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Next: A Linear Center-Fed Half-Wave Up: The Hansen Multipoles Previous: Green's Functions for the   Contents

Multipolar Radiation, revisited

We will now, at long last, study the complete radiation field including the scalar, longitudinal, and transverse parts. Recall that we wish to solve the two equations (in the Lorentz gauge):

$\displaystyle \big\{\nabla^2 + k^2 \big\} \Phi(\mbox{\boldmath$x$})$ $\textstyle =$ $\displaystyle -
\frac{\rho}{\epsilon_0}
({\bf r})$ (13.20)
$\displaystyle \big\{\nabla^2 + k^2 \big\} \mbox{\boldmath$A$}(\mbox{\boldmath$x$})$ $\textstyle =$ $\displaystyle - \mu_0 \mbox{\boldmath$J$}(\mbox{\boldmath$x$})$ (13.21)

with the Lorentz condition:
\begin{displaymath}
\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$A$}+ \frac{1}{c^2} \frac{\partial \Phi}{\partial t} = 0
\end{displaymath} (13.22)

which is connected (as we shall see) to the continuity equation for charge and current.

E and B are now (as usual) determined from the vector potential by the full relations, i. e. - we make no assumption that we are outside the region of sources:

$\displaystyle \mbox{\boldmath$E$}$ $\textstyle =$ $\displaystyle - \mbox{\boldmath$\nabla$}\Phi - \frac{\partial \mbox{\boldmath$A$}}{\partial t}$ (13.23)
$\displaystyle \mbox{\boldmath$B$}$ $\textstyle =$ $\displaystyle \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$A$},$ (13.24)

Using the methods discussed before (writing the solution as an integral equation, breaking the integral up into the interior and exterior of the sphere of radius $r$, and using the correct order of the multipolar expansion of the Green's function in the interior and exterior regions) we can easily show that the general solution to the IHE's above is:

\begin{displaymath}
\Phi(\mbox{\boldmath$r$}) = i k \sum_L \left\{ p_L^{\rm ext...
...$}) +
p_L^{\rm int}(r) H_L^+(\mbox{\boldmath$r$}) \right\}
\end{displaymath} (13.25)

where
$\displaystyle p_L^{\rm ext}(r)$ $\textstyle =$ $\displaystyle \int_r^\infty h_\ell^+(kr') Y_L^\ast(\hat{\mbox{\boldmath$r$}}')
\rho(\mbox{\boldmath$r$}') d^3r'$ (13.26)
$\displaystyle p_L^{\rm int}(r)$ $\textstyle =$ $\displaystyle \int_0^r j_\ell(kr') Y_L^\ast(\hat{\mbox{\boldmath$r$}}') \rho(\mbox{\boldmath$r$}') d^3r'$ (13.27)

Outside the (bounding sphere of the) source, the exterior coefficient is zero and the interior coefficient is the scalar multipole moment $p_L =
p_L^{\rm int}(\infty)$ of the charge source distribution, so that:

\begin{displaymath}
\Phi(\mbox{\boldmath$r$}) = \frac{ik}{\epsilon_0} \sum_L p_L H_L^+(\mbox{\boldmath$r$})
\end{displaymath} (13.28)

This is an important relation and will play an significant role in the implementation of the gauge condition below.

Similarly we can write the interior and exterior multipolar moments of the current in terms of integrals over the various Hansen functions to obtain a completely general expression for the vector potential $\mbox{\boldmath$A$}(\mbox{\boldmath$r$})$. To simplify matters, I am going to only write down the solution obtained outside the current density distribution, although the integration volume can easily be split into $r_<$ and $r_>$ pieces as above and an exact solution obtained on all space including inside the charge distribution. It is:

\begin{displaymath}
\mbox{\boldmath$A$}(\mbox{\boldmath$r$}) = i k \mu_0 \sum_L...
...$}) + l_L \mbox{\boldmath$L$}_L(\mbox{\boldmath$r$}) \right\}
\end{displaymath} (13.29)

where
$\displaystyle m_L$ $\textstyle =$ $\displaystyle \int \mbox{\boldmath$J$}(\mbox{\boldmath$r$}') \cdot \mbox{\boldmath$M$}_L^0(\mbox{\boldmath$r$}')^\ast d^3r'$ (13.30)
$\displaystyle n_L$ $\textstyle =$ $\displaystyle \int \mbox{\boldmath$J$}(\mbox{\boldmath$r$}') \cdot \mbox{\boldmath$N$}_L^0(\mbox{\boldmath$r$}')^\ast d^3r'$ (13.31)
$\displaystyle l_L$ $\textstyle =$ $\displaystyle \int \mbox{\boldmath$J$}(\mbox{\boldmath$r$}') \cdot \mbox{\boldmath$L$}_L^0(\mbox{\boldmath$r$}')^\ast d^3r'$ (13.32)

Note well that the action of the dot product within the dyadic form for the Green's function (expanded in Hansen solutions) reduces the dyadic tensor to a vector again.

