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When considering radiation from a moving charge, we have to think about how far the observer is from the charge and the finite propagation speed of the electromagnetic fields. This results in us solving Maxwell’s equations with a ‘retarded time’
\[t = t_\mathrm{ret} - \frac{r(t_\mathrm{ret})}{c}\]Using these and doing some math (Green’s functions or Fourier transformations, you take your pick) we get the Lienard-Wiechert potentials
\[\phi = \left[ \frac{q}{\kappa r} \right]\mathrm{ret}\] \[\vec{A} = \left[ \frac{q\vec{v}}{c\kappa r} \right]\mathrm{ret}\]where $\kappa$ contains the effect of line-of-sight:
\[\kappa_\mathrm{ret} = 1 - \frac{ \hat{r}_\mathrm{ret} \cdot \vec{v}_\mathrm{ret} }{c} = 1 - \frac{v_\mathrm{ret} }{c} \cos \phi_\mathrm{ret}\]The concept is “where was the emitting particle one light travel time ago?” That defines the retarded time and allows us to calculate the electromagnetic fields at the observer generated by the source.
Assuming $v \ll c$ and large $r$, the electric and magnetic fields are
\[\vec{E} = \frac{q}{c^2 r}\hat{k}\times \left( \hat{k} \times \vec{a} \right)\] \[\vec{B} = \frac{q}{c^2 r} \vec{a}\times \hat{k}\]Both of these have the same magnitude:
\[\| \vec{B} \| = \| \vec{E} \| = \frac{q}{c^2 r} a \sin \theta\]where $\theta$ is the angle between the observer in direction $\hat{k}$ and the particle’s acceleration $\vec{a}$.
From these expressions, we know the particle generates a pointing flux
\[\vec{S} = \frac{c}{4\pi} \vec{E} \times \vec{B} = \frac{q^2 a^2 }{4\pi c^3 r^2} \sin^2 \theta \, \hat{k}\]The pointing flux is power per unit area. The power per unit solid angle is then
\[\frac{dW}{dt \, d\Omega} = \frac{dW}{dt \, dA} \frac{dA}{d\Omega } = \|\vec{S}\| \frac{dA}{d\Omega } = \frac{q^2 a^2 }{4\pi c^3 r^2} \sin^2 \theta \cdot r^2 = \frac{q^2 a^2 }{4\pi c^3 } \sin^2 \theta\]Integrating over solid angle gives the Larmor Formula
\[P = \int d\Omega \left[ \frac{dW}{dt \, d\Omega} \right]= \frac{q^2 a^2 }{4\pi c^3 } 2\pi \int_0^{\pi} d\theta \left[ \sin^3 \theta \right]\] \[P = \frac{q^2 a^2 }{2 c^3 } (-1)\int_{1}^{-1} d\mu\left[ 1-\mu^2 \right] = \frac{2q^2 a^2 }{3c^3 }\]For dipole radiation, we define the dipole moment of a source as
\[\vec{d} = \sum_i q_i \vec{r}_i\]This allows us to generalize the Larmor formulat for objects with zero net charge and nonzero dipole moment:
\[P = \frac{2\ddot{d}^2 }{3c^3 }\]where the dots above $d$ denote time derivatives.
For relativistic particles, the Larmor formula depends on the acceleration parallel and perpendicular to the velocity of the particle. Specifically, the relativistic Larmor formula is
\[P = \frac{2q^2 }{3c^3 } \gamma^4 \left(a_\perp^2 + \gamma^2 a_\parallel^2 \right)\]where $\gamma$ is the Lorentz factor and $a_\parallel$ is the acceleration component parallel to the particle’s velocity.
Also, we can fourier transform the Larmor formula such that
\[\frac{dW}{d\nu \, d\Omega} = \left( \frac{2\pi}{c}\right)^3 \nu^4 \| d(\nu)\|^2 \sin^2 \theta\]where the dipole moment $d(\nu)$ is defined by the fourier transform of the dipole moment:
\[\vec{d}(\nu) = \frac{-1}{(2\pi \nu)^2} \mathcal{F}\left[ \ddot{\vec{d}}\right]\]