bloch equations feb 2010 ol
TRANSCRIPT
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Bloch equations
Ravinder Reddy
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Outline
Magnetic moment and Larmor precession Bloch equations
With and without relaxation With RF and Relaxation
Transformation into rotating frame Steady state solutions
Absorption and dispersion
Steady state solutions Pulse responses
Advantages and Limitations of BE
Single pulse and spin-echo sequence
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Angular momentum and magnetic moment
Consider a charge q moving anticlockwise
On a circle of radius r with velocity v
A charge whose region of circulation is smallcompared to the distance at which the field ismeasured is called a magnetic dipole with dipolemoment having absolute value
=iA
Where A is the area of the current loop and i is the
current.
The period of rotation of such particle is t
!=Magnetic moment
vr
q
t=circumference/speed= 2"r/v
For a circulating charge q, this gives anequivalent current of
i=q/t=qv/2!r
= (q/2m)/(mvr) m=mass of charge
mvr = L, angular momentum of circulatingcharge
=(q/2m) L (magnetic moment is proportionalto angular momentum)
=(q/2m) L =L
=I for the nucleus
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Nuclear magnetic dipole moment
Associated with each rotatingobject there will be a angularmomentum
Associated with each nuclear spin isa magnetic momentarising from theangular momentum of the nucleus
The magnetic moment is a vectorperpendicular to the current loop
In a magnetic field (B) the magneticmoment will behave like magneticdipole
will experience a torque "=xB
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Larmor PrecessionArclength =(diameter).!.[(d")/360]
Arclength =radius.(d")
Sin!= r /
r = .Sin! r
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Larmor precession!=
"I
"t=
Isin#"$
"t
1 Tesla Magnetic Field !42.578 MHz
!Larmor
= "B
! =
!
B =
Bsin
" =
#IBsin
"
# =g e
2mp
!Larmor
=
!"
dt= #B
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Larmor Precession
Precession of themagnetization
vector around thez-axis of themagnetic field
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Boltzman factor
Boltzman factor b=(N+-N-)/N=E/2kT
K=1.3805x10-23 J/Kelvin
T=300 k E= h#=6.626x10-34Js x100 MHz b= 7.99 x10-6
How is the Boltzman factor changes with Boand or T?
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Bloch equations
In terms of total angular momentum of a
sample
d
dt="xB
M= i
i
!
Total magnetic moment of a sample
Interaction of magnetic moment with magnetic field gives atorque on the system and changes the angular momentum ofthe system
M=!L
! =dL
dt= M" B
dM
dt=!dL
dt=!M" B
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Bloch equations
In terms of individual componentsd
x
dt="(M
yBz#Mz
By )
dy
dt="(M
zBx#Mx
Bz )
dz
dt="(M
xBy# My
Bx)
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BE Static magnetic
field is applied alongz-direction. Bzisnon zero andBx=By=0 and spinsprecess around B
z
For free precession inthe absence ofrelaxation and RF,
these equations haveto be solved. Thesolutions to theseequations are:
M
x
'= M
x
cos(!t) + M
y
sin(!t)
My
'= M
ycos(!t)" M
xsin(!t)
dMx (t)
dt=!My (t)Bz
dMy (t)
dt="!Mx (t)Bz
dMz(t)
dt=
0
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Including relaxation
Spin-lattice and spin-spin relaxation can betreated as first orderprocesses with
characteristic times T1and T2respectively.
dMx
dt=!(M
yB
z" M
zB
y) "
Mx
T2
dMy
dt=!(M
zB
x" M
xB
z)"
My
T2
dMzdt
=!(MxBy" My
Bx) " "(Mz
"Mo)
T1
Including relaxation termsalone and treating that the Bo
applied along z-axis (Bx=By=0,Bo=Bz)) the Bloch equationsare given by:
dMx
dt=!M
yBo"
Mx
T2
dMy
dt="
!MxBo"
My
T2
dMz
dt="
(Mz" M
o)
T1
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T1
Solving for Mz Mz(t) = Mzequ +[Mz (0)! Mzequ]exp(!t/T1)
Mz(t)
Mz(0)nullt = 1T ln(1!
