Physics · Electricity and magnetism · Grade 11

Radiation from a dipole antenna

How does an oscillating charge produce a wave that travels far away?

The situation

The Tashkent TV Tower (375 m) and radio masts on the edge of the city send electromagnetic waves into the air. An antenna is a current that changes direction millions of times a second: the electrons in it oscillate up and down. How does an oscillating charge produce a wave that breaks away from it and travels far? And why does a drone flying right above the top of the antenna pick up no signal at all — and how is this connected with the blueness of the sky?

Topic

Maxwell's equations. A wave spreading from an antenna — a 3D field

Key ideas

Maxwell's equations
Hertzian dipole
Poynting vector

Formulas

\nabla \cdot \vec{E} = \frac{ρ}{ε_{0}}, \nabla \cdot \vec{B} = 0

Gauss's laws: an electric field begins on a charge, while a magnetic “charge” does not exist — the $\vec{B}$ lines are always closed.

\nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}, \nabla \times \vec{B} = μ_{0} \vec{j} + μ_{0} ε_{0} \frac{\partial \vec{E}}{\partial t}

Faraday's and Ampère–Maxwell laws: a changing magnetic field produces an electric field, and a changing electric field produces a magnetic field — this is a wave.

c = \frac{1}{\sqrt{μ_{0} ε_{0}}} \approx 3 \cdot 10^{8} m/s, λ = \frac{c}{f}

The wave speed that follows from Maxwell's equations is the speed of light.

E_{θ} \approx - \frac{p_{0} k^{2} \sin θ}{4 π ε_{0} r} \cos (ω t - k r), B_{φ} = \frac{E_{θ}}{c}

Far zone ( $r ≫ λ$ ): $E$ and $B$ are in phase, perpendicular to each other and to the direction of propagation, and fall off as $1 / r$ .

\overline{S} = \frac{μ_{0} p_{0}^{2} ω^{4} \sin^{2} θ}{32 π^{2} c r^{2}}, P = \frac{μ_{0} p_{0}^{2} ω^{4}}{12 π c}

The mean value of the Poynting vector and the total radiated power of the dipole.

Theory

Maxwell's equations

Four equations describe all of electromagnetism. The most important are the last two: a changing $\vec{B}$ produces a curling $\vec{E}$ , and a changing $\vec{E}$ (through the displacement current $ε_{0} \partial \vec{E} / \partial t$ added by Maxwell) produces a curling $\vec{B}$ . Supporting each other, the fields break away from the charges and propagate at the speed $c = 1 / \sqrt{μ_{0} ε_{0}}$ . In integral form: $\oint \vec{E} \cdot d \vec{l} = - d Φ_{B} / d t$ , $\oint \vec{B} \cdot d \vec{l} = μ_{0} (I + ε_{0} d Φ_{E} / d t)$ , $\oint \vec{E} \cdot d \vec{S} = Q / ε_{0}$ , $\oint \vec{B} \cdot d \vec{S} = 0$ ; the differential form is obtained from them with the Stokes and Gauss theorems. Together with the Lorentz force on a charge $\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})$ this is the complete Maxwell–Lorentz system. Its first pair ( $\nabla \cdot \vec{B} = 0$ , $\nabla \times \vec{E} = - \partial \vec{B} / \partial t$ ) is satisfied automatically once potentials are introduced; the second pair, with sources, follows from the principle of least action for the field action integral $S = \int (\frac{1}{2} ε_{0} E^{2} - \frac{1}{2 μ_{0}} B^{2} - ρ φ + \vec{j} \cdot \vec{A}) d V d t$ .

Near and far zones

The violet lines on the stage are the electric field lines in the meridian plane of the antenna; they are the contour lines of the stream function of the exact solution $ψ = \sin^{2} θ (\cos u / k r - \sin u)$ , $u = ω t - k r$ . In the near zone ( $r ≪ λ / 2 π$ ) the lines are tied to the charges, and the field falls off as $1 / r^{3}$ , like that of a static dipole. Every half period, when the charges swap places, the lines break off into loops — in the far zone they become a wave; the distance between neighbouring loops of the same colour is $λ$ . The blue loops are magnetic field lines; they go round the antenna axis.

