(a) Does a particle in hyperbolic motion (Eq. 10.52) radiate? (Use the exact formula (Eq. 11.75) to calculate the power radiated.) (b) Does a particle in hyperbolic motion experience a radiation reaction? (Use the exact formula (Prob. 11.33) to determine the reaction force.) [Comment: These fam

Question

(a) Does a particle in hyperbolic motion (Eq. 10.52) radiate? (Use the exact formula (Eq. 11.75) to calculate the power radiated.)

[latex]w (t)=\sqrt{b^{2}+(c t)^{2}} \hat{ x } \quad(-\infty(10.52)

[latex]P=\frac{\mu_{0} q^{2} a^{2} \gamma^{6}}{6 \pi c}[/latex] (11.75)

(b) Does a particle in hyperbolic motion experience a radiation reaction? (Use the exact formula (Prob. 11.33) to determine the reaction force.)

[Comment: These famous questions carry important implications for the principle of equivalenc[latex] ext { e }^{27}[/latex]].

Accepted Answer

[latex] ext { (a) } P=\frac{\mu_{0} q^{2} a^{2} \gamma^{6}}{6 \pi c} ext { (Eq. 11.75). } \quad w=\sqrt{b^{2}+c^{2} t^{2}}( ext { Eq. } 10.52) ; v=\dot{w}=\frac{c^{2} t}{\sqrt{b^{2}+c^{2} t^{2}}}[/latex];

[latex]a=\dot{v}=\frac{c^{2}}{\sqrt{b^{2}+c^{2} t^{2}}}-\frac{c^{2} t\left(c^{2} t ight)}{\left(b^{2}+c^{2} t^{2} ight)^{3 / 2}}=\frac{c^{2}}{\left(b^{2}+c^{2} t^{2} ight)^{3 / 2}}\left(b^{2}+c^{2} t^{2}-c^{2} t^{2} ight)=\frac{b^{2} c^{2}}{\left(b^{2}+c^{2} t^{2} ight)^{3 / 2}}[/latex];

[latex]\gamma^{2}=\frac{1}{1-v^{2} / c^{2}}=\frac{1}{1-\left[c^{2} t^{2} /\left(b^{2}+c^{2} t^{2} ight) ight]}=\frac{b^{2}+c^{2} t^{2}}{b^{2}+c^{2} t^{2}-c^{2} t^{2}}=\frac{1}{b^{2}}\left(b^{2}+c^{2} t^{2} ight)[/latex] . So

[latex]P=\frac{\mu_{0} q^{2}}{6 \pi c} \frac{b^{4} c^{4}}{\left(b^{2}+c^{2} t^{2} ight)^{3}} \frac{\left(b^{2}+c^{2} t^{2} ight)^{3}}{b^{6}}=\frac{q^{2} c}{6 \pi \epsilon_{0} b^{2}}[/latex] . Yes, it radiates (in fact, at a constant rate).

[latex] ext { (b) } F_{ rad }=\frac{\mu_{0} q^{2} \gamma^{4}}{6 \pi c}\left(\dot{a}+\frac{3 \gamma^{2} a^{2} v}{c^{2}} ight) ; \quad \dot{a}=-\frac{3}{2} \frac{b^{2} c^{2}\left(2 c^{2} t ight)}{\left(b^{2}+c^{2} t^{2} ight)^{5 / 2}}=-\frac{3 b^{2} c^{4} t}{\left(b^{2}+c^{2} t^{2} ight)^{5 / 2}} ; \quad\left(\dot{a}+\frac{3 \gamma^{2} a^{2} v}{c^{2}} ight)=[/latex]

[latex]-\frac{3 b^{2} c^{4} t}{\left(b^{2}+c^{2} t^{2} ight)^{5 / 2}}+\frac{3}{c^{2}} \frac{\left(b^{2}+c^{2} t^{2} ight)}{b^{2}} \frac{b^{4} c^{4}}{\left(b^{2}+c^{2} t^{2} ight)^{3}} \frac{c^{2} t}{\sqrt{b^{2}+c^{2} t^{2}}}=0 . \quad F_{ rad }=0[/latex] . No, the radiation reaction is zero.

Question 11.34: (a) Does a particle in hyperbolic motion (Eq. 10.52) radiate...

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