Mz (0)
Mzeq
)
Mzeq
http://www.cis.rit.edu/htbooks/nmr/inside.htm
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Solutions for Mxand My
Solutions for MxandMyare:
Mx(t) =[M
x(0)cos(!Bt) " M
y(0)sin(!Bt)]exp("t/T2)
My(t) =["Mx (0)sin(!Bt) +My (0)cos(!Bt)]exp("t/T2)
Mx(t) = M0(0)cos(!Bt" #o)exp("t/T2)
My(t)="
M0(0)sin(!Bt"
#o)exp("
t/T2)
These are complicated to visualize
If the magnitude of the
transverse magnetizationat time zero is Moand theangle between Moand thex-axis at time zero is $o,then
NMR signal (~kHz)
Larmor precession frequency (Mhz)
NMR signal (~kHz)
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Frequency or Amplitudemodulation
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BE with relaxation and RF
Static field B isapplied along z-direction B
z
=Bo
and RF field at%with anamplitude %1isapplied in the
transverse planewith components
Bx = B1cos(!t),By = B1 sin(!t),Bz =Bo
dMx
dt=!(M
yBo +Mz
B1sin("t)) #M
xT2
dMy
dt=!(M
zB1cos("t)#Mx
Bo) #
My
T2
dMzdt
=#
!(MxB1sin("t)+MyB1cos(
"t))#
(Mz# M
o)
T1
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Rotating frame of reference In the laboratory frame
the magnetization vectorprecess at the Larmorfrequecny.
To probe the changesinduced in the frequencyit is convenient to workin a rotating frame ofreference
In the rotating framethe magnetization vector,at equilibrium appearsstationery.
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Magnetization vector
Bo is along z-axis
Equilibriummagnetization Mo isalong z-axis
Mx and My = 0
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Rotating frame In the rotating
frame the
magnetizationappear stationery.
It can only changein the magnitude in
response toperturbations
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Rotating frame vs Laboratory frame
Magnetization trajectory following RF field in
Rotating frameLaboratory frame
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Transformation into rotating frame
Additional time dependence introduced inthe form of time dependent RF field
complicates the analysis.
Two new variables u and v are defined as
u =Mxcos(!t)"M
ysin(!t)
v =Mxsin(!t)+M
ycos(!t)
Mx=
ucos(!
t)+
vsin(!
t)M
y=ucos(!t) " vsin(!t)
=(!o" !)v "
u
T2
Expression for dv/dt can also be obtained similarly
du
dt=
dMx
dtcos(!t)"
dMy
dtsin(!t)"![M
xsin(!t) +M
ycos(!t)]
=#B0v" uT
2
"!vUsing Eqn: I
Eqn: I
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Rotating frame solutions
Usually u and v arewritten as Mx and
My respectively. Wewill drop the primeshere and now on wewill work in therotating frame with
the followingequations
du
dt=(!
o"!)v"
u
T2
dv
dt="(!
o"!)u +!1Mz
"
v
T2
dMz
dt="!1v"
(Mz" M
o)
T1
dMx
dt=(!o"!)My
"
Mx
T2
dMydt
="(!o"!)Mx+!1Mz
"
MyT2
dMz
dt="!1My
"
(Mz" M
o)
T1
Full Bloch equations in rotating frame
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Steady state solutions
If we assume that thesystem has been allowed tosoak in this combination of
of static and time varyingfields a steady stateultimately will be reachedin which none of thecomponents change withtime:
!"=("o#")
"1 =$B1
now on we will work in therotating frame with thefollowing eqautions
Mx
=
!1T2
2"!
1+!1
2T1T2+(T
2"!)
2M
o
My
=!1T2
1+!1
2T1T2+(T
2"!)
2M
o
Mz
=
1+T2
2"!
2
1+!1
2T1T2+(T
2"!)
