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Retarded potentials of infinitesimal dipole

Short answer

Retarded potential uses source values at the earlier time t − R/v; in phasor form the delay is e^(−jkR). For an infinitesimal dipole in the Fraunhofer region, the far-field vector potential is Az = μ I0 l e^(−jkr)/(4πr), and the scalar potential is V = (I0 l cosθ/4πε)(1/(c r) + 1/(jω r²)) e^(−jkr), which reduces to the 1/r term for r ≫ λ.

What is retarded potential? Write the expression of retarded scalar potential and vector potential of infinitesimal dipole antenna in Fraunhofer far field region.

Retarded potential is the potential (scalar V or vector A) at a point computed from the source value at the earlier time t − R/v, since the effect of a time-varying source takes R/v seconds to arrive. In phasor form the delay appears as a factor e^(−jkR).

V = (1/4πε) ∫ ρ e^(−jkR)/R dv
A = (μ/4π)  ∫ J e^(−jkR)/R dv

Infinitesimal dipole

Length l ≪ λ along the z-axis, uniform current I0, end charges ±q with I0 = jωq.

Retarded vector potential (exact for small l):

Az = μ I0 l e^(−jkr) / (4π r)
Ar = Az cosθ = μ I0 l cosθ e^(−jkr)/(4πr)
Aθ = −Az sinθ = −μ I0 l sinθ e^(−jkr)/(4πr)

Retarded scalar potential (from the two end charges, or from the Lorenz gauge):

V = (q l cosθ / 4πε r²) (1 + jkr) e^(−jkr)
  = (I0 l cosθ / 4πε) [1/(c r) + 1/(jω r²)] e^(−jkr)

In the Fraunhofer (far-field) region

For r ≫ λ (kr ≫ 1, r > 2D²/λ) only the 1/r terms remain:

Az ≈ μ I0 l e^(−jkr) / (4π r)
Aθ ≈ −μ I0 l sinθ e^(−jkr) / (4π r)
V  ≈ I0 l cosθ e^(−jkr) / (4π ε c r)
   = η I0 l cosθ e^(−jkr) / (4π r)

(using 1/(εc) = η = 120π Ω).

From these the far fields are

Eθ ≈ −jω Aθ = jη k I0 l sinθ e^(−jkr)/(4πr)
Hφ ≈ Eθ / η

which give the sinθ (doughnut) pattern of the dipole.

✓ Answer checked by Claude Opus 5.5 (high effort)

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