta, t) = - \frac{\mu_0 p_0 \omega}{4 \pi r} \sin [\omega(t - r/c)] \vu z

Great, we've got the potentials! What are the fields in the radiation zone (far-field)?

V=Vrr^+1rVθθ^=p0ω4πϵ0c[cosθ(1r2sin[ω(tr/c)]ωrccos[ω(tr/c)])r^sinθr2sin[ω(tr/c)]θ^]p0ω24πϵ0c2(cosθr)cos[ω(tr/c)]r^ \begin{aligned} \grad V & = & \pdv{V}{r} \vu r + \frac{1}{r} \pdv{V}{\theta} \vu \theta \\ & = & - \frac{p_0 \omega}{4 \pi \epsilon_0 c} \left[ \cos \theta \left( - \frac{1}{r^2} \sin [\omega (t - r/c)] - \frac{\omega}{rc} \cos [\omega(t - r/c)] \right) \vu r \right. \\ & & \qquad \left. - \frac{\sin \theta}{r^2} \sin [\omega(t - r/c)] \vu \theta \right] \\ & \approx & \frac{p_0 \omega^2}{4 \pi \epsilon_0 c^2} \left( \frac{\cos \theta}{r} \right) \cos [\omega(t - r/c)] \vu r \end{aligned}

where we've dropped the first and last terms in accordance with our far-field approximation. Similarly,

At=μ0p0ω24πrcos[ω(tr/c)](cosθr^sinθθ^) \pdv{\vec A}{t} = - \frac{\mu_0 p_0 \omega^2}{4 \pi r} \cos [\omega(t - r/c)] (\cos \theta \vu r - \sin \theta \vu \theta)

therefore

E=VAt=μ0p0ω24π(sinθr)cos[ω(tr/c)]θ^ \vec E = - \grad V - \pdv{\vec A}{t} = - \frac{\mu_0 p_0 \omega^2}{4 \pi} \left( \frac{\sin \theta}{r} \right) \cos [ \omega( t - r/c)] \vu \theta

What about the magnetic field?

×A=1r[r(rAθ)Arθ]ϕ^=μ0p0ω4πr[sinθcos[ω(tr/c)]+sinθr[ω(tr/c)]]ϕ^ \curl \vec A = \frac{1}{r} \left[ \pdv{}{r} (r A_\theta) - \pdv{A_r}{\theta} \right] \vu \phi \\ = - \frac{\mu_0 p_0 \omega}{4 \pi r} \left[ \sin \theta \cos [ \omega(t - r/c)] + \frac{\sin \theta}{r} [\omega(t - r/c)] \right] \vu \phi

The second term can be eliminated in the far-field, so

B=×A=μ0p0ω24πc(sinθr)cos[ω(tr/c)]ϕ^ \vec B = \curl \vec A = - \frac{\mu_0 p_0 \omega^2}{4 \pi c} \left( \frac{\sin \theta}{r} \right) \cos [\omega(t - r/c)]\vu \phi

These fields are in phase, mutually perpendicular, and transverse to the direction of propagation, and the ratio of their amplitudes is E0/B0=c E_0 / B_0 = c , exactly as we expect from electromagnetic waves. These are actually spherical waves, not plane waves, and their amplitude decreases like 1/r 1/r as they progress. But for large r, they are approximately plane over small regions.

The energy flux is determined by the Poynting vector

S(r,t)=1μ0E×B=μ0c[p0ω24π(sinθr)cos[ω(tr/c)]]2r^ \vec S(\vec r, t) = \frac{1}{\mu_0} \vec E \cross \vec B = \frac{\mu_0}{c} \left[ \frac{p_0 \omega^2}{4 \pi} \left( \frac{\sin \theta}{r} \right) \cos [\omega(t - r/c)] \right]^2 \vu r

The intensity is the time average over a cycle

I=S=(μ0p02ω432π2c)sin2θr2r^ I = \langle \vec S \rangle = \left( \frac{\mu_0 p_0 ^2 \omega^4}{32 \pi^2 c} \right) \frac{\sin^2 \theta}{r^2} \vu r

and the total power radiated is found by integrating S \langle \vec S \rangle over a sphere of radius r r :

P=Sda=μ0p02ω432π2csin2θr2r2sinθdθdϕ=μ0p02ω412πc \langle P \rangle = \int \langle \vec S \rangle \cdot \dd \vec a = \frac{\mu_0 p_0 ^2 \omega^4}{32 \pi^2 c} \int \frac{\sin^2 \theta}{r^2} r^2 \sin \theta \dd \theta \dd \phi = \frac{\mu_0 p_0 ^2 \omega^4}{12 \pi c}