What is the origin of the nuclear magnetic dipole
moment (m) and how is it oriented relative to an
external magnetic field B0?
m  iA  qA
m
q
A
29-May-16
From Quantum Mechanics
(I = ½-particle; 1H, 13C,31P,…)
z
cos  
m


x
29-May-16
m
I ( I  1)
m   I ,..0,..I
y
I  1/ 2
3
   54.7 0
3
 : not det er min ed
cos  
Characterizing the motion of m in B0
m = gL
L; angular momentum (spin)
µ
=mxB
; torque
dμ
 gμ  B
dt
Exercise 1.1: Derive the above equation (qualitatively)
29-May-16
Solution
Larmor Equation - the basic NMR equation
w = -gB
m
29-May-16
Conclusion
•THE LARMOR EQUATION (w  gB) IS DERIVED FROM A CLASSICAL
MECHANICAL APPROACH
THE SPIN MOTION IS WITHIN THE MHz-RANGE (Radio-frequencies)
29-May-16
MOTION OF m IN A ROTATING FRAME OF REFERENCE
y
m
wt
y
d 
d 

 dt 
 dt   wx
fixed
rot
x
u
v
x
w; rotation frequency of the reference frame relative to the static frame
 dμ 
 gμ  B 0 ;
 dt 
  fixed
 dμ 
 dμ 
 ωμ
 dt    dt 
  rot   fixed
 dμ 
 dt   gm  B0  w  μ  gμ  (B0  ω / g )  gμ  Beff
rot
29-May-16

Exercise 1.4. Find the solution of the above equation in the rotating frame,
(Note w0= -gB0)
 dμ 
 dt   μ  (ω0  ω)
rot
mU  cos(w-w0)t
mV  sin(ww0)t
29-May-16
One dipole (mi) → Many dipoles ( M   mi )
i
M    m0 cos  e i i  wt 
z
i
M z   m0 sin 
i
y
y
x
x
29-May-16
Mxy = 0 !!!!!
LINEAR POLARIZED FIELD [B1= B1cos(wt)]
Rf-irradiation
e iwt  eiwt
2
 2 cos(wt )
2
Conclusion:
A LPF is composed of two opposite rotating fields
29-May-16
RFR
Application of an rf-pulse along the u-direction
(Why and what effect ?)
Beff
Bo (-Bw/g
w /g )
dM
 gM  B eff
dt
0
Beff
M
u
29-May-16
B1
v

 B1u  ( B0  w / g )k
Excersize 1.5: What effect will B1have on the magnetization
when on resonance (w0 = w) ? Discuss
dM U
 (w0  w ) M V  0
dt
dM V
 (w0  w ) M U  gB1M z  gB1M z
dt
dM z
 gB1M V
dt
29-May-16
zM
0
z
v
a
v
B1
u
u
M V  M 0 sin( w1t )  M 0 sin( gB1t )
a  gB1t
The magnetization is rotated an angle a during the duration t of the rf-pulse
Show experiment !
29-May-16
Decaying FID? – show experiment
29-May-16
Including (phenomenologicaly) relaxation terms in the
Bloch equation (RFR):
M
M
M  M0
M
 gMxB eff  u u  v v  z
k
t
T2
T2
T1
29-May-16
Bloch Equation (on resonance, after magnetization is rotated onto the v-axis)
dM v
M
 v
dt
T2
dM z M 0  M z

dt
T1
(3a)
(3b)
M u (0)  0, M v (0)  M 0 , M z (0)  0
Solution (after the magnetization is rotated into the uv-plane (w1 = 0); Exercise 2.2)
M v (t )  M 0  exp  t / T2 
M z  M 0  1  exp( t / T1 )
29-May-16
Water Confined between Solid Glass beads - Experiment
1
4000
H-FID
3500
Intensity (a.u.)
3000
.
2
Gaussian; I0 exp -at-bt
(
2500
2000
1500
1000
500
0
Exponential; I0exp(-ct)
-500
0,0000
0,0005
0,0010
Time/s
29-May-16
0,0015
0,0020
)
Effect of various types of magnetic field inhonogeneities on the FID decay
Let N(B) represent the number of spins experiencing a magnetic field B.
Since w  gB, we may write dN = N(w)dw
What is the observable signal intensity M if ?
N (w )  N 0
B1
B0
B-1
B1 > B0 > B-1
29-May-16
M  e t / T2
a
1
 a2  w 2

 N (w ) cos(wt )dw

N 0 ( a 1/ T2 )t

e
2
1 / T2,eff  a  1 / T2
What if N(B) = dN  N ( B)dB  N (w )dw  N 0
 N0
1
 2
M  e t / T2
29-May-16
 2
1 w w 2
 (
)
2

e
d
w

 N (w ) cos(wt )dw


1
N0
2 
e
 ( 2t 2 / 2  t / T2 )
1 BB 2
 (
)
2

e
dB
Why doesn’t the FID decay exponentially with time
as expected from theory?
(Perform a CPMG pulse sequence)
1
H-FID
10000
1000
100
10
1
0
0,0005
0,001
0,0015
0,002
0,0025
Time/s
CPMG
10000
1000
100
10
1
0
0,2
0,4
Time/s
29-May-16
0,6
0,8
Why does the FID decay faster (1/T2*) than what is predicted by the true spinspin relaxation rate 1/T2?
FID
lnM
B1
Homogeneous field
Inhomogeneous field
B0
B-1
Time
B1 > B0 > B-1
29-May-16
Principles for Measuring T2
Spin gymnastics: Controlled sequence of rf-pulses
Hahn spin-echo: /2----deteksjon
z
z
v
u
u
(/2)-puls
29-May-16
v
The basic (/2)x – -x spin-echo experiment
29-May-16
Magnetization in the rotating frame

v
v
u
u
“100 meter”
AVSTAND
29-May-16
v
v
-pulse (along u)
u
u
29-May-16
v

v
u
u
EKKO
AVSTAND
29-May-16
1000M z/M 0
SPIN-ECHO-EKSPERIMENT
1/T2
100
1/T2+1/T2INH
1
0
5
10
15
20
TIME (2)
29-May-16
25
30
CPMG-pulse sequence
/2--(-deteksjon)n
/2



TID
CPMG
I (t )  I (n  2 )  I (0)  exp( n2 / T2 )  I (0)  exp( t / T2 )
29-May-16
T1-experiment
M ( )  M (0)  1  2 exp( / T1 )
29-May-16