BMI1 FS05 Class09 MRI Theory

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Published on November 14, 2007

Author: brod

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Biomedical Imaging I:  Biomedical Imaging I Class 9 – Magnetic Resonance Imaging (MRI) Physical Theory 11/09/05 Magnetic Resonance in a Nutshell:  Magnetic Resonance in a Nutshell Hydrogen Nuclei (Protons) Axis of Angular Momentum (Spin), Magnetic Moment Magnetic Resonance in a Nutshell:  Magnetic Resonance in a Nutshell Magnetic Resonance in a Nutshell:  Magnetic Resonance in a Nutshell Irradiating with a (radio frequency) field of frequency f0, causes spins to precess coherently, or in phase Magnetic Field I:  Magnetic Field I N S magnetic 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 II:  Magnetic Field II N S Magnetic Field III:  Magnetic Field III Operationally, magnetic field is defined in terms of q, v and F, according to the formula F = qvB. Notice, the force is the cross-product, or vector product of qv and B. Thus F is  both v and B. Recall that ab = |a||b|sin n, where  is the angle between a and b and the direction of n is determined by the right-hand rule. Magnetic Field IV:  Magnetic Field IV F[N] = q[A-s]v[m-s-1]B For 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 I:  Magnetic Pole Strength, Magnetic Moment I qm  |F|/|B| = magnetic pole strength. Units are N/N-(A-m)-1 = A-m  = ×F = ×qmB = qm ×B m  qm = magnetic moment, or magnetic dipole moment [A-m2]. So,  = m×B Magnetic Pole Strength, Magnetic Moment II:  Magnetic Pole Strength, Magnetic Moment II 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 III:  Magnetic 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]. Magnetic Moment III:  Magnetic 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×B m = IAn, magnetic moment For a N-turn coil, m = NIAn Magnetization: Definition, Relation to Magnetic Moment:  Magnetization: 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 Moment:  Angular Momentum  Magnetic Moment 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 Ratio:  Magnetogyric 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 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 precesses about an axis parallel to field lines, but with what frequency? Magnetogyric Ratio and Precession Frequency:  Magnetogyric Ratio and Precession Frequency  = 2f |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 I:  Angular 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. Angular Momentum (Spin) of Atomic Nuclei II:  Angular Momentum (Spin) of Atomic Nuclei II Alignment of 1H Nuclei in a Magnetic Field:  Alignment of 1H Nuclei in a Magnetic Field Protons 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? What is the precession frequency? Orientational Distribution of 1H Nuclei:  Orientational Distribution of 1H Nuclei What fraction of nuclei are in the “up” state and what fraction are “down”? Protons 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. Transitions Between Spin States (Orientations) I:  Transitions 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 II:  Transitions 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: Transitions Between Spin States III:  Transitions 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. Saturation:  Saturation 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 I:  Radiation ↔ Rotating Magnetic Field I N S B0 Imagine that we replace the EM field with… Radiation ↔ Rotating Magnetic Field II:  S S Radiation ↔ 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 III:  Radiation ↔ 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 I:  Rotating Reference Frame I y x z B0 (1-10 T) y x z, z’ y’ x’  Instead of a constant rotation angle , let  = 2f0t = 0t Original (laboratory) coordinate system Coordinate system rotated about z axis x’ = ysin + xcos = -ysin0t + xcos0t y’ = ycos - xsin = ycos0t + xsin0t Rotating Reference Frame II:  Rotating Reference Frame II B0 (1-10 T) y x z, z’ y’ x’  Rotating coordinate system, observed from laboratory frame B0 z’ y’ x’ Rotating coordinate system, observed from within itself Excursion: Bloch Equations I:  Excursion: 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. Excursion: Bloch Equations II:  Excursion: 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 = -2My|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: Excursion: Bloch Equations III:  Excursion: Bloch Equations III Bloch equations: dMx/dt = My|B0| - Mx/T2, dMy/dt = (2Mz|B1|cosω0t - Mx|B0|) – My/T2, dMz/dt = -2My|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 |B0| in the reference frame rotating at the Larmor frequency (“on-resonance” rotation)? Effective Field I:  Effective 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 II:  Effective 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 I:  Relaxation 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 I:  Spin-Lattice Relaxation I Spin-lattice interactions occur whenever a physical process causes the magnetic field at a 1H nucleus to fluctuate Spin-lattice interactions cause the perturbed distribution of magnetic moments (i.e., tipped bulk magnetization) to return to equilibrium more rapidly Types of spin-lattice interaction Magnetic dipole-dipole interactions Electric quadrupole interactions Chemical shift anisotropy interactions Scalar-coupling interactions Spin-rotation interactions What is the T1 time constant associated with these processes? Spin-Lattice Relaxation II:  Spin-Lattice Relaxation II What is the T1 time constant associated with spin-lattice interactions? Spin-Lattice Relaxation III:  Spin-Lattice Relaxation III What is the T1 time constant associated with spin-lattice interactions? Spin-Lattice Relaxation IV:  Spin-Lattice Relaxation IV What is the T1 time constant associated with spin-lattice interactions? x׳ y׳ z׳ B0 The z component of M, Mz, grows back into its equlibrium value, exponentially: Mz = |M|(1 - e-t/T1) Mz M Relaxation II:  Relaxation 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 I:  Spin-Spin Relaxation I What is the T2 time constant associated with spin-spin interactions? x׳ y׳ z׳ B0 M Mz 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 II:  Spin-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 Mz But 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 III:  Spin-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 Mz When 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 III:  Relaxation III In this example, T1 = 0.5 s In this example, T2 = 0.2 s Effect of B0 Field Heterogeneity:  Effect 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

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