Chemical Sciences: A Manual for CSIR-UGC National Eligibility Test for Lectureship and JRF/Bloch equations

In physics and chemistry, specifically in NMR (nuclear magnetic resonance) or MRI (magnetic resonance imaging), or ESR (electron spin resonance) the Bloch equations are a set of macroscopic equations that are used to calculate the nuclear magnetization M = (Mx, My, Mz) as a function of time when relaxation times T1 and T2 are present. These are phenomenological equations that were introduced by Felix Bloch in 1946. Sometimes they are called the equations of motion of nuclear magnetization.

Bloch equations in laboratory (stationary) frame of reference
Let M(t) = (Mx(t), My(t), Mz(t)) be the nuclear magnetization. Then the Bloch equations read:


 * $$\frac {d M_x(t)} {d t} = \gamma ( \mathbf {M} (t) \times \mathbf {B} (t) ) _x - \frac {M_x(t)} {T_2}$$
 * $$\frac {d M_y(t)} {d t} = \gamma ( \mathbf {M} (t) \times \mathbf {B} (t) ) _y - \frac {M_y(t)} {T_2}$$
 * $$\frac {d M_z(t)} {d t} = \gamma ( \mathbf {M} (t) \times \mathbf {B} (t) ) _z - \frac {M_z(t) - M_0} {T_1}$$

where γ is the gyromagnetic ratio and B(t) = (Bx(t), By(t), B0 + Bz(t)) is the magnetic flux density experienced by the nuclei. The z component of the magnetic flux density B is sometimes composed of two terms: M(t) &times; B(t) is the cross product of these two vectors. M0 is the steady state nuclear magnetization (that is, for example, when t → ∞); it is in the z direction.
 * one, B0, is constant in time,
 * the other one, Bz(t), is time dependent. It is present in magnetic resonance imaging and helps with the spatial decoding of the NMR signal.

Physical background
With no relaxation (that is both T1 and T2 → ∞) the above equations simplify to:


 * $$\frac {d M_x(t)} {d t} = \gamma ( \mathbf {M} (t) \times \mathbf {B} (t) ) _x$$
 * $$\frac {d M_y(t)} {d t} = \gamma ( \mathbf {M} (t) \times \mathbf {B} (t) ) _y$$
 * $$\frac {d M_z(t)} {d t} = \gamma ( \mathbf {M} (t) \times \mathbf {B} (t) ) _z$$

or, in vector notation:


 * $$\frac {d \mathbf {M}(t)} {d t} = \gamma \mathbf {M} (t) \times \mathbf {B} (t)  $$

This is the equation for Larmor precession of the nuclear magnetization M in an external magnetic flux density B.

The relaxation terms,


 * $$\left ( -\frac {M_x} {T_2}, -\frac {M_y} {T_2}, -\frac {M_z - M_0} {T_1} \right ) $$

represent an established physical process of transverse and longitudinal relaxation of nuclear magnetization M.

Bloch equations are macroscopic equations
These equations are not microscopic: they do not describe the equation of motion of individual nuclear magnetic moments. These are governed and described by laws of quantum mechanics.

Bloch equations are macroscopic: they describe the equations of motion of macroscopic nuclear magnetization that can be obtained by summing up all nuclear magnetic moment in the sample.

Alternative form of Bloch equations
The above form is simplified assuming


 * $$M_{xy} = M_x + iM_y \text{ and  } B_{xy} = B_x + iB_y $$

where i = &radic;(-1). After some algebra one obtains:


 * $$\frac {d M_{xy}(t)} {d t} = -i \gamma \left ( M_{xy} (t) B_z (t) - M_z (t) B_{xy} (t) \right ) -

\frac {M_{xy}} {T_2} $$.
 * $$\frac {d M_z(t)} {d t} = i \gamma \left ( M_{xy} (t) \overline{B_{xy} (t)} -

\overline {M_{xy}} (t) B_{xy} (t) \right ) - \frac {M_z - M_0} {T_1}$$

where


 * $$\overline {M_{xy}} = M_x - i M_y $$.

The real and imaginary parts of Mxy correspond to Mx and My respectively. Mxy is sometimes called transverse nuclear magnetization.

