a)Find the radiation reaction force on a particle moving with arbitrary velocity in a straight line, by reconstructing the argument in Sect. 11.2.3 without assuming $\mathbf{\upsilon }\mathbf{\left(}{\mathbf{t}}_r\mathbf{\right)}\mathbf{=}\mathbf{0}$. [Answer: $\mathbf{(}{\mathbf{\mu }}_0{\mathbf{q}}^2{\mathbf{\gamma }}^4\mathbf{/}\mathbf{6}\mathbf{\pi c}\mathbf{)}\mathbf{}\mathbf{(}\dot{\mathbf{a}}\mathbf{+}\mathbf{3}{\mathbf{\gamma }}^2{\mathbf{a}}^2\mathbf{\upsilon }\mathbf{/}{\mathbf{c}}^2\mathbf{)}$]

(b) Show that this result is consistent (in the sense of Eq. 11.78) with the power radiated by such a particle (Eq. 11.75).

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.

${\mathbf{F}}_{\mathrm{self}}=\frac{{\mathbf{q}}^2}{\mathbf{4}\mathrm{\pi }{\mathbf{\epsilon }}_0}\left[-{\mathrm{\gamma }}^3\frac{a}{\mathbf{4}{\mathbf{c}}^2\mathbf{d}}+\frac{{\mathrm{\gamma }}^4}{\mathbf{4}{\mathbf{c}}^2}\mathbf{(}\frac{\mathbf{a}}{\mathbf{3}}\mathbf{+}\frac{{\mathbf{\nu \gamma }}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{)}+\left(\right)\mathbf{d}+...\right]\widehat{\mathbf{x}}$ …… (1)

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

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

${\mathbf{\int }}_{{\mathbf{t}}_1}^{{\mathbf{t}}_{2\text{'}}}{\mathbf{F}}_{\mathrm{rad}}\mathbf{\cdot }\mathbf{vdt}$ …… (2)

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

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

${\text{E}}_{\text{1}}=\frac{\left(\frac{\mathrm{q}}{2}\right)}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{r}}{{(\mathrm{r}\cdot \mathrm{u})}^3}[({\mathrm{c}}^2-{\mathrm{v}}^2)\mathrm{u}+(\mathrm{r}\cdot \mathrm{a})\mathrm{u}-(\mathrm{r}\cdot \mathrm{u})\mathrm{a}]$

Substitute $\mathrm{cr}-\mathrm{v}$ for $\mathrm{u}$, $(\mathrm{l}\widehat{\mathrm{x}}+\mathrm{d}\widehat{\mathrm{y}})$ for $\mathrm{r}$, $\nu \widehat{x}$ for $v$, role="math" localid="1658740581012" $\mathrm{a}\widehat{\mathrm{x}}$ for $\mathrm{a}$, $\mathrm{cr}-\mathrm{lv}$for $\mathrm{r}\cdot \mathrm{u}$ and $(\mathrm{cl}-\mathrm{\nu r})/\mathrm{r}$ for $u_x$into above equation ${\mathrm{E}}_1$.

$\begin{array}{c}{\mathrm{E}}_{1_{\mathrm{x}}}=\frac{\mathrm{q}}{8{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{r}}{{(\mathrm{cr}-\mathrm{lv})}^3}\left[\frac{1}{\mathrm{r}}(\mathrm{cl}-\mathrm{\nu r})({\mathrm{c}}^2-{\mathrm{v}}^2+\mathrm{la})-\mathrm{a}(\mathrm{cr}-\mathrm{lv})\right]\\ =\frac{\mathrm{q}}{8{\mathrm{\pi \epsilon }}_0}\frac{1}{{(\mathrm{cr}-\mathrm{lv})}^3}[(\mathrm{cl}-\mathrm{\nu r})({\mathrm{c}}^2-{\mathrm{v}}^2)+{\mathrm{cl}}^2\mathrm{a}-\mathrm{\nu da}-{\mathrm{acr}}^2+\mathrm{al\nu r}]\mathrm{a}\end{array}$

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

${\mathrm{E}}_{1_{\mathrm{x}}}=\frac{\mathrm{q}}{8{\mathrm{\pi \epsilon }}_0}\frac{1}{{(\mathrm{cr}-\mathrm{lv})}^3}[(\mathrm{cl}-\mathrm{\nu r})({\mathrm{c}}^2-{\mathrm{\nu }}^2)-{\mathrm{acd}}^2]$

Determine the self-radiation reaction force.

