bmi i fs05 – class 9 “mri physic” slide 1 biomedical imaging i class 9 – magnetic resonance...
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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
BMI I FS05 – Class 9 “MRI Physics” Slide 3
Magnetic Resonance in a NutshellMagnetic Resonance in a Nutshell
Spins PRECESS at a single frequency (f0), but incoherently − they are not in phase
External Magnetic Field
BMI I FS05 – Class 9 “MRI Physics” Slide 4
Magnetic Resonance in a NutshellMagnetic Resonance in a Nutshell
Irradiating with a (radio frequency) field of frequency f0, causes spins to precess coherently, or in phase
BMI I FS05 – Class 9 “MRI Physics” Slide 5
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?
BMI I FS05 – Class 9 “MRI Physics” Slide 6
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!)
BMI I FS05 – Class 9 “MRI Physics” Slide 7
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
BMI I FS05 – Class 9 “MRI Physics” Slide 8
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
BMI I FS05 – Class 9 “MRI Physics” Slide 9
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
BMI I FS05 – Class 9 “MRI Physics” Slide 10
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|).
BMI I FS05 – Class 9 “MRI Physics” Slide 11
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)
BMI I FS05 – Class 9 “MRI Physics” Slide 12
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
BMI I FS05 – Class 9 “MRI Physics” Slide 13
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]
BMI I FS05 – Class 9 “MRI Physics” Slide 14
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)
BMI I FS05 – Class 9 “MRI Physics” Slide 15
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?
BMI I FS05 – Class 9 “MRI Physics” Slide 16
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.
BMI I FS05 – Class 9 “MRI Physics” Slide 17
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
BMI I FS05 – Class 9 “MRI Physics” Slide 18
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
BMI I FS05 – Class 9 “MRI Physics” Slide 19
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?
BMI I FS05 – Class 9 “MRI Physics” Slide 20
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
BMI I FS05 – Class 9 “MRI Physics” Slide 21
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
BMI I FS05 – Class 9 “MRI Physics” Slide 22
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)
BMI I FS05 – Class 9 “MRI Physics” Slide 23
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)
BMI I FS05 – Class 9 “MRI Physics” Slide 24
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.
BMI I FS05 – Class 9 “MRI Physics” Slide 25
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
BMI I FS05 – Class 9 “MRI Physics” Slide 26
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
BMI I FS05 – Class 9 “MRI Physics” Slide 27
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
BMI I FS05 – Class 9 “MRI Physics” Slide 28
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
BMI I FS05 – Class 9 “MRI Physics” Slide 29
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
BMI I FS05 – Class 9 “MRI Physics” Slide 30
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
BMI I FS05 – Class 9 “MRI Physics” Slide 31
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?
BMI I FS05 – Class 9 “MRI Physics” Slide 32
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)?
BMI I FS05 – Class 9 “MRI Physics” Slide 33
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
BMI I FS05 – Class 9 “MRI Physics” Slide 34
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
BMI I FS05 – Class 9 “MRI Physics” Slide 35
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
BMI I FS05 – Class 9 “MRI Physics” Slide 36
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!
BMI I FS05 – Class 9 “MRI Physics” Slide 37
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׳
BMI I FS05 – Class 9 “MRI Physics” Slide 38
x׳
y׳
z׳
B0
Spin-Lattice Relaxation IIISpin-Lattice Relaxation III
What is the T1 time constant associated with spin-lattice interactions?
Now impose a transverse magnetic field
…and tip the magnetization towards the x׳-y׳ plane
Then turn the transverse field off
BMI I FS05 – Class 9 “MRI Physics” Slide 39
Spin-Lattice Relaxation IVSpin-Lattice Relaxation IV
What is the T1 time constant associated with spin-lattice interactions?
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
BMI I FS05 – Class 9 “MRI Physics” Slide 40
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?
BMI I FS05 – Class 9 “MRI Physics” Slide 41
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?
BMI I FS05 – Class 9 “MRI Physics” Slide 42
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.
BMI I FS05 – Class 9 “MRI Physics” Slide 43
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.
BMI I FS05 – Class 9 “MRI Physics” Slide 44
Relaxation IIIRelaxation III
In this example, T1 = 0.5 s
In this example, T2 = 0.2 s
BMI I FS05 – Class 9 “MRI Physics” Slide 45
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