# Bloch Equations

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### Introduction

In 1946 Felix Bloch, co-discoverer of NMR, proposed a set of equations to described the time dependence of the net magnetization during the course of the NMR experiment. These equations are known as the Bloch equations and give insights into many processes in NMR. The Bloch equations follow first order kinetics and the derivations are first-order differentials.

### Bloch Equations in the Lab Frame

It has been shown that nuclei in placed in a magnetic field precess with a characteristic Larmor frequency, $$\omega$$.

$\omega_0=-\gamma (1-\sigma) B_0$

which can be made time-dependent by considering a time-dependent magnetic field B(t).

In addition we have also seen that there is a bulk magnetization, based on Boltzmann statistics, which lies along the direction of the applied magnetic field, $$B_0$$. This bulk magnetization can then be thought of precessing at the Larmor frequency. Assuming that the $$B_0$$ is along the Z-axis, we can describe the time dependence of the magnetization $$M$$, by

$\frac{dM}{dt}=\omega (t) x M(t)-[R][M(t)-M_eq]$

where R is a rotation matrix, and $$M_{eq}$$ is the magnetization at equilibrium along the z-axis. The rotation matrix accounts for relaxation processes and is given by

$R= \begin{pmatrix} \frac{1}{T_2} & 0 & 0 \\ 0 & \frac{1}{T_2} &0 \\ 0& 0 & \frac{1}{T_1} \end{pmatrix}$

The $$[R][M(t)-M_{eq}]$$ term then describes the decay of the magnetization in the x-y plane and the growth of the equilibrium magnetization, due to T2 and T1 effects, repectively.

Expansion of the equation yeilds the magnetization in each direction

$[\begin{pmatrix} \frac{dM^*_x(t)}{dt} \\ \frac{dM^*_y(t)}{dt} \\ \frac{dM^*_z(t)}{dt} \end{pmatrix}= \begin{pmatrix} [M_x(t) \omega _y(t)-M_y\omega _z(t)]- \frac{M_x(t)}{T_2} \\ [M_x(t) \omega _z(t)-M_z\omega_x(t)]- \frac{M_y(t)}{T_2} \\ [M_y(t) \omega _x(t)-M_x\omega _y(t)]-\frac{[M_z(t)-M_{eq}]}{T_1} \end{pmatrix}$

Assuming that we have a static magnetic field these equations simplify to

$[\begin{pmatrix} \frac{dM^*_x(t)}{dt} \\ \frac{dM^*_y(t)}{dt} \\ \frac{dM^*_z(t)}{dt} \end{pmatrix} = \begin{pmatrix} -\omega _0M_y(t)-\frac{M_x(t)}{T_2} \\ \omega _0M_x(t)-\frac{M_y(t)}{T_2} \\ -\frac{[M_z(t)-M_{eq}]}{T_1} \end{pmatrix}$

As the magnetization is precessing about the Z axis as it recovers from the x-y plane it is realized that the time dependance will follow an oscilitory pattern which can be described using trigonmetric functions. Therefore at a given time the magnetization in any direction will be

$\begin{pmatrix} M_x(t) \\ M_y(t) \\ M_z(t) \end{pmatrix} = \begin{pmatrix} [M_x(0)cos \omega _0t - M_y(0)sin \omega _0t]e^{\frac{-t}{T_2}} \\ [M_y(0)cos \omega _0t - M_x(0)sin \omega _0t]e^{\frac{-t}{T_2}} \\M_z(0)e^{\frac{-t}{T_1}} + M_{eq}(1-e^{\frac{-t}{T_1}})\end{pmatrix}$

Where $$M_{x,y,z}(0)$$ is the magnetization at t=0.

#### What Do The Equations Describe?

From this derivation we are able to describe the motion of the bulk magnetization of the nuclei after a pulse as a function of time accounting for relaxation.