${\mathrm{F}}_{\mathrm{self}}=\frac{\mathrm{q}}{8{\mathrm{\pi \epsilon }}_0}\frac{1}{{(\mathrm{cr}-\mathrm{lv})}^3}[(\mathrm{cl}-\mathrm{\nu r})({\mathrm{c}}^2-{\mathrm{\nu }}^2)-{\mathrm{acd}}^2]\widehat{\mathrm{x}}$

Now

role="math" localid="1658740798843" $\begin{array}{c}\mathrm{x}\left(\mathrm{t}\right)-\mathrm{x}\left({\mathrm{t}}_{\mathrm{r}}\right)=\mathrm{l}\\ =\mathrm{\nu T}+\frac{1}{2}{\mathrm{aT}}^2+\frac{1}{6}\dot{\mathrm{a}}{\mathrm{T}}^3+\mathrm{...},\end{array}$

Here, $\mathrm{T}=\mathrm{t}-{\mathrm{t}}_{\mathrm{r}}$, and $\nu$, $a$, and $\dot{a}$ are all evaluate at the retarded time $t_r$.

$\begin{array}{c}{\left(\mathrm{cT}\right)}^2={\mathrm{r}}^2\\ ={\mathrm{l}}^2+{\mathrm{d}}^2\\ ={\mathrm{d}}^2+{(\mathrm{\nu T}+\frac{1}{2}{\mathrm{aT}}^2+\frac{1}{6}\dot{\mathrm{a}}{\mathrm{T}}^3)}^2\\ ={\mathrm{d}}^2+{\mathrm{\nu }}^2\mathrm{T}+{\mathrm{\nu aT}}^3+\frac{1}{3}\mathrm{\nu }\dot{\mathrm{a}}{\mathrm{T}}^4+\frac{1}{4}{\mathrm{a}}^2{\mathrm{T}}^4\end{array}$

Solve further as

$\begin{array}{c}{\mathrm{c}}^2{\mathrm{T}}^2\left(\frac{1-{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)=\frac{{\mathrm{c}}^2{\mathrm{T}}^2}{{\mathrm{\gamma }}^2}\\ ={\mathrm{d}}^2+{\mathrm{\nu aT}}^3+\left(\frac{1}{3}\mathrm{\nu }\dot{\mathrm{a}}+\frac{1}{4}{\mathrm{a}}^2\right){\mathrm{T}}^4\end{array}$

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

$\begin{array}{c}\mathrm{T}=\frac{\mathrm{\gamma d}}{\mathrm{c}}(1+\mathrm{Ad}+{\mathrm{Bd}}^2+...)\\ \Rightarrow \frac{{\mathrm{c}}^2}{{\mathrm{\gamma }}^2}\frac{{\mathrm{\gamma }}^2{\mathrm{d}}^2}{{\mathrm{c}}^2}(1+2\mathrm{Ad}+2{\mathrm{Bd}}^2+{\mathrm{A}}^2{\mathrm{d}}^2)\\ ={\mathrm{d}}^2+\mathrm{\nu a}\frac{{\mathrm{\gamma }}^3{\mathrm{d}}^3}{{\mathrm{c}}^3}(1+3\mathrm{Ad})+\left(\frac{\mathrm{\nu }\dot{\mathrm{a}}}{3}+\frac{{\mathrm{a}}^2}{4}\right)\frac{{\mathrm{\gamma }}^4}{{\mathrm{c}}^4}{\mathrm{d}}^4\end{array}$

