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Oscillating dipoles radiate $P = \mu_0p_0^2\omega^4/12\pi c$ in a $\sin^2\theta$ pattern; radiation resistance of antennas and Rayleigh scattering.
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By the end of this lesson you will be able to compute the power and angular pattern of dipole radiation, find the radiation resistance of short antennas, and apply the $\omega^4$ law to scattering.
You know the retarded potentials, the fields of static dipoles, the Poynting vector and plane waves. You know that radiation fields fall as $1/r$. This lesson applies the retarded potentials to the simplest radiating system, an oscillating dipole, and finds how much power it sends out and in which directions.
| Term | What it means |
|---|---|
| Oscillating dipole | A dipole moment $p(t) = p_0\cos\omega t$, such as charge sloshing along an antenna. |
| Radiation zone | Distances much greater than the wavelength, where the $1/r$ fields dominate. |
| Radiation pattern | The dependence of radiated intensity on direction; $\sin^2\theta$ for a dipole. |
| Larmor-like power | $P = \mu_0p_0^2\omega^4/12\pi c$ for an oscillating dipole. |
| Radiation resistance | The resistance $R$ with $P = \tfrac{1}{2}I_0^2R$ that represents power lost to radiation. |
| Rayleigh scattering | Scattering by particles much smaller than the wavelength, proportional to $\omega^4$. |
| Directivity | How strongly an antenna concentrates power in its best direction compared with an isotropic source. |
Take a dipole $\vec{p}(t) = p_0\cos\omega t\,\hat{z}$, small compared with the wavelength, and observe far away, $r \gg \lambda$. The retarded potentials give fields in the radiation zone
$$\vec{E} = -\frac{\mu_0p_0\omega^2}{4\pi}\frac{\sin\theta}{r}\cos[\omega(t - r/c)]\,\hat{\theta}, \qquad \vec{B} = \frac{E}{c}\,\hat{\phi},$$
falling as $1/r$, transverse, in phase, like a plane wave locally. The time-averaged Poynting vector points outward:
$$\langle\vec{S}\rangle = \frac{\mu_0p_0^2\omega^4}{32\pi^2c}\frac{\sin^2\theta}{r^2}\,\hat{r}.$$
No energy goes along the dipole's axis; the most goes broadside. Integrating over a sphere gives the total radiated power,
$$P = \frac{\mu_0p_0^2\omega^4}{12\pi c},$$
independent of $r$: the energy genuinely leaves. The steep $\omega^4$ dependence comes from the field's proportionality to the charges' acceleration, $\omega^2p_0$. For a short antenna of length $L$ carrying current $I_0$, $p_0 = I_0L/\omega$, and the power is $\tfrac{1}{2}I_0^2R$ with radiation resistance $R = 80\pi^2(L/\lambda)^2$ Ω.
Another way: picture
Picture charge sloshing up and down a vertical antenna. Looking at it from the side, you see the full motion and receive the strongest wave. Looking straight down its axis, you see the charge moving toward and away from you, which produces no transverse field and no wave. The radiation pattern is a doughnut around the antenna: fat at the equator, pinched to nothing at the poles.
Another way: steps
Checks. The power must not depend on distance. The pattern must vanish along the axis and peak at $90°$. Units: $\mu_0p_0^2\omega^4/c$ gives watts. And for antennas, doubling the length quadruples the radiation resistance, so short antennas are poor radiators, which is why efficient antennas are a sizable fraction of a wavelength long.
Near a static dipole the field falls as $1/r^3$. An oscillating dipole has, in addition, fields that fall as $1/r^2$ and $1/r$, from differentiating the retarded potentials: the retarded time $t - r/c$ inside $\cos\omega t$ brings a factor $\omega/c$ each time it is differentiated with respect to $r$. Far from the source, where $r \gg c/\omega = \lambda/2\pi$, the $1/r$ terms dominate.
Only $1/r$ fields carry energy to infinity: the Poynting flux goes as $1/r^2$, and multiplied by the sphere's area $4\pi r^2$ gives a constant. The $1/r^2$ and $1/r^3$ fields store energy near the source and give it back each cycle. Antenna engineers call these the near field and far field; wireless chargers and RFID tags work in the near field, radio broadcasting in the far field.
