====== Radiation from acceleration ======
An accelerated charge particle creates waves on the [[em|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 $E_\theta=0$.

{{:un:charge-sphere.jpg?nolink&500|}}

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.

{{:un:charge-cone.jpg?nolink&500|}}

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.

{{:un:charge-S.jpg?nolink&500|}}

  - $P$ gives the total radiation in all directions.
  - $S$ 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.
  - 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.