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Next: Vector Spherical Harmonics and Up: Vector Multipoles Previous: Angular momentum and spherical   Contents

Magnetic and Electric Multipoles Revisited

As we have now seen repeatedly from Chapter J6 on, in a source free region of space, harmonic electromagnetic fields are divergenceless and have curls given by:

$\displaystyle \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$E$}$ $\textstyle =$ $\displaystyle i\omega \mbox{\boldmath$B$}= ikc \mbox{\boldmath$B$}$ (12.23)
$\displaystyle \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$B$}$ $\textstyle =$ $\displaystyle -i\frac{k}{c}\mbox{\boldmath$E$}.$ (12.24)

By massaging these a little bit (recall $\mbox{\boldmath$\nabla$}\times (\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$...{\boldmath$\nabla$}\cdot \mbox{\boldmath$X$}) - \nabla^2 \mbox{\boldmath$X$}$ and $\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$X$} = 0$ for $\mbox{\boldmath$X$} = \mbox{\boldmath$E$},\mbox{\boldmath$B$})$ we can easily show that both $\mbox{\boldmath$E$}$ and $\mbox{\boldmath$B$}$ must be divergenceless solutions to the HHE:
(\nabla^2+ k^2) \mbox{\boldmath$X$} = 0
\end{displaymath} (12.25)

If we know a solution to this equation for $\mbox{\boldmath$X$} = \mbox{\boldmath$E$}$ we can obtain $\mbox{\boldmath$B$}$ from its curl from the equation above:
\mbox{\boldmath$B$}= - \frac{i}{\omega} \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$E$}
\end{displaymath} (12.26)

and vice versa. However, this is annoying to treat directly, because of the vector charactor of $\mbox{\boldmath$E$}$ and $\mbox{\boldmath$B$}$ which complicate the description (as we have seen - transverse electric fields are related to magnetic multipoles and vice versa). Let's eliminate it.

By considering the action of the Laplacian on the scalar product of $\mbox{\boldmath$r$}$ with a well-behaved vector field $X$,

\nabla^2 (\mbox{\boldmath$r$}\cdot \mbox{\boldmath$X$}) = \...
...h$X$}) + 2
\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$X$}
\end{displaymath} (12.27)

and using the divergenceless of $\mbox{\boldmath$E$}$ and $\mbox{\boldmath$B$}$, we see that the scalars $(\mbox{\boldmath$r$}\cdot \mbox{\boldmath$E$})$ and $(\mbox{\boldmath$r$}\cdot \mbox{\boldmath$B$})$ also satisfy the HHE:
$\displaystyle (\nabla^2 + k^2) (\mbox{\boldmath$r$}\cdot \mbox{\boldmath$E$}) = 0$     (12.28)
$\displaystyle (\nabla^2 + k^2) (\mbox{\boldmath$r$}\cdot \mbox{\boldmath$B$}) = 0$     (12.29)

We already know how to write a general solution to either of these equations in terms of the spherical bessel, neumann, and hankel functions times spherical harmonics.

Recall, that when we played around with multipole fields, I kept emphasizing that electric n-pole fields were transverse magnetic and vice versa? Well, transverse electric fields have $(\mbox{\boldmath$r$}\cdot \mbox{\boldmath$E$}) = 0$ by definition, right? So now we define a magnetic multipole field of order L by

$\displaystyle \mbox{\boldmath$r$}\cdot \mbox{\boldmath$B$}_L^{(M)} = \frac{\ell(\ell+1)}{k} g_\ell (kr)
Y_L(\hat{r})$     (12.30)
$\displaystyle \mbox{\boldmath$r$}\cdot \mbox{\boldmath$E$}_L^{(M)}$ $\textstyle =$ $\displaystyle 0.$ (12.31)

Similarly, a electric multipole field of order L (which must be transverse magnetic) is any solution such that
$\displaystyle \mbox{\boldmath$r$}\cdot \mbox{\boldmath$E$}_L^{(E)} = - \frac{\ell(\ell+1)}{k} f_\ell (kr)
Y_L(\hat{r})$     (12.32)
$\displaystyle \mbox{\boldmath$r$}\cdot \mbox{\boldmath$B$}_L^{(E)}$ $\textstyle =$ $\displaystyle 0.$ (12.33)

In these two definitions, $g_\ell (kr)$ and $f_\ell(kr)$ are arbitrary linear combinations of spherical bessel functions12.1, two at a time. Jackson uses the two hankel functions in (J9.113)k, but this is not necessary.

Now, a little trickery. Using the curl equation for $\mbox{\boldmath$B$}$ we get:

k (\mbox{\boldmath$r$}\cdot \mbox{\boldmath$B$}_L^{(M)}) = ...
...ath$E$}_L^{(M)} =
{\bf L} \cdot \mbox{\boldmath$E$}_L^{(M)}
\end{displaymath} (12.34)

so that ${\bf L} \cdot \mbox{\boldmath$E$}_L^(M)$ is a scalar solution to the HHE for magnetic multipolar fields. Ditto for ${\bf L} \cdot \mbox{\boldmath$B$}_L^{(E)}$ in the case of electric multipolar fields. Thus,
{\bf L} \cdot \mbox{\boldmath$E$}_L^{(M)} = \ell(\ell + 1) g_\ell(kr) Y_L(\hat{r})
\end{displaymath} (12.35)

etc. for ${\bf L} \cdot \mbox{\boldmath$B$}_L^{(E)}$.

Now we get really clever. Remember that $\mbox{\boldmath$r$}\cdot {\bf L} = 0$. Also, $L^2 = {\bf L} \cdot {\bf L}$. We have arranged things just so that if we write:

$\displaystyle \mbox{\boldmath$E$}_L^{(M)}$ $\textstyle =$ $\displaystyle g_\ell(kr) {\bf L} Y_L(\hat{r})$ (12.36)
$\displaystyle \mbox{\boldmath$B$}_L^{(M)}$ $\textstyle =$ $\displaystyle - \frac{i}{\omega} \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$E$}_L^{(M)}$ (12.37)

we exactly reconstruct the solutions above. Neato! This gives us a completely general TE, MM EMF. A TM, EM EMF follows similarly with $g \to f$ and $\mbox{\boldmath$E$}\leftrightarrow \mbox{\boldmath$B$}$ (and a minus sign in the second equation).

This is good news and bad news. The good news is that this is a hell of a lot simpler than screwing around with symmetric and antisymmetric vector decompositions and integrations by parts ad nauseam. The radial part of the solutions is straightforward, and the angular part is written in a concise notation. The bad news is we have never seen that notation, good or bad, ever before. We have two choices. Either we can laboriously crank out the operator products and curls for each problem as we need to (which is really just as bad as what we have been doing) or we have to work out the algebra of these new objects once and for all so we can plug and chug out the most difficult of answers with comparative ease.

Guess which one we're about to do.

next up previous contents
Next: Vector Spherical Harmonics and Up: Vector Multipoles Previous: Angular momentum and spherical   Contents
Robert G. Brown 2013-01-04