Electric dipole radiation: angular distribution, polarization, and total radiated power

Intuition Beginner

When an electric charge oscillates back and forth, it radiates electromagnetic waves. The simplest radiating system is the oscillating electric dipole — two equal and opposite charges separated by a small distance that varies sinusoidally with time.

The radiation pattern from an oscillating dipole is not uniform. The radiated power is strongest in the directions perpendicular to the oscillation (the "equator" of the dipole) and zero along the axis of oscillation (the "poles"). The angular dependence goes as ${\sin }^2\theta$ where $\theta$ is the angle from the dipole axis. Think of a radio antenna: it radiates most strongly broadside to the antenna and not at all along its length.

The total power radiated by an oscillating dipole with dipole moment $p=p_0\cos (\omega t)$ is:

$$P=\frac{{\mu }_0{p}_0^2{\omega }^4}{12\pi c}$$

The power scales as ${\omega }^4$: doubling the frequency increases the radiated power by a factor of 16. This strong frequency dependence explains why higher-frequency radiation (UV, X-rays) is produced by atomic-scale dipoles while lower-frequency radiation (radio waves) requires macroscopic antennas.

Visual Beginner

Direction from dipole	Radiated power
Along axis ($\theta =0$)	Zero (no radiation)
45 degrees ($\theta =45^{\circ }$)	Half maximum
Perpendicular ($\theta =90^{\circ }$)	Maximum

Worked example Beginner

A short radio antenna of length $\ell =1$ m carries an oscillating current $I=I_0\cos (\omega t)$ with $I_0=1$ A and frequency $f=100$ MHz. Find the total radiated power.

The antenna is an electric dipole with $p_0=I_0\ell /\omega$. (The current $I_0$ flowing over the length $\ell$ creates a dipole moment $p=q\ell$ where $dq/dt=I_0$.) The angular frequency is $\omega =2\pi \times 10^8$ rad/s.

$p_0=I_0\ell /\omega =1\times 1/(2\pi \times 10^8)=1.59\times 10^{-9}$ C$\cdot$m.

$P={\mu }_0{p}_0^2{\omega }^4/(12\pi c)={\mu }_0{\omega }^2(I_0\ell {)}^2/(12\pi c)$. Substituting: $P=4\pi \times 10^{-7}\times (2\pi \times 10^8{)}^2\times 1/(12\pi \times 3\times 10^8)=4\pi \times 10^{-7}\times 3.95\times 10^{17}\times 1/(12\pi \times 3\times 10^8)\approx 4.39$ W. About 4.4 watts of radiation from a 1-meter antenna at 100 MHz with 1 amp of current.

Check your understanding Beginner

Formal definition Intermediate+

An oscillating electric dipole has dipole moment $\mathbf{p}(t)=p_0\cos (\omega t)\hat{\mathbf{z}}$ (or more generally $\mathbf{p}(t)={\mathbf{p}}_0{e}^{-i\omega t}$, taking the real part).

Retarded potential approach. In the Lorenz gauge, the retarded vector potential is:

$$\mathbf{A}(\mathbf{r},t)=\frac{{\mu }_0}{4\pi }\int \frac{\mathbf{J}({\mathbf{r}}^{\prime },t_r)}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}{d}^3{r}^{\prime }$$

For a source of size $d$ much smaller than the wavelength ($d\ll \lambda =2\pi c/\omega$) and the observation distance ($d\ll r$), expand the retarded time: $t_r\approx t-r/c+\hat{\mathbf{r}}\cdot {\mathbf{r}}^{\prime }/c$. Keeping the leading correction (the dipole term):

$$\mathbf{A}(\mathbf{r},t)=\frac{{\mu }_0}{4\pi r}\dot{\mathbf{p}}(t_r)=\frac{{\mu }_0}{4\pi r}(-\omega p_0\sin (\omega t_r))\hat{\mathbf{z}}$$

where $t_r=t-r/c$ is the retarded time.

