bmi i fs05 – class 9 “mri physic” slide 1 biomedical imaging i class 9 – magnetic resonance...

BMI I FS05 – Class 9 “MRI Physic” Slide 1

Biomedical Imaging IBiomedical Imaging I

Class 9 – Magnetic Resonance Imaging (MRI)

Physical Theory

11/09/05

BMI I FS05 – Class 9 “MRI Physics” Slide 2

Magnetic Resonance in a NutshellMagnetic Resonance in a Nutshell

Hydrogen Nuclei (Protons)

Axis of Angular Momentum (Spin), Magnetic Moment



Spins PRECESS at a single frequency (f0), but incoherently − they are not in phase

External Magnetic Field



Irradiating with a (radio frequency) field of frequency f0, causes spins to precess coherently, or in phase


Magnetic Field IMagnetic Field I

N

Smagnetic field lines

By staying in the interior region of the field, we can ignore edge effects.

But how do we describe magnetic fields and field strengths quantitatively?


Magnetic Field IIMagnetic Field II

N

S

q

v

An electric charge q moves between the N and S poles with velocity v.

If the charge is crossing magnetic field lines, it experiences a force F.F

B

(Perhaps better to put it the other way: if the charge experiences a force, then a magnetic field B is present!)


Magnetic Field IIIMagnetic Field III

N

S

qvF

B

Operationally, magnetic field is defined in terms of q, v and F, according to the formula

F = qvB.

Notice, the force is the cross-product, or vector product of qv and B. Thus F is both v and B.

Recall that ab = |a||b|sin n, where is the angle between a and b andthe direction of n is determined by theright-hand rule.

Alternatively, x y z

x y z

y z z y z x x z x y y x

a a a

b b b

a b a b a b a b a b a b

i j k

a b

i j k


Magnetic Field IVMagnetic Field IV

F[N] = q[A-s]v[m-s-1]BFor consistency, units of B must be N-(A-m)-1

1 N-(A-m)-1 1 T (tesla)

If a current of 1 A flows in a direction perpendicular to the field lines of a 1 T magnetic field, each one-meter length of moving charges will experience a magnetic force of 1 N

B goes by several different names in physics literature:Magnetic field

Magnetic induction

Magnetic induction vector

Magnetic flux density


Magnetic Pole Strength, Magnetic Moment IMagnetic Pole Strength, Magnetic Moment I

N

S

N

S

B

θ

F1

F2

qm |F|/|B| = magnetic pole strength.

Units are N/N-(A-m)-1 = A-m

F1 and F2 are a force couple, and as such exert a net torque on the bar magnet

= ×F = ×qmB = qm ×B

m qm = magnetic moment, or magnetic dipole moment [A-m2].

So, = m×B


Magnetic Pole Strength, Magnetic Moment IIMagnetic Pole Strength, Magnetic Moment II

N

S

N

S

B

θ

F1

F2

Recall also that, in general, [N-m] = dL/dt

L = angular momentum [kg-m2-s-1]

(Analogy to Newton’s second law: F [N] = dp/dt, where p [kg-m-s-1] = linear momentum)

Definition of magnetic moment as product of distance and pole strength is analogous to electric dipole moment definition (i.e., product of separated charge and distance). But it is somewhat fictitious, given that there are no magnetic monopoles.

Note that we could define m without invoking the intermediate concept of magnetic pole strength: m (|F|/|B|).


Magnetic Moment IIIMagnetic Moment III

N

S

B

I

a

b

A loop carrying current I is placed in a uniform magnetic field.

There is no magnetic force on the loop segments in which the current flows || the field lines. There is a magnetic force on the segments in which the current is the field lines.

The magnitude of the force can be computed from F = qv×B, when we recall that charge [A-s] times velocity [m-s-1] equals current [A] times length [m].

|F1| = |F2| = Ia|B|,

a force couple that generates a net torque:

F1

F2

|| = (Ia|B|)(bsin) (force times distance)

= IA|B|sin (A = ab = loop area)


Magnetic Moment IIIMagnetic Moment III

N

S

B

A loop carrying current I is placed in a uniform magnetic field.

There is no magnetic force on the loop segments in which the current flows || the field lines. There is a magnetic force on the segments in which the current is the field lines.

