416 Chapter 14. Wave-particle nonlinearities
so
v ̃ 1 =−eE/m ̄
(p+iω)(p+ikv 0 ). (14.95)
The inverse Laplace transform gives
̃v 1 =−1
2 πieE ̄
m∫b+i∞b−i∞ept
(p+iω)(p+ikv 0 )dp. (14.96)Analytic continuation allows the contour to be completed on the left hand side andit is
seen that there are two poles, one atp=−iωand the other atp=−ikv.Evaluation of the
residues for the two poles gives
̃v 1 = −1
2 πieE ̄
m
2 πi lim
p→−iω
(p+iω)[
ept
(p+iω)(p+ikv 0 )]
+2πi lim
p→−ikv
(p+ikv)[
ept
(p+iω)(p+ikv 0 )]
= −
eE ̄
m{
e−iωt
(−iω+ikv 0 )+
e−ikv^0 t
(−ikv 0 +iω)}
(14.97)
which is the same as Eq.(14.92). The ballistic term∼e−ikv^0 tthus results from a pole
originating from the Laplace transform of the source term, whereas the homogeneous term
∼e−iωtresults from the pole originating from the factor(p+ikv 0 )that appeared in the
left hand side of Eq.(14.94).
14.3.2Phase mixing of the ballistic term for multiple beams
Now suppose that instead of just one electron beam, there are multiple beams where
the density of each beam is proportional toexp(−v^20 /v^2 T); this would be one way of
characterizing a Maxwellian velocity distribution. Superposition of the ballistic terms from
all these beams leads to a vanishing sum, because the ballistic terms each have a velocity-
dependent phase and so the superposition would give destructive interference dueto phase
mixing. In particular, the superposition would involve integrals of the form
∫
dv 0 e−v
(^20) /vT (^2) +ikv 0 t
=e−k
(^2) v (^2) Tt (^2) / 4
∫
dv 0 e−(v^0 /vT+ikvTt/2)2
=√
πvTe−k(^2) v (^2) Tt (^2) / 4
(14.98)
which would quickly become extremely small. This suggests the ballisticterm has no en-
during macroscopic physical importance and, in fact, this is true for linear problems where
the ballistic term is typically ignored. However, the ballistic term can assume importance
when nonlinearities are considered.
14.3.3Beam echoes
The linearized continuity equation corresponding to Eq.(14.84) is
∂n 1
∂t
+v 0
∂n 1
∂x
+n 0
∂v 1
∂x
=0 (14.99)
and so, invoking the assumedeikxdependence, this becomes
∂n 1
∂t
+ikv 0 n 1 +ikn 0 v 1 =0. (14.100)