Geoscience Reference
In-Depth Information
The free molecule flux is now expressed in terms of ‰ as follows:
Z
J.a/
D
a
2
n
1
v
T
e
x
‰.x/.‰.x//dx;
(3.112)
0
where
v
T
D
p
8kT=m is the thermal velocity. From Eq.
3.112
we can find the
enhancement factor:
Z
e
x
‰.x/.‰.x//dx;
".a/
D
(3.113)
0
because the multiplier before the integral on the right-hand side of Eq.
3.112
is equal
to the free molecule flux of neutral molecules on a neutral particle (see following).
Now we prove that
r
2
a
2
:
d‰
dx
D
(3.114)
Indeed, differentiating both sides of Eq.
3.112
over
x
gives
x
ˇU .r
/
2
r
U
0
.r
/
:
r
2
a
2
C
2rr
0
x
a
2
d‰
dx
D
1
The expression in the parentheses on the right-hand side of this equation is zero
because of Eq.
3.110
.
Integrating by parts in Eq.
3.113
and using Eq.
3.114
leads to the final result:
0
1
Z
@
a
2
‰
0
C
A
;
e
x
r
2
J.a/
D
n
1
v
T
.x/ .‰.x// dx
(3.115)
0
where ‰
0
D
‰.x
D
0/ if ‰.x/ >0 at x>0and ‰
0
D
0 otherwise.
3.5.2.3
Ion Density Profile
Equations
3.27
,
3.105
,and
3.107
allow us to find the ion density profile n.r/
n.r/
D
n
1
ŒF.r/
C
G.r/;
(3.116)
where n
1
is the ion density far away from the particle,
L
2
m
.r/
Z
Z
dL
2
p
L
2
m
.r/
L
2
;
e
ˇE
dE
F.r/
D
A.r/
(3.117)
0
E
0
.E
0
/