Radiation pattern

The energy flow is the Poynting vector $\vec{S} = \vec{E} \times \vec{B} / μ_{0}$ . Its mean value is proportional to $\sin^{2} θ$ : greatest in the equatorial plane, and zero along the antenna axis ( $θ = 0 °$ and $180 °$ ). The 3D “doughnut” shows this pattern. That is why radio towers mount their antennas vertically: the energy spreads along the horizon, towards people, not into the sky.

Potentials and retardation

It is convenient to write the fields through potentials: $\vec{B} = \nabla \times \vec{A}$ , $\vec{E} = - \nabla φ - \partial \vec{A} / \partial t$ . The potentials are not unique: under the substitution $\vec{A} \to \vec{A} + \nabla χ$ , $φ \to φ - \partial χ / \partial t$ the fields do not change (gauge, i.e. gradient invariance). If the Lorenz condition $\nabla \cdot \vec{A} + c^{- 2} \partial φ / \partial t = 0$ is chosen, the equations become the d'Alembert equations: $□ \vec{A} = - μ_{0} \vec{j}$ , $□ φ = - ρ / ε_{0}$ . Their solution is the retarded potentials: the potential at the point $r$ is determined by the state of the source at the moment $t - r / c$ . For our dipole $\vec{A} = μ_{0} \dot{\vec{p}} (t - r / c) / (4 π r)$ — all the fields on the stage are derived from it, and the $k r$ in $u = ω t - k r$ is exactly the delay (see $r / c$ at the drone's position in the results window). For an arbitrarily moving point charge the same idea gives the Liénard–Wiechert potentials: $φ = q / (4 π ε_{0} (R - \vec{R} \cdot \vec{v} / c))$ , where $R$ and $\vec{v}$ are taken at the retarded time.

Every accelerating charge radiates

At a large distance the field of a charge depends on its acceleration at the retarded time and falls off as $1 / r$ ; the radiated power is given by the Larmor formula $P = q^{2} a^{2} / (6 π ε_{0} c^{3})$ , from which $P \propto ω^{4}$ follows for a dipole. Hertz's oscillator — two conductors separated by a spark gap — was such a dipole, and the receiver was a wire loop with a small spark gap. A loop antenna is a magnetic dipole: $P = μ_{0} m_{0}^{2} ω^{4} / (12 π c^{3})$ ; for the same size it is weaker than an electric dipole by roughly a factor of ${(k ℓ)}^{2}$ ; quadrupole radiation is proportional to $ω^{6}$ and is weaker still. Because energy is carried away, a radiation reaction force acts on the charge: the electron oscillator is damped, and its spectral line has a natural width — in the classical calculation $Δ λ = \frac{4 π}{3} r_{e} \approx 1.2 \cdot 10^{- 5}$ nm, independent of the wavelength. A wave makes a free electron oscillate, turning it into a tiny antenna — Thomson scattering, with cross-section $σ_{T} = \frac{8 π}{3} r_{e}^{2} \approx 6.65 \cdot 10^{- 29}$ m². Charges moving close to the speed of light emit light forward in a narrow cone — synchrotron radiation (in the “Charged-particle accelerator” experiment). The field energy density $w$ , the energy flux $\vec{S}$ and the momentum density $\vec{S} / c^{2}$ together form one energy–momentum tensor.

Why is the sky blue?

Power $P \propto ω^{4}$ : if the frequency increases 2 times, the radiation becomes 16 times stronger. Sunlight makes the electrons in air molecules oscillate, and each molecule becomes a small dipole antenna scattering the light in all directions (Rayleigh scattering). Blue light (≈ 450 nm) has a frequency 1.56 times that of red (≈ 700 nm), so it is scattered ${1.56}^{4} \approx 6$ times more strongly. Model: antenna length $ℓ = 1$ m ≪ $λ$ (Hertz dipole), reflection from the ground is ignored (free space), time is slowed down by a factor of $5 \cdot 10^{7}$ , the drone is enlarged.

Did you know?

In 1865, in his paper “A Dynamical Theory of the Electromagnetic Field”, James Clerk Maxwell showed that electromagnetic waves must travel at the speed of light, and concluded that light itself is an electromagnetic wave.
In 1887 Heinrich Hertz produced and detected electromagnetic waves in an experiment, using a spark oscillator — in fact a dipole antenna — and a loop receiver. The unit of frequency, the hertz (Hz), is named after him.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

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