It turns out that these four sets of numbers: $p_L, m_L, n_L, l_L$ are not independent. They are related by the requirement that the solutions satisfy the Lorentz gauge condition, which is a constraint on the admissible solutions. If we substitute these forms into the gauge condition itself and use the differential relations given above for the Hansen functions to simplify the results, we obtain:

$\displaystyle \mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$A$}+ \frac{1}{c^2} \frac{\partial \Phi}{\partial t}$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle i k \sum_L \left\{ \mu_0 l_L \mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$L$}_L^+ -
\frac{i \omega }{c^2 \epsilon_0} p_L H_L^+ \right\}$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle i k \sum_L \left\{ l_L \mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$L$}_L^+ - i k c p_L H_L^+ \right\}$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle -k^2 \sum_L \left\{ l_L - c p_L\right\} H_L^+$ $\textstyle =$ $\displaystyle 0$ (13.33)

where we used $\mbox{\boldmath$\nabla$}\cdot {\bf L}_L^+ = ik H_L^+$ in the last step. If we multiply from the left by $Y_{\ell',m'}^\ast$ and use the fact that the $Y_L$ form a complete orthonormal set, we find the relation:
\begin{displaymath}
l_L - cp_L = 0
\end{displaymath} (13.34)

or
\begin{displaymath}
l_L = cp_L
\end{displaymath} (13.35)

This tells us that the effect of the scalar moments and the longitudinal moments are connected by the gauge condition. Instead of four relevant moments we have at most three. In fact, as we will see below, we have only two!

Recall that the potentials are not unique - they can and do vary according to the gauge chosen. The fields, however, must be unique or we'd get different experimental results in different gauges. This would obviously be a problem!

Let us therefore calculate the fields. There are two ways to proceed. We can compute $\mbox{\boldmath$B$}$ directly from $vA$:

$\displaystyle \mbox{\boldmath$B$}$ $\textstyle =$ $\displaystyle \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$A$}$  
  $\textstyle =$ $\displaystyle ik \mu_0 \sum_L \left\{m_l (\mbox{\boldmath$\nabla$}\times \mbox{...
...}_L^+) +
l_l (\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$L$}_L^+) \right\}$  
  $\textstyle =$ $\displaystyle ik\mu_0 \sum_L \left\{m_l (-ik \mbox{\boldmath$N$}_L^+) + n_l (ik \mbox{\boldmath$M$}_L^+) \right\}$  
  $\textstyle =$ $\displaystyle k^2\mu_0 \sum_L \left\{ m_L \mbox{\boldmath$N$}_L^+ - n_L \mbox{\boldmath$M$}_L^+ \right\}$ (13.36)

and use Ampere's Law, $\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$B$}= \mu_0\epsilon_0\frac{\partial \mbox{\boldmath$E$}}{\partial t} =
-i \omega \mu_0 \epsilon_0 E$ to find $\mbox{\boldmath$E$}$:
$\displaystyle \mbox{\boldmath$E$}$ $\textstyle =$ $\displaystyle \frac{i c^2}{kc} \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$B$}$  
  $\textstyle =$ $\displaystyle i k c \mu_0 \sum_L \left\{ m_L (\mbox{\boldmath$\nabla$}\times \m...
...}_L^+) - n_L (\mbox{\boldmath$\nabla$}\times
\mbox{\boldmath$M$}_L^+) \right\}$  
  $\textstyle =$ $\displaystyle i k \sqrt{\frac{1}{\mu_0\epsilon_0}} \mu_0
\sum_L \left\{ m_L (ik \mbox{\boldmath$M$}_L^+) - n_L (-ik \mbox{\boldmath$N$}_L^+) \right\}$  
  $\textstyle =$ $\displaystyle - k^2 \sqrt{\frac{\mu_0}{\epsilon_0}} \sum_L
\left\{ m_L \mbox{\boldmath$M$}_L^+ + n_L \mbox{\boldmath$N$}_L^+ \right\}$  
  $\textstyle =$ $\displaystyle - k^2 Z_0 \sum_L \left\{ m_L \mbox{\boldmath$M$}_L^+ + n_L \mbox{\boldmath$N$}_L^+ \right\}.$ (13.37)

where $Z_0 = \sqrt{\frac{\mu_0 }{\epsilon_0 }}$ is the usual impedance of free space, around 377 ohms.

