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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

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