### Bloch Equations in the Rotating Frame

One can imagine that application of a $$\frac{\pi}{2}$$ pulse in which the B- field now becomes time dependent would be challenging to calculate using the Bloch equations in the lab frame. Therefore we now shift our discussion to the rotating frame. It should be stated here that the insertion of the $$^*$$ to denote the magnetization projection on the rotating frame.

The frequency of rotation $$\omega _{rot}$$, is equivalent to spins precession frequency, which will be the Larmor frequency. From this we can describe the change of the magnetization as

$\frac{d^*M}{dt}=\frac{dM}{dt}-\omega _{rot}xM$

We know that $$\frac{dM}{dt}$$ is given by the bloch equations we derived for the lab frame that is:

$\frac{dM}{dt}=\omega (t) x M(t)-[R][M(t)-M_eq]$

Substituting this into our equation, we obtain

$\frac{d^*M}{dt}=\omega (t)XM-\omega _{rot}xM-[R][M-M_{eq}]$

which simplifies (using cross product relations) to

$\frac{d^*M}{dt}=\omega _{eff}(t)xM-[R][M-M_{eq}]$

where

$\omega _{eff}(t)=\omega - \omega _{rot}$

Noting that the relation between the lab frame frame and the rotating frame is the common z-axis, we can define the Bloch equations in the rotating frame

$\begin{pmatrix} \frac{dM^*_x(t)}{dt} \\ \frac{dM^*_y(t)}{dt} \\ \frac{dM^*_z(t)}{dt} \end{pmatrix}= \begin{pmatrix} \gamma [M^*_y(t)B_z(t)-M^*_z(t)B_y(t)]-\frac{M^*_x(t)}{T_2} \\ \gamma [M^*_z(t)B_x(t)-M^*_x(t)B_z(t)]-\frac{M^*_y(t)}{T_2} \\ \gamma [M^*_x(t)B_y(t)-M^*_y(t)B_x(t)]-\frac{[M^*_z(t)-M^*_{eq}]}{T_2} \end{pmatrix}$

Which can be simplified to give

$\begin{pmatrix} \frac{dM^*_x(t)}{dt} \\ \frac{dM^*_y(t)}{dt} \\ \frac{dM^*_z(t)}{dt} \end{pmatrix}= \begin{pmatrix} -\Omega M^*_y(t)- \gamma M^*_z(t)B^*_y(t)]-\frac{M^*_x(t)}{T_2} \\ -\gamma [M^*_x(t)+ \gamma M^*_z(t)B^*_x(t)]-\frac{M^*_y(t)}{T_2} \\ \gamma [M^*_x(t)B^*_y(t)-M^*_y(t)B^*_x(t)]-\frac{[M^*_z(t)-M^*_{eq}]}{T_2} \end{pmatrix}$

where

$\Omega=\omega _{0}-\omega _{rot}$

### Effects of RF pulses

During the NMR experiment, we can choose which direction we can apply our magnetic field during the pulse. For simplicity we assume that we apply the pulse along the +x axis. The nutation frequency is then defined as

$\omega_1=\mid{\gamma B_1(t)} \mid$

where $$B_1$$ is the magnetic field applied along rhe x-axis. Application of this to the Bloch equations in the rotating frame we obtain

$\begin{pmatrix} M^*_x(t) \\ M^*_y(t) \\ M^*_z(t) \end{pmatrix}= \begin{pmatrix} \omega_1 M^*_zsin\phi - \frac{-M^*_x}{T_2} \\ -\omega_1 M^*_zcos\phi - \frac{-M^*_y}{T_2} \\-\omega_1 M^*_zsin\phi + \omega_1 M^*_zcos\phi-\frac{(M^*_z-M^*_eq)}{T_1} \end{pmatrix}$

where $$\phi$$ is the phase of the pulse (0=x axis and 270=-y axis). The RF pulse is much shorter than T1, T2, or $$\omega_0$$. Therefore we can neglect any terms containing these values.THe equations then simplify to

$\begin{pmatrix} M^*_x(t) \\ M^*_y(t) \\ M^*_z(t) \end{pmatrix}= \begin{pmatrix} \omega_1 M^*_zsin\phi \\ -\omega_1 M^*_zcos\phi \\-\omega_1 M^*_zsin\phi + \omega_1 M^*_zcos\phi \end{pmatrix}$