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

$\begin{array}{c}2\mathrm{B}+{\mathrm{A}}^2=\frac{3{\mathrm{\nu a\gamma }}^3}{{\mathrm{c}}^3}\mathrm{A}+\left(\frac{\mathrm{\nu }\dot{\mathrm{a}}}{3}+\frac{{\mathrm{a}}^2}{4}\right)\frac{{\mathrm{\gamma }}^4}{{\mathrm{c}}^4}\\ 2\mathrm{B}=\frac{3{\mathrm{\nu a\gamma }}^3}{{\mathrm{c}}^3}\frac{1}{2}\mathrm{\nu a}\frac{{\mathrm{\gamma }}^3}{{\mathrm{c}}^3}-\frac{1}{4}{\mathrm{\nu }}^2{\mathrm{a}}^2\frac{{\mathrm{\gamma }}^6}{{\mathrm{c}}^6}+\frac{\mathrm{\nu }\dot{\mathrm{a}}}{3}\frac{{\mathrm{\gamma }}^4}{{\mathrm{c}}^4}+\frac{{\mathrm{a}}^2{\mathrm{\gamma }}^4}{4{\mathrm{c}}^4}\\ 2\mathrm{B}=\frac{\mathrm{\nu }\dot{\mathrm{a}}}{3}\frac{{\mathrm{\gamma }}^4}{{\mathrm{c}}^4}+\frac{{\mathrm{\gamma }}^4{\mathrm{a}}^2}{4{\mathrm{c}}^4}\left(\frac{1}{{\mathrm{\gamma }}^2}-\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right)+\frac{3}{2}\frac{{\mathrm{\nu }}^2{\mathrm{a}}^2{\mathrm{\gamma }}^6}{{\mathrm{c}}^6}\\ 2\mathrm{B}=\frac{{\mathrm{\gamma }}^4}{{\mathrm{c}}^4}\left[\frac{\mathrm{\nu }\dot{\mathrm{a}}}{3}+\frac{{\mathrm{a}}^2{\mathrm{\gamma }}^2}{4}\left(1-\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}-\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}+6\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right)\right]\end{array}$

Solve further as

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

Now solve again $\mathrm{T}$.

$\mathrm{T}=\frac{\mathrm{\gamma d}}{\mathrm{c}}\left\{1+\frac{\mathrm{\nu a}}{2}\frac{{\mathrm{\gamma }}^3}{{\mathrm{c}}^3}\mathrm{d}+\frac{{\mathrm{\gamma }}^4}{2{\mathrm{c}}^4}\left[\frac{\mathrm{\nu }\dot{\mathrm{a}}}{3}+\frac{{\mathrm{\gamma }}^2{\mathrm{a}}^2}{4}\left(1+4\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right)\right]]{\mathrm{d}}^2\right\}+\left(\right){\mathrm{d}}^4+...$

Determine the current $I$.

role="math" localid="1658741573624" $\begin{array}{c}\mathrm{I}=\mathrm{\nu T}+\frac{1}{2}{\mathrm{aT}}^2+\frac{1}{6}\dot{\mathrm{a}}{\mathrm{T}}^3+...\\ =\left\{\begin{array}{c}\frac{\mathrm{\nu \gamma d}}{\mathrm{c}}\left\{1+\frac{\mathrm{\nu a}}{2}\frac{{\mathrm{\gamma }}^3}{{\mathrm{c}}^3}\mathrm{d}+\frac{{\mathrm{\gamma }}^4}{2{\mathrm{c}}^4}\left[\frac{\mathrm{\nu }\dot{\mathrm{a}}}{3}+\frac{{\mathrm{\gamma }}^2{\mathrm{a}}^2}{4}\left(1+4\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right)\right]{\mathrm{d}}^2\right\}\\ +\frac{1}{2}\mathrm{a}\frac{{\mathrm{\gamma }}^2{\mathrm{d}}^2}{{\mathrm{c}}^2}\left[1+\mathrm{\nu a}\frac{{\mathrm{\gamma }}^3}{{\mathrm{c}}^3}\mathrm{d}\right]+\frac{1}{6}\dot{\mathrm{a}}\frac{{\mathrm{\gamma }}^3}{{\mathrm{c}}^3}{\mathrm{d}}^3\end{array}\right\}\\ =\left(\frac{\mathrm{\nu \gamma }}{\mathrm{c}}\right)\mathrm{d}+\frac{\mathrm{a}}{2}\frac{{\mathrm{\gamma }}^4}{{\mathrm{c}}^2}\left(1-\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}+\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right){\mathrm{d}}^2\\ =\left\{\frac{\mathrm{\nu \gamma }}{\mathrm{c}}\frac{{\mathrm{\gamma }}^4}{{\mathrm{c}}^4}\left[\frac{\mathrm{\nu }\dot{\mathrm{a}}}{3}+\frac{{\mathrm{\gamma }}^2{\mathrm{a}}^2}{4}\left(1+4\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right)\right]+\frac{1}{2}\mathrm{a}\frac{{\mathrm{\gamma }}^2}{{\mathrm{c}}^2}\mathrm{\nu a}\frac{{\mathrm{\gamma }}^3}{{\mathrm{c}}^3}+\frac{1}{6}\dot{\mathrm{a}}\frac{{\mathrm{\gamma }}^3}{{\mathrm{c}}^3}\right\}{\mathrm{d}}^3\end{array}$