Radiation fields (far field, $r\gg \lambda$). The electric and magnetic fields in the radiation zone are:

$$\mathbf{E}=-\frac{{\mu }_0{p}_0{\omega }^2}{4\pi }\frac{\sin \theta }{r}\cos (\omega (t-r/c))\hat{\mathbf{\theta }}$$ $$\mathbf{B}=\frac{{\mu }_0{p}_0{\omega }^2}{4\pi c}\frac{\sin \theta }{r}\cos (\omega (t-r/c))\hat{\mathbf{\varphi }}$$

Both fields fall off as $1/r$ and are transverse (perpendicular to $\hat{\mathbf{r}}$). They satisfy $\mathbf{B}=\hat{\mathbf{r}}\times \mathbf{E}/c$.

Time-averaged Poynting vector.

$$⟨\mathbf{S}⟩=\frac{{\mu }_0{p}_0^2{\omega }^4}{32{\pi }^{2}c}\frac{{\sin }^2\theta }{r^2}\hat{\mathbf{r}}$$

Total radiated power.

$$P=\oint ⟨S⟩r^{2}d\Omega =\frac{{\mu }_0{p}_0^2{\omega }^4}{12\pi c}$$

This is the electric dipole radiation formula. Using $p_0=q_{0}d$ (charge times separation) or $p_0=I_0\ell /\omega$ (current times length divided by angular frequency):

$$P=\frac{{\mu }_0{q}_0^2{d}^2{\omega }^4}{12\pi c}=\frac{{\mu }_0{I}_0^2{\ell }^2{\omega }^2}{12\pi c}$$

Key derivation Intermediate+

Derivation (Electric dipole radiation fields from the retarded potentials).

Theorem. An oscillating electric dipole $\mathbf{p}(t)=p_0{e}^{-i\omega t}\hat{\mathbf{z}}$ radiates fields (in the far zone $r\gg \lambda$):

$$\mathbf{E}=\frac{k^2{p}_0}{4\pi {ϵ}_0}\frac{e^{i(kr-\omega t)}}{r}\sin \theta \hat{\mathbf{\theta }},\mathbf{B}=\frac{{\mu }_0{k}^2\omega p_0}{4\pi }\frac{e^{i(kr-\omega t)}}{r}\sin \theta \hat{\mathbf{\varphi }}$$

The total radiated power is $P={\mu }_0{p}_0^2{\omega }^4/(12\pi c)$.

Proof. For a localized source with $\mathbf{p}(t)=\int {\mathbf{r}}^{\prime }\rho ({\mathbf{r}}^{\prime },t)d^3{r}^{\prime }$, the retarded scalar and vector potentials can be expanded in powers of $k{r}^{\prime }$ (multipole expansion). The leading (dipole) term for the vector potential is:

$${\mathbf{A}}_{\text{dip}}=\frac{{\mu }_0}{4\pi r}(-i\omega {\mathbf{p}}_0)e^{i(kr-\omega t)}$$

From $\mathbf{A}$, compute $\mathbf{B}=\nabla \times \mathbf{A}$. In the far zone ($kr\gg 1$), the gradient acts mainly on the phase factor $e^{ikr}$: $\nabla e^{ikr}\approx ik\hat{\mathbf{r}}{e}^{ikr}$. So:

$$\mathbf{B}=\nabla \times \mathbf{A}\approx ik\hat{\mathbf{r}}\times \mathbf{A}=\frac{{\mu }_0{k}^2\omega p_0}{4\pi }\frac{e^{i(kr-\omega t)}}{r}\sin \theta \hat{\mathbf{\varphi }}$$

From Faraday's law in the frequency domain ($\nabla \times \mathbf{E}=i\omega \mathbf{B}$), or directly from $V_{\text{dip}}$ and ${\mathbf{A}}_{\text{dip}}$:

$$\mathbf{E}\approx -c\hat{\mathbf{r}}\times \mathbf{B}=\frac{k^2{p}_0}{4\pi {ϵ}_0}\frac{e^{i(kr-\omega t)}}{r}\sin \theta \hat{\mathbf{\theta }}$$

