Deduce Eq. 11.100 from Eq. 11.99. Here are three methods:

(a) Use the Abraham-Lorentz formula to determine the radiation reaction on each end of the dumbbell; add this to the interaction term (Eq. 11.99).

(b) Method (a) has the defect that it uses the Abraham-Lorentz formula—the very thing that we were trying to derive. To avoid this,$\mathit{F}\left(q\right)$ let be the total d-independent part of the self-force on a charge q. Then

$\mathit{F}\left(q\right)\mathbf{=}{\mathit{F}}^{\mathbf{i}\mathbf{n}\mathbf{t}}\left(q\right)\mathbf{+}\mathbf{2}\mathit{F}\left(\frac{q}{2}\right)$

Where${\mathit{F}}^{\mathbf{i}\mathbf{n}\mathbf{t}}$ is the interaction part (Eq. 11.99), and$\mathit{F}\left(\frac{q}{2}\right)$ is the self-force on each end. Now,$\mathit{F}\left(q\right)$ must be proportional to ${\mathit{q}}^{\mathit{2}}$, since the field is proportional to q and the force is qE. So$\mathit{F}\left(\frac{q}{2}\right)\mathbf{=}\left(\frac{1}{4}\right)\mathit{F}\left(q\right)$ Take it from there.

(c) Smear out the charge along a strip of length L oriented perpendicular to the motion (the charge density, then, is $\mathit{\lambda }\mathbf{=}\frac{\mathbf{q}}{\mathbf{L}}$); find the cumulative interaction force for all pairs of segments, using Eq. 11.99 (with the correspondence $\frac{\mathbf{q}}{\mathbf{2}}\mathbf{\to }\mathit{\lambda }\mathit{d}\mathit{y}$, at one end and at the other). Make sure you don’t count the same pair twice.

Write the expression for the Abraham-Lorentz formula.

${\mathit{F}}_{\mathbf{r}\mathbf{a}\mathbf{d}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}^{\mathbf{2}}}{\mathbf{6}\mathbf{\Pi }\mathbf{c}}\dot{\mathbf{a}}$

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

(a)

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

$$\begin{gathered} F_{rad}^{end}=\frac{{\mu }_0{\left({\displaystyle \frac{q}{2}}\right)}^2}{6\pi c}\dot{a} \\ {F}_{rad}^{end}=\frac{{\mu }_0{q}^2}{24\pi c}\dot{a} \end{gathered}$$

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

$F_{rad}^{int}=\frac{{\mu }_0{q}^2}{12\pi c}\dot{a}$

Hence, the total radiation force in dumbbell will be,

$$\begin{gathered} F_{rad}=2F_{rad}^{end}+F_{rad}^{int} \\ {F}_{rad}=2\left(\frac{{\mu }_0{q}^2}{24\pi c}\dot{a}\right)+\frac{{\mu }_0{q}^2}{12\pi c}\dot{a} \\ {F}_{rad}=\frac{{\mu }_0{q}^2}{12\pi c}\dot{a}\left(1+1\right) \\ {F}_{rad}=\frac{{\mu }_0{q}^2}{6\pi c}\dot{\dot{a}} \end{gathered}$$

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$F_{rad}=\frac{{\mu }_0{q}^2}{6\pi c}\dot{a}$ and$F\left(\frac{q}{2}\right)=\frac{1}{4}F\left(q\right)$ in equation (2).

$$\begin{gathered} F\left(q\right)=\frac{{\mu }_0{q}^2}{12\pi c}\dot{a}+2\left(\frac{1}{4}F\left(q\right)\right)\left\{F^{int}\left(q\right)=F_{rad}^{int}\right\} \\ F\left(q\right)-\left(\frac{1}{2}F\left(q\right)\right)=\frac{{\mu }_0{q}^2}{12\pi c}\dot{a} \\ F\left(q\right)=\frac{{\mu }_0{q}^2}{6\pi c}\dot{a} \end{gathered}$$

Therefore, equation 11.100 is deduced.

(c)

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

$F=\frac{{\mu }_0}{12\pi c}\dot{a}\left[{\int }_0^{y_1}2\lambda d{y}_2\right]2d{y}_1$

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

role="math" localid="1653989418237" $$\begin{gathered} F=\frac{{\mu }_0}{12\pi c}\dot{a}\left(4{\lambda }^2\right){\int }_0^L{y}_{1}d{y}_1 \\ F=\frac{{\mu }_0}{12\pi c}\dot{a}\left(4{\lambda }^2\right)\left(\frac{L^2}{2}\right) \\ F=\frac{{\mu }_0}{6\pi c}\dot{a}{\lambda }^2{L}^2 \end{gathered}$$

Here, $\lambda =\frac{q}{L}$.

Hence, the above equation becomes,

$$\begin{gathered} F=\frac{{\mu }_0}{6\pi c}\dot{a}{\left(\frac{q}{L}\right)}^2{L}^2 \\ F=\frac{{\mu }_0}{6\pi c}\dot{a}{q}^2 \end{gathered}$$

Therefore, equation 11.100 is deduced.