13.2 Statistical argument for the development of the Fokker-Planck equation 389in which case
〈
∆v
∆t
〉
=
q^2 TqF^2 lnΛ
4 πε^20 m^2 T∂hF
∂v
〈
∆v∆v
∆t〉
=
q^2 TqF^2 lnΛ
4 πε^20 m^2 T∂^2 gF
∂v∂v. (13.45)
The Fokker-Planck equation, Eq.(13.20) thus becomes∂fT
∂t=
∑
F=i,eq^2 TqF^2 lnΛ
4 πε^20 m^2 T[
−
∂
∂v·
(
fT∂hF
∂v)
+
1
2
∂
∂v∂
∂v:
(
fT∂^2 gF
∂v∂v)]
. (13.46)
The first term on the right hand side is a friction term which slows downthe mean velocity
associated withfTwhile the second term is an isotropization term which spreads out (i.e.,
diffuses) the velocity distribution described byfT.The right hand side of Eq.(13.46) thus
gives the rate of change of the distribution function due to collisions and sois the correct
quantity to put on the right hand side of Eq.(2.12).
13.2.1Slowing down
Mean velocity is defined by
u=∫
vfdv
n(13.47)
wheren=
∫
fdv.The rate of change of the mean velocity of speciesTis thus found by
taking the first velocity moment of Eq.(13.46). Integration by parts on the right hand side
terms respectively give
−
∫
v∂
∂v·
(
fT∂hF
∂v)
dv=∫
fT∂hF
∂vdv (13.48)and ∫
v∂
∂v∂
∂v:
(
fT∂^2 gF
∂v∂v)
dv=−∫
∂
∂v·
(
fT∂^2 gF
∂v∂v)
dv=0 (13.49)where Gauss’s theorem has been used in Eq.(13.49). The first velocity momentof Eq.(13.46)
is therefore
∂uT
∂t
=
∑
F=i,eq^2 Tq^2 FlnΛ
4 πε^20 nTm^2 T∫
fT∂hF
∂vdv. (13.50)Let us now suppose that the test particles consist of a mono-energetic beam so that
fT(v)=nTδ(v−u 0 ). (13.51)In this case Eq.(13.50) becomes
∂uT
∂t=
∑
F=i,eq^2 TqF^2 lnΛ
4 πε^20 m^2 T(
∂hF
∂v)
v=u 0. (13.52)
Let us further suppose that the field particles have a Maxwellian distribution so that
fF(v)=nF(
mF
2 πκTF) 3 / 2
exp(
−mFv^2 / 2 κTF