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core39 min studyLesson 18 in Physics

Retarded potentials and radiation

Use causal source times, distinguish near fields from radiation, and estimate the power emitted by a small oscillating dipole.

Before you begin
On this page

The question for today

What you’ll understand

  • Interpret retarded source time in an extended-source integral.
  • Separate static, induction, and radiative distance scaling.
  • Calculate dipole radiation power with peak moments and explicit validity conditions.

An antenna does not update the electric field everywhere at once. Changes in its charges and currents propagate outward, so an observer samples source behavior at earlier times. Retarded potentials encode this delay while preserving Maxwell's equations and charge conservation. Their distant fields also explain why a small alternating current can carry energy far beyond the region where its static field would be appreciable.

Every source point has its own delay

In Lorenz gauge, the scalar and vector potentials obey sourced wave equations. Choosing the retarded solution implements a boundary condition with no incoming radiation supplied from infinity in addition to the specified sources. Each source point contributes at time t−R/ct-R/c, where R is its distance to the observation point. An extended antenna therefore cannot generally be represented by one common delayed time without an approximation.

Retardation is not achieved by evaluating Coulomb's law at the delayed position and stopping there. Differentiating time-dependent potentials also produces terms involving changing sources, including the time derivative of vector potential. Their combination is essential for the correct field. The source distributions themselves must obey continuity; prescribing an oscillating charge that appears and disappears without a corresponding current is not a legitimate isolated antenna model.

ϕ(r,t)=14πϵ0∫ρ(r′,t−R/c)R d3r′,A(r,t)=μ04π∫J(r′,t−R/c)R d3r′\phi(\mathbf r,t)=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\mathbf r',t-R/c)}{R}\,d^3r',\qquad \mathbf A(\mathbf r,t)=\frac{\mu_0}{4\pi}\int\frac{\mathbf J(\mathbf r',t-R/c)}{R}\,d^3r'
R = |r − r′|. These are retarded vacuum potentials in Lorenz gauge for suitable localized conserved sources; additional homogeneous solutions would represent independently specified incoming fields.↔ Scroll wide equations horizontally.

A far field is a controlled limit

For a source of size a with wavelength much larger than a, the leading time-dependent electric multipole is often a dipole. Its near electric field decreases as inverse distance cubed, an induction contribution as inverse distance squared, and radiation as inverse distance. In the radiation zone the observer is far from both source size and the reduced wavelength. A small source alone does not guarantee that a nearby observer lies in that zone.

The radiative field is transverse to the observation direction and vanishes along the oscillating dipole axis. Its energy flux decreases as inverse distance squared, so integrating across a sphere gives a finite distance-independent outgoing power. The faster-decaying terms mainly describe local storage and exchange. They can affect near-field flux but cannot be interpreted as the same asymptotic power independently of distance.

Erad,0(r,θ)=p0ω2sin⁡θ4πϵ0c2r,⟨dPdΩ⟩=p02ω4sin⁡2θ32π2ϵ0c3E_{\mathrm{rad},0}(r,\theta)=\frac{p_0\omega^2\sin\theta}{4\pi\epsilon_0c^2r},\qquad \left\langle\frac{dP}{d\Omega}\right\rangle=\frac{p_0^2\omega^4\sin^2\theta}{32\pi^2\epsilon_0c^3}
These magnitudes describe a nonrelativistic localized electric dipole p(t) = p₀ cos(ωt) along a fixed axis, with ka ≪ 1 and kr ≫ 1. θ is measured from that axis; p₀ is a peak moment, not an RMS value.↔ Scroll wide equations horizontally.

Radiation power must be supplied

Integrating the angular pattern over a sphere produces the dipole's average emitted power. The strong fourth-power frequency dependence holds at fixed peak dipole moment within the small-source approximation. It cannot be extrapolated indefinitely: changing frequency also changes antenna currents, material behavior, and electrical size. A driven source must supply the radiated energy in addition to any resistive losses.

