Topics. The concept of spin Precession of magnetic spin Relaxation Bloch Equation. Bioengineering 280A Principles of Biomedical Imaging
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1 Bioengineering 280A Principles of Biomedical Imaging Fall Quarter 2006 MRI Lecture 1 Topics The concept of spin Precession of magnetic spin Relaxation Bloch Equation 1
2 Spin Intrinsic angular momentum of elementary particles -- electrons, protons, neutrons. Spin is quantized. Key concept in Quantum Mechanics. The History of Spin 1921 Stern and Gerlach observed quantization of magnetic moments of silver atoms 1925 Uhlenbeck and Goudsmit introduce the concept of spin for electrons Stern and Gerlach measure the effect of nuclear spin Rabi predicts and observes nuclear magnetic resonance. 2
3 Classical Magnetic Moment I A r µ = IAˆ n Energy in a Magnetic Field Maximum Energy State B E = " r µ B = "µ z B Lorentz Force Minimum Energy State 3
4 Stern-Gerlach Experiment Image from Force in a Field Gradient F = "#E = µ z $B z $z Deflected up Increasing vertical B-field. Deflected down 4
5 Stern-Gerlach Experiment Image from Quantization of Magnetic Moment The key finding of the Stern- Gerlach experiment is that the magnetic moment is quantized. That is, it can only take on discrete values. In the experiment, the finding was that the component of magnetization along the direction of the applied field was quantized: µ z = + µ 0 OR - µ 0 5
6 Magnetic Moment and Angular Momentum A charged sphere spinning about its axis has angular momentum and a magnetic moment. This is a classical analogy that is useful for understanding quantum spin, but remember that it is only an analogy! Relation: µ = γ S where γ is the gyromagnetic ratio and S is the spin angular momentum. Quantization of Angular Momentum Because the magnetic moment is quantized, so is the angular momentum. In particular, the z-component of the angular momentum Is quantized as follows: S z = m s h m s "{#s,#(s #1),...s } s is an integer or half intege 6
7 Nuclear Spin Rules Number of Protons Number of Neutrons Spin Examples Even Even 0 12 C, 16 O Even Odd j/2 17 O Odd Even j/2 1 H, 23 Na, 31 P Odd Odd j 2 H Hydrogen Proton Spin 1/2 # S z = $ +h/2 %& "h/2 # +'h /2 µ z = $ %& "'h /2 7
8 Boltzmann Distribution Ε = µ z Β 0 B 0 ΔΕ = γhβ 0 Ε = µ z Β 0 Number Spins Down Number Spins Up = exp(-δe/kt) Ratio = at 1.5T!!! Corresponds to an excess of about 10 up spins per million Equilibrium Magnetization # M 0 = N µ z = N n up µ z % $ ( ) " n down ( µ z ) N = Nµ eµ z B /( kt ) " e "µ z B /( kt ) ( ) + e "µ z B /( kt ) e µ z B / kt ) Nµ z 2 B /(kt) = N* 2 h 2 B /(4kT) & ( ' N = number of nuclear spins per unit volume Magnetization is proportional to applied field. 8
9 Bigger is better 3T Human imager at UCSD. 7T Rodent Imager at UCSD 7T Human imager at U. Minn. 9.4T Human imager at UIC Gyromagnetic Ratios Nucleus Spin Magnetic γ/(2π) Abundance (MHz/Tesla) Moment 1H 1/ M 23Na 3/ mm 31P 1/ mm Source: Haacke et al., p. 27 9
10 Torque B µ For a non-spinning magnetic moment, the torque will try to align the moment with magnetic field (e.g. compass needle) N Torque N = µ x B Precession Torque ds dt N = µ x B = N ds dt = µ x B dµ dt = µ x γb Change in Angular momentum µ = γ S Relation between magnetic moment and angular momentum 10
11 dµ dt = µ x γb B Precession Analogous to motion of a gyroscope Precesses at an angular frequency of ω = γ Β This is known as the Larmor frequency. dµ µ Larmor Frequency ω = γ Β f = γ Β / (2 π) Angular frequency in rad/sec Frequency in cycles/sec or Hertz, Abbreviated Hz For a 1.5 T system, the Larmor frequency is MHz which is million cycles per second. For comparison, KPBS-FM transmits at 89.5 MHz. Note that the earth s magnetic field is about 50 µτ, so that a 1.5T system is about 30,000 times stronger. 11
12 Notation and Units 1 Tesla = 10,000 Gauss Earth's field is about 0.5 Gauss 0.5 Gauss = 0.5x10-4 T = 50 µt " = radians/second/gauss " = " /2# = 4258 Hz/Gauss = MHz/Tesla Recap Spins: angular momentum and magnetic moment are quantized. Spins precess about a static field at the Larmor frequency. In MRI we work with the net magnetic moment. In the presence of a static field and non-zero temperature, the equilibirum net magnetic moment is aligned with the field (longitudinal), since transverse components cancel out. We will use an radiofrequency pulse to tip this longitudinal component into the transverse plane. 12
13 Vector sum of the magnetic moments over a volume. M = 1 V Magnetization Vector For a sample at equilibrium in a magnetic field, the transverse components of the moments cancel out, so that there is only a longitudinal component. Equation of motion is the same form as for individual moments. dm dt " protons in V µ i = "M # B Simplified Drawing of Basic Instrumentation. Body lies on table encompassed by coils for static field B o, gradient fields (two of three shown), and radiofrequency field B 1. Image, caption: copyright Nishimura, Fig
