390 Chapter 13. Fokker-Planck theory of collisions
and so, using Eq.(13.44),
hF(v)=nFmT
μ(
mF
2 πκTF) 3 / 2 ∫
exp(
−mFv′^2 / 2 κTF)
|v−v′|dv′. (13.54)The velocity integral in Eq.(13.54) can be evaluated using standard means (see assign-
ments) to obtain
hF(v)=nFmT
μverf(√
mF
2 κTFv)
(13.55)
where
erf(x)=2
√
π∫x0exp(−w^2 )dw (13.56)is the Error Function.
Thus, Eq.(13.52) becomes
∂uT
∂t=
niq^2 Tq^2 ilnΛ
4 πε^20 m^2 TmT
μi{
∂
∂v[
v−^1 erf(√
mi
2 κTiv)]}
v=u 0+neq^2 Tqe^2 lnΛ
4 πε^20 m^2 TmT
μe{
∂
∂v[
v−^1 erf(√
me
2 κTev)]}
v=u 0(13.57)
whereμ−i,e^1 =m−i,e^1 +m−T^1.
This can be further simplified by noting (i) quasi-neutrality implies
niZqi+neqe=0 (13.58)whereZis the charge of the ions, (ii) the masses are related by
mT
μi,e=1+
mT
mi,e, (13.59)
and (iii) the velocity gradient of the Error Function must be in the direction ofu 0 because
that is the only direction there is in the problem. Using these relationships and realizing
that both the left and right sides are in the direction ofu,Eq.(13.57) becomes
∂u
∂t=
nee^2 lnΛ
4 πε^20q^2 T
m^2 T
Z
(
1+
mT
mi)
d
du
erf(√
mi
2 κTiu)
u
+
(
1+
mT
me)
d
du
erf(√
me
2 κTeu)
u
. (13.60)
Let us define
ξi,e=√
mi,e
2 κTi,eu (13.61)