91Ó°ÊÓ

Write the expression for the Abraham-Lorentz formula.

${\mathit{F}}_{\mathbf{r}\mathbf{a}\mathbf{d}}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}{\mathbf{q}}^{\mathbf{2}}}{\mathbf{6}\mathbf{Π}\mathbf{c}}\stackrel{\mathbf{˙}}{\mathbf{a}}$

Here, q is the point charge, c is the speed of light, and$\stackrel{Ë™}{a}$ is the time rate of the acceleration.

(a)

Write the expression for the radiation reaction force at the end of the dumbbell.

$$\begin{gathered} F_{rad}^{end}=\frac{{\mathrm{μ}}_0{\left({\displaystyle \frac{q}{2}}\right)}^2}{6\mathrm{π}c}\stackrel{˙}{a} \\ {F}_{rad}^{end}=\frac{{\mathrm{μ}}_0{q}^2}{24\mathrm{π}c}\stackrel{˙}{a} \end{gathered}$$

Write the expression for the interaction force (using equation 11.99).

$F_{rad}^{int}=\frac{{\mathrm{μ}}_0{q}^2}{12\mathrm{π}c}\stackrel{˙}{a}$

Hence, the total radiation force in dumbbell will be,

$$\begin{gathered} F_{rad}=2F_{rad}^{end}+F_{rad}^{int} \\ {F}_{rad}=2\left(\frac{{\mathrm{μ}}_0{q}^2}{24\mathrm{π}c}\stackrel{˙}{a}\right)+\frac{{\mathrm{μ}}_0{q}^2}{12\mathrm{π}c}\stackrel{˙}{a} \\ {F}_{rad}=\frac{{\mathrm{μ}}_0{q}^2}{12\mathrm{π}c}\stackrel{˙}{a}\left(1+1\right) \\ {F}_{rad}=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{π}c}\stackrel{˙}{\stackrel{˙}{a}} \end{gathered}$$

Therefore, the equation 11.100 is deduced.

(b)

Consider is the total$F\left(q\right)$d-independent part of the self-force that acts on the charge q, then,

$F\left(q\right)=F^{int}\left(q\right)+2F\left(\frac{q}{2}\right)$ …… (2)

Here,$F\left(\frac{q}{2}\right)$ is the self-force on each end of the dumbbell and$F^{int}\left(\left(q\right)\right)$ is the interaction force part.

Write the relation between$F\left(q\right)$ and $F\left(\frac{q}{2}\right)$.

$F\left(\frac{q}{2}\right)=\frac{1}{4}F\left(q\right)$

Substitute$F_{rad}=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{π}c}\stackrel{˙}{a}$ and$F\left(\frac{q}{2}\right)=\frac{1}{4}F\left(q\right)$ in equation (2).

$$\begin{gathered} F\left(q\right)=\frac{{\mathrm{μ}}_0{q}^2}{12\mathrm{π}c}\stackrel{˙}{a}+2\left(\frac{1}{4}F\left(q\right)\right)\left\{F^{int}\left(q\right)=F_{rad}^{int}\right\} \\ F\left(q\right)-\left(\frac{1}{2}F\left(q\right)\right)=\frac{{\mathrm{μ}}_0{q}^2}{12\mathrm{π}c}\stackrel{˙}{a} \\ F\left(q\right)=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{π}c}\stackrel{˙}{a} \end{gathered}$$

Therefore, equation 11.100 is deduced.

(c)

Write the cumulative interaction force expression on all the pairs of segments.

$F=\frac{{\mathrm{μ}}_0}{12\mathrm{π}c}\stackrel{˙}{a}\left[{∫}_0^{y_1}2\mathrm{λ}d{y}_2\right]2d{y}_1$

Integrate from 0 to L and divide the result with L.

role="math" localid="1653989418237" $$\begin{gathered} F=\frac{{\mathrm{μ}}_0}{12\mathrm{π}c}\stackrel{˙}{a}\left(4{\mathrm{λ}}^2\right){∫}_0^L{y}_{1}d{y}_1 \\ F=\frac{{\mathrm{μ}}_0}{12\mathrm{π}c}\stackrel{˙}{a}\left(4{\mathrm{λ}}^2\right)\left(\frac{L^2}{2}\right) \\ F=\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}\stackrel{˙}{a}{\mathrm{λ}}^2{L}^2 \end{gathered}$$

Here, $\mathrm{λ}=\frac{q}{L}$.

Hence, the above equation becomes,

$$\begin{gathered} F=\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}\stackrel{˙}{a}{\left(\frac{q}{L}\right)}^2{L}^2 \\ F=\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}\stackrel{˙}{a}{q}^2 \end{gathered}$$

Therefore, equation 11.100 is deduced.