The factor $\sin\theta$ in the radiation field comes from projecting the dipole's acceleration onto the plane perpendicular to the line of sight: only the transverse part radiates. That gives the doughnut pattern, with half the power within $\pm 45°$ of the equatorial plane.
Broadcast towers exploit it. An AM radio tower is a vertical antenna, radiating mostly horizontally toward listeners and little straight up, where it would be wasted. Arrays of several antennas, fed with chosen phases, add their fields to concentrate power further, which is how stations protect other areas from interference and how phased-array radars and $5$G base stations steer beams electronically.
Radiated power acts like a resistance in the antenna's circuit: to push current $I_0$ into the antenna, the transmitter must supply $\tfrac{1}{2}I_0^2R$ that leaves as radiation. For a short dipole, $R = 80\pi^2(L/\lambda)^2$, only about $8$ Ω at a tenth of a wavelength. The antenna's ordinary ohmic resistance competes with it, and if the two are comparable, much of the power becomes heat instead of radio waves.
A half-wave dipole, too long for the short-dipole formula, has a radiation resistance of $73$ Ω, conveniently matched to common cables. Very small antennas, such as those in phones or hearing aids, have tiny radiation resistances and use matching networks and careful design to radiate efficiently. Every antenna datasheet lists the radiation resistance or its close relative, the input impedance.
Sunlight's electric field drives the electrons in air molecules, making them oscillating dipoles that re-radiate — scatter — some of the light. Visible frequencies are far below the molecules' resonances, so the induced dipole moment is nearly the same for all colors, and the scattered power goes as $\omega^4$. Violet light at $400$ nm is scattered about nine times as strongly as red light at $700$ nm.
Looking away from the Sun, we see this scattered light, dominated by blue (violet is weaker in sunlight, and our eyes are less sensitive to it). At sunset, sunlight crosses much more air to reach us, and the blue is scattered out of the direct beam, leaving it red and orange. The scattered light is also polarized, strongly so at $90°$ from the Sun, as the dipole pattern predicts — which bees and some birds use to navigate.
A current loop whose current oscillates is a magnetic dipole, and it radiates too, with the same $\sin^2\theta$ pattern but electric and magnetic fields swapped. Its power, $\mu_0m_0^2\omega^4/12\pi c^3$, is smaller than an electric dipole's of comparable size by a factor of order $(\text{size}/\lambda)^2$, so small loops radiate weakly. Quadrupoles are weaker still.
This hierarchy is why atomic transitions that change the electric dipole moment are "allowed" and fast, while those that can only proceed by magnetic dipole or quadrupole radiation are "forbidden" and slow, as the quantum lesson on selection rules found. Classical radiation theory and quantum transition rates are two views of the same physics.
A radiating charge loses energy, so something must slow it down. The radiated power acts like a damping force on the oscillating charge, the radiation reaction. For an antenna it appears as the radiation resistance; for an electron bound in an atom, classical physics predicts it would spiral into the nucleus in about $10^{-11}$ s.
That catastrophe was one of the failures of classical physics that quantum mechanics resolved: stationary states do not radiate. But radiation reaction is real for free charges. It sets the natural width of classical oscillators, limits the energy of electrons in circular accelerators, and, in its quantum form, gives excited atoms their spontaneous emission rates.
Astronomers read the radiation of oscillating charges across the cosmos. Pulsars emit beamed radio from charges accelerated in their magnetic fields; the $21$ cm line comes from a magnetic dipole transition in hydrogen; the cosmic microwave background is thermal radiation from the early universe's plasma. Every radio telescope receives these signals with antennas that are, in effect, dipoles run in reverse.
Reciprocity guarantees that an antenna's pattern for receiving matches its pattern for transmitting. The Very Large Array's dishes in New Mexico and the planned next-generation arrays focus radio waves onto small feed antennas at their focus, whose dipole-like patterns determine how the dish is illuminated. The theory of this lesson, reversed, describes how every radio photon from space is caught.
