91Ó°ÊÓ

For the problem refer to the figure in textbook.

Consider the dumbbell move along the x-axis. The resultant electric force on the charge will be in the same direction, so we only need the x-component of the electric fields.

Write the formula radiation reaction force on a moving particle.

${\mathbf{F}}_{\mathrm{self}}=\frac{{\mathbf{q}}^2}{\mathbf{4}\mathrm{Ï€}{\mathbf{Î\mu }}_0}\left[−{\mathrm{γ}}^3\frac{a}{\mathbf{4}{\mathbf{c}}^2\mathbf{d}}+\frac{{\mathrm{γ}}^4}{\mathbf{4}{\mathbf{c}}^2}\mathbf{(}\frac{\mathbf{a}}{\mathbf{3}}\mathbf{+}\frac{{\mathbf{νγ}}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{)}+\left(\right)\mathbf{d}+...\right]\widehat{\mathbf{x}}$ …… (1)

Here, ${\mathbf{Î\mu }}_0$ is absolute permittivity, role="math" localid="1658740081686" ${\mathbf{a}}_{}$is function of $\mathrm{t}$, role="math" localid="1658740380724" $\mathbf{ν}$and is function of $\mathrm{t}$.

Write the formula ofradiation reaction force of adding first integral to the second one.

${\mathbf{∫}}_{{\mathbf{t}}_1}^{{\mathbf{t}}_{2\text{'}}}{\mathbf{F}}_{\mathrm{rad}}\mathbf{â‹\dots }\mathbf{vdt}$ …… (2)

Here, ${\mathbf{F}}_{\mathrm{rad}}$ radiation reaction force and is velocity.

As we know that electric field due to a moving charge.

${\text{E}}_{\text{1}}=\frac{\left(\frac{\mathrm{q}}{2}\right)}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{r}}{{(\mathrm{r}â‹\dots \mathrm{u})}^3}[({\mathrm{c}}^2−{\mathrm{v}}^2)\mathrm{u}+(\mathrm{r}â‹\dots \mathrm{a})\mathrm{u}−(\mathrm{r}â‹\dots \mathrm{u})\mathrm{a}]$

Substitute $\mathrm{cr}−\mathrm{v}$ for $\mathrm{u}$, $(\mathrm{l}\widehat{\mathrm{x}}+\mathrm{d}\widehat{\mathrm{y}})$ for $\mathrm{r}$, $\mathrm{ν}\widehat{x}$ for $v$, role="math" localid="1658740581012" $\mathrm{a}\widehat{\mathrm{x}}$ for $\mathrm{a}$, $\mathrm{cr}−\mathrm{lv}$for $\mathrm{r}â‹\dots \mathrm{u}$ and $(\mathrm{cl}−\mathrm{ν°ù})/\mathrm{r}$ for $u_x$into above equation ${\mathrm{E}}_1$.

$\begin{array}{c}{\mathrm{E}}_{1_{\mathrm{x}}}=\frac{\mathrm{q}}{8{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{r}}{{(\mathrm{cr}−\mathrm{lv})}^3}\left[\frac{1}{\mathrm{r}}(\mathrm{cl}−\mathrm{ν°ù})({\mathrm{c}}^2−{\mathrm{v}}^2+\mathrm{la})−\mathrm{a}(\mathrm{cr}−\mathrm{lv})\right]\\ =\frac{\mathrm{q}}{8{\mathrm{Ï€Î\mu }}_0}\frac{1}{{(\mathrm{cr}−\mathrm{lv})}^3}[(\mathrm{cl}−\mathrm{ν°ù})({\mathrm{c}}^2−{\mathrm{v}}^2)+{\mathrm{cl}}^2\mathrm{a}−\mathrm{ν»å²¹}−{\mathrm{acr}}^2+\mathrm{alν°ù}]\mathrm{a}\end{array}$

But ${\mathrm{r}}^2={\mathrm{l}}^2+{\mathrm{d}}^2$

${\mathrm{E}}_{1_{\mathrm{x}}}=\frac{\mathrm{q}}{8{\mathrm{Ï€Î\mu }}_0}\frac{1}{{(\mathrm{cr}−\mathrm{lv})}^3}[(\mathrm{cl}−\mathrm{ν°ù})({\mathrm{c}}^2−{\mathrm{ν}}^2)−{\mathrm{acd}}^2]$

Determine the self-radiation reaction force.

