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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

$\displaystyle \int_{-\infty}^\infty \int_0^\infty \int _0^{2\pi} T^\xi_{~\tau}\,\xi d\tau
\, rdr \,d\theta$ $\textstyle =$ $\displaystyle \frac{1}{4\pi}\int_{-\infty}^\infty \int_0^\infty \int _0^{2\pi}
\hat B_r \times\hat E_\theta \xi \,\xi d\tau \, rdr \,d\theta$  
  $\textstyle =$ $\displaystyle \frac{(2\pi a^2)^2 2\pi}{4\pi}
\int_{-\infty}^\infty \int_0^\inft...
...} )(\alpha \dot q +\beta \ddot q +\gamma
\stackrel{...}{q})
\xi^2d\tau\, r\, dr$ (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

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

and

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

Thus there remain only three non-zero terms in the integral,
\begin{displaymath}
\int_{-\infty}^\infty \int_0^\infty \int _0^{2\pi} T^\xi_{~\...
...alpha\epsilon ) {\ddot q}^2 \, \right)
\xi^2 d\tau\, r\, dr~.
\end{displaymath} (87)

The coefficients of the squared terms are
$\displaystyle \epsilon\gamma$ $\textstyle =$ $\displaystyle \mp \frac{4r^2}{(2\xi\xi')^4}
\left(
\frac{1}{\xi^2}
\frac{u}{(u^2+1)^{5/2}}-
\frac{1}{\xi\xi'}
\frac{1}{(u^2+1)^{5/2}}
\right)$ (88)
$\displaystyle \beta\delta$ $\textstyle =$ $\displaystyle \mp\frac{4r^2}{(2\xi\xi')^4}
\left(
-\frac{1}{\xi^2} \frac{(2-u^2)u}{(u^2+1)^{7/2}}
-\frac{3}{\xi\xi'} \frac{u^2}{(u^2+1)^{7/2}}
\right)$ (89)
$\displaystyle \alpha\epsilon$ $\textstyle =$ $\displaystyle \mp \frac{4r^2}{(2\xi\xi')^4}
\left(
-\frac{3}{\xi^2} \frac{u}{(u^2+1)^{7/2}}+\frac{1}{\xi\xi'} \frac{1-2u^2}{(u^2+1)^{7/2}}
\right)$ (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

\begin{displaymath}
1\ll\frac{\xi}{\xi'}
\end{displaymath}

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

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

as the new integration variable. With this scheme one has
$\displaystyle \int_0^\infty \cdots \frac{r^2r\,dr}{(2\xi\xi')^2}$ $\textstyle =$ $\displaystyle \frac{1}{2} \int_{(\xi^2-\xi'^2)/2\xi\xi'}^{-\infty} \cdots \left(
\frac{\xi}{2\xi'}-\frac{\xi'}{2\xi}-u
\right)(-)du$ (91)

The to-be-used integrands have the form

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

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
$\displaystyle \int_0^\infty \cdots \frac{r^2r\,dr}{(2\xi\xi')^2}$ $\textstyle \rightarrow$ $\displaystyle \frac{\xi}{4\xi'}\int_{-\infty}^{\infty} \cdots du
\quad\textrm{whenever} \quad \xi'\ll\xi ~.$ (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
$\displaystyle \int_{-\infty}^\infty \int_0^\infty \int _0^{2\pi} T^\xi_{~\tau}\,\xi d\tau
\, rdr \,d\theta$ $\textstyle =$ $\displaystyle \frac{(2\pi a^2)^2 2\pi}{4\pi}
\frac{\pm 4}{(2\xi')^2}
\frac{1}{4...
...)^{5/2}}+
\ddot q^2 \int_{-\infty}^\infty \frac{du}{(u^2+1)^{5/2}}
\right)d\tau$  
  $\textstyle =$ $\displaystyle (\pi a^2)^2 \frac{\pm 1}{2\xi'^4}\int_{-\infty}^\infty
\left( \st...
...cos^3 \phi~d\phi+
\ddot q^2\int_{-\pi/2}^{\pi/2}\cos^3 \phi ~d\phi
\right)d\tau$ (93)

The value of the integral

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

implies that the final result is
\begin{displaymath}
\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~,
\end{displaymath} (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)




next up previous
Next: About this document ... Up: APPENDIX: POTENTIAL, FIELD AND Previous: Field Components
Ulrich Gerlach 2001-10-09