The magnitude of the force can be computed from F = qv×B, when we recall that charge [A-s] times velocity [m-s-1] equals current [A] times length [m].

|F1| = |F2| = Ia|B|,

a force couple that generates a net torque:

|| = (Ia|B|)(bsin) (force times distance)

= IA|B|sin (A = ab = loop area)

= IAn×B m×Bm = IAn, magnetic moment

For a N-turn coil, m = NIAn

a

bI

F1

F2

n


Magnetization: Definition, Relation to Magnetic MomentMagnetization: Definition, Relation to Magnetic Moment

A material of volume V [m3] has magnetic moment m [A-m2]

Its magnetization M is its magnetic moment per unit volume:

M m/V [A-m-1]


Angular Momentum Magnetic MomentAngular Momentum Magnetic Moment

r

Particle of charge q and mass m (do not confuse with m!), moving at speed |v| in a circular orbit of radius |r|.

Orbital period: T = 2|r|/|v|

Current: I = q/T = q|v|/(2|r|)

Magnetic moment:

|m| = IA = [q|v|/(2|r|)](|r|2)

= ½q|v||r|

Angular momentum:

|L| = m|v||r| = m(½q|v||r|)/(½q)

= 2m|m|/q

(Recall that L = mr×v)


Magnetogyric RatioMagnetogyric Ratio

|m|/|L| = (½q|v||r|)/(m|v||r|) = q/(2m)

is called the magnetogyric ratio

[A-s-kg-1] is inversely proportional to particle’s

mass-to-charge ratio r

Notice that units of can be rearranged to:

A-s-kg-1 = A-m-s2-kg-1-m-1-s-1 = (A-m)-N-1-s-1

= s-1-[N-(A-m)-1]-1 = Hz-T-1

If a charged particle has non-zero angular momentum, then it also has a magnetic moment (and vice versa), and m || L.

m B0

Now rotate plane of current loop, and place it in a uniform magnetic field:

m precesses about an axis parallel to field lines, but with what frequency?


Magnetogyric Ratio and Precession FrequencyMagnetogyric Ratio and Precession Frequency

m B0

= 2f |B0|

Proportionality constant is the magnetogyric ratio!

f = |B0|

(Some authors define so that = |B0|; be aware!)

Thus for macroscopic, or classical, cycling currents, precession frequency is inversely proportional to mass-to-charge ratio.

For quantum mechanical cycling currents (e.g., electrons, protons, neutrons, many types of atomic nuclei), relationship is more complicated, but same qualitative trend is seen. Among atomic nuclei, precession frequency trends downward as atomic number Z increases.


Angular Momentum (Spin) of Atomic Nuclei IAngular Momentum (Spin) of Atomic Nuclei I

Every atomic nucleus has a spin quantum number, s

Permissible values of s depend on mass number A.Odd A: s may be 1/2, 3/2, 5/2, …

Even A: s may be 0, 1, 2, …

The magnitude of the intrinsic angular momentum, or spin, S that corresponds to a given value of s is |S| =

, where h is Planck’s constant

The direction of S can not be precisely defined. The most we can say is that the component of S in a given direction is equal to , where permissible values of ms are -s, -s+1,…,s-1,s.

1s s

2h

sm


Angular Momentum (Spin) of Atomic Nuclei IIAngular Momentum (Spin) of Atomic Nuclei II

1 , z ss s S mS

1 312

3 1 12 2 2

So, if (e.g., H, H):

, or +z

s

SS

B0, z

0

+0.5

-0.5

1 1cos

354.74

2And if 1 (e.g., H):

2 , ,0, or +z

s

SS

B0, z

0

+1

-1

1

11

2

1cos

245

90


Alignment of 1H Nuclei in a Magnetic FieldAlignment of 1H Nuclei in a Magnetic Field

mmz

mmz

B0Protons must orient themselves such that the z-components of their magnetic moments lie in one of the two permissible directions

What about direction of m?

mzCorrect quantum mechanical description is that m does not have an orientation, but is delocalized over all directions that are consistent with fixed value of mz.

For the purpose of predicting/interpreting the interaction of m with radiation, we can think of m as a well-defined vector rapidly precessing about z-direction.

mz

What is the precession frequency?


