Question 11.17: (a) A particle of charge q moves in a circle of radius R at ...

(a) A particle of charge q moves in a circle of radius R at a constant speed υ. To sustain the motion, you must, of course, provide a centripetal force 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 amplitude A and angular frequency \omega: w (t)=A \cos (\omega t) \hat{ 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.

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(a) To counteract the radiation reaction (Eq. 11.80), you must exert a force F _{e}=-\frac{\mu_{0} q^{2}}{6 \pi c} \dot{ a }.

F _{ rad }=\frac{\mu_{0} q^{2}}{6 \pi c} \dot{ a }                          (11.80)

For circular motion, r (t)=R[\cos (\omega t) \hat{ x }+\sin (\omega t) \hat{ y }], v (t)=\dot{ r }=R \omega[-\sin (\omega t) \hat{ x }+\cos (\omega t) \hat{ y }];

a (t)=\dot{ v }=-R \omega^{2}[\cos (\omega t) \hat{ x }+\sin (\omega t) \hat{ y }]=-\omega^{2} r ; \dot{ a }=-\omega^{2} \dot{ r }=-\omega^{2} v . S o \quad F _{e}=\frac{\mu_{0} q^{2}}{6 \pi c} \omega^{2} v.

P_{e}= F _{e} \cdot v =\frac{\mu_{0} q^{2}}{6 \pi c} \omega^{2} v^{2} . This is the power you must supply.

Meanwhile, the power radiated is (Eq. 11.70) P_{ rad }=\frac{\mu_{0} q^{2} a^{2}}{6 \pi c}, \text { and } a^{2}=\omega^{4} r^{2}=\omega^{4} R^{2}=\omega^{2} v^{2} , so

P_{ rad }=\frac{\mu_{0} q^{2}}{6 \pi c} \omega^{2} v^{2} , and the two expressions agree.

P=\frac{\mu_{0} q^{2} a^{2}}{6 \pi c}                           (11.70)

(b) For simple harmonic motion, r (t)=A \cos (\omega t) \hat{ z } ; v =\dot{ r }=-A \omega \sin (\omega t) \hat{ z } ; a =\dot{ v }=-A \omega^{2} \cos (\omega t) \hat{ z }=-\omega^{2} r ; \quad \dot{ a }=-\omega^{2} \dot{ r }=-\omega^{2} v . So F _{e}=\frac{\mu_{0} q^{2}}{6 \pi c} \omega^{2} v ; P_{e}=\frac{\mu_{0} q^{2}}{6 \pi c} \omega^{2} v^{2} . But this time a^{2}=\omega^{4} r^{2}=\omega^{4} A^{2} \cos ^{2}(\omega t), whereas \omega^{2} v^{2}=\omega^{4} A^{2} \sin ^{2}(\omega t) , so

P_{ rad }=\frac{\mu_{0} q^{2}}{6 \pi c} \omega^{4} A^{2} \cos ^{2}(\omega t) \neq P_{e}=\frac{\mu_{0} q^{2}}{6 \pi c} \omega^{4} A^{2} \sin ^{2}(\omega t);

the power you deliver is not equal to the power radiated. However, since the time averages of \sin ^{2}(\omega t) and  \cos ^{2}(\omega t) are equal (to wit: 1/2), over a full cycle the energy radiated is the same as the energy input. (In the mean time energy is evidently being stored temporarily in the nearby fields.)

\text { (c) In free fall, } v (t)=\frac{1}{2} g t^{2} \hat{ y } ; v =g t \hat{ y } ; a =g \hat{ y } ; \dot{ a }=0 . \text { So } F _{e}= 0 ; the radiation reaction is zero, and hence P_{e}=0 . \text { But there is radiation: } P_{ rad }=\frac{\mu_{0} q^{2}}{6 \pi c} g^{2} \text {. } Evidently energy is being continuously extracted from the nearby fields. This paradox persists even in the exact solution (where we do not assume v \ll c, as in the Larmor formula and the Abraham-Lorentz formula)—see Prob. 11.34.

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