${\mathrm{F}}_{\mathrm{self}}=\frac{\mathrm{q}}{8{\mathrm{Ï€Î\mu }}_0}\frac{1}{{(\mathrm{cr}−\mathrm{lv})}^3}[(\mathrm{cl}−\mathrm{ν°ù})({\mathrm{c}}^2−{\mathrm{ν}}^2)−{\mathrm{acd}}^2]\widehat{\mathrm{x}}$

Now

role="math" localid="1658740798843" $\begin{array}{c}\mathrm{x}\left(\mathrm{t}\right)−\mathrm{x}\left({\mathrm{t}}_{\mathrm{r}}\right)=\mathrm{l}\\ =\mathrm{ν°Õ}+\frac{1}{2}{\mathrm{aT}}^2+\frac{1}{6}\stackrel{Ë™}{\mathrm{a}}{\mathrm{T}}^3+\mathrm{...},\end{array}$

Here, $\mathrm{T}=\mathrm{t}−{\mathrm{t}}_{\mathrm{r}}$, and $\mathrm{ν}$, $a$, and $\stackrel{˙}{a}$ are all evaluate at the retarded time $t_r$.

$\begin{array}{c}{\left(\mathrm{cT}\right)}^2={\mathrm{r}}^2\\ ={\mathrm{l}}^2+{\mathrm{d}}^2\\ ={\mathrm{d}}^2+{(\mathrm{ν°Õ}+\frac{1}{2}{\mathrm{aT}}^2+\frac{1}{6}\stackrel{Ë™}{\mathrm{a}}{\mathrm{T}}^3)}^2\\ ={\mathrm{d}}^2+{\mathrm{ν}}^2\mathrm{T}+{\mathrm{ν²¹T}}^3+\frac{1}{3}\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}{\mathrm{T}}^4+\frac{1}{4}{\mathrm{a}}^2{\mathrm{T}}^4\end{array}$

Solve further as

$\begin{array}{c}{\mathrm{c}}^2{\mathrm{T}}^2\left(\frac{1−{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)=\frac{{\mathrm{c}}^2{\mathrm{T}}^2}{{\mathrm{γ}}^2}\\ ={\mathrm{d}}^2+{\mathrm{ν²¹T}}^3+\left(\frac{1}{3}\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}+\frac{1}{4}{\mathrm{a}}^2\right){\mathrm{T}}^4\end{array}$

Solve for $\mathrm{T}$ as a power series in $\mathrm{d}$:

$\begin{array}{c}\mathrm{T}=\frac{\mathrm{γ»å}}{\mathrm{c}}(1+\mathrm{Ad}+{\mathrm{Bd}}^2+...)\\ ⇒\frac{{\mathrm{c}}^2}{{\mathrm{γ}}^2}\frac{{\mathrm{γ}}^2{\mathrm{d}}^2}{{\mathrm{c}}^2}(1+2\mathrm{Ad}+2{\mathrm{Bd}}^2+{\mathrm{A}}^2{\mathrm{d}}^2)\\ ={\mathrm{d}}^2+\mathrm{ν²¹}\frac{{\mathrm{γ}}^3{\mathrm{d}}^3}{{\mathrm{c}}^3}(1+3\mathrm{Ad})+\left(\frac{\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{a}}^2}{4}\right)\frac{{\mathrm{γ}}^4}{{\mathrm{c}}^4}{\mathrm{d}}^4\end{array}$

Comparing like powers of $\mathrm{d}:\mathrm{A}=\frac{1}{2}\mathrm{ν²¹}\frac{{\mathrm{γ}}^3}{{\mathrm{c}}^3}$;

