(a) A particle of charge $q$moves in a circle of radius$R$at a constant speed$v$. To sustain the motion, you must, of course, provide a centripetal force$\frac{m{v}^2}{R}$what additional force ($F_e$) must you exert, in order to counteract the radiation reaction? [It's easiest to express the answer in terms of the instantaneous velocity$v$.] What power ($P_e$) does this extra force deliver? Compare$P_e$with the power radiated (use the Larmor formula).

(b) Repeat part (a) for a particle in simple harmonic motion with amplitudeand angular frequency:$\omega$.$\omega \left(t\right)=\mathbf{Acos}\mathbf{(}\omega t\mathbf{)}\stackrel{⌢}{z}$ Explain the discrepancy.

(c) Consider the case of a particle in free fall (constant acceleration$g$). What is the radiation reaction force? What is the power radiated? Comment on these results.

When a charged particle or any matter rotates in a circular path, that matter experiences two forces.

One force of the forces out of them is directed towards the rotation axis called centripetal force. Another force acts outside the charged particle called centrifugal force.

The given data can be listed as:

• The charged particle is$q$
• The radius of the circle is $R$.
• The constant velocity is $v$.
• The force exerted over the particle is $F_e$.
• Thepower is $P_e$.

We know that to counteract the radiation reaction we must have to exert a force of

$F_e=-\frac{{\mu }_0{q}^2}{6\pi c}\dot{a}$ … (i)

Here, $F_e$is the force of radiation,${\mu }_0$is the permeability of free space,$c$is the velocity of light in a vacuum, $q$is the charged particle,$\dot{a}=\frac{da}{dt}$is the rate of change ofacceleration.

For circular motion

$r\left(t\right)=R[\text{cos}\left(\omega t\right)\widehat{x}+\text{sin}\left(\omega t\right)\widehat{y}]$ … (ii)

We can write from equation (ii) that

$\begin{array}{c}v\left(t\right)=\dot{r}\\ =\frac{dr}{dt}\\ =\frac{d}{dt}\left(R[\cos \left(\omega t\right)\widehat{x}+\sin \left(\omega t\right)\widehat{y}]\right)\\ =R\omega [-\sin \left(\omega t\right)\widehat{x}+\cos \left(\omega t\right)\widehat{y}]\end{array}$

Here,$R$is the radius of the circular path,$\omega$is the angular frequency, $\widehat{x}$and$\widehat{y}$are the unit vectors in the direction of the horizontal and vertical axis respectively,$v$is the velocity.

Now, from equation (ii) we can write

$\begin{array}{c}a\left(t\right)=\dot{v}\\ =\frac{d}{dt}v\\ =\frac{d}{dt}\left(R\omega [-\sin (\omega t)\widehat{x}+\cos (\omega t\left)\widehat{y}\right]\right)\\ =-R{\omega }^2\left[\cos \right(\omega t)\widehat{x}+\sin (\omega t\left)\widehat{y}\right]\end{array}$

It is further solved as

$\begin{array}{c}a\left(t\right)=-{\omega }^{2}r\\ a=\dot{v}\end{array}$

Here,a is the acceleration of the body.

Now,the rate of change of acceleration

$\begin{array}{c}\dot{a}=\frac{da}{dt}\\ =\frac{d}{dt}(-{\omega }^{2}r)\\ =-{\omega }^2\frac{dr}{dt}\\ =-{\omega }^2\dot{r}\end{array}$

It is further solved as

$\dot{a}=-{\omega }^{2}v$ … (iii)

Therefore, putting equation (iii) in equation (i), we get:

$\begin{array}{c}{F}_e=-\frac{{\mu }_0{q}^2}{6\pi c}(-{\omega }^{2}v)\\ =\frac{{\mu }_0{q}^2}{6\pi c}{\omega }^{2}v\end{array}$

Hence, the force of radiation is $\frac{{\mu }_0{q}^2}{6\pi c}{\omega }^{2}v$.

Now, as we know that

$\text{Power}=\text{Force}\times \text{velocity}$

Therefore

$\begin{array}{c}{P}_e=F_e\times v\\ =\frac{{\mu }_0{q}^2}{6\pi c}{\omega }^{2}v\times v\\ =\frac{{\mu }_0{q}^2}{6\pi c}{\omega }^2{v}^2\end{array}$

… (iv)

Here,$P_e$is the power which extra force delivers.

Thus, the power delivered is $\frac{{\mu }_0{q}^2}{6\pi c}{\omega }^2{v}^2$.

Meanwhile, we know that

$P_{rad}=\frac{{\mu }_0{q}^2}{6\pi c}{a}^2$

Here, $P_{rad}$is the power radiated.

We know that

$a=-{\omega }^{2}r$

Therefore

$\begin{array}{c}{a}^2={\left(-{\omega }^{2}r\right)}^2\\ ={\omega }^4{r}^2\\ ={\omega }^4{r}^2\\ ={\omega }^2{v}^2\end{array}$

Hence, we can write putting the value in equation (iv) as

$\begin{array}{c}{P}_e=\frac{{\mu }_0{q}^2}{6\pi c}{a}^2\\ =P_{rad}\end{array}$

Thus, the power delivered is equal to power radiated.

