Chemistry Reference
In-Depth Information
because we may take
(0)
=
0, we have Eqs. (94) and (95), in the noninertial case
for a step field gradient
γ
ζ
G
(
t
)
dt
dt
2
γλ
(
t
1
)
ζ
G
(
t
)
dt
dt
1
2
t
0
t
1
λ
(
t
2
)
t
2
0
t
1
2
(
t
)
=
(98)
0
0
t
t
1
λ
(
t
2
)
t
2
0
G
(
t
)
dt
dt
2
λ
(
t
1
)
t
1
0
2
γ
2
ζ
2
G
(
t
)
dt
dt
1
=
0
0
t
t
1
t
1
G
(
t
)
dt
t
2
0
2
γ
2
ζ
2
G
(
t
)
dt
λ
(
t
1
)
λ
(
t
2
)
dt
1
dt
2
=
(99)
0
0
0
Recalling the definition of
λ
(
t
) and the fact that
G
(
t
) is not stochastic,
t
t
1
t
1
G
(
t
)
dt
t
2
0
2
2
γ
2
ζ
2
G
(
t
)
dt
λ
(
t
1
)
λ
(
t
2
)
dt
1
dt
2
=
(100)
0
0
0
2
kTζ
t
0
t
1
t
1
G
(
t
)
dt
t
2
0
2
γ
2
ζ
2
G
(
t
)
dt
δ
(
t
2
−
=
t
1
)
dt
1
dt
2
0
0
Now the one-sided delta function has a sifting property such that,
t
2
f
(
t
2
)
2
f
(
t
1
)
δ
(
t
2
−
t
1
)
dt
1
=
0
Thus, for a rectangular field gradient pulse of strength
G
,
ζ
2
kTζ
t
dt
2
t
1
0
G
(
t
)
dt
t
2
0
4
γ
2
2
G
(
t
)
dt
=
0
t
t
2
dt
2
2
γ
2
G
2
kT
ζ
=
dt
2
0
0
Hence,
2
2
1
3
γ
2
G
2
Dt
3
=
(101)
Thus we have the Carr-Purcell-Torrey Eq. (60) result for the dephasing [29]
following the application of a single-step gradient,
exp
3
Dγ
2
G
2
t
3
e
i
=
1
e
−
2
/
2
=
−
(102)
The gradient-echo result, Eq.(64), may be obtained in a similar manner. For
the spin-echo case, Eq.(69) may be obtained by writing the right-hand side of