Lord Rayleigh explained the blue sky in 1871 using exactly this physics: molecules much smaller than the wavelength act as dipoles driven by sunlight, scattering power proportional to $\omega^4$. Violet at $400$ nm scatters about nine times as strongly as red at $700$ nm, and the sky's scattered light is dominated by blue, the brightest short-wavelength part of sunlight that our eyes see well.
The same law reddens sunsets and the Moon during a lunar eclipse, when the only light reaching it has passed through Earth's atmosphere edgewise. It sets how far visible light travels through clean air, and, applied to tiny fluctuations in glass density, it sets the minimum loss of optical fibers, which is why telecommunications use infrared wavelengths near $1550$ nm, where Rayleigh scattering is weaker by $(1550/400)^4 \approx 225$.
An AM radio station at $1$ MHz has a wavelength of $300$ m, so even its tall tower is a modest fraction of a wavelength. Stations typically use a vertical tower a quarter wavelength tall, $75$ m, standing on a network of buried copper radials that act as a ground mirror, making the tower and its image a half-wave dipole with a radiation resistance near $37$ Ω.
The vertical tower radiates in a doughnut pattern, sending ground waves out along Earth's surface toward listeners. Stations that must protect distant stations on the same frequency, as the Federal Communications Commission requires at night when signals travel farther, use arrays of several towers fed with chosen phases to shape the pattern, sending less power in protected directions. Each array is designed with the dipole fields of this lesson, added with their phases.
It is easy to picture radio waves spreading uniformly from an antenna like ripples from a stone. A dipole antenna sends no energy along its axis and the most perpendicular to it, with intensity proportional to $\sin^2\theta$. That is why a vertical car antenna receives stations on the horizon well and why broadcast towers are vertical: their doughnut pattern sends energy toward listeners, not into the sky.
A second misconception is that radiated power grows gently with frequency. At a fixed dipole moment it grows as $\omega^4$: ten times the frequency gives ten thousand times the power. This steep dependence explains both the blue sky and why radio circuits radiate far more than power-line circuits of the same size.
Charges of $\pm 1.0$ nC oscillate with separation amplitude $1.0$ cm at $100$ MHz. Find $p_0$.
$p_0 = qd = 10^{-9} \times 0.010 = 10^{-11}\ \text{C m}$
Charge times separation.
Find the angular frequency.
$\omega = 2\pi \times 10^{8} = 6.28 \times 10^{8}\ \text{rad/s}$
Radio frequency.
Raise it to the fourth power.
$\omega^4 = 1.56 \times 10^{35}\ \text{s}^{-4}$
The dominant factor.
Evaluate the power.
$P = \dfrac{4\pi \times 10^{-7} \times 10^{-22} \times 1.56 \times 10^{35}}{12\pi \times 3.00 \times 10^8} = 1.7 \times 10^{-3}\ \text{W}$
About two milliwatts.
Find the power at $1$ GHz.
$P' = 10^4 \times 1.7 \times 10^{-3} = 17\ \text{W}$
Ten times the frequency, ten thousand times the power.
A current $I_0\cos\omega t$ flows in a short antenna of length $L$. Find the charge amplitude at the ends.
$q_0 = \dfrac{I_0}{\omega}$
Integrating the current over time.
Write the dipole moment.
$p_0 = q_0L = \dfrac{I_0L}{\omega}$
Charge times length.
Substitute into the power formula.
$P = \dfrac{\mu_0I_0^2L^2\omega^2}{12\pi c}$
Two powers of $\omega$ cancel.
Write the power as a resistance.
$P = \tfrac{1}{2}I_0^2R \quad\Rightarrow\quad R = \dfrac{\mu_0L^2\omega^2}{6\pi c}$
Time-averaged power in a resistor.
Express it with the wavelength.
$R = \dfrac{2\pi\mu_0c}{3}\left(\dfrac{L}{\lambda}\right)^2 = 80\pi^2\left(\dfrac{L}{\lambda}\right)^2\ \Omega$
Using $\omega = 2\pi c/\lambda$ and $\mu_0c = 120\pi$ Ω.
