DOC.
178
SEPTEMBER
1909
133
Thus,
ßnx
is
the number of
effective collisions in the
layer
(x,
x
+
1)
But this
number
n
is
also
-
e
0.
Hence,
by
equating
we
obtain
o
ß
=
1
0
X
(3)
o
Now
we
begin
with
(1)
and
seek
the
mean
value
of
Ax+1
in the
cross
section
x
+
1.
If
the collisions in
the
layer
under consideration
were
to
occur
in
a
fluctuation-free
manner,
we
would have:
Ax+1
equal
to
(1
+
ß)Ax
(fluctuation
a).
But
a
fluctuation
Ax+1
would be
present
even
in
the
absence
of
a
fluctuation
Ax
in
the
layer x,
because of
the
fluctuation of
the effective collisions in this
layer.
The number
n
of
these
collisions
is
·,
hence the fluctuation
Xo
/ \
A2 equal
to
e ·o
\
1
-
e
0
n
X
(fluctuation b)
o
The
squares
of the
two
fluctuations
(a
and
b)
add,
so
that
one
obtains
/ \
Al,
=
(i
+
ß)2K
+
n
X0
k
0
\
1
-
e
0
/
If
we
set
/
A*
=
J,
2ß
=
a,
Xq
0
\
1
-
e
x0
=
b,
then
we
obtain[5]
dydx
^-
=
ay + beßx,
and
for
x
=
0,
we
must
have
y
=
0.
dx
We
obtain
y
=
a
-
ß
(ew-
eßx)
(4)