Geoscience Reference
In-Depth Information
3.3.3
Flux Matching for
S
p
¤ 1
Now we find
R
from the condition Eq.
3.39
dn
fm
dr
D
n
a
S
p
n
R
1
S
p
b.R/
db
dr
(3.56)
ˇ
ˇ
ˇ
ˇ
R
D
a
2
R
3
p
1
a
2
=R
2
db
dr
(3.57)
The equation for
R
is
a
2
R
p
R
2
˛
fm
2DR
D
(3.58)
a
2
or
v
T
2D
D
1
p
R
2
(3.59)
a
2
From here we have
q
a
2
R
D
C
R
o
(3.60)
with
2D
v
T
R
o
D
(3.61)
We had already
˛.a;R/
1
C
˛
D
(3.62)
S
p
˛
.
a;R
/
4DR
On substituting thus ˛.a;R/and
R
we find
˛
fm
1
S
p
b.R/
C
˛
D
(3.63)
S
p
˛
fm
4DR
After some transformation one finally gets
˛
fm
q
1
C
a
v
T
1
˛.a/
D
(3.64)
2D
2
S
p
2
1
C
