Resonant particles

Since there is a $1 / \alpha^2$ factor in the integrand of the above integral, the important contribution to the integral must come from the vicinity of $\alpha = 0$ (i.e. resonant particles). Therefore we expand $g (\alpha)$ as

$$g (\alpha) = g (0) + g' (0) \alpha + g'' (0) \frac{\alpha^2}{2} + \ldots$$ (37)

Since $\sin (\alpha t) / \alpha^2$ is odd in $\alpha$, only terms that are also odd need to be retained in the above expansion. Using these, expression (36) is written as

$$- \frac{1}{\vert k\vert} \frac{(q E)^2}{2 m} \int_{- \infty}^{\infty} \frac{\omega}{\alpha^2} \sin (\alpha t) g (\alpha) d \alpha \approx - \frac{1}{\vert k\vert} \frac{(q E)^2}{2 m} \int_{- \infty}^{\infty} \frac{\omega}{\alpha^2} \sin (\alpha t) (g' (0) \alpha) d \alpha$$

$$= - \frac{1}{\vert k\vert} \frac{(q E)^2}{2 m} g' (0) \omega \int_{- \infty}^{\infty} \frac{1}{\alpha} \sin (\alpha t) d \alpha$$

$$= - \frac{1}{\vert k\vert} \frac{(q E)^2}{2 m} g' (0) \omega \pi$$

$$= - \frac{\pi \omega}{\vert k\vert k} \frac{(q E)^2}{2 m} \left[ \frac{d f (v_0)}{d v_0} \right]_{v_0 = \omega / k} .$$ (38)

Using these results, Eq. (35) is written as

$$\left\langle \left\langle \frac{d}{d t} \left( \frac{1}{2} m v^2 ... ...\frac{(q E)^2}{2 m} \left[ \frac{d f (v_0)}{d v_0} \right]_{v_0 = \omega / k} .$$ (39)

which agrees with Eq. (16) in Chapter 8 of Stix's book[4]. Equation (39) indicates that the time rate of change of the averaged kinetic energy of resonant particles is proportional to the derivative of the initial distribution function at the phase velocity of the wave.