Wow! Recall that the M waves are transverse, so the $m_L$ and $n_L$ are the magnetic (transverse electric) and electric (transverse magnetic) multipole moments respectively. The field outside of the source is a pure expansion in elementary transverse multipoles. (Later we will show that the (approximate) definitions we have used to date as "multipoles" are the limiting forms of these exact definitions.)

Note well that the actual fields require only two of the basic hansen solutions - the two that are mutually transverse. Something happened to the longitudinal part and the dependence of the field on the scalar potential. To see just what, let us re-evaluate the electric field from:

$\displaystyle \mbox{\boldmath$E$}$ $\textstyle =$ $\displaystyle - \mbox{\boldmath$\nabla$}\Phi - \frac{\partial \bf A}{\partial t}$  
  $\textstyle =$ $\displaystyle - \mbox{\boldmath$\nabla$}\left(\frac{ik}{\epsilon_0} \sum_L p_L ...
...boldmath$r$}) + l_L \mbox{\boldmath$L$}_L(\mbox{\boldmath$r$}) \right\} \right)$  
  $\textstyle =$ $\displaystyle -\frac{i k}{\epsilon_0} \sum_L \left\{ p_L (\mbox{\boldmath$\nabl...
...um_L \left\{ m_l \mbox{\boldmath$M$}_L^+ + n_L \mbox{\boldmath$N$}_L^+ \right\}$  
  $\textstyle =$ $\displaystyle \frac{k^2}{\epsilon_0} \sum_L \left\{ p_L -
\frac{1}{c} l_L \ri...
...um_L \left\{ m_l \mbox{\boldmath$M$}_L^+ + n_L \mbox{\boldmath$N$}_L^+ \right\}$ (13.38)

(Note that we used $\omega = kc$ and $\mbox{\boldmath$\nabla$}H_L^+ = ik {\bf L}_L^+$.) From this we see that if the gauge condition:
\begin{displaymath}
l_L = cp_L
\end{displaymath} (13.39)

is satisfied, the scalar and longitudinal vector parts of the electric field cancel exactly! All that survives are the transverse parts:
\begin{displaymath}
\mbox{\boldmath$E$}= - k^2 Z_0 \sum_L \left\{ m_L \mbox{\boldmath$M$}_L^+ + n_L \mbox{\boldmath$N$}_L^+ \right\}
\end{displaymath} (13.40)

as before. The Lorentz gauge condition is thus intimately connected to the vanishing of a scalar or longitudinal contribution to the $\mbox{\boldmath$E$}$ field! Also note that the magnitude of $\mbox{\boldmath$E$}$ is greater than that of $\mbox{\boldmath$B$}$ by $c$, the velocity of light.

Now, we are interested (as usual) mostly in obtaining the fields in the far zone, where this already simple expression attains a clean asymptotic form. Using the $kr \to \infty$ form of the hankel function,

\begin{displaymath}
\lim_{ kr \rightarrow \infty} h_\ell^+(kr) = \frac{e^{ikr -
(\ell+1)\frac{i\pi}{2}}}{kr}
\end{displaymath} (13.41)

we obtain the limiting forms (for $kr \to \infty$):
\begin{displaymath}
\mbox{\boldmath$M$}_L^+ \sim \frac{e^{ikr - (\ell+1)\frac{i...
...ll+1} \frac{e^{ikr}}{kr} \mbox{\boldmath$Y$}_{\ell \ell}^{m}
\end{displaymath} (13.42)


\begin{displaymath}
\mbox{\boldmath$N$}_L^+ \sim \frac{e^{ikr - \ell \frac{i\pi...
...ll}{2\ell + 1}} \mbox{\boldmath$Y$}_{\ell,\ell+1}^{m} \right]
\end{displaymath} (13.43)

