Lower order approximation

In Eq. (23), the time change rate of the kinetic energy of a particle (the absorbed power by the particle) is approximated by

$$\frac{d}{d t} \left( \frac{1}{2} m v^2 \right) = v^{(1)} m \frac{d v^{(2)}}{d t},$$ (47)

which uses a high order approximation of the velocity $v^{(2)}$. Next, we consider lower order approximations of $d (m v^2 / 2) d t$ and check whether Landau damping can be recovered in these lower order approximations. If we approximate $d (m v^2 / 2) d t$ as

$$\frac{d}{d t} \left( \frac{1}{2} m v^2 \right) \approx v^{(0)} m \frac{d v^{(1)}}{d t}$$

$$= v_0 q E \cos (k z_0 + k v_0 t - \omega t),$$ (48)

then it is obvious that $d (m v^2 / 2) d t$ will reduce to zero when it is averaged over initial position $z_0$ in one wavelength. Therefore, Landau damping is missed in this approximation. If we approximate $d (m v^2 / 2) d t$ as

$$\frac{d}{d t} \left( \frac{1}{2} m v^2 \right) \approx v^{(1)} m \frac{d v^{(1)}}{d t}$$

$$= \left[ v_0 + \frac{q E}{m} \frac{\sin (k z_0 + \alpha t) - \sin (k z_0)}{\alpha} \right] q E \cos (k z_0 + k v_0 t - \omega t),$$ (49)

then, according to the derivation given in the above section, we have

$$\left\langle \frac{d}{d t} \left( \frac{1}{2} m v^2 \right) \right\rangle_{z_0} = \frac{(q E)^2}{2 m} \frac{1}{\alpha} \sin (\alpha t)$$ (50)

and

$$\left\langle \left\langle \frac{d}{d t} \left( \frac{1}{2} m v^2 \right) \right\rangle_{z_0} \right\rangle_{v_0} = \frac{1}{\vert k\vert} \frac{(q E)^2}{2 m} \int_{- \infty}^{\infty} \frac{1}{\alpha} \sin (\alpha t) g (\alpha) d \alpha .$$ (51)

The right-hand side of Eq. (51) do not approach zero for large $t$ (I have verified this for the case that $g (\alpha) = 1 / (1 + \alpha^2)$). Since there is a $1 / \alpha$ factor in the integrand of the above integral, the important contribution to the integral must come from the vicinity of $\alpha = 0$. Therefore we expand $g (\alpha)$ as

$$g (\alpha) = g (0) + g' (0) \alpha + g'' (0) \frac{\alpha^2}{2} + \ldots$$ (52)

Since $\sin (\alpha t) / \alpha$ is even in $\alpha$, only terms that are also even need to be retained in the above expansion. Using these, Eq. (51) is written as

$$\left\langle \left\langle \frac{d}{d t} \left( \frac{1}{2} m v^2 \right) \right\rangle_{z_0} \right\rangle_{v_0} = \frac{1}{\vert k\vert} \frac{(q E)^2}{2 m} \int_{- \infty}^{\infty} \frac{1}{\alpha} \sin (\alpha t) g (\alpha) d \alpha$$

$$\approx \frac{1}{\vert k\vert} \frac{(q E)^2}{2 m} \int_{- \infty}^{\infty} \frac{1}{\alpha} \sin (\alpha t) g (0) d \alpha$$

$$= \frac{\pi}{\vert k\vert} \frac{(q E)^2}{2 m} g (0),$$ (53)

which is obviously not Landau damping. Then what is this contribution? The answer is that it is an ``error term'' often encountered in the iterative method. In the above section, we saw that the term $\sin (\alpha t) / \alpha$ was canceled when we go higher order approximation (refer to the derivation between Eqs. (28) and (29)). The iterative method has the undesirable feature that in the early iteration it gives erroneous values to the higher order terms. One can only check that a term is correct by making one more iteration, which of course is usually convincing but no rigorous proof. Therefore, strictly speaking, the result given here does not prove the existence of Landau damping, but only suggests that the Landau damping is very likely to exist.