Pulsed-field gradient nuclear magnetic resonance as a tool for ...
Pulsed-field gradient nuclear magnetic resonance as a tool for ...
Pulsed-field gradient nuclear magnetic resonance as a tool for ...
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316<br />
PRICE<br />
ing displacement in the direction of the <strong>gradient</strong>,<br />
Ž .<br />
which can be written most generally <strong>as</strong> 79<br />
² Ž . Ž .:<br />
z t z t a<br />
a b<br />
HHH 1 0 z 2 0 z 0 0 1 a<br />
Ž r r . Ž r r . Ž r . PŽ r ,r ,t .<br />
PŽ r ,r ,t t . dr dr dr . 60 1 2 b a 0 1 2<br />
It should be noted that Eq. 60 holds only when<br />
t t . We will now evaluate Eq. 60 b a<br />
<strong>for</strong> the<br />
Ž particularly simple. c<strong>as</strong>e of PŽ r , r , t. 0 1 <strong>as</strong> given<br />
by Eq. 27 Ž i.e., free diffusion . . Since we are only<br />
interested in motion in one dimension, we can<br />
simplify our t<strong>as</strong>k by using the one-dimensional<br />
version of Eq. 27 ,<br />
12<br />
PŽ z , z ,t. Ž 4Dt. exp <br />
0 1 ž<br />
2<br />
Ž z z . 1 0<br />
4Dt /<br />
61 <br />
and making obvious changes to Eq. 60 thus<br />
obtain<br />
² Ž . Ž .:<br />
z t z t a<br />
a b<br />
<br />
H H H 0 1 0 2 0<br />
<br />
Ž z .Ž z z .Ž z z .<br />
12<br />
Ž .<br />
4Dt a<br />
ž /<br />
ž<br />
Ž z z .<br />
/<br />
4DŽ t t .<br />
2<br />
1 0 12<br />
Ž z z .<br />
exp<br />
Ž4DŽ t t ..<br />
b a<br />
4Dta 2 1<br />
exp<br />
dz dz dz .<br />
2<br />
b a<br />
0 1 2<br />
Now we let Z z z and Z z z , and<br />
1 1 0 2 2 0<br />
thus,<br />
<br />
H Ž z . 0 dz0<br />
<br />
ž /<br />
<br />
2<br />
12 Z1 H Z Ž 4Dt .<br />
1 a exp dZ1<br />
4Dta<br />
<br />
H Z24D tbta <br />
Ž Ž ..<br />
ž /<br />
12<br />
Ž Z . 2Z1 exp<br />
dZ 2 .<br />
4DŽ t t .<br />
b a<br />
2<br />
By noting Eq. 29 ,<br />
we can remove the integral<br />
over z , and making the substitution Z 0 2Z2 Z 1,<br />
we then get<br />
ž /<br />
2<br />
<br />
12 Z1 H Z Ž 4Dt .<br />
1 a exp dZ1<br />
4Dta<br />
<br />
H Z2Z1 4D tbta <br />
Z 2<br />
ž b a /<br />
Ž .Ž Ž ..<br />
12<br />
2 <br />
exp<br />
dZ . 62 2<br />
4DŽ t t .<br />
<br />
We now consider the integral over Z in Eq. 62 <br />
2 ,<br />
which we rewrite <strong>as</strong><br />
12<br />
Ž4DŽ t t ..<br />
b a<br />
½<br />
2<br />
<br />
Z2 Z1Hexp dZ<br />
4DŽ t t . b a<br />
ž /<br />
2<br />
Z2 ž 4DŽ t t . /<br />
<br />
H 5<br />
2<br />
b a<br />
<br />
Z exp dZ . 63 The first integral in Eq. 63 can be evaluated with<br />
the standard integral e.g.,<br />
integral 3.323 2. in Ref.<br />
Ž 53 .,<br />
H e dxe . 64<br />
p<br />
2 2 2 2 <br />
p x qx q 4p <br />
<br />
'<br />
Ž Ž .. 12<br />
2 a b<br />
by setting x Z , p 4D t t and<br />
Ž Ž .. 12<br />
q0, to give 4D tbt a . The second inte-<br />
gral in Eq. 63 can be evaluated using the standard<br />
integral Eq. 3.462 6. in Ref. Ž 53 .,<br />
H (<br />
<br />
2 q 2 2<br />
px 2qx q p <br />
xe dx e Re p 0.<br />
p p<br />
<br />
65<br />
<br />
In our c<strong>as</strong>e, x Z , p 4DŽ t t. 2 b a and q 0,<br />
and so this integral equals 0. Hence, Eq. 63 reduces to simply Z and now Eq. 62 becomes<br />
1<br />
<br />
2<br />
12<br />
² zŽ t . zŽ t .: Z Ž 4Dt .<br />
a b 1 a<br />
<br />
Z2 1<br />
1 ž 4Dt / a<br />
a H<br />
exp dZ . 66