112 Chapter 3. Motion of a single plasma particle
The average velocity for each of these four representative particles will now be evalu-
ated:
ParticleA– Let the distance between maximum and minimum potential bed.Letx= 0
be the location of the minimum so the injection point is atx=−d.Thus the trajectory on
the downslope is
x(t) =−d+v 0 t+at^2 / 2 (3.206)
and the time for particleAto go from its injection point to the potential minimum is found
by settingx(t) = 0giving
tAdown=
v 0
a(
−1 +
√
1 + 2δ)
(3.207)
whereδ=ad/v^20 is the normalized acceleration. When particleAreaches the next poten-
tial peak, it again has velocityv 0 and if the time and space origins are re-set to be at the
new peak, the trajectory will be
x(t) =v 0 t−at^2 / 2. (3.208)The negative time when the particle is at the preceding potential minimum is found from
−d=v 0 t−at^2 / 2. (3.209)Solving for this negative time and then calculating the time increment to go from the mini-
mum to the maximum shows that this time is the same as going from the maximum to the
minimum, i.e.,tAdown=tAup.Thus the average velocity for particleAis
vAavg=da/v 0
−1 +√
1 + 2δ. (3.210)
The average velocity of particleAis thus alwaysfasterthan its injection velocity.
ParticleC– Now letx= 0be the location of maximum potential andx=−dbe the
point of injection so the particle trajectory is
x(t) =−d+v 0 t−at^2 / 2 (3.211)and the time to get tox= 0is
tCup=v 0
a(
1 −
√
1 − 2 δ)
. (3.212)
From symmetry it is seen that the time to go from the maximum to the minimum will be
the same so the average velocity will be
vCavg=ad/v 0
1 −√
1 − 2 δ. (3.213)
Because particleBis always on a potential hill relative to its injection position, its average
velocity is alwaysslowerthan its injection velocity.
ParticlesBandD- ParticleBcan be considered as first traveling in a potential well and
then in a potential hill, while the reverse is the case for particleD.For the potential well