2M
o
(!o"!)My"
Mx
T2
=0
"(!o"!)M
x+!
1Mz"
My
T2
=0
"!1My
"
(Mz"M
o)
T1=0
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Steady state solutions in the limiting case
In the limiting case
of small RF limit,
!1
2T1T2
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Absorption and dispersion lines
In the limiting case
of small RF limit,
Signal due to Myand
Mxcomponents are
known as absorption
and dispersion,
respectively.
(!T2)-1
Frequency (Hz)--
Absorption mode
Dispersion mode
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Mx and My
Laboratory frame solutions
dMx
dt=!M
yBo"
Mx
T2
dMy
dt="!M
xBo"
My
T2
dMz
dt="
(Mz"
Mo)
T1
dMx
dt=(!o"!)My
"
Mx
T2
dMy
dt="(!o"!)Mx
+!1Mz"
My
T2
dMz
dt
="!1My"
(Mz" M
o)
T1
Rotating frame solutions
Mx(t) = M0(0)cos(!Bt" #o)exp("t/T2)
My(t) ="M0(0)sin(!Bt" #o)exp("t/T2)
Mx(t)Cos!
or
cos"#t
My(t)Sin!
or
Sin"#t
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Pulsed NMR
Flip angel &&(rad)= (rad sec-1G-1)B1(G)tw(sec)
http://www.cis.rit.edu/htbooks/nmr/inside.htm
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Pulse response
Response of apulse duration (tw)
and amplitudew
1applied along x-axis,
Evolution with achemical shift (%
during acquisition
Mx(+) =M
x(!)
My(+) =M
y(!) cos("
1tw
) +Mz
(!)sin("1tw)
Mz
(+) =Mz
(!) cos("1tw
)!My(!)sin("
1tw)
Mx(+) = M
x(!)cos("#t) !M
y(!)sin("#t)exp(!t/T
2)
My(+) = M
y(!)cos("#t) +M
x(!)sin("#t)exp(!t/T
2)
Mz(+) = M
0(1!exp(!t/T
2))
dMx
dt=(!o"!)My
"
Mx
T2
dMy
dt="(!o"!)Mx
+!1M
z"
My
T2
dMz
dt="
!1My"
(Mz" M
o)
T1
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Pulse response
Following the pulse
My(t)
M0Sin!
o
Mx(t)M
0cos!
o
Mx(+) = M
x(!)
My(+) = M
y(!) cos("
1tw
) +Mz
(!)sin("1tw)
Mz
(+) = Mz
(!) cos("1tw
) !My(!)sin("
1tw)
Mx(+) = M
x(!)cos("#t)!M
y(!)sin("#t)exp(!t/T
2)
My(+) = M
y(!)cos("#t) +M
x(!)sin("#t)exp(!t/T
2)
Mz(+) = M0(1!exp(!t/T2 ))
After evolution with the CS
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Saturation If applied B1 is too high the
peak height in theabsorption spectrum isdecreased due to thereduction in the difference
in population between thetwo levels.
This happens when the rateof energy absorption iscomparable to or greaterthan the rate of relaxationbetween energy levels, 1/T1.
!1
2T1T2
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Nuclear receptivity In Steady state NMR
experiment the spin systemcan absorb energy from RFat a rate R is depends on
transition probability p,energy level separation, (Eand population difference(no
NMR detected signal ~dMy/dt =R/B1
R = p!E!no" #
4
B0
2
NB1
2
g($) / kT
S! R /B1! "
4
B0
2
NB1g(#) / kT
!no= N!E/2kT
p" #2
B12
g($)!E= #!B
o
B1(opt) =(!
2
T1T
2)"1/ 2
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Limitations of BE
Can not be used toanalyze otherinteractions such as J-,dipolar and quadrupolar
2D NMR Higher spin (I>1/2)
interactions
MQ coherences can notbe explained
Molecular motionalprocesses (T1 and T2) andfield inhomogeneity
Gradient manipulation andMR imaging