Bloch equations in rotating frame of reference (outline)
In rotating frame of reference the it is easier to understand the behaviour of nuclear magnetization M. This is the motivation:

Solution of Bloch equations with T1, T2 → &infin;
Assume that:
 * at t = 0 the transverse nuclear magnetization Mxy(0) experiences a constant magnetic flux density B(t) = (0, 0, B0);
 * B0 is positive;
 * there are no longitudinal and transverse relaxations (that is T1 and T2 → &infin;).

Then the Bloch equations are simplified to:


 * $$\frac {d M_{xy}(t)} {d t} = -i \gamma M_{xy} (t) B_{0}$$,
 * $$\frac {d M_z(t)} {d t} = 0 $$.

These are two (not coupled) linear differential equations. Their solution is:


 * $$M_{xy}(t) = M_{xy} (0) e^{-i \gamma B_{0} t}$$,
 * $$M_z(t) = M_0 = \text{const} \,$$.

Thus the transverse magnetization, Mxy, rotates around the z axis with angular frequency ω0 = γB0 in counterclockwise direction (this due to the negative sign in the exponent). The longitudinal magnetization, Mz remains constant in time. This is also how the transverse magnetization appears to an observer in the laboratory frame of reference (that is to a stationary observer).

Mxy(t) is translated in the following way into observable quantities of Mx(t) and My(t): Since


 * $$M_{xy}(t) = M_{xy} (0) e^{-i \gamma B_{z0} t} = M_{xy} (0) \left [ \cos (\omega _0 t) - i \sin (\omega_0 t) \right ]$$

then


 * $$M_{x}(t) = \text {Re} \left (M_{xy} (t) \right ) = M_{xy} (0) \cos (\omega _0 t)$$,
 * $$M_{y}(t) = \text {Im} \left (M_{xy} (t) \right ) = -M_{xy} (0) \sin (\omega _0 t)$$,

where Re(z) and Im(z) are functions that return the real and imaginary part of complex number z. In this calculation it was assumed that Mxy(0) is a real number.

Transformation to rotating frame of reference
This is the conclusion of the previous section: in a constant magnetic flux density B0 along z axis the transverse magnetization Mxy rotates around the this axis in counterclockwise direction with angular frequency ω0. If the observer were around the same axis in  counterclockwise direction with angular frequency  Ω, Mxy it would appear to him

It is often more convenient to describe the physics and mathematics of nuclear magnetization in a rotating frame of reference: Let (x&prime;, y&prime;, z&prime;) be a Cartesian coordinate system that is rotating around the static magnetic field B0 (given in the laboratory reference system (x, y, z)) with angular frequency Ω. This is equivalent to assuming that:


 * $$M_{xy}(t) = e^{-i \Omega t} M_{xy}'$$
 * $$M_{z}(t) = M_{z}'$$

What are the equations of motion of Mxy(t)&prime; and Mz(t)&prime;?


 * $$\frac {d M_{xy}'(t)} {d t} = \frac {d M_{xy}(t) e^{+i \Omega t}} {d t} =

e^{+i \Omega t} \frac {d M_{xy}(t) } {d t} + i \Omega e^{+i \Omega t} M_{xy} = e^{+i \Omega t} \frac {d M_{xy}(t) } {d t} + i \Omega M_{xy}' $$

Substitute from the Bloch equation in laboratory frame of reference:


 * $$\begin{align} \frac {d M_{xy}'(t)} {d t} & = e^{+i \Omega t} \left [-i \gamma \left ( M_{xy} (t) B_z (t) - M_z (t) B_{xy} (t) \right ) -

\frac {M_{xy}} {T_2} \right ] + i \Omega M_{xy}' = \\

& = \left [-i \gamma \left ( M_{xy} (t) e^{+i \Omega t} B_z (t) - M_z (t) B_{xy} (t) e^{+i \Omega t}\right ) - \frac {M_{xy} e^{+i \Omega t} } {T_2} \right ] + i \Omega M_{xy}' = \\

& = -i \gamma \left ( M_{xy}' (t) B_z' (t) - M_z' (t) B_{xy}' (t) \right ) + i \Omega M_{xy}' - \frac {M_{xy}'} {T_2} =

\end{align} $$