For simple harmonic motion, it isgiven that

$r=A\cos \left(\omega t\right)\widehat{z}$ … (v)

Here,$\widehat{z}$ is the unit vector along thez-direction

Now, to calculate velocity, we have

role="math" localid="1658752719009" $\begin{array}{c}v\left(t\right)=\dot{\omega }\\ =\frac{dr}{dt}\\ =\frac{d}{dt}\left(A\cos \right(\omega t\left)\right)\widehat{z}\\ =-A\omega \sin \left(\omega t\right)\widehat{z}\end{array}$ … (vi)

Now, to calculate acceleration

$\begin{array}{c}a\left(t\right)=\dot{v}\\ =\frac{dv}{dt}\\ =\frac{d}{dt}(-A\omega \sin (\omega t\left)\right)\widehat{z}\\ =-A{\omega }^2\cos \left(\omega t\right)\widehat{z}\end{array}$ … (vii)

Now, from the equation (vii) and (v), we can write

$a=-{\omega }^{2}r$

Now, the rate of change of acceleration

$\begin{array}{c}\dot{a}=\frac{da}{dt}\\ =\frac{d}{dt}(-{\omega }^{2}r)\\ =-{\omega }^2\dot{r}\\ =-{\omega }^{2}v\end{array}$

Therefore, from equation(i) we can write

$\begin{array}{c}{F}_e=-\frac{{\mu }_0{q}^2}{6\pi c}\dot{a}\\ =-\frac{{\mu }_0{q}^2}{6\pi c}(-{\omega }^{2}v)\\ =\frac{{\mu }_0{q}^2}{6\pi c}{\omega }^{2}v\end{array}$

Thus, the radiation reaction force is$\frac{{\mu }_0{q}^2}{6\pi c}{\omega }^{2}v$.

Hence, power can be calculated as

$\begin{array}{c}{P}_e=\frac{{\mu }_0{q}^2}{6\pi c}{\omega }^{2}v\times v\\ =\frac{{\mu }_0{q}^2}{6\pi c}{\omega }^2{v}^2\end{array}$

Thus, the power is $\frac{{\mu }_0{q}^2}{6\pi c}{\omega }^2{v}^2$.

Now, as we know from part (a) that

$a^2={\omega }^4{r}^2$

Substitute equation (v) in the above expression.

Therefore

$\begin{array}{c}{a}^2={\omega }^4{r}^2\\ ={\omega }^4{\left(A\cos \left(\omega t\right)\right)}^2\\ ={\omega }^4{A}^2{\cos }^2\left(\omega t\right)\end{array}$

${\omega }^2{v}^2={\omega }^4{A}^2{\sin }^2\left(\omega t\right)$

Hence, the radiated power is

$\begin{array}{c}{P}_{rad}=\frac{{\mu }_0{q}^2}{6\pi c}{a}^2\\ =\frac{{\mu }_0{q}^2}{6\pi c}{A}^2{\omega }^4{\cos }^2\left(\omega t\right)\end{array}$ … (viii)

$\begin{array}{c}{P}_e=\frac{{\mu }_0{q}^2}{6\pi c}{\omega }^2{v}^2\\ =\frac{{\mu }_0{q}^2}{6\pi c}{\omega }^4{A}^2{\sin }^2\left(\omega t\right)\end{array}$

Therefore,$P_{rad}$ is not equal to $P_e$when the body moves in SHM.This occurs because we know that over a full cycle the time average of${\sin }^2\left(\omega t\right)$ and${\cos }^2\left(\omega t\right)$are equal.Hence, the energy which is radiating is the same as that of the energy input.

We know that for a free-falling body with constant acceleration g, the displacement of that particle will be

$s=\frac{1}{2}g{t}^2\widehat{y}$

Here,$\widehat{y}$is the unit vector along the vertical direction.

Therefore, velocity is given as

$\begin{array}{c}v\left(t\right)=\dot{s}\\ v=\frac{ds}{dt}\\ =\frac{d}{dt}\left(\frac{1}{2}g{t}^2\right)\\ =gt\widehat{y}\end{array}$

Here,$\widehat{y}$is the unit vector along the vertical direction. is the acceleration due to gravity, and $t$is the time.

Therefore, the acceleration is

$\begin{array}{c}a\left(t\right)=\dot{v}\\ =\frac{dv}{dt}\\ =\frac{d}{dt}\left(gt\right)\widehat{y}\\ =g\widehat{y}\end{array}$

Hence, the change in acceleration is

$\begin{array}{c}\dot{a}=\frac{da}{dt}\\ =\frac{d}{dt}g\\ =0\end{array}$

Here, a is equal to g.

Substitute the value in equation (1).

$\begin{array}{l}{F}_e=-\frac{{\mu }_0{q}^2}{6\pi c}\dot{a}\\ F_e=0\end{array}$

Therefore,the radiation reaction force is equal to zero.

Substitute the value in the equation$P_e=F_e\times v$ from equation (i).

$\begin{array}{l}{P}_e=F_e\times v\\ P_e=0\end{array}$

Hence, thereaction power is zero.

Now, put$a=g$ in the expression of radiated power from equation (viii).

$P_{rad}=\frac{{\mu }_0{q}^2}{6\pi c}{g}^2$

Hence, due to this reason, energy will be taken from the nearby fields.So, in this exact solution the paradox stays.