Dipole Radiation

Let us evaluate this term. It is (c. f. J9.13):

$$\vA(\vr) = \frac{\mu_0 e^{ikr}}{4\pi r} \int_0^r \vJ(\vr') d^3r'$$ (13.45)

(note: $Y_{00}(\hat{r}) = Y_{00}(\hat{r})^\ast = 1/\sqrt{4 \pi}$ ). If we integrate this term by parts (a surprisingly difficult chore that will be an exercise) and use the continuity equation and the fact that the source is harmonic we get:

$$\vA(\vr) = -\frac{i \mu_0 \omega}{4 \pi} \vp \frac{e^{ikr}}{r}$$ (13.46)

where

$$\vp = \int \vr' \rho(\vr') d^3r'$$ (13.47)

is the electric dipole moment (see J4.8). Note that if we define $\rho(\vr)$ to be a ``probability density'' for the electrons during a transition this expression is still valid.

This is wonderfully simple. If only we could quit with the vector potential. Alas, no. We must reconstruct the electromagnetic field being radiated away from the source from the expressions previously given

$${\bf B} = \del \times \vA$$

and

$${\bf E} = \frac{ic}{k} \del \times {\bf B}.$$

After a tremendous amount of straightforward but nonetheless difficult algebra that you will do and hand in next week (see problems) you will obtain:

$${\bf H} = \frac{ck^2}{4\pi}({\bf n} \times \vp) \frac{e^{ikr}}{r} \left(1 - \frac{1}{ikr} \right )$$ (13.48)

and E = 14&pi#pi;&epsi#epsilon;_0 {k^2(n × ) ×n e^ikrr + [3n(n ·) - ] ( 1r^3 - ikr^2 ) e^ikr} The magnetic field is always transverse to the radial vector. Electric dipole radiation is therefore also called transverse magnetic radiation. The electric field is transverse in the far zone, but in the near zone it will have a component (in the $\vp$ direction) that is not generally perpendicular to n.