Evaluate for $L = \lambda/10$.
$R = 789.6 \times 0.01 = 7.9\ \Omega$
Small: short antennas radiate inefficiently.
Model an air molecule as a dipole driven by the light's field.
$p_0 = \alpha E_0$
The polarizability is nearly constant across visible light.
Write the scattered power.
$P = \dfrac{\mu_0\alpha^2E_0^2\omega^4}{12\pi c}$
Dipole radiation.
Find the ratio for violet and red.
$\dfrac{P_{400}}{P_{700}} = \left(\dfrac{700}{400}\right)^4 = 9.4$
At equal incident intensity.
Find the ratio for blue and red.
$\left(\dfrac{700}{450}\right)^4 = 5.9$
Blue still dominates the scattered light.
Explain why the sky is not violet.
$\text{less violet in sunlight; eyes less sensitive to it}$
The spectrum and our vision tilt the balance to blue.
Explain the red sunset.
$\text{long path through air removes blue from the direct beam}$
What remains is red and orange.
Predict the polarization at $90°$ from the Sun.
$\text{strongly polarized}$
Dipoles driven perpendicular to the line of sight radiate linearly polarized light.
Test it with polarized sunglasses.
$\text{rotate the lenses: the sky darkens and brightens}$
An everyday demonstration of dipole radiation.
Write the scaling.
$P \propto \dfrac{1}{\lambda^4}$
Rayleigh scattering.
Form the ratio.
$\left(\dfrac{700}{350}\right)^4 = 2^4$
Twice the frequency.
Evaluate the ratio.
An oscillating dipole radiates $28$ mW. If its dipole moment amplitude stays the same but its frequency doubles, how much does it radiate?
Complete the worked solution: a dipole radiates $11$ μW at frequency $f$. With the same dipole moment, find the power at $2f$ and at $3f$, and the difference between them, in μW.
Scale by $2^4$.
$P(2f) = 16 \times 11 =$ a
Twice the frequency.
Scale by $3^4$.
$P(3f) = 81 \times 11 =$ b
Three times the frequency.
Subtract the two powers.
$P(3f) - P(2f) =$ c
Sixty-five times the original.
Match each property of dipole radiation to its expression.
| $\mu_0p_0^2\omega^4/12\pi c$ | $\propto\sin^2\theta$ | $\propto 1/r$ | $80\pi^2(L/\lambda)^2$ | |
|---|---|---|---|---|
| total power | ||||
| angular pattern | ||||
| field falloff | ||||
| radiation resistance |
At a fixed distance from an oscillating dipole, fill in the intensity relative to its maximum at angles $0°$, $30°$, $45°$ and $90°$ from the dipole's axis.
| $I/I_{\max}$ | |
|---|---|
| $0°$ | |
| $30°$ | |
| $45°$ | |
| $90°$ |
A short dipole antenna is $18$ cm long and operates at a wavelength of $1$ m. What is its radiation resistance, in ohms? Use $80\pi^2 = 789.6$.
Answer: Ω
A short dipole antenna $5$ cm long radiates at a wavelength of $1$ m, driven by a current of amplitude $1$ A. How much power does it radiate, in watts?
Answer: W
Air molecules scatter sunlight as tiny oscillating dipoles driven by the light's field. How many times more strongly is $400$ nm light scattered than $600$ nm light?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Complete the worked solution: a dipole radiates $12$ μW at frequency $f$. With the same dipole moment, find the power at $2f$ and at $3f$, and the difference between them, in μW.
Scale by $2^4$.
$P(2f) = 16 \times 12 =$ a
Twice the frequency.
Scale by $3^4$.
$P(3f) = 81 \times 12 =$ b
Three times the frequency.
Subtract the two powers.
$P(3f) - P(2f) =$ c
Sixty-five times the original.
You can analyze dipole radiation. Explain to someone why the sky is blue.
19. Your turn: violet light at $350$ nm and red light at $700$ nm strike the same molecules. How much more strongly is the violet scattered?, step 3
$16$
Sixteen times as strong.