Bloch Equations Feb 2010 Ol

8/21/2019 Bloch Equations Feb 2010 Ol

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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


16/32

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


20/32

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


26/32

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

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Bloch Equations Feb 2010 Ol