Solve further as

Determine the $r$.

Determine the$r\cdot u$.

$\begin{array}{c}\mathrm{r}\cdot \mathrm{u}=\mathrm{c\gamma d}+\frac{{\mathrm{\nu a\gamma }}^4}{2{\mathrm{c}}^2}{\mathrm{d}}^2+\frac{{\mathrm{\gamma }}^5}{2{\mathrm{c}}^3}{\mathrm{d}}^2+\frac{{\mathrm{\gamma }}^5}{2{\mathrm{c}}^3}\left[\frac{\mathrm{\nu }\dot{\mathrm{a}}}{3}+{\mathrm{\gamma }}^2{\mathrm{a}}^2\left(\frac{1}{4}+\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right)\right]{\mathrm{d}}^3\\ -\frac{{\mathrm{\nu }}^2\mathrm{\gamma }}{\mathrm{c}}\mathrm{d}-\frac{{\mathrm{a\nu \gamma }}^4}{2{\mathrm{c}}^2}{\mathrm{d}}^2-\frac{{\mathrm{\gamma }}^5\mathrm{\nu }}{2{\mathrm{c}}^3}\left[\frac{\dot{\mathrm{a}}}{3}+\frac{5}{4}\frac{{\mathrm{\nu \gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right]{\mathrm{d}}^3+...\\ =\mathrm{c\gamma d}\left(1-\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right)+\frac{{\mathrm{\gamma }}^5}{2{\mathrm{c}}^3}\left[\frac{\mathrm{\nu }\dot{\mathrm{a}}}{3}+{\mathrm{\gamma }}^2{\mathrm{a}}^2\left(\frac{1}{4}+\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right)-\frac{\mathrm{\nu }\dot{\mathrm{a}}}{3}-\frac{5}{4}\frac{{\mathrm{\nu }}^2{\mathrm{\gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right]{\mathrm{d}}^3+...\\ =\frac{\mathrm{c}}{\mathrm{\gamma }}\mathrm{d}+\frac{{\mathrm{\gamma }}^5{\mathrm{a}}^2}{8{\mathrm{c}}^3}{\mathrm{d}}^3+\left(\right){\mathrm{d}}^4+...\end{array}$

Determine the $\mathrm{cl}-\mathrm{\nu r}$.