The time-averaged Poynting vector is $⟨S⟩=\frac{1}{2{\mu }_0}|\mathbf{E}\times {\mathbf{B}}^{\ast }|=\frac{{\mu }_0{p}_0^2{\omega }^4}{32{\pi }^{2}c}\frac{{\sin }^2\theta }{r^2}$. Integrating over solid angle: $\int {\sin }^2\theta d\Omega =2\pi {\int }_0^{\pi }{\sin }^3\theta d\theta =8\pi /3$. So $P=\frac{{\mu }_0{p}_0^2{\omega }^4}{12\pi c}$. $\square$

Bridge. Electric dipole radiation is the dominant radiation mechanism for most radiating systems: radio antennas, oscillating atoms, and molecular transitions. The foundational insight is that the radiation fields are produced by the acceleration of charges (since $\ddot{\mathbf{p}}=\sum q_i{\mathbf{a}}_i$), which connects to the Larmor formula 10.07.01. The central message is that the dipole radiation power scales as ${\omega }^4$ (or equivalently ${\lambda }^{-4}$), the angular distribution goes as ${\sin }^2\theta$, and the fields are transverse with $|\mathbf{E}|=c|\mathbf{B}|$. Putting these together, the dipole radiation formula is the leading term in the multipole expansion 10.07.03, and the higher multipoles (magnetic dipole, electric quadrupole) are suppressed by additional powers of $kd$ (where $d$ is the source size). The dipole radiation also provides the classical model for atomic transitions 10.07.06, where the oscillating charge distribution in an atom produces radiation at the transition frequency.

Exercises Intermediate+

Lean formalization Intermediate+

Mathlib lacks the retarded potential formalism, the multipole expansion, and the derivation of radiation fields. The spherical harmonic expansion of Green's functions and the near/far field transition are absent. lean_status: none.

Advanced results Master

The Hertz vector. The electric and magnetic fields of a dipole radiator can be expressed compactly in terms of the electric Hertz vector ${\Pi }_e$:

$$\mathbf{E}=\nabla (\nabla \cdot {\Pi }_e)-{\mu }_0{ϵ}_0\frac{{\partial }^2{\Pi }_e}{\partial t^2},\mathbf{B}={\mu }_0{ϵ}_0\nabla \times \frac{\partial {\Pi }_e}{\partial t}$$

For an oscillating dipole: ${\Pi }_e=\frac{1}{4\pi {ϵ}_{0}r}\mathbf{p}(t_r)\hat{\mathbf{z}}$. This formulation automatically satisfies Maxwell's equations and is the basis for the multipole expansion.

The multipole expansion. The full radiation field is expanded in multipoles: the electric dipole (E1) is the leading term, followed by the magnetic dipole (M1) and electric quadrupole (E2), each suppressed by an additional factor of $kd$ (where $d$ is the source size). The expansion is analogous to the Taylor expansion of the retarded time.

Radiation from an arbitrary current distribution. For a general current $\mathbf{J}(\mathbf{r},t)$, the far-field radiation is:

$$\mathbf{E}(\mathbf{r},t)=\frac{{\mu }_0}{4\pi r}\hat{\mathbf{r}}\times (\hat{\mathbf{r}}\times {\ddot{\mathbf{p}}}_{\perp }(t_r))$$

where ${\mathbf{p}}_{\perp }$ is the component of the dipole moment perpendicular to $\hat{\mathbf{r}}$. This generalises the single-dipole formula to arbitrary source distributions.

Synthesis. Electric dipole radiation is the universal leading-order description of electromagnetic radiation from any charge distribution. The foundational insight is that radiation is produced by the second time derivative of the dipole moment ($\ddot{\mathbf{p}}$), which corresponds to charge acceleration. The central message is that the dipole radiation pattern (${\sin }^2\theta$), the ${\omega }^4$ power scaling, and the $1/r$ field decay are universal features that appear whenever the source is small compared to the wavelength. Putting these together, electric dipole radiation connects the microscopic (Larmor formula 10.07.01 for a single charge) to the macroscopic (antenna radiation resistance, Rayleigh scattering 10.07.06), and provides the foundation for the full multipole expansion 10.07.03 that describes radiation from extended sources.

