# 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 = (M$_x$, M$_y$, M$_z$) as a function of time when relaxation times T$_1$ and T$_2$ 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) = (M$_x$(t), M$_y$(t), M$_z$(t)) be the nuclear magnetization. Then the Bloch equations read:
    ${\frac {dM_{x}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{x}-{\frac {M_{x}(t)}{T_{2}}}$
    ${\frac {dM_{y}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{y}-{\frac {M_{y}(t)}{T_{2}}}$
    ${\frac {dM_{z}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{z}-{\frac {M_{z}(t)-M_{0}}{T_{1}}}$
where γ is the gyromagnetic ratio and B(t) = (B$_x$(t), B$_y$(t), B$_0$ + B$_z$(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:
* one, B$_0$, is constant in time,
* the other one, B$_z$(t), is time dependent. It is present in magnetic resonance imaging and helps with the spatial decoding of the NMR signal.
M(t) × B(t) is the cross product of these two vectors. M$_0$ is the steady state nuclear magnetization (that is, for example, when t → ∞); it is in the z direction.

### Physical background

With no relaxation (that is both T$_1$ and T$_2$ → ∞) the above equations simplify to:
    ${\frac {dM_{x}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{x}$
    ${\frac {dM_{y}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{y}$
    ${\frac {dM_{z}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{z}$
or, in vector notation:
    ${\frac {d\mathbf {M} (t)}{dt}}=\gamma \mathbf {M} (t)\times \mathbf {B} (t)$
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 = √(-1). After some algebra one obtains:
    ${\frac {dM_{xy}(t)}{dt}}=-i\gamma \left(M_{xy}(t)B_{z}(t)-M_{z}(t)B_{xy}(t)\right)-{\frac {M_{xy}}{T_{2}}}$.
    ${\frac {dM_{z}(t)}{dt}}=i\gamma \left(M_{xy}(t){\overline {B_{xy}(t)}}-{\overline {M_{xy}}}(t)B_{xy}(t)\right)-{\frac {M_{z}-M_{0}}{T_{1}}}$
where
    ${\overline {M_{xy}}}=M_{x}-iM_{y}$.
The real and imaginary parts of M$_{xy}$ correspond to M$_x$ and M$_y$ respectively. M$_{xy}$ 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 T$_1$, T$_2$ → ∞

Assume that:
* at t = 0 the transverse nuclear magnetization M$_{xy}$(0) experiences a constant magnetic flux density B(t) = (0, 0, B$_0$);
* B$_0$ is positive;
* there are no longitudinal and transverse relaxations (that is T$_1$ and T$_2$ → ∞).
Then the Bloch equations are simplified to:
    ${\frac {dM_{xy}(t)}{dt}}=-i\gamma M_{xy}(t)B_{0}$,
    ${\frac {dM_{z}(t)}{dt}}=0$.
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, M$_{xy}$, rotates around the z axis with angular frequency ω$_0$ = γB$_0$ in counterclockwise direction (this due to the negative sign in the exponent). The longitudinal magnetization, M$_z$ 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).
M$_{xy}$(t) is translated in the following way into observable quantities of M$_x$(t) and M$_y$(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 M$_{xy}$(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 B$_0$ along z axis the transverse magnetization M$_{xy}$ rotates around the this axis in counterclockwise direction with angular frequency ω$_0$. If the observer were around the same axis in counterclockwise direction with angular frequency Ω, M$_{xy}$ 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′, y′, z′) be a Cartesian coordinate system that is rotating around the static magnetic field B$_0$ (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 M$_{xy}$(t)′ and M$_z$(t)′?
    ${\frac {dM_{xy}'(t)}{dt}}={\frac {dM_{xy}(t)e^{+i\Omega t}}{dt}}=e^{+i\Omega t}{\frac {dM_{xy}(t)}{dt}}+i\Omega e^{+i\Omega t}M_{xy}=e^{+i\Omega t}{\frac {dM_{xy}(t)}{dt}}+i\Omega M_{xy}'$
Substitute from the Bloch equation in laboratory frame of reference:
    ${\begin{aligned}{\frac {dM_{xy}'(t)}{dt}}&=e^{+i\Omega t}\left[-i\gamma \left(M_{xy}(t)B_{z}(t)-M_{z}(t)B_{xy}(t)\right)-{\frac {M_{xy}}{T_{2}}}\right]+i\Omega M_{xy}'=\\&=\left[-i\gamma \left(M_{xy}(t)e^{+i\Omega t}B_{z}(t)-M_{z}(t)B_{xy}(t)e^{+i\Omega t}\right)-{\frac {M_{xy}e^{+i\Omega t}}{T_{2}}}\right]+i\Omega M_{xy}'=\\&=-i\gamma \left(M_{xy}'(t)B_{z}'(t)-M_{z}'(t)B_{xy}'(t)\right)+i\Omega M_{xy}'-{\frac {M_{xy}'}{T_{2}}}=\end{aligned}}$

## References

* ↑ F Bloch, Nuclear Induction, Physics Review 70, 460-473 (1946)