Next: Connection to Old (Approximate)
Up: The Hansen Multipoles
Previous: Multipolar Radiation, revisited
Contents
Suppose we are given a centerfed dipole antenna with length
(halfwave antenna). We will assume further that the antenna is aligned
with the z axis and centered on the origin, with a current given by:

(13.58) 
Note that in ``real life'' it is not easy to arrange for a given current
because the current instantaneously depends on the ``resistance'' which is a
function of the radiation field itself. The current itself thus comes out of
the solution of an extremely complicated boundary value problem.
For atomic or nuclear radiation, however, the ``currents'' are generally
matrix elements associated with transitions and hence are known.
In any event, the current density corresponding to this current is

(13.59) 
for
and

(13.60) 
for .
When we use the Hansen multipoles, there is little incentive to convert
this into a form where we integrate against the charge density in the
antenna. Instead we can easily and directly calculate the multipole
moments. The magnetic moment is
(where we have done the integral over ). Now,

(13.62) 
(Why? Consider
...) and yet
Consequently, we can conclude (
) that

(13.65) 
All magnetic multipole moments of this linear dipole vanish. Since the
magnetic multipoles should be connected to the rotational part of the current
density (which is zero for linear flow) this should not surprise you.
The electric moments are
If we look up the definition of the v.s.h.'s on the handout table, the z
components are given by:



(13.67) 



(13.68) 



(13.69) 



(13.70) 
so the electric multipole moments vanish for , and

(13.71) 
Examining this equation, we see that all the even terms vanish!
However, all the odd , terms do not vanish, so we can't
quit yet. We use the following relations:

(13.72) 
(the fundamental recursion relation),

(13.73) 
(true fact) and

(13.74) 
for any two spherical bessel type functions (a valuable thing to know that
follows from integration by parts and the recursion relation). From these we
get

(13.75) 
Naturally, there is a wee tad of algebra involved here that I have skipped.
You shouldn't. Now, let's figure out the power radiated from this source.
Recall from above that:
Now this also equals (recall)
, from which we
can find the radiation resistance of the half wave antenna:

(13.77) 
We are blessed by this having manifest units of resistance, as we
recognize our old friend
(the impedance of free space) and a bunch of dimensionless
numbers! In terms of this:

(13.78) 
We can obtain a good estimate of the magnitude by evaluating the first few
terms. Noting that
and doing some arithmetic, you should be able to show that
.
Note that the ratio of the first (dipole) term to the third (octupole) term is
That means that this is likely to be a good approximation (the
answer is very nearly unchanged by the inclusion of the extra term).
Even if the length of the antenna is on the order of , the
multipole expansion is an extremely accurate and rapidly converging
approximation. That is, after all, why we use it so much in all kinds
of localized source wave theory.
However, if we plug in the ``long wavelength'' approximation we previously
obtained for a short dipole antenna (with ) we get:

(13.81) 
which is off by close to a factor of 50%. This is not such a good
result. Using this formula with a long wavelength approximation for the
dipole moment (only) of

(13.82) 
yields
, still off by 11%.
Next: Connection to Old (Approximate)
Up: The Hansen Multipoles
Previous: Multipolar Radiation, revisited
Contents
Robert G. Brown
20150823