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Next: Concluding Remarks About Multipoles Up: The Hansen Multipoles Previous: Connection to Old (Approximate)   Contents

Angular Momentum Flux

Let us consider the angular momentum radiated away with the electromagnetic field. The angular momentum flux density is basically $vr$ crossed into the momentum density $\mbox{\boldmath$S$}/c$ or:

\mbox{\boldmath$\cal L$} = \frac{1}{2} {\rm Re}\left\{\frac{...{\boldmath$E$}\times
\end{displaymath} (13.95)

Into this expression we must substitute our expressions for $\mbox{\boldmath$E$}$ and $\mbox{\boldmath$H$}$:
$\displaystyle \mbox{\boldmath$E$}$ $\textstyle =$ $\displaystyle - k^2 Z_0 \sum_L \left\{ m_L \mbox{\boldmath$M$}_L^+ + n_L \mbox{\boldmath$N$}_L^+ \right\}$ (13.96)
$\displaystyle \mbox{\boldmath$H$}$ $\textstyle =$ $\displaystyle k^2 \sum_L \left\{ m_L \mbox{\boldmath$N$}_L^+ - n_L \mbox{\boldmath$M$}_L^+ \right\}.$ (13.97)

If we try to use the asymptotic far field results:

$\displaystyle \mbox{\boldmath$E$}$ $\textstyle =$ $\displaystyle - k Z_0 \frac{e^{ikr}}{r} \sum_L (-i)^{\ell+1} \left\{
m_L \mbox{...
...\mbox{\boldmath$r$}}\times \mbox{\boldmath$Y$}_{\ell \ell}^{m}
\right) \right\}$ (13.98)
$\displaystyle \mbox{\boldmath$H$}$ $\textstyle =$ $\displaystyle - k \frac{e^{ikr}}{r}
\sum_L (-i)^{\ell+1} \left\{ m_L \left(\hat...
...$Y$}_{\ell \ell}^{m}
\right) + n_L \mbox{\boldmath$Y$}_{\ell \ell}^{m} \right\}$ (13.99)

we get:
$\displaystyle \mbox{\boldmath$E$}\times \mbox{\boldmath$H$}^\ast$ $\textstyle =$ $\displaystyle \frac{k^2 Z_0}{r^2}\sum_L\sum_{L'} i^{\ell -
\ell'} \left\{m_L \m...
...boldmath$Y$}_{\ell \ell}^{m}(\hat{\mbox{\boldmath$r$}}) \right) \right\} \times$  
    $\displaystyle \quad \quad \left\{ m_{L'}^\ast \left(\hat{\mbox{\boldmath$r$}}\t...
... \mbox{\boldmath$Y$}_{\ell' \ell'}^{m'\ast}(\hat{\mbox{\boldmath$r$}}) \right\}$  
  $\textstyle =$ $\displaystyle \frac{k^2 Z_0}{r^2}\sum_L\sum_{L'} i^{\ell - \ell'}
\left\{ m_L m...
...\boldmath$Y$}_{\ell' \ell'}^{m'\ast}(\hat{\mbox{\boldmath$r$}}) \right) \right.$  
    $\displaystyle \quad \quad
+  m_L n_{L'}^\ast \mbox{\boldmath$Y$}_{\ell \ell}^{...
...) \times \mbox{\boldmath$Y$}_{\ell' \ell'}^{m'
    $\displaystyle \quad \quad -  n_L m_{L'}^\ast \left( \hat{\mbox{\boldmath$r$}}\...
... \mbox{\boldmath$Y$}_{\ell' \ell'}^{m'
\ast}(\hat{\mbox{\boldmath$r$}}) \right)$  
    $\displaystyle \quad \quad \left. -  n_L n_{L'}^\ast \left( \hat{\mbox{\boldmat...
...\mbox{\boldmath$Y$}_{\ell' \ell'}^{m'\ast}(\hat{\mbox{\boldmath$r$}}) \right\}.$ (13.100)

With some effort this can be shown to be a radial result - the Poynting vector points directly away from the source in the far field to leading order. Consequently, this leading order behavior contributes nothing to the angular momentum flux. We must keep at least the leading correction term to the asymptotic result.