Orientational Distribution of 1H NucleiOrientational Distribution of 1H Nuclei

What fraction of nuclei are in the “up” state and what fraction are “down”?

mmz

mmz

B0Protons must orient themselves such that the z-components of their magnetic moments lie in one of the two permissible directions

The orientation with mz aligned with B0 has lower potential energy, and is favored (North pole of nuclear magnet facing South pole of external field).

The fractional population of the favored state increases with increasing |B0|, and increases with decreasing (absolute) temperature T.

Boltzmann distribution: 0

23 -1down

up34

, 1.381 10 J - K

6.626 10 J - s

h

kTN

e kN

h

B


Transitions Between Spin States (Orientations) ITransitions Between Spin States (Orientations) I

QM result: energy difference between the “up” and “down” states of mz is ΔE0 = h|B0|

As always, frequency of radiation whose quanta (photons) have precisely that amount of energy is f0 = ΔE0/h

So, f0 = |B0|

Different nuclei have different values of . (Units of are MHz/T.)

1H: = 42.58; 2H: = 6.53; 3H: = 45.41

13C: = 10.71

31P: = 17.25

23Na: = 11.27

39K: = 1.99

19F: = 40.08


Transitions Between Spin States IITransitions Between Spin States II

The frequency f0 that corresponds to the energy difference between the spin states is called the Larmor frequency.

The Larmor frequency f0 is the (apparent) precession frequency for m about the magnetic field direction.

(In QM, the azimuthal part of the proton’s wave function precesses at frequency f0, but this is not experimentally observable.)

Three important processes occur:

+

+

+

+

+

+

hf0 hf0

2hf0

Absorption Stimulated emission

Spontaneous emission

(Relaxation)


Transitions Between Spin States IIITransitions Between Spin States III

The number of 1H nuclei in the low-energy “up” state is slightly greater than the number in the high-energy “down” state.

Irradiation at the Larmor frequency promotes the small excess of low-energy nuclei into the high-energy state.

When the nuclei return to the low-energy state, they emit radiation at the Larmor frequency.

The radiation emitted by the relaxing nuclei is the NMR signal that is measured and later used to construct MR images.

+

+

+

+

+

+

hf0 hf0

2hf0

Absorption Stimulated emission

Spontaneous emission

(Relaxation)


SaturationSaturation

Suppose the average time required for an excited nucleus to return to the ground state is long (low relaxation rate, long excited-state lifetime)

If the external radiation is intense or is kept on for a long time, ground-state nuclei may be promoted to the excited state faster than they can return to the ground state.

Eventually, an exact 50/50 distribution of nuclei in the ground and excited states is reached

At this point the system is saturated. No NMR signal is produced, because the rates of “up”→“down” and “down”→“up” transitions are equal.


Radiation ↔ Rotating Magnetic Field IRadiation ↔ Rotating Magnetic Field I

N

S

B0

Static magnetic field

Sinusoidal EM field

Imagine that we replace the EM

field with…

y

x

z


S

S

Radiation ↔ Rotating Magnetic Field IIRadiation ↔ Rotating Magnetic Field II

N

S

B0 …two more magnets, whose fields are B0, that rotate, in opposite

directions, at the Larmor frequency

N

N


Radiation ↔ Rotating Magnetic Field IIIRadiation ↔ Rotating Magnetic Field III

Simplified bird’s-eye view of counter-rotating magnetic field vectors

t = 0 1/(8f0) 1/(4f0) 3/(8f0) 1/(2f0) 5/(8f0) 3/(4f0) 7/(8f0) 1/f0

So what does resulting B vs. t look like?

This time-dependent field is called B1


Rotating Reference Frame IRotating Reference Frame I

y

x

zB0

(1-10 T)

y

x

z, z’

y’x’

Instead of a constant rotation angle , let = 2f0t = 0t

Original (laboratory) coordinate system

Coordinate system rotated about z axis

counter-rotating magnetic fields

resultant field, sinusoidally varying

in x direction

x’ = ysin + xcos = -ysin0t + xcos0t

y’ = ycos - xsin = ycos0t + xsin0t


Rotating Reference Frame IIRotating Reference Frame II

B0

(1-10 T)

y

x

z, z’

y’x’

Rotating coordinate system, observed from laboratory frame

These axes are rotating in the xy plane, with frequency f0

B0

z’

y’

x’

Rotating coordinate system, observed from within itself

But what is the magnitude of B0 in this reference frame?

