ABSTRACT
E = −∇φ. (8.2) where
∇2φ = − e 0
( n −
∫ fe d3v
) . (8.3)
Here, n is the number density of ions (which is the same as the equilibrium number density of electrons).
Physics: An
Because we are dealing with small amplitude waves, it is appropriate to linearize the Vlasov equation. Suppose that the electron distribution function is written
fe(r, v, t) = f0(v) + f1(r, v, t). (8.4)
Here, f0 represents the equilibrium electron distribution, whereas f1 represents the small perturbation due to the wave. Of course,
∫ f0 d3v = n, otherwise the equi-
librium state would not be quasi-neutral. The electric field is assumed to be zero in the unperturbed state, so that E(r, t) can be regarded as a small quantity. Thus, linearization of Equations (8.1) and (8.3) yields
∂ f1 ∂t + v · ∇ f1 − eme E · ∇v f0 = 0, (8.5)
and ∇2φ = e
∫ f1 d3v, (8.6)
respectively. Let us now follow the standard procedure for analyzing small amplitude waves,
by assuming that all perturbed quantities vary with r and t like exp[ i (k · r − ω t)]. Equations (8.5) and (8.6) reduce to
−i (ω − k · v) f1 + i eme φ k · ∇v f0 = 0, (8.7)
and −k2 φ = e
∫ f1 d3v, (8.8)
respectively. Solving the first of these equations for f1, and substituting into the integral in the second, we conclude that if φ is non-zero then we must have
1 + e2
∫ k · ∇v f0 ω − k · v d
3v = 0. (8.9)
We can interpret Equation (8.9) as the dispersion relation for electrostatic plasma waves, relating the wavevector, k, to the frequency,ω. However, in doing so, we run up against a serious problem, because the integral has a singularity in velocity space, where ω = k · v, and is, therefore, not properly defined.