$\begin{array}{c}2\mathrm{B}+{\mathrm{A}}^2=\frac{3{\mathrm{ν²¹Î³}}^3}{{\mathrm{c}}^3}\mathrm{A}+\left(\frac{\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{a}}^2}{4}\right)\frac{{\mathrm{γ}}^4}{{\mathrm{c}}^4}\\ 2\mathrm{B}=\frac{3{\mathrm{ν²¹Î³}}^3}{{\mathrm{c}}^3}\frac{1}{2}\mathrm{ν²¹}\frac{{\mathrm{γ}}^3}{{\mathrm{c}}^3}−\frac{1}{4}{\mathrm{ν}}^2{\mathrm{a}}^2\frac{{\mathrm{γ}}^6}{{\mathrm{c}}^6}+\frac{\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}}{3}\frac{{\mathrm{γ}}^4}{{\mathrm{c}}^4}+\frac{{\mathrm{a}}^2{\mathrm{γ}}^4}{4{\mathrm{c}}^4}\\ 2\mathrm{B}=\frac{\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}}{3}\frac{{\mathrm{γ}}^4}{{\mathrm{c}}^4}+\frac{{\mathrm{γ}}^4{\mathrm{a}}^2}{4{\mathrm{c}}^4}\left(\frac{1}{{\mathrm{γ}}^2}−\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right)+\frac{3}{2}\frac{{\mathrm{ν}}^2{\mathrm{a}}^2{\mathrm{γ}}^6}{{\mathrm{c}}^6}\\ 2\mathrm{B}=\frac{{\mathrm{γ}}^4}{{\mathrm{c}}^4}\left[\frac{\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{a}}^2{\mathrm{γ}}^2}{4}\left(1−\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}−\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}+6\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right)\right]\end{array}$

Solve further as

$\mathrm{B}=\frac{{\mathrm{γ}}^4}{2{\mathrm{c}}^4}\left[\frac{\mathrm{ν²¹}}{3}+\frac{{\mathrm{γ}}^2{\mathrm{a}}^2}{4}\left(1+4\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right)\right]$

Now solve again $\mathrm{T}$.

$\mathrm{T}=\frac{\mathrm{γ»å}}{\mathrm{c}}\left\{1+\frac{\mathrm{ν²¹}}{2}\frac{{\mathrm{γ}}^3}{{\mathrm{c}}^3}\mathrm{d}+\frac{{\mathrm{γ}}^4}{2{\mathrm{c}}^4}\left[\frac{\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{γ}}^2{\mathrm{a}}^2}{4}\left(1+4\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right)\right]]{\mathrm{d}}^2\right\}+\left(\right){\mathrm{d}}^4+...$

Determine the current $I$.

role="math" localid="1658741573624" $\begin{array}{c}\mathrm{I}=\mathrm{ν°Õ}+\frac{1}{2}{\mathrm{aT}}^2+\frac{1}{6}\stackrel{Ë™}{\mathrm{a}}{\mathrm{T}}^3+...\\ =\left\{\begin{array}{c}\frac{\mathrm{νγ»å}}{\mathrm{c}}\left\{1+\frac{\mathrm{ν²¹}}{2}\frac{{\mathrm{γ}}^3}{{\mathrm{c}}^3}\mathrm{d}+\frac{{\mathrm{γ}}^4}{2{\mathrm{c}}^4}\left[\frac{\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{γ}}^2{\mathrm{a}}^2}{4}\left(1+4\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right)\right]{\mathrm{d}}^2\right\}\\ +\frac{1}{2}\mathrm{a}\frac{{\mathrm{γ}}^2{\mathrm{d}}^2}{{\mathrm{c}}^2}\left[1+\mathrm{ν²¹}\frac{{\mathrm{γ}}^3}{{\mathrm{c}}^3}\mathrm{d}\right]+\frac{1}{6}\stackrel{Ë™}{\mathrm{a}}\frac{{\mathrm{γ}}^3}{{\mathrm{c}}^3}{\mathrm{d}}^3\end{array}\right\}\\ =\left(\frac{\mathrm{νγ}}{\mathrm{c}}\right)\mathrm{d}+\frac{\mathrm{a}}{2}\frac{{\mathrm{γ}}^4}{{\mathrm{c}}^2}\left(1−\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}+\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right){\mathrm{d}}^2\\ =\left\{\frac{\mathrm{νγ}}{\mathrm{c}}\frac{{\mathrm{γ}}^4}{{\mathrm{c}}^4}\left[\frac{\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{γ}}^2{\mathrm{a}}^2}{4}\left(1+4\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right)\right]+\frac{1}{2}\mathrm{a}\frac{{\mathrm{γ}}^2}{{\mathrm{c}}^2}\mathrm{ν²¹}\frac{{\mathrm{γ}}^3}{{\mathrm{c}}^3}+\frac{1}{6}\stackrel{Ë™}{\mathrm{a}}\frac{{\mathrm{γ}}^3}{{\mathrm{c}}^3}\right\}{\mathrm{d}}^3\end{array}$

Solve further as

Determine the $r$.

