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 /
 
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