# Bloch Model

Introduction

The phenomological equations proposed by Felix Bloch in 1946 have had a profound effect on the development of magnetic resonance, on the ways in which the experiments are described and on the analysis of line widths and saturation behavior. Here we will describe the phenomological model, derive the Bloch equations and solve them for steady-state conditions.

Bloch Equations

When a magnetic field is applied to a spin, the spin quantization axis is defined by the field direction. Spin magnetic moments aligned with the field are only slightly lower in energy than those aligned opposed to the field. If we consider ensemble of spins, the vector sum of all the spin magnetic moment or macroscopic magnetization:

$\displaystyle \vec{M}=\Sigma_i \vec{\mu}_i \ \ \ \ \ (1)$

At equilibrium ${\vec{M}}$ is in the direction of the field ${\vec{B}}$. If somehow ${\vec{M}}$ is tilted away from ${\vec{B}}$ there will be a torque that causes ${\vec{M}}$ to precess about ${\vec{B}}$ with the equation of motion

$\displaystyle {d\vec{M} \over dt}= \gamma \vec{B}\times \vec{M} \ \ \ \ \ (2)$

where ${\gamma=g\mu_B \hbar}$. In addition to the precessional motion , there are two relaxation effects.

If ${M_0}$ is the equilibrium magnetization along ${\vec{B}}$ and ${M_z}$ is the z-component magnetization under non-equilibrium conditions, then we assume ${M_z}$ approach ${M_0}$ with first-order kinetics:

$\displaystyle {dM_z \over dt}=-{M_z-M_0 \over T_1} \ \ \ \ \ (3)$

where ${T_1}$ is the chararcteristic time to reach the equilibrium. Since this process involve transfer of energy from the spin system to the surrounding lattices, ${T_1}$ is called the spin-relaxation time.

There is a second kind of relaxation process besides ${T_1}$ process. Suppose that ${\vec{M}}$ is somehow tilted down from the z-axis toward the xy-axis and the precessional motion is started. Each individual magnetic moment undergoes this precessional motion, but the individual spins may precess at slightly different rates that can be caused by either local shielding that makes small variations in ${\vec{B}}$ or the effective g-factor may vary slightly through the sample. Thus an ensemble of spins that all start out in phase will gradually lose phase coherence. The characteristic time for this process is called transverse relaxation time ${T_2}$, and that the transverse magnetization components decay to the equilibrium value of zero accordingly:

$\displaystyle {dM_x \over dt}=-{M_x \over T_2},\quad \quad {dM_y \over dt}=-{M_y \over T_2} \ \ \ \ \ (4)$

We have to notice that the dephasing of the transverse magnetization does not affect ${M_z}$; a ${T_2}$ process involves no energy transfer but, being a spontaneous process, does involve an increase in the entropy of the spin system.

The approach to equilirium by a ${T_1}$ process, in which ${M_z}$ approaches ${M_0}$, also causes ${M_x}$ and ${M_y}$ to approach zero. Thus, the ${T_2}$ of eq. (4) must include both of the effects of spin-lattice relaxation as well as the dephasing of the transverse magnetization. Transfer relaxation is often much faster than spin-lattice relaxation and ${T_2}$ is then determined mostly by spin dephasing. However, in general we should write:

$\displaystyle {1 \over T_2}={1 \over T_1}+{1 \over T_{2'}} \ \ \ \ \ (5)$

where ${T_{2'}}$ is the spin dephasing relaxation time, and ${T_2}$ is the observed transverse relaxation time.

Derivation of Bloch Equation

Combining eqs. (2)-(4), we get:

$\displaystyle {d\vec{M} \over dt}= \gamma \vec{B}\times \vec{M}-\hat{i}{M_x \over T_2}-\hat{j}{M_y \over T_2}-\hat{k}{M_z-M_0 \over T_1} \ \ \ \ \ (6)$

In a magnetic resonance experiment, we apply not only a static field ${B_0}$ in the ${z}$-direction but also an oscillating radiation field in ${B_1}$ in the ${xy}$-plane, so that the total field is:

$\displaystyle \vec{B}=\hat{i}B_1\text{cos} \omega t + \hat{j} B_1 \text{sin}\omega t+ \hat{k}B_0 \ \ \ \ \ (7)$

Note there are other possible ways to impose a time-dependent ${B_1}$. The one described in eq. (7) corresponds to a circularly polarized field initially aligned along the ${x}$-axis and rotating about the ${z}$-axis in a counterclockwise direction.

