Use the Larmor formula (Eq. 11.70) and special relativity to derive the Lienard formula (Eq. 11. 73).

$\begin{array}{l}\mathbf{P}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}}{\mathbf{6}\mathbf{\pi }\mathbf{c}}\mathbf{(}\mathbf{11}\mathbf{.}\mathbf{70}\mathbf{)}\\ \mathbf{P}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}^{\mathbf{2}}{\mathbf{\gamma }}^{\mathbf{6}}}{\mathbf{6}\mathbf{\pi }\mathbf{c}}\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{\left|}\frac{\mathbf{\upsilon }\mathbf{\times }\mathbf{a}}{\mathbf{c}}\mathbf{\right|}}^{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{11}\mathbf{.}\mathbf{73}\mathbf{)}\end{array}$

The Larmor formula for the power radiated by a point charge is$P=\frac{{\mu }_0{q}^2{a}^2}{6\pi c}$,.

Here,$q$ is the charge of the particle,$a$ is the acceleration of the particle at the retarded time$\tau$ , $c$is the speed of light and${\mu }_0$ is the permittivity of vacuum.

The Lienard formula for the power radiated by a point charge is,.

$P=\frac{{\mu }_0{q}^2}{6\pi c}{\gamma }^6[a^2-{\left|\frac{\upsilon \times a}{c}\right|}^2]$

Here, $\gamma \equiv \sqrt{1-\frac{{\upsilon }^2}{c^2}}$is the factor of time dilation and $\upsilon$is the velocity of the charge at the retarded time.

Consider a point charge of certain magnitude, moves in a uniform electric and magnetic field then the power radiated by the point charge is represented using the Larmor formula.

The value of the power radiated by a point charge increases with the increase in the acceleration of the point charge.

Using the Minkowski version of Newton’s second law, the proper power as the zeroth component of a 4-vector can be written as,

$\begin{array}{c}\frac{dW}{d\tau }=K^0\times c\\ K^0=\frac{1}{c}\frac{dW}{d\tau }\end{array}$

Here,$K$ is the proper force.

According to the Larmor formula, for $\upsilon =0$, the rate at which energy left the charge is given by,

$\begin{array}{l}\frac{dW}{d\tau }=\frac{\frac{dW}{dt}}{\frac{\partial \tau }{\partial t}}\\ \frac{dW}{d\tau }=\frac{1}{4\pi {\epsilon }_0}\frac{2}{3}\frac{q^2}{c^3}{a}^2\end{array}$

When $\upsilon =0$, the value of acceleration $a^2$can be written as,.$a^2={\alpha }^{\nu }{\alpha }_{\nu }$ Also,the value of a 4-vector$K^{\mu }$ in terms of${\eta }^{\mu }$, whose zeroth component is just $c$is given by,

$K^{\mu }=\frac{1}{4\pi {\epsilon }_0}\frac{2}{3}\frac{q^2}{c^3}\left({\alpha }^{\nu }{\alpha }_{\nu }\right){\eta }^{\mu }$

Reducing the above expression to the Larmor formula when$\upsilon =0$ ,

$\begin{array}{c}\frac{dW}{dt}=\frac{1}{\gamma }\frac{dW}{d\tau }\\ \frac{dW}{dt}=\frac{1}{\gamma }c{K}^0\\ \frac{dW}{dt}=\frac{1}{\gamma }c\frac{{\mu }_0{q}^2}{6\pi c^3}\left({\alpha }^{\nu }{\alpha }_{\nu }\right){\eta }^0\end{array}$

Here, ,${\eta }^0=c\gamma$ so,

$\frac{dW}{dt}=\frac{{\mu }_0{q}^2}{6\pi c}\left({\alpha }^{\nu }{\alpha }_{\nu }\right)$ ….. (1)

The value of the term$\left({\alpha }^{\nu }{\alpha }_{\nu }\right)$ in terms of ordinary velocity and acceleration is given by

role="math" localid="1658915579537" $\begin{array}{c}{\alpha }^{\nu }{\alpha }_{\nu }={\gamma }^4[a^2+\frac{{(\upsilon \cdot a)}^2}{(c^2-{\upsilon }^2)}]\\ {\alpha }^{\nu }{\alpha }_{\nu }={\gamma }^6[a^2{\gamma }^{-2}+\frac{1}{c^2}{(\upsilon \cdot a)}^2]\\ {\alpha }^{\nu }{\alpha }_{\nu }={\gamma }^6[a^2(1-\frac{{\upsilon }^2}{c^2})+\frac{1}{c^2}{(\upsilon \cdot a)}^2]\\ {\alpha }^{\nu }{\alpha }_{\nu }={\gamma }^6[a^2-\frac{1}{c^2}\{{\upsilon }^2{a}^2-{(\upsilon \cdot a)}^2\}]\end{array}$

Here,$\upsilon \cdot a=\upsilon a\cos \theta$, where$\theta$ is the angle between $\upsilon$and $a$, so,

$\begin{array}{c}{\upsilon }^2{a}^2-{(\upsilon \cdot a)}^2={\upsilon }^2{a}^2(1-{\cos }^2\theta )\\ {\upsilon }^2{a}^2-{(\upsilon \cdot a)}^2={\upsilon }^2{a}^2{\sin }^2\theta \\ {\upsilon }^2{a}^2-{(\upsilon \cdot a)}^2={|\upsilon \times a|}^2\end{array}$

Substituting this in the expression,

$\begin{array}{c}{\alpha }^{\nu }{\alpha }_{\nu }={\gamma }^6[a^2-\frac{1}{c^2}{|\upsilon \times a|}^2]\\ {\alpha }^{\nu }{\alpha }_{\nu }={\gamma }^6[a^2-{\left|\frac{\upsilon \times a}{c}\right|}^2]\end{array}$

From equation (1),

$\begin{array}{c}\frac{dW}{dt}=\frac{{\mu }_0{q}^2}{6\pi c}{\gamma }^6[a^2-{\left|\frac{\upsilon \times a}{c}\right|}^2]\\ P=\frac{{\mu }_0{q}^2}{6\pi c}{\gamma }^6[a^2-{\left|\frac{\upsilon \times a}{c}\right|}^2]\end{array}$

Hence, the above expressionrepresents the Lienard’s generalization of the Larmor formula.