The bracket in the second equation can be simplified, using the results of the table I handed out previously. Note that

\begin{displaymath}
\left[ \sqrt{\frac{\ell + 1}{2\ell+1}} \mbox{\boldmath$Y$}_...
...box{\boldmath$r$}}\times \mbox{\boldmath$Y$}_{\ell \ell}^{m})
\end{displaymath} (13.44)

so that (still in the far zone limit)
\begin{displaymath}
\mbox{\boldmath$N$}_L^+ \sim - \frac{e^{ikr - (\ell+1)\frac...
...\boldmath$r$}}
\times \mbox{\boldmath$Y$}_{\ell \ell}^{m}) .
\end{displaymath} (13.45)

Let us pause to admire this result before moseying on. This is just

$\displaystyle \mbox{\boldmath$B$}$ $\textstyle =$ $\displaystyle - k^2 \mu_0 \frac{e^{ikr}}{kr} \sum_L (-i)^{\ell+1}
\left\{ m_L ...
...Y$}_{\ell \ell}^{m} \right) + n_L
\mbox{\boldmath$Y$}_{\ell \ell}^{m} \right\}$  
  $\textstyle =$ $\displaystyle - k \mu_0 \frac{e^{ikr}}{r} \sum_L (-i)^{\ell+1}
\left\{ m_L \le...
...Y$}_{\ell \ell}^{m} \right) + n_L
\mbox{\boldmath$Y$}_{\ell \ell}^{m} \right\}$ (13.46)
$\displaystyle \mbox{\boldmath$E$}$ $\textstyle =$ $\displaystyle - k^2 Z_0 \frac{e^{ikr}}{kr} \sum_L (-i)^{\ell+1} \left\{
m_L \m...
...mbox{\boldmath$r$}}\times \mbox{\boldmath$Y$}_{\ell \ell}^{m}
\right) \right\}$  
  $\textstyle =$ $\displaystyle - k Z_0 \frac{e^{ikr}}{r} \sum_L (-i)^{\ell+1} \left\{
m_L \mbox...
...box{\boldmath$r$}}\times \mbox{\boldmath$Y$}_{\ell \ell}^{m}
\right) \right\}.$ (13.47)

If I have made a small error at this point, forgive me. Correct me, too. This is a purely transverse outgoing spherical wave whose vector character is finally translucent, if not transparent.

The power flux in the outgoing wave is still not too easy to express, but it is a damn sight easier than it was before. At least we have the satisfaction of knowing that we can express it as a general result. Recalling (as usual)

\begin{displaymath}
\mbox{\boldmath$S$}= \frac{1}{2} \mbox{\rm Re}(\mbox{\boldmath$E$}\times \mbox{\boldmath$H$}^\ast)
\end{displaymath} (13.48)

and that the power distribution is related to the flux of the Poynting vector through a surface at distance $r$ in a differential solid angle $d\Omega$:
\begin{displaymath}
\frac{dP}{d\Omega} = \frac{1}{2} \mbox{\rm Re}[r^2 \hat{\mb...
...}\cdot (\mbox{\boldmath$E$}\times
\mbox{\boldmath$H$}^\ast )]
\end{displaymath} (13.49)

we get
$\displaystyle \mbox{\boldmath$S$}$ $\textstyle =$ $\displaystyle \frac{k^2}{2r^2} Z_0 \mbox{\rm Re} \left[ \sum_L \sum_{L'}
i^{\e...
...r$}}\times
\mbox{\boldmath$Y$}_{\ell \ell}^{m} \right) \right\} \hfill \right.$  
    $\displaystyle \quad \quad \times \quad \left. \left\{ m_{L'}^\ast \left( \hat{\...
...ht) + n_{L'}^\ast \mbox{\boldmath$Y$}_{\ell' \ell'}^{m' \ast}
\right\} \right]$ (13.50)

(Note: Units here need to be rechecked, but they appear to be consistent at first glance).

