Larmor Power

Larmor radiation is produced by an accelerating particle. The power of the radiation depends only on the magnitude of the charge and the acceleration of the particle. both of these increasing leads to a quadratic increase in the radiation.

$P = \frac{q^2 a^2}{6 \pi \varepsilon_0 c^3}$

When an electric charge accelerates, the change in the direction of its electric field can only travel at the speed of light, so an oscillation of the electric field travels outward from the original position of the charge. This oscillation releases radiation.

As the red charged particle accelerates to the right, the oscillation of its electric field moves outward at the speed of light.

Since the disturbance will move radially outward a distance $c \Delta t$ in the same time than the field will oscillate tangentially a distance $\Delta v t \sin(\theta).$ So the ratio of the respective components of the electric field is

$\frac{E_{\text{tangentially}} }{E_{\text{radial} }} = \frac{ \Delta v t \sin(\theta) }{c \Delta t} = \frac{a r \sin(\theta) }{c^2} .$

Since $E = \frac{1}{4\pi \varepsilon_0}\frac{q}{r^2},$

$E_{\text{tangentially}} = \frac{q a \sin(\theta) }{4\pi \varepsilon_0 c^2 r}.$

The intensity is given by the Poynting vector.

$\begin{aligned} \vec{S} &= \frac{E^2}{2\mu_0} \hat{r} \\ &= \frac{q^2 a^2 \sin^2(\theta) }{16\pi^2 \varepsilon_0 c^3 R^2} \hat{r} \end{aligned}$

In order to calculate the power, integrate over all possible angles for $\theta.$ Since $\int \sin^2 (\theta) d\theta = \frac{8\pi}{3},$

$P = \frac{q^2 a^2}{6 \pi \varepsilon_0 c^3}.$

In the relativistic doppler effect, the radiation does not move out at a constant rate, but instead is subject to the Doppler effect. This is summarized by Liénard's generalization.

Liénard's generalization $P= \frac{\mu_0 q^2 \lambda^6}{6\pi c} \Bigg( a^2 - \bigg| \frac{\vec{v} \times \vec{a} }{c} \bigg|^2 \Bigg)$

Show that Liénard's generalization reduces to the Larmor formula when $v<<c.$

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Two things happen as a result of $v<<c:$

$\begin{aligned} \lambda &= \frac{1}{\sqrt{1 - \frac{v^2}{c^2} }} \\ &\rightarrow \frac{1}{1-0} = 1 \end{aligned}$

and

$\bigg| \frac{\vec{v} \times \vec{a} }{c} \bigg|^2 \rightarrow 0.$

Thus,

$\begin{aligned} P &= \frac{\mu_0 q^2 \lambda^6}{6\pi c} \Bigg( a^2 - \bigg| \frac{\vec{v} \times \vec{a} }{c} \bigg|^2 \Bigg) \\ &\rightarrow \frac{\mu_0 q^2 (1)^6}{6\pi c} \Bigg( a^2 - 0 \Bigg) \\ &= \frac{q^2 a^2}{6 \pi \varepsilon_0 c^3} \end{aligned}$