Determine the$râ‹\dots u$.

$\begin{array}{c}\mathrm{r}â‹\dots \mathrm{u}=\mathrm{cγ»å}+\frac{{\mathrm{ν²¹Î³}}^4}{2{\mathrm{c}}^2}{\mathrm{d}}^2+\frac{{\mathrm{γ}}^5}{2{\mathrm{c}}^3}{\mathrm{d}}^2+\frac{{\mathrm{γ}}^5}{2{\mathrm{c}}^3}\left[\frac{\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}}{3}+{\mathrm{γ}}^2{\mathrm{a}}^2\left(\frac{1}{4}+\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right)\right]{\mathrm{d}}^3\\ −\frac{{\mathrm{ν}}^2\mathrm{γ}}{\mathrm{c}}\mathrm{d}−\frac{{\mathrm{²¹Î½Î³}}^4}{2{\mathrm{c}}^2}{\mathrm{d}}^2−\frac{{\mathrm{γ}}^5\mathrm{ν}}{2{\mathrm{c}}^3}\left[\frac{\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{5}{4}\frac{{\mathrm{νγ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right]{\mathrm{d}}^3+...\\ =\mathrm{cγ»å}\left(1−\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right)+\frac{{\mathrm{γ}}^5}{2{\mathrm{c}}^3}\left[\frac{\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}}{3}+{\mathrm{γ}}^2{\mathrm{a}}^2\left(\frac{1}{4}+\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right)−\frac{\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}}{3}−\frac{5}{4}\frac{{\mathrm{ν}}^2{\mathrm{γ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right]{\mathrm{d}}^3+...\\ =\frac{\mathrm{c}}{\mathrm{γ}}\mathrm{d}+\frac{{\mathrm{γ}}^5{\mathrm{a}}^2}{8{\mathrm{c}}^3}{\mathrm{d}}^3+\left(\right){\mathrm{d}}^4+...\end{array}$

Determine the $\mathrm{cl}−\mathrm{ν°ù}$.

$\begin{array}{c}\mathrm{cl}−\mathrm{ν°ù}=\mathrm{ν°ùd}+\frac{{\mathrm{²¹Î³}}^4}{2\mathrm{c}}{\mathrm{d}}^2+\frac{{\mathrm{γ}}^5}{2{\mathrm{c}}^2}\left(\frac{\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{5}{4}\frac{{\mathrm{νγ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right){\mathrm{d}}^3−\mathrm{vγ»å}\\ −\frac{{\mathrm{ν}}^2{\mathrm{²¹Î³}}^4}{2{\mathrm{c}}^3}{\mathrm{d}}^2−\frac{{\mathrm{νγ}}^3}{2{\mathrm{c}}^4}\left[\frac{\mathrm{ν}\stackrel{Ë™}{\mathrm{a}}}{3}+{\mathrm{γ}}^2{\mathrm{a}}^2\left(\frac{1}{4}+\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right)\right]{\mathrm{d}}^3\\ =\frac{{\mathrm{²¹Î³}}^4}{2\mathrm{c}}\left(1−\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right){\mathrm{d}}^2+\frac{{\mathrm{γ}}^5}{2{\mathrm{c}}^2}\left[\begin{array}{l}\frac{\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{5}{4}\frac{{\mathrm{νγ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}−\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\frac{\stackrel{Ë™}{\mathrm{a}}}{3}\\ −\frac{{\mathrm{νγ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\left(\frac{1}{4}+\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right)\end{array}\right]{\mathrm{d}}^3+\left(\right){\mathrm{d}}^4+...\\ =\left(\frac{{\mathrm{²¹Î³}}^4}{2\mathrm{c}}\right){\mathrm{d}}^2+\frac{{\mathrm{γ}}^5}{2{\mathrm{c}}^2}\left[\frac{\stackrel{Ë™}{\mathrm{a}}}{3{\mathrm{γ}}^2}+\frac{{\mathrm{νγ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\left(\frac{5}{4}−\frac{1}{4}−\frac{{\mathrm{ν}}^2}{{\mathrm{c}}^2}\right)\right]{\mathrm{d}}^3+\left(\right){\mathrm{d}}^4+...\\ =\left(\frac{{\mathrm{²¹Î³}}^4}{2\mathrm{c}}\right){\mathrm{d}}^2+\frac{{\mathrm{γ}}^5}{2{\mathrm{c}}^2}\left(\frac{\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{νγ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right){\mathrm{d}}^3+\left(\right){\mathrm{d}}^4+...\end{array}$

