frequency,
rL
and
r
1
are
the
transverse
and
longitudinal
relaxation
rates
respectively,
and
n
the
density
of
the
atoms.
Let
us
limit
our
considerations
to
fast
transverse
relaxation,
which
means
that
the
polarization
follows
the
electric
field
adiabatically
(adiabatic
elimination
method)
i.e.,
aa;ar
~
o, we
have
a=
iJ,LNE/(i.t.n-
r_L).
(7.3-2)
Substituting
Eq.(7.3-2)
into
Eq.(7.3-la),
then
the
electric
field
can
be
written
in
the
integral
form.
E(r+zjc,z)
=
E(r,O)exp[28W(r,z)(i.t.fl
+
r_L)/(.t.fl'+rL')],
(7.3-3)
where
8 =
27TnkJ,L'
and
W(r,z)
=
foz
N(r
+
z'jc,z')dz'.
(7.3-4)
Substituting
Eqs.(7.3-3)
and
(7.3-4)
into
Eq.
(7.3-lb)
and
then
integrating
over
z,
one
has
aw(r,z);ar
=-
r
1
cw
+
z/2)
-
J.L'IE(r,O)
I'
(exp[4SrLW/(.t.n'
+
rL')]
-1)/48.
(7.3-5)
Introducing
the
dimensionless
quantities
E(t,z)
=
J.L
E(t,z)jzjrLr
1
(1
+
.t.'
),
x =
tr
1
,
cp(t) =
W(t-
J;./c,L)/L,
(7.3-6)
.t.
= .t.n;rL.
Combining
Eqs.(7.3-3)
and
(7.3-5)
together
with
the
boundary
conditions
with
the
dimensionless
quantities
in
(7.3-6)
we
have
the
following
set
of
equations
which
do
not
involve
the
optical
coordinates:
€(X,O)
=
)T€
1
(X)
+
RE(x-k,O)exp(aLcf>(x))exp(i(aL.t.(cp(x)-~)-6
0
},
(7.3-7a)
dcp(x)jdx
=
-(cp(x)+~)
-
2IE(x-k,O)
I'
[exp(2aLcf>(x))-1]/aL,
(7.3-7b)
€r
=
)TE(x-k,O)exp(aLcf>(x))exp(i(aL.t.(cp(x)+~)-(6
0
+kl)
},
where
and
is
the
Er(x)
=
J.LEr(t-
l/c)/2(rLr
1
(1
+ .t.')}v.,
€
1
(X)
=
J,LE
1
(t)/2(rLr
1
(1
+
.t.
2
)
}%,
a =
2Srv(.t.fl
2
+
rL')
I
effective
absorption
coefficient,
and
344
(7.3-8)