Question

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

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

Short answer

The Larmor formula is$P=\frac{{\mu }_0{q}^2{a}^2}{6\pi c}$ and the Lienard formula is$P=\frac{{\mu }_0{q}^2}{6\pi c}{\gamma }^6[a^2-{\left|\frac{\upsilon \times a}{c}\right|}^2]$ .

Step-by-step solution

Given information

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.

Larmor formula

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.

Derive the Lienard formula

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.