we calculated the energy per unit time radiated by a (non-relativistic) point charge- the Larmor formula. In the same spirit:

(a) Calculate the momentum per unit time radiated.

(b) Calculate the angular momentum per unit time radiated.

The Larmor formula is as follows:

$\frac{dp}{d{t}_r}=\frac{{\mu }_0{q}^2}{6\pi c^3}{a}^{2}v$

The expression of equation of motion of accelerated charge is given as follows:

$\mathit{m}\mathit{a}\mathbf{=}\mathit{E}\mathit{e}$ …… (1)

Here,$\mathit{E}$is the electric field strength, $\mathit{e}$is the charge of electron, $\mathit{m}$is the mass of the electron and is the acceleration.

The electric field is varying simple harmonically is given as follows”

$\mathit{E}\mathbf{=}{\mathit{E}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}\mathbf{j}\mathbf{\omega }\mathbf{t}}$ …… (2)

The relation between the linear momentum and angular momentum is given as follows:

role="math" localid="1658122582746" $\mathit{L}\mathbf{=}\mathit{r}\mathbf{\times }\mathit{p}$…… (3)

Here,$\mathit{L}$is the linear momentum, $\mathit{p}$is the angular momentum, $\mathit{r}$is the radius.

(a)

From the equation (1)

$\begin{array}{l}ma=Ee\\ a=\frac{Ee}{m}\end{array}$…… (4)

Now substitute equation (2) in equation (4) and simplify,

$\frac{d^{2}r}{d{t}^2}=\frac{E_0{e}^{-j\omega t}q}{m}$

Integrating above equation twice and simplify.

$\begin{array}{l}\int \frac{d^{2}r}{d{t}^2}=\int \frac{E_0{e}^{-j\omega t}q}{m}\\ \frac{dr}{dt}=\frac{q{E}_0}{m}\int e^{-j\omega t}\\ \frac{dr}{dt}=\frac{q{E}_0}{-mj\omega }\left[e^{-j\omega t}\right]\end{array}$

Again integrate,

$\begin{array}{l}\int \frac{dr}{dt}=\frac{q{E}_0}{-mj\omega }\int \left[e^{-j\omega t}\right]\\ r=\frac{q{E}_0}{m{i}^2{\omega }^2}\left[e^{-j\omega t}\right]\\ r=-\frac{q{E}_0}{m{\omega }^2}\left[e^{-j\omega t}\right]\end{array}$

The dipole moment of an oscillating electric dipole is,

$P=qr$

Here, $q$is the magnitude of each charge and $r$is the separation between the two charges.

Then,

$\left|P\right|=|-\frac{q^2{E}_0}{m{\omega }^2}|\left[e^{-j\omega t}\right]$…… (5)

Now,

$\begin{array}{c}P=\left[q_{0}dl\right]e^{-j\omega t}\\ P=P_0{e}^{-j\omega t}\end{array}$…… (6)

Comparing the equations (5) and (6).

$P_0=\frac{q^2{E}_0}{m{\omega }^2}$

If $a$is the acceleration of the charged particle, under the action of electric field of frequency $\omega$at$t=0$ , then

$\begin{array}{l}ma=q{E}_0{e}^{-j\omega t}\\ ma=q{E_0|}_{t=0}\\ a=\frac{q{E_0|}_{t=0}}{m}\end{array}$

The time averaged power radiated is given by:

$\begin{array}{l}⟨P⟩=\frac{1}{4\pi {\epsilon }_0}\frac{\frac{q^2{a}^2{\omega }^4}{{\omega }^4}}{3c^3}\\ ⟨P⟩=\frac{1}{4\pi {\epsilon }_0}\frac{q^2{a}^2}{3c^3}\end{array}$

Then:

$⟨P⟩=\frac{1}{4\pi {\epsilon }_0}\frac{q^2{a}^2}{3c^3}$

If $p$is the instantaneous power, then average power

$\begin{array}{l}⟨P⟩=⟨Pco{s}^2\omega (t-\frac{r}{c})⟩\\ ⟨P⟩=P⟨co{s}^2\omega (t-\frac{r}{c})⟩\end{array}$

The average value of$⟨co{s}^2\omega (t-\frac{r}{c})⟩=\frac{1}{2}$

Then,

$\begin{array}{l}⟨P⟩=\frac{P}{2}\\ P=2⟨P⟩\end{array}$

Substitute$⟨P⟩=\frac{1}{4\pi {\epsilon }_0}\frac{q^2{a}^2}{3c^3}$ in the above equation and simplify,

$\begin{array}{c}P=2\left[\frac{1}{4\pi {\epsilon }_0}\frac{q^2{a}^2}{3c^3}\right]\\ =\frac{1}{6\pi {\epsilon }_0}\frac{q^2{a}^2}{c^3}\end{array}$

Momentum per unit time in radiation is the amount of force.

The relativistic velocity is,

$\begin{array}{c}{v}^2=\frac{1}{{\mu }_0{\epsilon }_0}\\ {\epsilon }_0=\frac{1}{{\mu }_0{v}^2}\end{array}$

Power is the product of force and velocity.

$\begin{array}{l}P=Fv\\ \frac{1}{6\pi \left(\frac{1}{{\mu }_0{v}^2}\right)}\frac{q^2{a}^2}{c^3}=Fv\\ F=\frac{{\mu }_0}{6\pi }\frac{q^2{a}^2}{c^3}v\end{array}$

Therefore, the momentum per unit is radiated is$\boxed{\frac{{\mu }_0}{6\pi }\frac{q^2{a}^2}{c^3}v}$

(b)

The angular momentum per unit time is

$\frac{L}{t}=\frac{r\times p}{t}$

But force is the rate of change of momentum.

$F=\frac{p}{t}$

Then, solve for the angular momentum per unit time as:

$\begin{array}{c}\frac{L}{t}=r\times F\\ \frac{L}{t}=r\times \left[\frac{{\mu }_0}{6\pi }\frac{q^2\dot{a}}{c}\right]v\\ \frac{L}{t}=\left[\frac{{\mu }_0}{6\pi }\frac{q^{2}a}{c}\right]v\\ \frac{L}{t}=\left[\frac{{\mu }_0}{6\pi }\frac{q^2}{c}\right][v\times a]\end{array}$

Therefore, angular momentum per unit time is$\boxed{\left[\frac{{\mu }_0}{6\pi }\frac{q^2}{c}\right][v\times a]}$