$\begin{array}{c}\mathrm{cl}-\mathrm{\nu r}=\mathrm{\nu rd}+\frac{{\mathrm{a\gamma }}^4}{2\mathrm{c}}{\mathrm{d}}^2+\frac{{\mathrm{\gamma }}^5}{2{\mathrm{c}}^2}\left(\frac{\dot{\mathrm{a}}}{3}+\frac{5}{4}\frac{{\mathrm{\nu \gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right){\mathrm{d}}^3-\mathrm{v\gamma d}\\ -\frac{{\mathrm{\nu }}^2{\mathrm{a\gamma }}^4}{2{\mathrm{c}}^3}{\mathrm{d}}^2-\frac{{\mathrm{\nu \gamma }}^3}{2{\mathrm{c}}^4}\left[\frac{\mathrm{\nu }\dot{\mathrm{a}}}{3}+{\mathrm{\gamma }}^2{\mathrm{a}}^2\left(\frac{1}{4}+\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right)\right]{\mathrm{d}}^3\\ =\frac{{\mathrm{a\gamma }}^4}{2\mathrm{c}}\left(1-\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right){\mathrm{d}}^2+\frac{{\mathrm{\gamma }}^5}{2{\mathrm{c}}^2}\left[\begin{array}{l}\frac{\dot{\mathrm{a}}}{3}+\frac{5}{4}\frac{{\mathrm{\nu \gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}-\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\frac{\dot{\mathrm{a}}}{3}\\ -\frac{{\mathrm{\nu \gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\left(\frac{1}{4}+\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right)\end{array}\right]{\mathrm{d}}^3+\left(\right){\mathrm{d}}^4+...\\ =\left(\frac{{\mathrm{a\gamma }}^4}{2\mathrm{c}}\right){\mathrm{d}}^2+\frac{{\mathrm{\gamma }}^5}{2{\mathrm{c}}^2}\left[\frac{\dot{\mathrm{a}}}{3{\mathrm{\gamma }}^2}+\frac{{\mathrm{\nu \gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\left(\frac{5}{4}-\frac{1}{4}-\frac{{\mathrm{\nu }}^2}{{\mathrm{c}}^2}\right)\right]{\mathrm{d}}^3+\left(\right){\mathrm{d}}^4+...\\ =\left(\frac{{\mathrm{a\gamma }}^4}{2\mathrm{c}}\right){\mathrm{d}}^2+\frac{{\mathrm{\gamma }}^5}{2{\mathrm{c}}^2}\left(\frac{\dot{\mathrm{a}}}{3}+\frac{{\mathrm{\nu \gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right){\mathrm{d}}^3+\left(\right){\mathrm{d}}^4+...\end{array}$

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

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

Determine theradiation reaction force on a moving particle.

$\begin{array}{c}{\mathrm{F}}_{\mathrm{self}}=\frac{{\mathrm{q}}^2}{8{\mathrm{\pi \epsilon }}_0}{\left(\frac{\mathrm{\gamma }}{\mathrm{cd}}\right)}^3\left(1-3\frac{{\mathrm{\gamma }}^6{\mathrm{a}}^2}{8{\mathrm{c}}^4}{\mathrm{d}}^2\right)\left\{\left[\left(\frac{{\mathrm{a\gamma }}^2}{2\mathrm{c}}\right){\mathrm{d}}^2+\frac{{\mathrm{\gamma }}^3}{2{\mathrm{c}}^2}\left(\frac{\dot{\mathrm{a}}}{3}+\frac{{\mathrm{\nu \gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right){\mathrm{d}}^3\right]\frac{{\mathrm{c}}^2}{{\mathrm{\gamma }}^2}-{\mathrm{acd}}^2\right\}\widehat{\mathrm{x}}\\ =\frac{{\mathrm{q}}^2}{9{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{\gamma }}^3}{{\mathrm{c}}^3\mathrm{d}}\left(1-\frac{3}{8}\frac{{\mathrm{\gamma }}^6{\mathrm{a}}^2}{{\mathrm{c}}^4}{\mathrm{d}}^2\right)\left[-\frac{\mathrm{ac}}{2}+\frac{\mathrm{\gamma }}{2}\left(\frac{\dot{\mathrm{a}}}{3}+\frac{{\mathrm{\nu \gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right)\mathrm{d}\right]]\widehat{\mathrm{x}}\\ =\frac{{\mathrm{q}}^2}{8{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{\gamma }}^3}{{\mathrm{c}}^3\mathrm{d}}\frac{1}{2}\left[-\mathrm{ac}+\mathrm{\gamma }\left(\frac{\dot{\mathrm{a}}}{3}+\frac{{\mathrm{\nu \gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right)\mathrm{d}+\left(\right){\mathrm{d}}^2+...\right]\widehat{\mathrm{x}}\\ =\frac{{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0}\left[-{\mathrm{\gamma }}^3\frac{\mathrm{a}}{4{\mathrm{c}}^2\mathrm{d}}+\frac{{\mathrm{\gamma }}^4}{4{\mathrm{c}}^3}\left(\frac{\dot{\mathrm{a}}}{3}+\frac{{\mathrm{\nu \gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right)+\left(\right)\mathrm{d}+...\right]\widehat{\mathrm{x}}\end{array}$

Determine the radiation reaction force on a moving particle.