Determine the ${(\mathrm{cr}−\mathrm{lv})}^3$

$\begin{array}{c}{(\mathrm{cr}−\mathrm{lv})}^3={\left[\frac{\mathrm{cd}}{\mathrm{γ}}\left(1+\frac{{\mathrm{γ}}^6{\mathrm{a}}^2}{8{\mathrm{c}}^4}{\mathrm{d}}^2\right)\right]}^{−3}\\ ={\left(\frac{\mathrm{γ}}{\mathrm{cd}}\right)}^3\left(1−3\frac{{\mathrm{γ}}^6{\mathrm{a}}^2}{8{\mathrm{c}}^4}{\mathrm{d}}^2\right)+...\end{array}$

Determine theradiation reaction force on a moving particle.

$\begin{array}{c}{\mathrm{F}}_{\mathrm{self}}=\frac{{\mathrm{q}}^2}{8{\mathrm{Ï€Î\mu }}_0}{\left(\frac{\mathrm{γ}}{\mathrm{cd}}\right)}^3\left(1−3\frac{{\mathrm{γ}}^6{\mathrm{a}}^2}{8{\mathrm{c}}^4}{\mathrm{d}}^2\right)\left\{\left[\left(\frac{{\mathrm{²¹Î³}}^2}{2\mathrm{c}}\right){\mathrm{d}}^2+\frac{{\mathrm{γ}}^3}{2{\mathrm{c}}^2}\left(\frac{\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{νγ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right){\mathrm{d}}^3\right]\frac{{\mathrm{c}}^2}{{\mathrm{γ}}^2}−{\mathrm{acd}}^2\right\}\widehat{\mathrm{x}}\\ =\frac{{\mathrm{q}}^2}{9{\mathrm{Ï€Î\mu }}_0}\frac{{\mathrm{γ}}^3}{{\mathrm{c}}^3\mathrm{d}}\left(1−\frac{3}{8}\frac{{\mathrm{γ}}^6{\mathrm{a}}^2}{{\mathrm{c}}^4}{\mathrm{d}}^2\right)\left[−\frac{\mathrm{ac}}{2}+\frac{\mathrm{γ}}{2}\left(\frac{\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{νγ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right)\mathrm{d}\right]]\widehat{\mathrm{x}}\\ =\frac{{\mathrm{q}}^2}{8{\mathrm{Ï€Î\mu }}_0}\frac{{\mathrm{γ}}^3}{{\mathrm{c}}^3\mathrm{d}}\frac{1}{2}\left[−\mathrm{ac}+\mathrm{γ}\left(\frac{\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{νγ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right)\mathrm{d}+\left(\right){\mathrm{d}}^2+...\right]\widehat{\mathrm{x}}\\ =\frac{{\mathrm{q}}^2}{4{\mathrm{Ï€Î\mu }}_0}\left[−{\mathrm{γ}}^3\frac{\mathrm{a}}{4{\mathrm{c}}^2\mathrm{d}}+\frac{{\mathrm{γ}}^4}{4{\mathrm{c}}^3}\left(\frac{\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{νγ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right)+\left(\right)\mathrm{d}+...\right]\widehat{\mathrm{x}}\end{array}$

Determine the radiation reaction force on a moving particle.

Substitute $(1−3{\mathrm{ν²¹Î³}}^3\mathrm{d}/{\mathrm{c}}^3)(\mathrm{a}−\stackrel{Ë™}{\mathrm{a}}\mathrm{γ»å}/\mathrm{c})$for $a$ into equation (1)