Inserting the ${\vec{B}}$ into eq. (7) and separating the results into their components, then:

$\displaystyle {dM_x \over dt}=-\gamma B_0 M_y + \gamma B_1 M_z \text{sin} \omega t-{M_x \over T_2} \ \ \ \ \ (8)$

$\displaystyle {dM_y \over dt}=\gamma B_0 M_x - \gamma B_1 M_z \text{cos}\omega t-{M_y \over T_2} \ \ \ \ \ (9)$

$\displaystyle {dM_z \over dt}=\gamma B_1[M_y \text{cos}\omega t-M_x \text{sin}\omega t]-(M_z-M_0)/T_1 \ \ \ \ \ (10)$

It is convenient to write ${M_x}$ and ${M_y}$ as:

$\displaystyle M_x=\mu \text{cos}\omega t + \nu \text{sin} \omega t \ \ \ \ \ (11)$

$\displaystyle M_y=\mu \text{sin}\omega t-\nu \text{cos} \omega t \ \ \ \ \ (12)$

or

$\displaystyle \mu=M_x \text{cos} \omega t+ M_y \text{sin} \omega t \ \ \ \ \ (13)$

$\displaystyle \nu=M_x \text{sin} \omega t-M_y \text{cos} \omega t \ \ \ \ \ (14)$

This is equivalent to transformation into a coordinate system that rotates with the oscillating field; ${\mu}$ is that part of ${M_x}$ which is in phase with ${B_1}$ and ${\nu}$ is the part which is 90 out of phase. Differentiating eq. (13) and substituting eqs. (8) and (9), we get:

$\displaystyle {d\mu \over dt}=-[\omega-\gamma B_0]\nu-{\mu \over T_2} \ \ \ \ \ (15)$

Similarly, we obtain:

$\displaystyle {d\nu \over dt}=-[\omega-\gamma B_0]\mu-{\nu \over T_2}+\gamma B_1 M_z \ \ \ \ \ (16)$

$\displaystyle {dM_z \over dt}=-\gamma B_1 \nu-{M_z-M_0 \over T_1} \ \ \ \ \ (17)$

Equations (15-17) are the Bloch equations in the rotating coordinate frame.

In a continuous wave (CW) magnetic resonance experiment, the radiation field ${B_1}$ is continuous and ${B_0}$ is changed only slowly compared with the relaxation rates (slow passage conditions). Thus a steady-state solution to eqs (15-17) is appropriate. Setting the derivatives to zero and solving the three simultaneous equations, we get:

$\displaystyle u= {\gamma B_1 M_0 (\omega_0-\omega)T_2^2 \over 1+T_2^2 (\omega_0-\omega)^2+\gamma^2 B_1^2 T_1 T_2}, \ \ \ \ \ (18)$

$\displaystyle v= {\gamma B_1 M_0 T_2 \over 1+T_2^2 (\omega_0-\omega)^2+\gamma^2 B_1^2 T_1 T_2}, \ \ \ \ \ (19)$

$\displaystyle M_z= {M_0 \left[1+(\omega_0-\omega)T_2^2\right] \over 1+T_2^2 (\omega_0-\omega)^2+\gamma^2 B_1^2 T_1 T_2}, \ \ \ \ \ (20)$

where ${\omega_0=\gamma B_0}$ is Larmor frequency and corresponds to the frequency of the energy level transition.

With ${\chi={\partial M \over \partial H}}$ and ${\vec{B}=\mu_0 \vec{H}}$, we get the two components of ac susceptibilities:

$\displaystyle \chi '= {\gamma \mu_0 M_0 (\omega_0-\omega)T_2^2 \over 1+T_2^2 (\omega_0-\omega)^2+\gamma^2 B_1^2 T_1 T_2}, \ \ \ \ \ (21)$

$\displaystyle \chi ''= {\gamma \mu_0 M_0 T_2 \over 1+T_2^2 (\omega_0-\omega)^2+\gamma^2 B_1^2 T_1 T_2}. \ \ \ \ \ (22)$

${\chi'}$ is the ac susceptibility component in-phase with the driving field ${B_1}$. In general a response that is exactly in phase with a driving signal does not absorb power from the signal source and in spectroscopy corresponds to dispersion. In contrast, an out of phase corresponds to absorption. In magnetic resonance, it is usually the absorption, or ${\chi''}$, that is detected.

When the microwave or radiofrequency power, proportional to ${B_1^2}$, is small so that ${\gamma^2 B_1^2 T_1 T_2<<1}$, Eqs. (21) and (22) becomes

$\displaystyle \chi '= {\gamma \mu_0 M_0 (\omega_0-\omega)T_2^2 \over 1+T_2^2 (\omega_0-\omega)^2} \ \ \ \ \ (23)$

$\displaystyle \chi ''= {\gamma \mu_0 M_0 T_2 \over 1+T_2^2 (\omega_0-\omega)^2} \ \ \ \ \ (24)$

Equation (24) of ${\chi''}$ corresponds to the classical Lorentzian line shape function. The absorption curve will be a Lorentzian line with the half-width of the half height:

$\displaystyle \Delta\omega= {1 \over T_2} \quad \text{or} \quad \Delta B={\hbar \over g_s \mu_B T_2} \ \ \ \ \ (25)$