This magnetic field, rotating at 2f0, can be ignored; its frequency is too high to induce transitions between orientational states of the protons’ magnetic moments

This magnetic field, B1, is fixed in direction and has constant magnitude: ~0.01 T


Excursion: Bloch Equations IExcursion: Bloch Equations I

For an individual atomic nucleus, dL/dt = m×B

L – angular momentum, m – magnetic moment, B – magnetic field

dL/dt = m×B = dm/dt

Summing over all nuclei gives the corresponding equation for the bulk (macroscopic) magnetization: dM/dt = M×B

The net magnetic field B is the vector sum of the static longitudinal field and the counter-rotating transverse fields. In the laboratory frame, these sum to: (B1cosω0t + B1cosω0t + 0)i + (B1sinω0t - B1sinω0t + 0)j + (0 + 0 + B0)k.

x y z

x y z

y z z y z x x z x y y x

M M M

B B B

M B M B M B M B M B M B

i j k

M B

i j k


Excursion: Bloch Equations IIExcursion: Bloch Equations II

Combining the preceding equations, we have:

dM/dt = [(MyBz – MzBy)i + (MzBx – MxBz)j + (MxBy – MyBx)k],

and Bx = 2|B1|cosω0t, By = 0, Bz = |B0|

So the three components of dM/dt are:

dMx/dt = My|B0|,

dMy/dt = (2Mz|B1|cosω0t - Mx|B0|),

dMz/dt = -2My|B1|cosω0t

Then Bloch assumed that there are two relaxation processes (i.e., spin-lattice and spin-spin), and that these are first-order, with time constants T1 and T2. So the final form of the Bloch equations are:

What do these mean?


Excursion: Bloch Equations IIIExcursion: Bloch Equations III

Bloch equations:

dMx/dt = My|B0| - Mx/T2,

dMy/dt = (2Mz|B1|cosω0t - Mx|B0|) – My/T2,

dMz/dt = -2My|B1|cosω0t - (Mz – M0)/T1

These are three coupled ordinary linear differential equations.

Can be solved exactly, if laboriously

Tell us exactly how the magnetization responds to an EM field, of any duration, strength, and frequency

• The quantity ω0 in the equations can actually be any frequency (“off-resonance” rotation), doesn’t have to be the Larmor frequency.

Now we are able to answer question from Slide 29:

What is |B0| in the reference frame rotating at the Larmor frequency (“on-resonance” rotation)?


Effective Field IEffective Field I

M = Mxi + Myj + Mzk

dM/dt = (Mx/t)i + Mx(i/t) + (My/t)j + My(j/t) + (Mz/t)k + Mz(k/t)

= [(Mx/t)i + (My/t)j + (Mz/t)k] + [Mx(i/t) + My(j/t) + Mz(k/t)]

i/t = (ω×i)/(2), j/t = (ω×j)/(2), k/t = (ω×k)/(2)• ω is the angular frequency vector

dM/dt = (dM/dt)fixed = M/t + ω×(Mxi + Myj + Mzk)/(2) = (M/t)rot + (ω×M)/(2)

As shown previously, (dM/dt)fixed = M×B

So, (M/t)rot = M×B - (ω×M)/(2) = M×B + (M×ω)/(2)= M×(B + ω/(2)) M×Beff

The apparent, or effective field in a rotating reference frame is different from that in the laboratory frame


Effective Field IIEffective Field II

The apparent, or effective field in a rotating reference frame is different from that in the laboratory frame

Starting with a homogeneous static longitudinal field B0, add a transverse field B1 that rotates in the x-y plane with frequency f = /(2). In the frame that rotates at frequency f, the effective field is Beff = B0 + ω/(2) + B1

If = 0 (f = f0), then the effective longitudinal field is zero!

Beff = B1, the transverse field is all the field there is

Magnetization M precesses about B1 with frequency f1 = |B1|

If the B1 field is present for time tp, then the resulting tip angle is = 2 |B1|tp


Relaxation IRelaxation I

From Slide 31, what are spin-lattice relaxation and spin-spin relaxation?

What do time constants T1 and T2 mean?

