Electric dipole radiation

For a localized source whose size is small compared with its radiation wavelength, the radiation fields of its electric dipole moment $p$ areTheir total instantaneous power is

With all electric dipole moments and time derivatives evaluated at $t_0=t-r/c$, the retarded potentials of a point electric dipole give, for $r>0$,For a short verification, apply $\nabla t_0=-\mathbf{n}/c$ and ${\partial }_i{n}_j=({\delta }_{ij}-n_i{n}_j)/r$ when computing $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$. This produces the displayed $r^{-3}$, $r^{-2}$ and $r^{-1}$ terms. The radiation electric field is transverse and obeys ${\mathbf{B}}_{\mathrm{rad}}=\mathbf{n}\times {\mathbf{E}}_{\mathrm{rad}}/c$. The vector triple product then gives the leading Poynting vector $\mathbf{n}[|{\mathbf{p}}^{\prime \prime }{|}^2-(\mathbf{n}\cdot {\mathbf{p}}^{\prime \prime }{)}^2]/(16{\pi }^2{ϵ}_0{c}^3{r}^2)$, displaying the angular distribution of electric dipole radiation.

A permanent electric dipole moment of magnitude $p_0$ rotating perpendicular to a fixed axis emits $P=K{\omega }^4$ in the electric-dipole approximation for radiation, with $K={\mu }_0{p}_0^2/(6\pi c)$. For fixed mechanical moment of inertia $\mathcal{I}$, weak radiative loss gives $\mathcal{I}\omega \dot{\omega }=-K{\omega }^4$ and hence the displayed spin-down law. It assumes $|\dot{\omega }|\ll {\omega }^2$: the full dipole acceleration has squared magnitude $p_0^2({\omega }^4+{\dot{\omega }}^2)$, whose second term is beyond this leading adiabatic approximation.

For a source of characteristic size $a$ and angular frequency $\omega$, the electric-dipole approximation requiresso the source is much smaller than its radiation wavelength. Observing the pure $1/R$ radiation field additionally requires the radiation zone, $R\gg c/\omega$ and $R\gg a$.

If $\theta$ is the angle between the observation direction and $\ddot{p}$, thenIntegrating over solid angle uses $\int {\sin }^2\theta d\Omega =8\pi /3$ and gives the total electric dipole radiation power.

The angular distribution of electric dipole radiation measures squared projections of the dipole axis, so a few intensity measurements need not determine that axis or the total radiated power uniquely. For a real unit dipole axis $\mathbf{d}$, the radiation-zone formula is $I(\mathbf{n})=3P[1-(\mathbf{d}\cdot \mathbf{n}{)}^2]/(8\pi R^2)$. Equal intensities along $\hat{\mathbf{z}}$, $(\hat{\mathbf{x}}+\hat{\mathbf{z}})/\sqrt{2}$ and $(\hat{\mathbf{y}}+\hat{\mathbf{z}})/\sqrt{2}$ permit four axes, up to reversal:Three generally different radiated powers are compatible with the same intensity data. Additional directional measurements or polarization information can remove the ambiguity.

For an orbit of radius $b$ in the $xy$ plane, $\mathbf{a}=-b{\omega }^2(\cos \omega t,\sin \omega t,0)$. In spherical coordinates, at angle $\theta$ from the orbit normal, $⟨(\mathbf{n}\cdot \mathbf{a}{)}^2⟩=b^2{\omega }^4{\sin }^2\theta /2$. The acceleration covariance in nonrelativistic radiation therefore gives the displayed pattern. Its axial power per solid angle is twice its equatorial value; no observation direction has zero time-averaged power. Integrating gives $⟨P⟩=q^2{b}^2{\omega }^4/(6\pi {ϵ}_0{c}^3)$. The condition $b\omega \ll c$ is essential: relativistic synchrotron radiation has a different beamed pattern.

A charge $q$ undergoing simple harmonic motion of angular frequency $\omega$ and slowly varying amplitude $A$ radiates the averaged powerIts mechanical energy $E=m{A}^2{\omega }^2/2$ therefore has exponential decay rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$ and half-life