14 RF Excitation From Levitt, Spin Dynamics, 2001 RF Excitation At equilibrium, net magnetizaion is parallel to the main magnetic field. How do we tip the magnetization away from equilibrium? Image & caption: Nishimura, Fig. 3.2 B 1 radiofrequency field tuned to Larmor frequency and applied in transverse (xy) plane induces nutation (at Larmor frequency) of magnetization vector as it tips away from the z-axis. - lab frame of reference 14
15 RF Excitation a) Laboratory frame behavior of M b) Rotating frame behavior of M Images & caption: Nishimura, Fig
16 RF Excitation From Buxton 2002 Free Induction Decay (FID) 16
17 z RF Excitation z z M0 y y y x x x Doing nothing Excitation M 0 (1 e -t/t1 ) z Relaxation e -t/t2 y x T1 recovery T2 decay Credit: Larry Frank Relaxation An excitation pulse rotates the magnetization vector away from its equilibrium state (purely longitudinal). The resulting vector has both longitudinal M z and tranverse M xy components. Due to thermal interactions, the magnetization will return to its equilibrium state with characteristic time constants. T 1 spin-lattice time constant, return to equilibrium of M z T 2 spin-spin time constant, return to equilibrium of M xy 17
18 Longitudinal Relaxation dm z dt = " M z " M 0 T 1 After a 90 degree pulse M z (t) = M 0 (1" e "t /T 1 ) Due to exchange of energy between nuclei and the lattice (thermal vibrations). Process continues until thermal equilibrium as determined by Boltzmann statistics is obtained. The energy ΔE required for transitions between down to up spins, increases with field strength, so that T 1 increases with B. T1 Values Gray Matter muscle White matter kidney liver fat Image, caption: Nishimura, Fig
19 Transverse Relaxation dm xy dt = " M xy T 2 x y x y x y z z z Each spin s local field is affected by the z-component of the field due to other spins. Thus, the Larmor frequency of each spin will be slightly different. This leads to a dephasing of the transverse magnetization, which is characterized by an exponential decay. T 2 is largely independent of field. T 2 is short for low frequency fluctuations, such as those associated with slowly tumbling macromolecules. T2 Relaxation After a 90 degree excitation M xy (t) = M 0 e "t /T 2 19
20 T2 Relaxation Runners x x x xx x x x x x x xx x x x x x xx x x x x xx x x Net signal Credit: Larry Frank T2 Values Tissue T 2 (ms) gray matter 100 white matter 92 muscle 47 fat 85 kidney 58 liver 43 CSF 4000 Table: adapted from Nishimura, Table 4.2 Solids exhibit very short T 2 relaxation times because there are many low frequency interactions between the immobile spins. On the other hand, liquids show relatively long T 2 values, because the spins are highly mobile and net fields average out. 20
21 Example T 1 -weighted Density-weighted T 2 -weighted Questions: How can one achieve T2 weighting? What are the relative T2 s of the various tissues? Bloch Equation dm dt = M " #B $ M i + M j x y ( $ M $ M z 0)k T 2 T 1 Precession Transverse Relaxation Longitudinal Relaxation i, j, k are unit vectors in the x,y,z directions. 21
22 Free precession about static field dm dt = M " #B ˆ i ˆ j ˆ k = # M x M y M z B dμ Μ B x B y B z ( ) % i ˆ B z M y $ B y M ( ' z * = #' $ ˆ j ( B z M x $ B x M z )* ' k ˆ & ( B y M x $ B x M y ) * ) Free precession about static field " dm x $ $ dm y # $ dm z dt% " B z M y ) B y M z % ' $ ' dt' = ( $ B x M z ) B z M x ' dt& ' # $ B y M x ) B x M y & ' " 0 B z )B y %" $ ' $ = ( $ )B z 0 B x ' $ # $ B y )B x 0 & '# $ M x M y M z % ' ' &' 22
23 " dm x $ $ dm y # $ dm z Precession dt% " 0 B 0 0% " ' $ ' $ dt' = ( $ )B ' $ dt& ' # $ 0 0 0& '# $ M x M y M z % ' ' &' Useful to define M " M x + jm y dm dt = d dt( M x + im y ) = " j#b 0 M Solution is a time-varying phasor M(t) = M(0)e " j#b 0t = M(0)e " j$ 0t M x jm y Question: which way does this rotate with time? Matrix Form with B=B 0 " dm x $ $ dm y # $ dm z dt% "(1/T 2 )B 0 0 %" ' $ ' $ dt' = $ ()B 0 1/T 2 0 ' $ dt& ' # $ 0 0 (1/T 1 &' # $ M x M y M z % " 0 % ' $ ' ' + $ 0 ' &' # $ M 0 /T 1 &' 23
24 Z-component solution M z (t) = M 0 + (M z (0) " M 0 )e "t /T 1 Saturation Recovery If M z (0) = 0 then M z (t) = M 0 (1" e "t /T 1 ) Inversion Recovery If M z (0) = "M 0 then M z (t) = M 0 (1" 2e "t /T 1 ) M " M x + jm y Transverse Component ( ) dm dt = d dt M x + im y = " j (# 0 +1/T 2 )M M(t) = M(0)e " j# 0 t e "t /T 2 24
25 Summary 1) Longitudinal component recovers exponentially. 2) Transverse component precesses and decays exponentially. Fact: Can show that T 2 < T 1 in order for M(t) M 0 Physically, the mechanisms that give rise to T 1 relaxation also contribute to transverse T 2 relaxation. 25
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