Full proof set Master

Proposition (Dipole radiation fields). The far-field radiation from an oscillating dipole $\mathbf{p}(t)=p_0\cos (\omega t)\hat{\mathbf{z}}$ is:

$$\mathbf{E}(r,\theta ,t)=-\frac{{\mu }_0{p}_0{\omega }^2}{4\pi r}\sin \theta \cos (\omega t_r)\hat{\mathbf{\theta }},\mathbf{B}=\hat{\mathbf{r}}\times \mathbf{E}/c$$

Proof. The retarded vector potential for the dipole is $\mathbf{A}=\frac{{\mu }_0}{4\pi r}\dot{\mathbf{p}}(t_r)$. In the far zone ($kr\gg 1$), $\nabla \times \mathbf{A}\approx ik\hat{\mathbf{r}}\times \mathbf{A}$ (the gradient acts on the phase factor). This gives $\mathbf{B}=ik\hat{\mathbf{r}}\times \mathbf{A}$. The electric field follows from $\mathbf{E}=-c\hat{\mathbf{r}}\times \mathbf{B}$ (far-field relation from Maxwell's equations). Computing $\hat{\mathbf{r}}\times \hat{\mathbf{z}}=-\sin \theta \hat{\mathbf{\varphi }}$ and $\hat{\mathbf{r}}\times \hat{\mathbf{\varphi }}=-\hat{\mathbf{\theta }}$ gives the stated result. $\square$

Connections Master

• Larmor formula 10.07.01 is the single-charge version of dipole radiation; $P={\mu }_0{q}^2{a}^2/(6\pi c)$.
• Multipole radiation 10.07.03 extends the dipole treatment to magnetic dipole and electric quadrupole terms.
• Synchrotron radiation 10.07.05 is produced by relativistic charges; the dipole formula is the non-relativistic limit.
• Thomson scattering 10.07.06 is dipole radiation from a bound charge driven by an incident wave.
• Vector potential 10.02.02 provides the retarded potentials from which the radiation fields are derived.
• EM waves 10.04.02 are the general framework; dipole radiation is a specific solution.

Historical & philosophical context Master

Hertz's experiments (1886-1888) were the first demonstration of electromagnetic radiation from an oscillating dipole. His spark-gap transmitter was essentially an oscillating electric dipole, and he measured the radiation pattern, finding the ${\sin }^2\theta$ angular distribution predicted by Maxwell's theory. This was the decisive confirmation of Maxwell's equations.

The connection between dipole radiation and atomic transitions was made by Lorentz (1903), who modelled atoms as oscillating electrons (the "Lorentz oscillator") and showed that they radiate at the oscillation frequency. This classical model correctly predicts the ${\omega }^4$ scaling but fails to explain the discrete nature of atomic spectra, which required quantum mechanics.

Rayleigh scattering (1871) was explained by Lord Rayleigh as dipole radiation from the induced polarisation of air molecules, correctly predicting the ${\lambda }^{-4}$ wavelength dependence that makes the sky blue. This was one of the first quantitative applications of electromagnetic theory to a natural phenomenon.

Bibliography Master

• Hertz, H., "On electromagnetic waves in air and their reflection," Ann. Phys. 34, 610 (1888).
• Rayleigh, Lord, "On the light from the sky, its polarisation and colour," Phil. Mag. 41, 107-120, 274-279 (1871).
• Jackson, J. D., Classical Electrodynamics, 3rd ed. (Wiley, 1999).
• Griffiths, D. J., Introduction to Electrodynamics, 4th ed. (Cambridge, 2017).
• Landau, L. D. and Lifshitz, E. M., The Classical Theory of Fields, 4th ed. (Butterworth-Heinemann, 1975).