$\begin{array}{c}{\mathrm{F}}_{\mathrm{self}}=\frac{{\mathrm{q}}^2}{4{\mathrm{Ï€Î\mu }}_0}\left[−{\mathrm{γ}}^3\frac{(1−3{\mathrm{ν²¹Î³}}^3\mathrm{d}/{\mathrm{c}}^3)(\mathrm{a}−\stackrel{Ë™}{\mathrm{a}}\mathrm{γ»å}/\mathrm{c})}{4{\mathrm{c}}^2\mathrm{d}}+\frac{{\mathrm{γ}}^4}{4{\mathrm{c}}^3}\left(\frac{\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{νγ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right)+\left(\right)\mathrm{d}+...\right]\widehat{\mathrm{x}}\\ =\frac{{\mathrm{q}}^2}{4{\mathrm{Ï€Î\mu }}_0}\left[−\frac{{\mathrm{γ}}^3\mathrm{a}}{4{\mathrm{c}}^2\mathrm{d}}+\frac{{\mathrm{γ}}^4}{4{\mathrm{c}}^3}\left(\frac{\stackrel{Ë™}{\mathrm{a}}\mathrm{γ}}{\mathrm{c}}+3\frac{{\mathrm{ν²¹}}^2{\mathrm{γ}}^2}{{\mathrm{c}}^3}\right)+\frac{{\mathrm{γ}}^4}{4{\mathrm{c}}^3}\left(\frac{\stackrel{Ë™}{\mathrm{a}}}{3}+\frac{{\mathrm{νγ}}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right)+\left(\right)\mathrm{d}+...\right]\widehat{\mathrm{x}}\\ =\frac{{\mathrm{q}}^2}{4{\mathrm{Ï€Î\mu }}_0}\left[−\frac{{\mathrm{γ}}^3\mathrm{a}}{4{\mathrm{c}}^2\mathrm{d}}+\frac{{\mathrm{γ}}^4}{4{\mathrm{c}}^3}\left(\stackrel{Ë™}{\mathrm{a}}+\frac{\stackrel{Ë™}{\mathrm{a}}}{3}+3\frac{{\mathrm{ν²¹}}^2{\mathrm{γ}}^2}{{\mathrm{c}}^3}+\frac{{\mathrm{ν²¹}}^2{\mathrm{γ}}^2}{{\mathrm{c}}^3}\right)+\left(\right)\mathrm{d}+..\right]\widehat{\mathrm{x}}\\ =\frac{{\mathrm{q}}^2}{4{\mathrm{Ï€Î\mu }}_0}\left[−\frac{{\mathrm{γ}}^3\mathrm{a}}{4{\mathrm{c}}^2\mathrm{d}}+\frac{{\mathrm{γ}}^4}{3{\mathrm{c}}^3}\left(\stackrel{Ë™}{\mathrm{a}}+3\frac{{\mathrm{ν²¹}}^2{\mathrm{γ}}^2}{{\mathrm{c}}^2}\right)+\left(\right)\mathrm{d}+...\right]\widehat{\mathrm{x}}\end{array}$

Therefore, the value of first term is the electromagnetic mass is ${\mathrm{F}}_{\mathrm{rad}}^{\mathrm{int}}=\frac{{\mathrm{μ}}_0{\mathrm{q}}^2}{12\mathrm{Ï€³¦}}{\mathrm{γ}}^4\left(\mathrm{a}+3\frac{{\mathrm{ν²¹}}^2{\mathrm{γ}}^2}{{\mathrm{c}}^2}\right)$ and the radiation reaction itself is the second term is ${\mathrm{F}}_{\mathrm{rad}}=\frac{{\mathrm{μ}}_0{\mathrm{q}}^2}{6\mathrm{Ï€³¦}}{\mathrm{γ}}^4\left(\mathrm{a}+3\frac{{\mathrm{ν²¹}}^2{\mathrm{γ}}^2}{{\mathrm{c}}^2}\right)$.

The power radiated in the case of collinear velocity and acceleration is:

$\mathrm{P}=\frac{{\mathrm{μ}}_0{\mathrm{q}}^2}{6\mathrm{Ï€³¦}}{\mathrm{γ}}^6{\mathrm{a}}^2$

There is need to demonstrate from equation 11.75 that, on average, the power radiated by the dumbbell under force ${\mathrm{F}}_{\mathrm{rad}}$ equals the above-described $P$:

${∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{F}}_{\mathrm{rad}}â‹\dots \mathrm{vdt}=−{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\mathrm{Pdt}$

Assuming that the motion is periodic, $v$and an are assumed to have the identical values at $t_1$ and $t_2$, respectively. Now:

