Radiated Momentum

Being generated by a circular loop, the density of radiated momentum pointing into the $\pm \tau$ direction is independent of the polar angle $\theta$. Consequently, that density's spatial integral, Eq.(66), reduces to

$$\int_{-\infty}^\infty \int_0^\infty \int _0^{2\pi} T^\xi_{~\tau}\,\xi d\tau \, rdr \,d\theta = \frac{1}{4\pi}\int_{-\infty}^\infty \int_0^\infty \int _0^{2\pi} \hat B_r \times\hat E_\theta \xi \,\xi d\tau \, rdr \,d\theta$$

$$= \frac{(2\pi a^2)^2 2\pi}{4\pi} \int_{-\infty}^\infty \int_0^\inft... ...} )(\alpha \dot q +\beta \ddot q +\gamma \stackrel{...}{q}) \xi^2d\tau\, r\, dr$$ (86)

The $\tau$-integration affects only the dotted factors, and they vanish outside a sufficiently large $\tau$-interval, i.e. $\dot q(\pm\infty)=\ddot q(\pm\infty)=0$. Consequently, integration by parts yields

$$\int_{-\infty}^\infty\ddot q \dot q \,d\tau=\int_{-\infty}^\infty\stackrel{...}{q} \,\stackrel{..}{q}\,d\tau=0$$

and

$$\int_{-\infty}^\infty\stackrel{...}{q}\,\stackrel{.}{q} \,d\tau=-\int_{-\infty}^\infty{\ddot q}^2 d\tau \ne 0$$

Thus there remain only three non-zero terms in the integral,

$$\int_{-\infty}^\infty \int_0^\infty \int _0^{2\pi} T^\xi_{~\... ...alpha\epsilon ) {\ddot q}^2 \, \right) \xi^2 d\tau\, r\, dr~.$$ (87)

The coefficients of the squared terms are

$$\epsilon\gamma = \mp \frac{4r^2}{(2\xi\xi')^4} \left( \frac{1}{\xi^2} \frac{u}{(u^2+1)^{5/2}}- \frac{1}{\xi\xi'} \frac{1}{(u^2+1)^{5/2}} \right)$$ (88)

$$\beta\delta = \mp\frac{4r^2}{(2\xi\xi')^4} \left( -\frac{1}{\xi^2} \frac{(2-u^2)u}{(u^2+1)^{7/2}} -\frac{3}{\xi\xi'} \frac{u^2}{(u^2+1)^{7/2}} \right)$$ (89)

$$\alpha\epsilon = \mp \frac{4r^2}{(2\xi\xi')^4} \left( -\frac{3}{\xi^2} \frac{u}{(u^2+1)^{7/2}}+\frac{1}{\xi\xi'} \frac{1-2u^2}{(u^2+1)^{7/2}} \right)$$ (90)

We now take advantage of the fact that the integral, Eq.(87), is independent of the synchronous time $\xi$. This simplifies the evaluation of the integral considerably because one may assume

$$1\ll\frac{\xi}{\xi'}$$

without changing the value of the integral. The final outcome is that (i) in each of the expressions, Eqs.(88)-(90), only the last term contributes to the $r$-integral and (ii) the integral assumes a simple mathematical form if one introduces

$$u=\frac{\xi^2-\xi^{'2} -r^2}{2\xi \xi'},\quad\quad du=-\frac{2r\,dr}{2\xi\xi'}$$

as the new integration variable. With this scheme one has

$$\int_0^\infty \cdots \frac{r^2r\,dr}{(2\xi\xi')^2} = \frac{1}{2} \int_{(\xi^2-\xi'^2)/2\xi\xi'}^{-\infty} \cdots \left( \frac{\xi}{2\xi'}-\frac{\xi'}{2\xi}-u \right)(-)du$$ (91)

The to-be-used integrands have the form

$$\frac{1}{(u^2+1)^{n/2}},\quad \frac{u}{(u^2+1)^{n/2}}\quad\quad\quad\quad n=5,7,\cdots~,$$

both of which are always less than one in absolute value, even when they get multiplied by $u$. Consequently, one is perfectly justified in saying that

$$\int_0^\infty \cdots \frac{r^2r\,dr}{(2\xi\xi')^2} \rightarrow \frac{\xi}{4\xi'}\int_{-\infty}^{\infty} \cdots du \quad\textrm{whenever} \quad \xi'\ll\xi ~.$$ (92)

Taking note that only the last terms of Eqs.(88)-(90) give nonzero contribution, apply the limiting form, Eq.(92), to evaluate the integral, Eq.(87). One finds that

$$\int_{-\infty}^\infty \int_0^\infty \int _0^{2\pi} T^\xi_{~\tau}\,\xi d\tau \, rdr \,d\theta = \frac{(2\pi a^2)^2 2\pi}{4\pi} \frac{\pm 4}{(2\xi')^2} \frac{1}{4... ...)^{5/2}}+ \ddot q^2 \int_{-\infty}^\infty \frac{du}{(u^2+1)^{5/2}} \right)d\tau$$

$$= (\pi a^2)^2 \frac{\pm 1}{2\xi'^4}\int_{-\infty}^\infty \left( \st... ...cos^3 \phi~d\phi+ \ddot q^2\int_{-\pi/2}^{\pi/2}\cos^3 \phi ~d\phi \right)d\tau$$ (93)

The value of the integral

$$\int_{-\pi/2}^{\pi/2}\cos^3 \phi~d\phi=\frac{4}{3}$$

implies that the final result is

$$\int_{-\infty}^\infty \int_0^\infty \int _0^{2\pi} T^\xi_{~\... ...{2}{3}\stackrel{...}{q}^2 +\frac{2}{3}\ddot q^2 \right)d\tau~,$$ (94)

the total momentum into the $\tau$-direction radiated by a magnetic dipole accelerated in Rindler sector $I$ (upper sign) or in Rindler sector $II$ (lower sign). This is the result stated by Eq.(66)