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 Lorenz gauge):

(15.8) | |||

(15.9) |

with the Lorenz condition:

(15.10) |

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:

(15.11) | |||

(15.12) |

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
, 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:
&Phi#Phi;() = i k &sum#sum;_L { p_L^ext(r) J_L() +
p_L^int(r) H_L^+() }
where
p_L^ext(r) & = & &int#int;_r^&infin#infty;h_&ell#ell;^+(kr') Y_L^&ast#ast;(')
&rho#rho;(') d^3r'

p_L^int(r) & = & &int#int;_0^r j_&ell#ell;(kr') Y_L^&ast#ast;(') &rho#rho;(') d^3r'

Outside the (bounding sphere of the) source, the exterior coefficient is zero and the interior coefficient is the scalar multipole moment of the charge source distribution, so that:

(15.13) |

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
. 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
and
pieces as
above and an exact solution obtained on all space *including inside
the charge distribution*. It is:
() = i k &mu#mu;_0 &sum#sum;_L { m_L _L^+() +
n_L _L^+() + l_L _L() }
where
m_L & = & &int#int;(') ·_L^0(')^&ast#ast;d^3r'

n_L & = & &int#int;(') ·_L^0(')^&ast#ast;d^3r'

l_L & = & &int#int;(') ·_L^0(')^&ast#ast;d^3r'
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:
are not independent. They are related by the requirement that the
solutions satisfy the *Lorenz 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:
+ 1c^2 &Phi#Phi;t & = & 0

i k &sum#sum;_L { &mu#mu;_0 l_L _L^+ -
i &omega#omega;c^2 &epsi#epsilon;_0 p_L H_L^+ } & = & 0

i k &sum#sum;_L { l_L _L^+ - i k c p_L H_L^+ } & =
& 0

-k^2 &sum#sum;_L { l_L - c p_L} H_L^+ & = & 0
where we used
in the last step. If
we multiply from the left by
and use the fact that
the
form a complete orthonormal set, we find the relation:

(15.14) |

or

(15.15) |

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
directly from
:
& = &

& = & ik &mu#mu;_0 &sum#sum;_L {m_l (_L^+) + n_l (_L^+) +
l_l (_L^+) }

& = & ik&mu#mu;_0 &sum#sum;_L {m_l (-ik _L^+) + n_l (ik _L^+) }

& = & k^2&mu#mu;_0 &sum#sum;_L { m_L _L^+ - n_L _L^+ }
and use Ampere's Law,
to find
:
& = & i c^2kc

& = & i k c &mu#mu;_0 &sum#sum;_L { m_L (_L^+) - n_L (
_L^+) }

& = & i k 1&mu#mu;_0&epsi#epsilon;_0 &mu#mu;_0
&sum#sum;_L { m_L (ik _L^+) - n_L (-ik _L^+) }

& = & - k^2 &mu#mu;_0&epsi#epsilon;_0 &sum#sum;_L
{ m_L _L^+ + n_L _L^+ }

& = & - k^2 Z_0 &sum#sum;_L { m_L _L^+ + n_L _L^+ }.
where
is the usual impedance of
free space, around 377 ohms.

Wow! Recall that the **M** waves are transverse, so the
and
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:
& = & - &Phi#Phi;- **A**t

& = & - (ik&epsi#epsilon;_0 &sum#sum;_L p_L H_L^+ )
+ i &omega#omega;(ik&mu#mu;_0&sum#sum;_L { m_L _L^+() +
n_L _L^+() + l_L _L() } )

& = & -i k&epsi#epsilon;_0 &sum#sum;_L { p_L (H_L^+)
- ikc &mu#mu;_0&epsi#epsilon;_0 &sum#sum;_L l_L _L^+ } -
k^2&mu#mu;_0c &sum#sum;_L { m_l _L^+ + n_L _L^+ }

& = & k^2&epsi#epsilon;_0 &sum#sum;_L { p_L -
1c l_L }_L^+ -
k^2 Z_0 &sum#sum;_L { m_l _L^+ + n_L _L^+ }
(Note that we used
and
.)
From this we see that *if* the gauge condition:
l_L = c p_L
is satisfied, *the scalar and longitudinal vector parts of the
electric field cancel exactly!* All that survives are the transverse
parts:
= - k^2 Z_0 &sum#sum;_L { m_L _L^+ + n_L _L^+ }
as before. The Lorenz gauge condition is thus intimately connected to
the vanishing of a scalar or longitudinal contribution to the
field! Also note that the magnitude of
is greater than that of
by
, 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 form of the hankel function,

(15.16) |

we obtain the limiting forms (for ):

(15.17) |

(15.18) |

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

(15.19) |

so that (still in the far zone limit)

(15.20) |

Let us pause to admire this result before moseying on. This is just
& = & - k^2 &mu#mu;_0 e^ikrkr &sum#sum;_L (-i)^&ell#ell;+1
{ m_L (×&ell#ell;&ell#ell;m ) + n_L
&ell#ell;&ell#ell;m }

& = & - k &mu#mu;_0 e^ikrr &sum#sum;_L (-i)^&ell#ell;+1
{ m_L (×&ell#ell;&ell#ell;m ) + n_L
&ell#ell;&ell#ell;m }

& = & - k^2 Z_0 e^ikrkr &sum#sum;_L (-i)^&ell#ell;+1 {
m_L &ell#ell;&ell#ell;m - n_L (×&ell#ell;&ell#ell;m
) }

& = & - k Z_0 e^ikrr &sum#sum;_L (-i)^&ell#ell;+1 {
m_L &ell#ell;&ell#ell;m - n_L (×&ell#ell;&ell#ell;m
) }.
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)
= 12 Re(×^&ast#ast;)
and that the power distribution is related to the flux of the Poynting
vector through a surface at distance
in a differential solid angle
:
dPd&Omega#Omega; = 12 Re[r^2 ·(×
^&ast#ast;)]
we get
& = & k^22r^2 Z_0 Re [ &sum#sum;_L &sum#sum;_L'
i^&ell#ell;' - &ell#ell; { m_L &ell#ell;&ell#ell;m - n_L ( ×
&ell#ell;&ell#ell;m ) } .

& & × . { m_L'^&ast#ast;( ×
&ell#ell;' &ell#ell;'m' &ast#ast; ) + n_L'^&ast#ast;&ell#ell;' &ell#ell;'m' &ast#ast;
} ]
(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

(15.21) |

and

(15.22) |

to evaluate typical terms. Using these relations, it is not too difficult to show that

(15.23) |

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
to see how it
compares to the usual static results. Static results are obtained in
the
(long wavelength) limit. In this limit e.g.
and:
n_L &ap#approx;ic&ell#ell;+1&ell#ell; k^&ell#ell;(2&ell#ell;+1)! &int#int;
&rho#rho;() r^&ell#ell;Y_&ell#ell;,m() d^3r
The dipole term comes from
. For a simple dipole:
n_1,m & &ap#approx;& ic 23 k &int#int;&rho#rho;r Y_1,m d^3r

& &ap#approx;& i kc23 34&pi#pi; e<r>

& &ap#approx;& i kc636&pi#pi; e<r>

& &ap#approx;& - ie6&pi#pi;&omega#omega; <r>
where we use
.

In terms of this the average power radiated by a single electron dipole
is:
P = 12(e^26&pi#pi;&epsi#epsilon;_0 c^3) |r|^2
which compares well with the *Larmor Formula*:
P = 23 (e^24&pi#pi;&epsi#epsilon;_0 c^3) |r|^2
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 moments differ from the Hansen multipolar moments by factors of several powers of . 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.