The closely related Larmor formula expresses the instantaneous radiation power of a nonrelativistically accelerating point charge. It is a classical result, not a model of stable atomic electrons orbiting indefinitely. Radiation reaction and quantum effects require further treatment when they materially alter the source dynamics. Here the prescribed dipole is a controlled emitter model whose external drive is responsible for maintaining its motion.

⟨P⟩=p02ω412πϵ0c3,PLarmor=q2a26πϵ0c3(v≪c)\langle P\rangle=\frac{p_0^2\omega^4}{12\pi\epsilon_0c^3},\qquad P_{\mathrm{Larmor}}=\frac{q^2a^2}{6\pi\epsilon_0c^3}\quad(v\ll c)
The first formula is cycle averaged for a sinusoidal dipole. The second uses an instantaneous nonrelativistic point-charge acceleration. An oscillating moment p = qx connects the two after averaging the squared acceleration.↔ Scroll wide equations horizontally.

Work it through

Power from an ideal small dipole

A localized sinusoidal dipole has peak moment p0=1.0×10−9 C mp_0=1.0\times10^{-9}\,\mathrm{C\,m} at 1.0 MHz. Assume its size is much smaller than 300 m and its charge motion is nonrelativistic. Estimate its mean radiation power.

  1. Convert frequency to angular frequency: ω=2πf=6.283×106 s−1\omega=2\pi f=6.283\times10^6\,\mathrm{s^{-1}}. Using f directly in a formula written for ω would miss a factor (2π)4(2\pi)^4.

  2. The numerator is p02ω4=(10−9)2(6.283×106)4≃1.559×109p_0^2\omega^4=(10^{-9})^2(6.283\times10^6)^4\simeq1.559\times10^9 in the corresponding SI combination. Use c=3.0×108 m/sc=3.0\times10^8\,\mathrm{m/s} and ϵ0=8.85×10−12 F/m\epsilon_0=8.85\times10^{-12}\,\mathrm{F/m}.

  3. Divide by 12πϵ0c3≃9.01×101512\pi\epsilon_0c^3\simeq9.01\times10^{15} to obtain ⟨P⟩=1.73×10−7 W\langle P\rangle=1.73\times10^{-7}\,\mathrm W. Doubling f at fixed p₀ predicts sixteen times the power only while the same assumptions remain valid.

The result

The ideal dipole emits approximately 0.173 μW on average. This total is measured by integrating radiation-zone flux, and is not the power dissipated in a particular real antenna's resistance.

The living notebook

Explore the model

Interactive lab

Interactive lab · waves

Delay appears as phase for a monochromatic signal

Vary the phase difference between the two equal-frequency scalar waves. A time delay becomes a phase shift for a single frequency, helping interpret retarded sinusoidal sources. The lab shows co-propagating one-dimensional sums, not an antenna radiation pattern, a three-dimensional Green function, or radiation power.

Delay appears as phase for a monochromatic signalTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). 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 forA phase delay can represent one frequency's travel time, but a general transient requires its full causal time dependence.

Pause & reason

Check your understanding

Which distance scaling allows a radiative electric field to carry finite nonzero total power through increasingly large spheres?

Your turn

Try it yourself

For the same ideal sinusoidal dipole, reduce the peak moment by a factor of two and double the frequency. Find the new mean power relative to the original. State two assumptions needed before trusting that scaling.

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

Give me a hint

Power is quadratic in moment and quartic in angular frequency.

Compare with the solution

The factor is (1/2)2(2)4=4(1/2)^2(2)^4=4, giving about 6.92×10−7 W6.92\times10^{-7}\,\mathrm W. The source must remain electrically small, with ka≪1ka\ll1, and its charge motion must remain nonrelativistic. Its drive must also actually maintain the specified new moment; a fixed voltage or fixed input power does not automatically do so. A far-field detector must stay in the radiation zone at the new frequency.

Keep the idea

Take it with you

  • Retarded potentials combine conserved sources with a causal boundary condition.
  • Near-field and radiation terms have different distance scaling and physical roles.
  • Radiation scaling is meaningful only while the emitter and observation approximations remain valid.

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