Radiation Damping of an Oscillating Charge

The most important application of the Abraham-Lorentz force law is the radiation reaction of bound electrons in atoms as they radiate. This is the problem originally studied by Lorentz, in the context of a classical oscillator, and yes, we are returning to our discussion of dispersion but now with a physical model for why we expect there to be a damping term instead of a strictly phenomenological one.

To simplify life, we consider a Lorentz ``atom'' to be an electron on a spring with constant $k = m \omega_0^2$; a one-dimensional classical oscillator with a resonant frequency $\omega_0$. If the oscillator is displaced from equilibrium, it radiates energy away and is simultaneously damped. This is a classical analogue of the emission of a photon by a quantum atom, which is accompanied by the atom entering a lower energy level.

The equation of motion for the electron is (from the AL force law above, integrated as described for offset times):

$$\ddot{x}(t) + \omega_0^2 \int_0^\infty e^{-s} x(t + \tau s) ds = 0$$ (19.28)

where we have used Hooke's law. If we try the usual song and dance (assume that $x(t) = x_0 e^{-\alpha t}$ we get the characteristic equation

$$x_0 e^{-\alpha t} \left( \alpha^2 + \omega_0^2 \int_0^\infty e^{-(1 + \alpha \tau) s} ds \right) = 0 .$$ (19.29)

In order for the integral to exist, it must damp at infinity, so Re $(1 + \alpha \tau) > 0$. In that case, we get:

$$\alpha^2 + \frac{\omega_0^2}{-(1 + \alpha \tau)} \int_0^\infty e^{-x}dx = 0$$

$$\alpha^2 + \frac{\omega_0^2}{(1 + \alpha \tau)} = 0$$

$$\alpha^2(1 + \alpha \tau) + \omega_0^2 = 0$$

$$\tau \alpha^3 + \alpha^2 + \omega_0^2 = 0$$

$$(\tau \alpha)^3 + (\tau \alpha)^2 + (\omega_0\tau)^2 = 0$$

$$z^3 + z^2 + \omega_0^2\tau^2 = 0$$ (19.30)

where we've defined $z = \alpha\tau$.

This is the same cubic that would arise directly from the original AL equation of motion but the restriction on the integral eliminates the ``runaway'' solutions ( $\alpha = -(1 + \omega_0^2 \tau^2)/\tau$) at the expense of introducing preaccelerated ones. There is no point in giving the physical roots in closed form here, but you should feel free to crank up e.g. Mathematica and take a look.

If $\omega_0 \tau « 1$ (which is the physically relevant range), then the first order result is

$$\alpha = \frac{\Gamma}{2} \pm i(\omega_0 + \Delta \omega)$$ (19.31)

whith

$$\Gamma = \omega_0^2 \tau$$ (19.32)

and

$$\Delta \omega = - \frac{5}{8} \omega_0^3 \tau^2 .$$ (19.33)

The constant $\Gamma$ is the decay constant and the $\Delta \omega$ is the level shift. Note that the radiative force both introduces damping and shifts the frequency, just like it does for a classical damped oscillator. If we evaluate the electric field radiated by such an oscillator (neglecting the transient signal at the beginning) we find that the energy radiated as a function of frequency is

$$\frac{dI(\omega)}{d\omega} = I_0 \frac{\Gamma}{2 \pi} \frac{1}{(\omega - \omega_0 - \Delta \omega)^2 + (\Gamma/2)^2}$$ (19.34)

which is the characteristic spectrum of a broadened, shifted resonant line.

Figure 19.2: A typical broadened and shifted resonant line due to radiation reaction.

This concludes our discussion of the consequences of radiation reaction. You will note that the derivations we have seen are not particularly satisfying or consistent. Now we will examine the ``best'' of the derivations (Dirac's and Wheeler and Feynman's) and try to make some sense of it all.

The following sections are alas still incomplete but will be added shortly.