Substitute $(1-3{\mathrm{\nu a\gamma }}^3\mathrm{d}/{\mathrm{c}}^3)(\mathrm{a}-\dot{\mathrm{a}}\mathrm{\gamma d}/\mathrm{c})$for $a$ into equation (1)

$\begin{array}{c}{\mathrm{F}}_{\mathrm{self}}=\frac{{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0}\left[-{\mathrm{\gamma }}^3\frac{(1-3{\mathrm{\nu a\gamma }}^3\mathrm{d}/{\mathrm{c}}^3)(\mathrm{a}-\dot{\mathrm{a}}\mathrm{\gamma d}/\mathrm{c})}{4{\mathrm{c}}^2\mathrm{d}}+\frac{{\mathrm{\gamma }}^4}{4{\mathrm{c}}^3}\left(\frac{\dot{\mathrm{a}}}{3}+\frac{{\mathrm{\nu \gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right)+\left(\right)\mathrm{d}+...\right]\widehat{\mathrm{x}}\\ =\frac{{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0}\left[-\frac{{\mathrm{\gamma }}^3\mathrm{a}}{4{\mathrm{c}}^2\mathrm{d}}+\frac{{\mathrm{\gamma }}^4}{4{\mathrm{c}}^3}\left(\frac{\dot{\mathrm{a}}\mathrm{\gamma }}{\mathrm{c}}+3\frac{{\mathrm{\nu a}}^2{\mathrm{\gamma }}^2}{{\mathrm{c}}^3}\right)+\frac{{\mathrm{\gamma }}^4}{4{\mathrm{c}}^3}\left(\frac{\dot{\mathrm{a}}}{3}+\frac{{\mathrm{\nu \gamma }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}\right)+\left(\right)\mathrm{d}+...\right]\widehat{\mathrm{x}}\\ =\frac{{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0}\left[-\frac{{\mathrm{\gamma }}^3\mathrm{a}}{4{\mathrm{c}}^2\mathrm{d}}+\frac{{\mathrm{\gamma }}^4}{4{\mathrm{c}}^3}\left(\dot{\mathrm{a}}+\frac{\dot{\mathrm{a}}}{3}+3\frac{{\mathrm{\nu a}}^2{\mathrm{\gamma }}^2}{{\mathrm{c}}^3}+\frac{{\mathrm{\nu a}}^2{\mathrm{\gamma }}^2}{{\mathrm{c}}^3}\right)+\left(\right)\mathrm{d}+..\right]\widehat{\mathrm{x}}\\ =\frac{{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0}\left[-\frac{{\mathrm{\gamma }}^3\mathrm{a}}{4{\mathrm{c}}^2\mathrm{d}}+\frac{{\mathrm{\gamma }}^4}{3{\mathrm{c}}^3}\left(\dot{\mathrm{a}}+3\frac{{\mathrm{\nu a}}^2{\mathrm{\gamma }}^2}{{\mathrm{c}}^2}\right)+\left(\right)\mathrm{d}+...\right]\widehat{\mathrm{x}}\end{array}$

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

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

$\mathrm{P}=\frac{{\mathrm{\mu }}_0{\mathrm{q}}^2}{6\mathrm{\pi c}}{\mathrm{\gamma }}^6{\mathrm{a}}^2$

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 $P$:

${\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{F}}_{\mathrm{rad}}\cdot \mathrm{vdt}=-{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\mathrm{Pdt}$