Thereby application of a pulse of phase zero will result in the following result for the magnetization as a function of time

$\begin{pmatrix} \frac{dM^*_x(t)}{dt} \\ \frac{dM^*_y(t)}{dt} \\ \frac{dM^*_z(t)}{dt} \end{pmatrix}= \begin{pmatrix} 0 \\ \gamma B_1(t) M^*_z \\-\gamma B_1(t) M^*_z\end{pmatrix}$

Or more generally as

$\begin{pmatrix} M^*_x(t) \\ M^*_y(t) \\ M^*_z(t) \end{pmatrix}= \begin{pmatrix} M^*_x \\ M^*_ycos\omega _1t - M^*_zsin\omega _1t \\M^*_zcos\omega _1t - M^*_ysin\omega _1t \end{pmatrix}$

We can also describe the effect of a pulse of any phase on a Cartesian basis

$\begin{pmatrix} M^*_x(t) \\ M^*_y(t) \\ M^*_z(t) \end{pmatrix}=\begin{pmatrix}\frac{1}{2}M^*_x(1+cos\omega _1t)+\frac{1}{2}(M^*_x cos2\phi +M^*_ysin2\phi)(1-cos\omega _1t)+M^*_zsin\phi sin\omega _1t \\ \frac{1}{2}M^*_y(1+cos\omega _1t)-\frac{1}{2}(M^*_y cos2\phi +M^*_xsin2\phi)(1-cos\omega _1t)+M^*_zcos\phi sin\omega _1t \\ M^*_zcos\omega _1t-(M^*_xsin\phi -M^*_ycps\phi )sin\omega _1t \end{pmatrix}$

If the movement of the magnetization is slow, then

$\frac{dM^*_x}{dt}=\frac{dM^*_y}{dt}=\frac{dM_z}{dt}=0$

When can then solve for the magnetizations in each direction

$M^*_x=-\frac{2\pi \gamma\ B_1 M_0 T^2_2 \Omega}{1+4\pi ^2T^2_2 \Omega ^2+\gamma ^2 B_1^2T_1T_2}$

$M^*_y=-\frac{\gamma\ B_1 M_0 T_2}{1+4\pi ^2T^2_2 \Omega ^2+\gamma ^2 B_1^2T_1T_2}$

$M_z=\frac{M_0[1+4\pi ^2T^2_2\Omega ^2]}{1+4\pi ^2T^2_2 \Omega ^2+\gamma ^2 B_1^2T_1T_2}$

### Free Precession

In the absences of any magnetization applied in the x-y plane, the change in the magnetization is due to relaxation effects of T1 and T2 processes.
The Bloch equations in the rotating fram then reduce to

$\begin{pmatrix} \frac{dM^*_x(t)}{dt} \\ \frac{dM^*_y(t)}{dt} \\ \frac{dM^*_z(t)}{dt} \end{pmatrix}= \begin{pmatrix} -(\omega _{rot} - \omega _0)M^*_y(t)-\frac{M^*_x(t)}{T_2} \\ (\omega _{rot} - \omega _0)M^*_x(t)-\frac{M^*_y(t)}{T_2} \\ \frac{-[M^*_z(t)-M^*_{eq}]}{T_1} \end{pmatrix}$

which can be solved to give

$\begin{pmatrix} M^*_x(t) \\ M^*_y(t) \\ M^*_z(t) \end{pmatrix}= \begin{pmatrix} [M^*_x(0) cos\Omega t - M^*_y sin\Omega t] e^{\frac{-t}{T_2}} \\ [M^*_y(0) cos\Omega t + M^*_x sin\Omega t]e^{\frac{-t}{T_2}} \\ M^*_z(0)e^{\frac{-t}{T_1}} + M_{eq}(1-e^{\frac{-t}{T_1}}\end{pmatrix}$

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14:26, 17 Jan 2014

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