This is an extremely complicated result, but it has to be, since it expresses the most general possible angular distribution of radiation (in the far zone). The power distribution follows trivially. We can, however, evaluate the total power radiated, which is a very useful number. This will be an exercise. You will need the results

$\displaystyle \int d^2\Omega \hat{r} \cdot \mbox{\boldmath$Y$}_{\ell \ell}^{m} \times \bigg(\hat{r}
\times \mbox{\boldmath$Y$}_{\ell' \ell'}^{m' \ast} \bigg)$ $\textstyle =$ $\displaystyle \int d^2\Omega \mbox{\boldmath$Y$}_{\ell
\ell}^{m} \cdot \mbox{\boldmath$Y$}_{\ell' \ell'}^{m' \ast}$  
  $\textstyle =$ $\displaystyle \delta_{\ell \ell'} \delta_{m m'}$ (13.51)

and
\begin{displaymath}
\int d^2\Omega \hat{r} \cdot \bigg( \mbox{\boldmath$Y$}_{\e...
...times \mbox{\boldmath$Y$}_{\ell'
\ell'}^{m' \ast} \bigg) = 0
\end{displaymath} (13.52)

to evaluate typical terms. Using these relations, it is not too difficult to show that
\begin{displaymath}
P = \frac{k^2}{2}Z_0 \sum_L \left\{ \mid m_L \mid^2 + \mid n_L \mid^2
\right\}
\end{displaymath} (13.53)

which is the sum of the power emitted from all the individual multipoles (there is no interference between multipoles!).

Let us examine e.g. the electric multipolar moment $n_L$ to see how it compares to the usual static results. Static results are obtained in the $k \to 0$ (long wavelength) limit. In this limit e.g. $j_\ell(kr)
\sim k^\ell r^\ell$ and:

\begin{displaymath}
n_L \approx ic\sqrt{\frac{\ell+1}{\ell}} \frac{k^\ell}{(2\e...
...ldmath$r$}) r^\ell Y_{\ell,m}(\hat{\mbox{\boldmath$r$}}) d^3r
\end{displaymath} (13.54)

The dipole term comes from $\ell = 1$. For a simple dipole:
$\displaystyle n_{1,m}$ $\textstyle \approx$ $\displaystyle ic \frac{\sqrt{2}}{3} k \int \rho r Y_{1,m} d^3r$  
  $\textstyle \approx$ $\displaystyle i \frac{kc\sqrt{2}}{3} \sqrt{\frac{3}{4\pi}} e<r>$  
  $\textstyle \approx$ $\displaystyle i \frac{kc\sqrt{6}}{36\pi} e<r>$  
  $\textstyle \approx$ $\displaystyle - \frac{ie}{\sqrt{6\pi}\omega} <\ddot{r}>$ (13.55)

where we use $<\ddot{r}> = -\omega^2<r>$.

In terms of this the average power radiated by a single electron dipole is:

\begin{displaymath}
P = \frac{1}{2}\left(\frac{e^2}{6\pi\epsilon_0 c^3}\right) \vert\ddot{r}\vert^2
\end{displaymath} (13.56)

which compares well with the Larmor Formula:
\begin{displaymath}
P = \frac{2}{3} \left(\frac{e^2}{4\pi\epsilon_0 c^3}\right) \vert\ddot{r}\vert^2
\end{displaymath} (13.57)

The latter is the formula for the instantaneous power radiated from a point charge as it is accelerated. Either flavor is the death knell of classical mechanics - it is very difficult to build a model for a stable atom based on classical trajectories of an electron around a nucleus that does not involve acceleration of the electron in question.

While it is not easy to see, the results above are essentially those obtained in Jackson (J9.155) except that (comparing e.g. J9.119, J9.122, and J91.165 to related results above) Jackson's $a_{E,M}(\ell,m)$ moments differ from the Hansen multipolar moments by factors of several powers of $k$. If one works hard enough, though, one can show that the results are identical, and even though Jackson's algebra is more than a bit Evil it is worthwhile to do this if only to validate the results above (where recall there has been a unit conversion and hence they do need validation).

Another useful exercise is to recover our old friends, the dipole and quadrupole radiation terms of J9 from the exact definition of their respective moments. One must make the long wavelength approximation under the integral in the definition of the multipole moments, integrate by parts liberally, and use the continuity equation. This is quite difficult, as it turns out, unless you have seen it before, so let us look at an example. Let us apply the methods we have developed above to obtain the radiation pattern of a dipole antenna, this time without assuming that it's length is small w.r.t. a wavelength. Jackson solves more or less the same problem in his section 9.12, so this will permit the direct comparison of the coefficients and constants in the final expressions for total radiated power or the angular distribution of power.


next up previous contents
Next: A Linear Center-Fed Half-Wave Up: The Hansen Multipoles Previous: Green's Functions for the   Contents
Robert G. Brown 2014-08-19