Chapter 11 is radiation. We started talking about radiation in Chapter 10 when we noticed that the field of a point charge

\[\vec{E}(\vec{r},t) = \frac{q}{4\pi \epsilon_0}\frac{\mathscr{r}}{(\vec{\mathscr{r}}\cdot\vec{u})^3}\left[(c^2 - v^2)\vec{u} + \vec{\mathscr{r}}\times(\vec{u}\times\vec{a})\right]\]

has a term that drops off like \(1/\mathscr{r}\) rather than a \(1/\mathscr{r}^2\). That’s the term that can give us radiation to infinity. In chapter 10, however, we were more concerned with the ramifications of the deferred potential than we were with radiation.

In Chapter 11, by contrast, we use many of the equations from Chapter 10, but we throw away all the terms that don’t contribute to the radiative energy.

Every E&M book starts its official radiation section with dipole radiation. A dipole antenna is the simplest to make - you attach a paper clip to a signal generator and bam, you’ve got a radiating dipole.

Dipole radiation

Starting with the voltage of a simple static dipole

\[V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}\]

I made you guess what the intensity from a dipole radiator was. You humored me and got all the factors of \(\omega\) (the frequency of the dipole oscillation) and \(c\) and \(p_0\) (the amplitude of the dipole oscillation) and \(r\) the distance you are from the dipole. Here’s the actual formula:

\[<\vec{S}> = \frac{\mu_0 p_0^2 \omega^4}{32\pi^2 c} \frac{\sin^2\theta}{r^2} \hat{r}\]

The dependence on \(\theta\) is really important. You don’t get any radiation looking down on the dipole along its axis. The dependence on \(\omega^4\) is really important. Intensity is the 4th power of frequency. So if you make your signal generator go twice as fast you get 16 times as much intensity.

The approximations we made are (1) that the size of the dipole is much smaller than the distance you are away from it (\(d<<r\)).
(2) We have to assume that \(d<<\lambda\).
(3) We also have to assume that we’re farther away from the source than one wavelength of light, so \(r>>\lambda\). That’s what makes the Coulomb terms (the non-radiative terms) unimportant. This last requirement is called the “radiation zone” and all it means is that the radiation term is more important that the coulomb term.

The fields look like this:

\[\vec{E} = \frac{\mu_0 p_0 \omega^2}{4\pi} \frac{\sin\theta}{r} \cos[\omega(t - r/c)]\hat{\theta}\] \[\vec{B} = \nabla \times \vec{A} = \frac{\mu_0 p_0 \omega^2}{4\pi c} \frac{\sin\theta}{r} \cos[\omega(t - r/c)]\hat{\phi}\]

If you integrate intensity over the sphere you get the time-averaged total power radiated:

\[<P> = \frac{\mu_0 p_0^2 \omega^4}{12 \pi^2 c}\]

For magnetic dipoles it’s remarkably similar:

\[<P> = \frac{\mu_0 m_0^2 \omega^4}{12 \pi^2 c^3}\]

where \(m_0\) is the amplitude of the magnetic dipole moment oscillating at \(\omega\) rather than the amplitude of the electric dipole oscillation oscillating at \(\omega\).

The important thing about the polarization of the waves is that no matter where you’re looking, the polarization (the e-field) will be in line with the projection of the axis of the dipole. And the amplitude of the radiation is proportional to the projection of that dipole onto a plane that’s perpendicular to the propagation direction.

Radiation from a point charge

Radiation from a point charge is less prevalent in common applications, but very common in astrophysics. Also, if you know the radiation from a point charge, you can integrate any distribution of charges to get the radiation.

Here’s the total power:

\[P = \frac{\mu_0 q^2 a^2}{6\pi c}\]

where \(a\) is the magnitude of the acceleration of the charge. The angular structure of the radiation looks again like a donut, with no radiation along the direction of the acceleration. (See Griffiths figure on page 484, Figure 11.11)

Here’s what the formula for intensity as a function of angle looks like:

\[\vec{S}_{rad} = \frac{1}{\mu_0 c} \left(\frac{\mu_0 q}{4\pi\mathscr{r}} \right)^2\left[a^2 - (\hat{\mathscr{r}}\cdot\vec{a})^2\right]\hat{\mathscr{r}} = \frac{\mu_0q^2 a^2}{16\pi^2 c}\left(\frac{\sin^2\theta}{\mathscr{r}^2}\right)\hat{\mathscr{r}}\]

where \(\theta\) is the angle between \(\hat{\mathscr{r}}\) and \(\vec{a}\) (so it’s the angle between your line of site to the particle and the acceleration of the particle). So there’s the donut (punchline #2). If the particle is coming toward you, you’ll see nothing, because that angle is zero.

If you don’t assume velocity is zero, then you get a more interesting formula for power.

\[P = \frac{\mu_0 q^2 \gamma^6}{6\pi c}\left(a^2 - \Bigg|\frac{\vec{v}\times\vec{a}}{c}\Bigg|^2\right)\]

where \(\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}\)

Notice that the factor of gamma means that the power increases enormously as the velocity approaches the speed of light.