An accelerated charge particle creates waves on the electromagnetic field which is called radiation because it enables transfer of energy via radiation. Here we will derive Larmor’s formula, which describes the power of this radiation, using the intuition of J J Thomson.

A charge at rest or moving with constant velocity only exhibits radial electric field lines, so its $|\mathbf{E}|=E_r$ and its component perpendicular to the radial direction $E_\theta=0$.

Now if it is accelerated within the time $\Delta t$ with a small difference in velocity $\Delta v \ll c$, each line is disturbed with a perpendicular electric field. The sphere of radius $r=ct$ is already affected by the disturbance, but the regions outside are still unaware of the effect. Between the aware and unaware regions, there is a shell of thickness $c\Delta t$ where we have to join the inner and outer field lines.

Let us zoom into the shell. There is an electric field in the circumferential direction $\mathbf{i}_\theta$ which is like a pulse carrying away the lost energy of the charge at the speed of light.

The diagram imagines two cones, one at the rest position $O$ and the other a time $\Delta t$ later. The cones are different at $AB$, but join up at $CD$. Using the relative sizes of the sides of the rectangle $ABCD$, it can be seen that

$$ \frac{E_\theta}{E_r} = \frac{\Delta v t \sin\theta}{c\Delta t} $$

which is the ratio of the strengths of the two electric field components along $\mathbf{i}_r$ and $\mathbf{i}_\theta$. Remembering from Coulomb’s law that $E_r=q/r^2$ (in Gaussian units) we can write

$$ E_\theta = \frac{q}{r^2} \frac{\Delta v}{\Delta t} \frac{2\sin\theta}{c^2} = \frac{qa\sin\theta}{rc^2} $$

where $a$ is the average magnitude of the acceleration. This field mimics the acceleration directly as $E_\theta\propto a$.

Now to find the power, we have to use Poynting flux which gives power per unit area from a cross product of the electric and magnetic fields:

$$ S = \frac{c}{4\pi} |\mathbf{E}\times \mathbf{B}| = \frac{cE^2}{4\pi} $$

because $E=B$ in Gaussian units. So the flux

$$ S = \frac{q^2a^2}{4\pi c^3} \frac{\sin^2\theta}{r^2} $$

which has to be integrated over the surface of a sphere of radius $r$ in order to get the power:

$$ P = \int_{sphere} S dA = \frac{q^2a^2}{4\pi c^3} \int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi} \frac{\sin^2\theta}{r^2} r \sin\theta \ d\theta \ r\ d\phi $$

because the area element $dA=r \sin\theta \ d\theta \ r\ d\phi$. Looking up the result of the integral $\int\sin^3d\theta=4/3$ in the given limits, we finally find

$$ P = \frac{2q^2a^2}{3c^3} $$

in CGS units. The same can be done in SI units to yield

$$ P = \frac{q^2a^2}{6\pi\epsilon_0 c^3} $$

where $\epsilon_0$ is the permittivity of free space and the unit of power here will be W. This is Larmor’s formula. The formula demonstrates the 3 key properties of this radiation.

1. $P$ gives the total radiation in all directions.
2. $S$ gives the radiation per unit area of a spherical surface and this has a dipolar form, because $S\propto \sin^2\theta$: The radiation is zero along the acceleration and maximum perpendicular to it.
3. The radiation is polarized.

These results are valid only at the instantaneous rest frame of the particle and must be revised if the particle has a relativistic speed.