advanced32 min studyLesson 254 in Physics

Normalize electric dipole radiation

Derive the sin-squared pattern, its total power, and the distinction between peak and cycle-averaged radiation.

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What you’ll understand

  • State both compact-source and far-zone assumptions.
  • Integrate a dipole angular pattern with the solid-angle measure.
  • Distinguish instantaneous, peak, and cycle-averaged power.

A small oscillating antenna radiates most strongly sideways and not along its electric dipole axis. That pattern follows from transversality, not from an arbitrary preference built into the source. The electric dipole approximation replaces a compact conserved charge distribution by its changing first moment. We will derive the angular power normalization in SI units, then average a sinusoidal source carefully. The frequent missing factor of two comes from confusing the squared peak amplitude with its time average.

Compress a source without discarding conservation

Define the electric dipole moment as the integral of position times charge density. Its units are coulomb metres. Continuity connects its first time derivative to the volume integral of current for a localized source with no boundary current escaping the chosen region. The radiation field depends on the second time derivative of the moment. A constant dipole has an electrostatic field but no electric dipole radiation; constant first derivative likewise gives no leading radiation term.

Two independent scale comparisons are needed. The source size must be small compared with the reduced wavelength, so source elements have nearly equal retarded phase. The observer must also be in the radiation zone, far compared with the source size and reduced wavelength. A distant observer does not by itself make an electrically large antenna a dipole. Higher multipoles and angular phase structure remain important for a large source even if its individual charges move slowly.

p(t)=∫x′ρ(x′,t) d3x′,Erad=n×[n×p¨(t−r/c)]4πϵ0c2r,kℓ≪1,kr≫1\mathbf p(t)=\int\mathbf x^{\prime}\rho(\mathbf x^{\prime},t)\,d^3x^{\prime},\qquad \mathbf E_{\mathrm{rad}}=\frac{\mathbf n\times[\mathbf n\times\ddot{\mathbf p}(t-r/c)]}{4\pi\epsilon_0c^2r},\qquad k\ell\ll1,\quad kr\gg1
p is in C·m, ℓ is source size in metres, k=ω/c for a characteristic angular frequency in rad/s, and r≫ℓ. The displayed field keeps the leading electric dipole term in vacuum.↔ Scroll wide equations horizontally.

Integrate the angular pattern

Suppose the second derivative of the moment lies along a fixed axis. Its projection onto the plane perpendicular to the observation direction has magnitude proportional to sine of the polar angle. Squaring the transverse field and multiplying by epsilon zero times c and radius squared gives power per solid angle. This quantity has units of watts per steradian. It is not power per unit polar angle, because a band of fixed polar-angle width contains a different solid angle near the axis and equator.

Integrate over azimuth and include the sine from the solid-angle measure. The integral of sine squared over the sphere is eight pi over three. This factor turns the angular coefficient into the total dipole power. The pattern's axial zeros concern the radiation term, not the full near electric field. A detector located on the dipole axis can still sense a changing induction field nearby without contradicting the far-zone radiation null.

dPdΩ=∣p¨∣2sin⁡2θ16π2ϵ0c3,∫sin⁡2θ dΩ=8π3,P=∣p¨∣26πϵ0c3\frac{dP}{d\Omega}=\frac{|\ddot{\mathbf p}|^2\sin^2\theta}{16\pi^2\epsilon_0c^3},\qquad \int\sin^2\theta\,d\Omega=\frac{8\pi}{3},\qquad P=\frac{|\ddot{\mathbf p}|^2}{6\pi\epsilon_0c^3}
θ is measured from the dipole-acceleration axis, dΩ=sinθ dθ dφ, and P is instantaneous emitted power evaluated at the source time corresponding to the outgoing wave.↔ Scroll wide equations horizontally.

Averages and coherent addition

For a moment with cosine time dependence, the second derivative contributes angular frequency squared and a minus sign. The sign does not survive the square, while the squared cosine averages to one half. Average power therefore grows as frequency to the fourth power at fixed moment amplitude. This statement holds only while source size remains small compared with wavelength and the prescribed moment remains physically realizable. Increasing frequency indefinitely eventually violates the approximation.

Fields from coherent sources must be added before squaring. Two equal dipoles with identical orientation and phase can radiate four times one dipole's intensity in a direction where their path phases agree. Opposite phases can cancel the leading dipole moment and reveal higher multipoles. Adding independently computed powers would miss that interference. Random relative phases, averaged over suitable times or ensembles, are a different physical preparation and may justify an incoherent sum.

p(t)=p0cos⁡(ωt)z^,⟨P⟩=p02ω412πϵ0c3,⟨cos⁡2ωt⟩=12\mathbf p(t)=p_0\cos(\omega t)\hat{\mathbf z},\qquad \langle P\rangle=\frac{p_0^2\omega^4}{12\pi\epsilon_0c^3},\qquad \left\langle\cos^2\omega t\right\rangle=\frac12
p₀ is peak moment in C·m and ω is angular frequency in rad/s. This average assumes a fixed-axis sinusoid over a complete cycle in the compact-source vacuum radiation regime.↔ Scroll wide equations horizontally.

Normalize a measured dipole pattern

A cycle-averaged fixed-axis dipole pattern is measured as dP/dΩ=9 sin²θ mW/sr. Find total average power and its angular density at 30° and 90°. State why multiplying 9 mW/sr by 4π gives the wrong answer.

  1. The angular coefficient is the equatorial maximum, not an isotropic average. Integrate sin²θ over solid angle to obtain 8π/3 rather than 4π.

  2. Total power is 9×8π/3=24π mW, approximately 75.4 mW. The isotropic estimate 36π mW is larger by a factor 3/2.

  3. At 30° the squared sine is 1/4, giving 2.25 mW/sr. At 90° it is one, giving 9 mW/sr. These angular densities are distinct from power collected by a finite detector, which also needs its accepted solid angle.

The result

The total is 24π mW, with angular densities 2.25 and 9 mW/sr at 30° and 90°. The maximum angular density cannot be treated as uniform over the sphere.

Explore the model

Interactive lab

Interactive lab · vectors

Predict the dipole angular null

For independent runs Reset and set BOTH angle 0°, magnitude 1; then Reset and set angle 90°, magnitude 1. Identify y as the transverse projection and predict zero and unit squared projection. The display contains one vector, so the radiation pattern and its solid-angle integral must be calculated separately.

Predict the dipole angular nullPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe angular null comes from a transverse projection, and total power requires integration over directions.

Check your understanding

For a sinusoidal electric dipole with fixed peak moment, doubling angular frequency changes average power by what factor within the same dipole regime?

Try it yourself

A compact sinusoidal dipole has peak instantaneous total power 12 μW. Find its cycle average. If its moment amplitude is then halved at unchanged frequency, find the new peak and average powers.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Average the squared cosine, and remember that power is quadratic in moment amplitude.

Compare with the solution

The first average is 6 μW. Halving moment amplitude divides every instantaneous power by four, so the new peak is 3 μW and the new average is 1.5 μW. No additional averaging factor should be applied to the original average twice. These ratios assume the same frequency, angular pattern, and compact-source regime.

Take it with you

  • Electric dipole radiation requires a nonzero second derivative of the electric moment.
  • The sin²θ angular pattern integrates to 8π/3, not 4π.
  • A fixed-axis sinusoid has average total power equal to half its peak.

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