“Lattice” means the material (i.e., tissue) the 1H nuclei are embedded in

1H nuclei are not the only things around that have magnetic moments

• Other species of nuclei• Electrons

A 1H magnetic moment can couple (i.e., exchange energy) with these other moments


Spin-Lattice Relaxation ISpin-Lattice Relaxation I

Spin-lattice interactions occur whenever a physical process causes the magnetic field at a 1H nucleus to fluctuateSpin-lattice interactions cause the perturbed distribution of magnetic moments (i.e., tipped bulk magnetization) to return to equilibrium more rapidlyTypes of spin-lattice interaction

Magnetic dipole-dipole interactionsElectric quadrupole interactionsChemical shift anisotropy interactionsScalar-coupling interactionsSpin-rotation interactions

What is the T1 time constant associated with these processes?

Look ’em up!


x׳

y׳

z׳

B0

Spin-Lattice Relaxation IISpin-Lattice Relaxation II

What is the T1 time constant associated with spin-lattice interactions?

At equilibrium, M point in z׳ direction

Recall that static field direction defines z, z׳


x׳

y׳

z׳

B0

Spin-Lattice Relaxation IIISpin-Lattice Relaxation III


Now impose a transverse magnetic field

…and tip the magnetization towards the x׳-y׳ plane

Then turn the transverse field off


Spin-Lattice Relaxation IVSpin-Lattice Relaxation IV


x׳

y׳

z׳ B0In the laboratory frame, M takes a spiralling path back to its equilibrium orientation. But here in the rotating frame, it simply rotates in the y׳-z׳ plane.

The z component of M, Mz, grows back into its equlibrium value, exponentially:

Mz = |M|(1 - e-t/T1)

Mz M


Relaxation IIRelaxation II

From Slide 31, what are spin-lattice relaxation and spin-spin relaxation?

What do time constants T1 and T2 mean?

A 1H magnetic moment can couple (i.e., exchange energy with) the magnetic moments of other 1H nuclei in its vicinity

These are called “spin-spin coupling”Spin-spin interactions occur when the magnetic field at a given 1H nucleus fluctuates

Therefore, should the rates of these interaction depend on temperature? If so, do they increase or decrease with increasing temperature?


Spin-Spin Relaxation ISpin-Spin Relaxation I

What is the T2 time constant associated with spin-spin interactions?

x׳

y׳

z׳ B0

MMz

Mtr If there were no spin-spin coupling, the transverse component of M, Mtr, would decay to 0 at the same rate as Mz returns to its original orientation

What are the effects of spin-spin coupling?


Spin-Spin Relaxation IISpin-Spin Relaxation II

W hat are the effects of spin-spin coupling?

Because the magnetic fields at individual 1H nuclei are not exactly B0, their Larmor frequencies are not exactly f0.

x׳

y׳

z׳ B0

MzBut the frequency of the rotating reference frame is exactly f0. So in this frame M appears to separate into many magnetization vectors the precess about z׳.

Some of them (f < f0) precess counterclockwise (viewed from above), others (f > f0) precess clockwise.


Spin-Spin Relaxation IIISpin-Spin Relaxation III

W hat are the effects of spin-spin coupling?

Within a short time, M is completely de-phased. It is spread out over the entire cone defined by cosθ = Mz/|M|

x׳

y׳

z׳ B0

MzWhen M is completely de-phased, Mtr is 0, even though Mz has not yet grown back completely: Mtr = 0, Mz < |M|

Mtr decreases exponentially, with time constant T2:

Mtr = Mtr0 e-t/T2

This also shows why T2 can not be >T1. It must be the case that T2 T1. In practice, usually T2 << T1.


Relaxation IIIRelaxation III

In this example, T1 = 0.5 s

In this example, T2 = 0.2 s


Effect of B0 Field HeterogeneityEffect of B0 Field Heterogeneity

What is the common element in spin-spin and spin-lattice interactions?

They require fluctuations in the strength of the magnetic field in the immediate environment of a 1H nucleus

If the static B0 field itself is not perfectly uniform, its spatial heterogeneity accelerates the de-phasing of the bulk magnetization vector

The net, or apparent, decay rate of the transverse magnetization is 1/T2* 1/T2 + |B0|.

T2* (“tee-two-star”) has a spin-spin coupling contribution and a field inhomogeneity contribution

T2* < T2 always, and typically T2* << T2

bmi i fs05 – class 9 “mri physic” slide 1 biomedical imaging i class 9 – magnetic resonance...

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