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

$\begin{array}{c}{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{F}}_{\mathrm{rad}}\cdot \mathrm{vdt}=\frac{{\mathrm{\mu }}_0{\mathrm{q}}^2}{6\mathrm{\pi c}}{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{\gamma }}^4\left(\dot{\mathrm{a}}+3\frac{{\mathrm{\gamma }}^2{\mathrm{a}}^2\mathrm{\upsilon }}{{\mathrm{c}}^2}\right)\mathrm{\upsilon dt}\\ =\frac{{\mathrm{\mu }}_0{\mathrm{q}}^2}{6\mathrm{\pi c}}({\mathrm{I}}_1+{\mathrm{I}}_2)\end{array}$

Determine the first integral:

$\begin{array}{c}\mathrm{I}{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{\gamma }}^4\mathrm{\upsilon }\dot{\mathrm{a}}\mathrm{dt}={\left|\left({\mathrm{a\gamma }}^4\mathrm{\upsilon }\right)\right|}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}-{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{\gamma }}^4\mathrm{\upsilon }\right)\mathrm{adt}\\ =0-{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}(4{\mathrm{\gamma }}^3\dot{\mathrm{\gamma }}\mathrm{\upsilon }+{\mathrm{\gamma }}^4\mathrm{a})\mathrm{adt}\\ =-{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}(4{\mathrm{\gamma }}^3\dot{\mathrm{\gamma }}\mathrm{\upsilon a}+{\mathrm{\gamma }}^4{\mathrm{a}}^2)\mathrm{dt}\end{array}$

Substitute $\begin{array}{c}\dot{\mathrm{\gamma }}=-\frac{1}{2}\frac{-2\mathrm{\upsilon }/{\mathrm{c}}^2}{{(1-{\mathrm{\upsilon }}^2/{\mathrm{c}}^2)}^{3/2}}\\ ={\mathrm{\gamma }}^3\frac{\mathrm{\upsilon a}}{{\mathrm{c}}^2}\end{array}$ into above equation.

$\begin{array}{c}-{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\left(4{\mathrm{\gamma }}^6\frac{{\mathrm{\upsilon }}^2{\mathrm{a}}^2}{{\mathrm{c}}^2}+{\mathrm{\gamma }}^4{\mathrm{a}}^2\right)\mathrm{dt}=-{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{\gamma }}^6{\mathrm{a}}^2\left(4\frac{{\mathrm{\upsilon }}^2}{{\mathrm{c}}^2}+{\mathrm{\gamma }}^2\right)\mathrm{dt}\\ ={\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{\gamma }}^6{\mathrm{a}}^2\left(1+3\frac{{\mathrm{\upsilon }}^2}{{\mathrm{c}}^2}\right)\mathrm{dt}\end{array}$

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 $-\frac{{\mu }_0{q}^2}{6\pi c}{\gamma }^6{a}^2\left(1+3\frac{{\upsilon }^2}{c^2}\right)$for $F_{rad}$ and $3\frac{{\mathrm{\mu }}_0{\mathrm{q}}^2}{6\mathrm{\pi c}}{\mathrm{\gamma }}^6\frac{{\mathrm{a}}^2{\mathrm{\upsilon }}^2}{{\mathrm{c}}^2}$ for $vdt$into equation (2).

$\begin{array}{c}{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{F}}_{\mathrm{rad}}\cdot \mathrm{vdt}=\frac{{\mathrm{\mu }}_0{\mathrm{q}}^2}{6\mathrm{\pi c}}{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\left[-{\mathrm{\gamma }}^6{\mathrm{a}}^2\left(1+\frac{{\mathrm{\upsilon }}^2}{{\mathrm{c}}^2}\right)+3{\mathrm{\gamma }}^6\frac{{\mathrm{a}}^2{\mathrm{\upsilon }}^2}{{\mathrm{c}}^2}\right]\mathrm{dt}\\ =-\frac{{\mathrm{\mu }}_0{\mathrm{q}}^2}{6\mathrm{\pi c}}{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{\gamma }}^6{\mathrm{a}}^2\mathrm{dt}\\ =-{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\mathrm{Pdt}\end{array}$

Therefore, the value of radiation reaction force of adding first integral to the second one is ${\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{F}}_{\mathrm{rad}}\cdot \mathrm{vdt}=-{\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\mathrm{Pdt}$.