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
/