$\begin{array}{c}{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{F}}_{\mathrm{rad}}â‹\dots \mathrm{vdt}=\frac{{\mathrm{μ}}_0{\mathrm{q}}^2}{6\mathrm{Ï€³¦}}{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{γ}}^4\left(\stackrel{Ë™}{\mathrm{a}}+3\frac{{\mathrm{γ}}^2{\mathrm{a}}^2\mathrm{Ï\dots }}{{\mathrm{c}}^2}\right)\mathrm{Ï\dots »å³Ù}\\ =\frac{{\mathrm{μ}}_0{\mathrm{q}}^2}{6\mathrm{Ï€³¦}}({\mathrm{I}}_1+{\mathrm{I}}_2)\end{array}$

Determine the first integral:

$\begin{array}{c}\mathrm{I}{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{γ}}^4\mathrm{Ï\dots }\stackrel{Ë™}{\mathrm{a}}\mathrm{dt}={\left|\left({\mathrm{²¹Î³}}^4\mathrm{Ï\dots }\right)\right|}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}−{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{γ}}^4\mathrm{Ï\dots }\right)\mathrm{adt}\\ =0−{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}(4{\mathrm{γ}}^3\stackrel{Ë™}{\mathrm{γ}}\mathrm{Ï\dots }+{\mathrm{γ}}^4\mathrm{a})\mathrm{adt}\\ =−{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}(4{\mathrm{γ}}^3\stackrel{Ë™}{\mathrm{γ}}\mathrm{Ï\dots ²¹}+{\mathrm{γ}}^4{\mathrm{a}}^2)\mathrm{dt}\end{array}$

Substitute $\begin{array}{c}\stackrel{Ë™}{\mathrm{γ}}=−\frac{1}{2}\frac{−2\mathrm{Ï\dots }/{\mathrm{c}}^2}{{(1−{\mathrm{Ï\dots }}^2/{\mathrm{c}}^2)}^{3/2}}\\ ={\mathrm{γ}}^3\frac{\mathrm{Ï\dots ²¹}}{{\mathrm{c}}^2}\end{array}$ into above equation.

$\begin{array}{c}−{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\left(4{\mathrm{γ}}^6\frac{{\mathrm{Ï\dots }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}+{\mathrm{γ}}^4{\mathrm{a}}^2\right)\mathrm{dt}=−{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{γ}}^6{\mathrm{a}}^2\left(4\frac{{\mathrm{Ï\dots }}^2}{{\mathrm{c}}^2}+{\mathrm{γ}}^2\right)\mathrm{dt}\\ ={∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{γ}}^6{\mathrm{a}}^2\left(1+3\frac{{\mathrm{Ï\dots }}^2}{{\mathrm{c}}^2}\right)\mathrm{dt}\end{array}$

where the boundary term has been eliminated using the assumption of periodicity of motion. The result of adding the first integral to the second is:

Determine the radiation reaction force of adding first integral to the second one.

Substitute $−\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{Ï€}c}{\mathrm{γ}}^6{a}^2\left(1+3\frac{{\mathrm{Ï\dots }}^2}{c^2}\right)$for $F_{rad}$ and $3\frac{{\mathrm{μ}}_0{\mathrm{q}}^2}{6\mathrm{Ï€³¦}}{\mathrm{γ}}^6\frac{{\mathrm{a}}^2{\mathrm{Ï\dots }}^2}{{\mathrm{c}}^2}$ for $vdt$into equation (2).

$\begin{array}{c}{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{F}}_{\mathrm{rad}}â‹\dots \mathrm{vdt}=\frac{{\mathrm{μ}}_0{\mathrm{q}}^2}{6\mathrm{Ï€³¦}}{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\left[−{\mathrm{γ}}^6{\mathrm{a}}^2\left(1+\frac{{\mathrm{Ï\dots }}^2}{{\mathrm{c}}^2}\right)+3{\mathrm{γ}}^6\frac{{\mathrm{a}}^2{\mathrm{Ï\dots }}^2}{{\mathrm{c}}^2}\right]\mathrm{dt}\\ =−\frac{{\mathrm{μ}}_0{\mathrm{q}}^2}{6\mathrm{Ï€³¦}}{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{γ}}^6{\mathrm{a}}^2\mathrm{dt}\\ =−{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\mathrm{Pdt}\end{array}$

Therefore, the value of radiation reaction force of adding first integral to the second one is ${∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{F}}_{\mathrm{rad}}â‹\dots \mathrm{vdt}=−{∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\mathrm{Pdt}$.