It is convenient to use a radial/tangential decomposition of the Hansen solutions. The $\mbox{\boldmath$M$}_L$ are completely tangential (recall $\mbox{\boldmath$r$}\cdot
\mbox{\boldmath$M$}_L = 0$). For the $\mbox{\boldmath$N$}_L$ we have:

\mbox{\boldmath$N$}_L(\mbox{\boldmath$r$}) = \frac{1}{kr} \...
f_ell(kr) Y_L(\hat{\mbox{\boldmath$r$}})
\end{displaymath} (13.101)

Using our full expressions for $\mbox{\boldmath$E$}$ and $\mbox{\boldmath$H$}^\ast$:
$\displaystyle \mbox{\boldmath$E$}$ $\textstyle =$ $\displaystyle - k^2 Z_0 \sum_L \left\{ m_L \mbox{\boldmath$M$}_L^+ + n_L \mbox{\boldmath$N$}_L^+ \right\}$ (13.102)
$\displaystyle \mbox{\boldmath$H$}$ $\textstyle =$ $\displaystyle k^2 \sum_L \left\{ m_L \mbox{\boldmath$N$}_L^+ - n_L \mbox{\boldmath$M$}_L^+ \right\}$ (13.103)

with this form substituted for $\mbox{\boldmath$N$}_L$ and the usual form for $\mbox{\boldmath$M$}_L$ we get:
$\displaystyle \mbox{\boldmath$\cal L$}$ $\textstyle =$ $\displaystyle \frac{1}{2} {\rm Re}\left\{\frac{\mbox{\boldmath$r$}\times (\mbox{\boldmath$E$}\times
  $\textstyle =$ $\displaystyle - \frac{k^4 Z_0}{2 c} {\rm Re} \sum_L \sum_{L'} \mbox{\boldmath$r...
m_L h_\ell^+(kr)\mbox{\boldmath$Y$}_{\ell \ell}^{m}(\hat{\mbox{\boldmath$r$}})$  
    $\displaystyle \quad \quad + n_L \left[ \frac{1}{kr}\frac{d
...ll(\ell+1)}\frac{h_\ell^+(kr)}{kr}Y_L(\hat{\mbox{\boldmath$r$}})\right] \bigg\}$  
    $\displaystyle \times \bigg\{ m_{L'}^\ast \left[ \frac{1}{kr}\frac{d
    $\displaystyle \quad \quad + n_{L'}^\ast h_{\ell'}^-(kr) \mbox{\boldmath$Y$}_{\ell' \ell'}^{m' \ast}(\hat{\mbox{\boldmath$r$}})
\bigg\}$ (13.104)

All the purely radial terms in the outermost $\bigg\{\bigg\}$ under the sum do not contribute to the angular momentum flux density. The surviving terms are:
$\displaystyle \mbox{\boldmath$\cal L$}$ $\textstyle =$ $\displaystyle - \frac{k^4 Z_0}{2 c} {\rm Re} \sum_L \sum_{L'} \mbox{\boldmath$r...
...h$r$}}) \times
    $\displaystyle + n_L m_{L'}^\ast \frac{1}{kr}\frac{d(rh_\ell^+(kr))}{dr} \left((...
    $\displaystyle - n_L m_{L'}^\ast \frac{h_\ell^+(kr)}{kr}\sqrt{\ell(\ell+1)}
\mbox{\boldmath$Y$}_{\ell' \ell'}^{m' \ast}(\hat{\mbox{\boldmath$r$}}))\right)$  
    $\displaystyle - n_L n_{L'}^\ast \frac{h_\ell^+(kr)}{kr} \sqrt{\ell(\ell+1)}
... \mbox{\boldmath$Y$}_{\ell' \ell'}^{m' \ast}(\hat{\mbox{\boldmath$r$}}) \bigg\}$ (13.105)

The lowest order term in the asymptotic form for the spherical bessel functions makes a contribution in the above expressions. After untangling the cross products and substituting the asymptotic forms, we get:

$\displaystyle \mbox{\boldmath$\cal L$}$ $\textstyle =$ $\displaystyle \frac{k \mu_0}{2 r^2} {\rm Re} \sum_L \sum_{L'} \bigg\{
m_L m_{L'...
...{\boldmath$r$}}) \mbox{\boldmath$Y$}_{\ell
    $\displaystyle - n_L m_{L'}^\ast \sqrt{\ell(\ell+1)} i^{\ell'-\ell}
...$}}\times \mbox{\boldmath$Y$}_{\ell \ell}^{m}(\hat{\mbox{\boldmath$r$}})\right)$  
    $\displaystyle + n_L m_{L'}^\ast \sqrt{\ell(\ell+1)} i^{\ell'-\ell} Y_L(\hat{\mb...
...s \mbox{\boldmath$Y$}_{\ell' \ell'}^{m' \ast}(\hat{\mbox{\boldmath$r$}})\right)$  
    $\displaystyle + n_L n_{L'}^\ast \sqrt{\ell(\ell+1)} i^{\ell'-\ell} Y_L(\hat{\mb...
\mbox{\boldmath$Y$}_{\ell' \ell'}^{m' \ast}(\hat{\mbox{\boldmath$r$}}) \bigg\}$ (13.106)

The angular momentum about a given axis emitted per unit time is obtained by selecting a particular component of this and integrating its flux through a distant spherical surface. For example, for the $z$-component we find (noting that $r^2$ cancels as it should):

\frac{dL_z}{dt} = \frac{k\mu_0}{2}{\rm Re}\sum_L\sum_{L'} \i...
\bigg\{...\bigg\}\sin(\theta)d\theta d\phi
\end{displaymath} (13.107)

where the brackets indicate the expression above. We look up the components of the vector harmonics to let us do the dot product and find:
$\displaystyle \hat{\mbox{\boldmath$z$}}\cdot \mbox{\boldmath$Y$}_{\ell \ell}^{m}$ $\textstyle =$ $\displaystyle \frac{m}{\sqrt{\ell(\ell+1)}}Y_{\ell,m}$ (13.108)
$\displaystyle \hat{\mbox{\boldmath$z$}}\cdot (\hat{\mbox{\boldmath$r$}}\times \mbox{\boldmath$Y$}_{\ell \ell}^{m})$ $\textstyle =$ $\displaystyle -i
\left[ \sqrt{\frac{\ell+1}{2\ell+1}} \hat{\mbox{\boldmath$z$}}...
...1}} \hat{\mbox{\boldmath$z$}}\cdot \mbox{\boldmath$Y$}_{\ell \ell+1}^{m}\right]$  
  $\textstyle =$ $\displaystyle -i \left[
...{\frac{[(\ell+1)^2 - m^2]\ell}{(2\ell+1)(2\ell+3)(\ell+1)}}Y_{\ell+1,m}

Doing the integral is now simple, using the orthonormality of the spherical harmonics. One obtains (after still more work, of course):

\frac{dL_z}{dt} = \frac{k\mu_0}{2} \sum_L m\left(\vert m_L\vert^2 +
\vert n_L\vert^2\right)
\end{displaymath} (13.109)

Compare this to:
P = \frac{k^2}{2}Z_0 \sum_L \left\{ \mid m_L \mid^2 + \mid n_L \mid^2
\end{displaymath} (13.110)

term by term. For example:
$\displaystyle \frac{dL_z(m_L)}{dt}$ $\textstyle =$ $\displaystyle \frac{k\mu_0 m}{2} \left\{\frac{2}{k^2 \mu_0 c}
P(m_L) = \vert m_L\vert^2\right\}$  
  $\textstyle =$ $\displaystyle \frac{m}{\omega} P(m_L)$ (13.111)

(where $m$ in the fraction is the spherical harmonic $m$, not the multipole $m_L$). In other words, for a pure multipole the rate of angular momentum about any given axis transferred is $m/\omega$ times the rate of energy transferred, where $m$ is the angular momentum aligned with that axis. (Note that if we chose some other axis we could, with enough work, find an answer, but the algebra is only simple along the $z$-axis as the multipoles were originally defined with their $m$-index referred to this axis. Alternatively we could rotate frames to align with the new direction and do the entire computation over.)

This is quite profound. If we insist, for example, that energy be transferred in units of $\hbar \omega$, then angular momentum is also transferred in units of $m\hbar$!

next up previous contents
Next: Concluding Remarks About Multipoles Up: The Hansen Multipoles Previous: Connection to Old (Approximate)   Contents
Robert G. Brown 2007-12-28