Bloch equations

In physics and chemistry, specifically in nuclear magnetic resonance (NMR), magnetic resonance imaging (MRI), and electron spin resonance (ESR), the Bloch equations are a set of macroscopic equations that are used to calculate the nuclear magnetization Template:Math as a function of time when relaxation times Template:Math and Template:Math 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. They are analogous to the Maxwell–Bloch equations.

In the laboratory (stationary) frame of reference

Under the effect of the external field B, the magnetization vector M relaxes to its equilibrium configuration while precessing around the magnetic field.

Let Template:Math be the nuclear magnetization. Then the Bloch equations read:

$$\begin{array}{cc}\frac{d{M}_x(t)}{dt}& =\gamma {(𝐌(t)\times 𝐁(t))}_x-\frac{M_x(t)}{T_2}\\ \frac{d{M}_y(t)}{dt}& =\gamma {(𝐌(t)\times 𝐁(t))}_y-\frac{M_y(t)}{T_2}\\ \frac{d{M}_z(t)}{dt}& =\gamma {(𝐌(t)\times 𝐁(t))}_z-\frac{M_z(t)-M_0}{T_1}\end{array}$$

where Template:Mvar is the gyromagnetic ratio and Template:Math is the magnetic field experienced by the nuclei. The Template:Mvar component of the magnetic field Template:Math is sometimes composed of two terms:

• one, Template:Math, is constant in time,
• the other one, Template:Math, may be time dependent. It is present in magnetic resonance imaging and helps with the spatial decoding of the NMR signal.

Template:Math is the cross product of these two vectors. Template:Math is the steady state nuclear magnetization (that is, for example, when Template:Math); it is in the z direction.

Physical background

With no relaxation (that is both Template:Math and Template:Math) the above equations simplify to:

$$\begin{array}{cc}\frac{d{M}_x(t)}{dt}& =\gamma (𝐌(t)\times 𝐁(t){)}_x\\ \frac{d{M}_y(t)}{dt}& =\gamma (𝐌(t)\times 𝐁(t){)}_y\\ \frac{d{M}_z(t)}{dt}& =\gamma (𝐌(t)\times 𝐁(t){)}_z\end{array}$$

or, in vector notation:

$$\frac{d𝐌(t)}{dt}=\gamma 𝐌(t)\times 𝐁(t)$$

This is the equation for Larmor precession of the nuclear magnetization Template:Mvar in an external magnetic field Template:Math.

The relaxation terms,

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

represent an established physical process of transverse and longitudinal relaxation of nuclear magnetization Template:Math.

As macroscopic equations

Template:Main articles These equations are not microscopic: they do not describe the equation of motion of individual nuclear magnetic moments. Those 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 forms

Opening the vector product brackets in the Bloch equations leads to:

$$\begin{array}{cc}\frac{d}{dt}{M}_x(t)& =\gamma (M_y(t)B_z(t)-M_z(t)B_y(t))-\frac{M_x(t)}{T_2}\\ \frac{d}{dt}{M}_y(t)& =\gamma (M_z(t)B_x(t)-M_x(t)B_z(t))-\frac{M_y(t)}{T_2}\\ \frac{d}{dt}{M}_z(t)& =\gamma (M_x(t)B_y(t)-M_y(t)B_x(t))-\frac{M_z(t)-M_0}{T_1}\end{array}$$

The above form is further simplified assuming

$$M_{xy}=M_x+i{M}_y\text{ and }{B}_{xy}=B_x+i{B}_y$$

where Template:Math. After some algebra one obtains:

$$\begin{array}{cc}\frac{d{M}_{xy}(t)}{dt}& =-i\gamma (M_{xy}(t)B_z(t)-M_z(t)B_{xy}(t))-\frac{M_{xy}(t)}{T_2}\\ \frac{d{M}_z(t)}{dt}& =i\frac{\gamma }{2}(M_{xy}(t)\overline{B_{xy}(t)}-\overline{M_{xy}}(t)B_{xy}(t))-\frac{M_z(t)-M_0}{T_1}.\end{array}$$

where

$$\overline{M_{xy}}=M_x-i{M}_y.$$

is the complex conjugate of Template:Math. The real and imaginary parts of Template:Math correspond to Template:Math and Template:Math respectively. Template:Math is sometimes called transverse nuclear magnetization.

Matrix form

The Bloch equations can be recast in matrix-vector notation: $$\frac{d}{dt}\left(\begin{array}{c}{M}_x\\ M_y\\ M_z\end{array}\right)=\left(\begin{array}{ccc}-\frac{1}{T_2}& \gamma B_z& -\gamma B_y\\ -\gamma B_z& -\frac{1}{T_2}& \gamma B_x\\ \gamma B_y& -\gamma B_x& -\frac{1}{T_1}\end{array}\right)\left(\begin{array}{c}{M}_x\\ M_y\\ M_z\end{array}\right)+\left(\begin{array}{c}0\\ 0\\ \frac{M_0}{T_1}\end{array}\right)$$

In a rotating frame of reference

In a rotating frame of reference, it is easier to understand the behaviour of the nuclear magnetization M. This is the motivation:

Solution of Bloch equations with T₁, T₂ → ∞

Assume that:

• at Template:Math the transverse nuclear magnetization Template:Math experiences a constant magnetic field Template:Math;
• Template:Math is positive;
• there are no longitudinal and transverse relaxations (that is Template:Math and Template:Math).

Then the Bloch equations are simplified to:

$$\begin{array}{cc}\frac{d}{dt}{M}_{xy}(t)& =-i\gamma M_{xy}(t)B_0,\\ \frac{d}{dt}{M}_z(t)& =0.\end{array}$$

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

$$\begin{array}{cc}{M}_{xy}(t)& =M_{xy}(0)e^{-i\gamma B_{0}t},\\ M_z(t)& =M_0=\text{const}.\end{array}$$

Thus the transverse magnetization, Template:Math, rotates around the z axis with angular frequency Template:Math in clockwise direction (this is due to the negative sign in the exponent). The longitudinal magnetization, Template:Math 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).

Template:Math is translated in the following way into observable quantities of Template:Math and Template:Math: Since

$$\begin{array}{cc}{M}_{xy}(t)& =M_{xy}(0)e^{-i\gamma B_{z0}t}\\ & =M_{xy}(0)[\cos ({\omega }_{0}t)-i\sin ({\omega }_{0}t)]\end{array}$$

then

$$\begin{array}{cc}{M}_x(t)& =\mathrm{Re}(M_{xy}(t))=M_{xy}(0)\cos ({\omega }_{0}t),\\ M_y(t)& =\mathrm{Im}(M_{xy}(t))=-M_{xy}(0)\sin ({\omega }_{0}t),\end{array}$$

where Template:Math and Template:Math are functions that return the real and imaginary part of complex number Template:Mvar. In this calculation it was assumed that Template:Math is a real number.

Transformation to rotating frame of reference

This is the conclusion of the previous section: in a constant magnetic field Template:Math along z axis the transverse magnetization Template:Math rotates around this axis in clockwise direction with angular frequency Template:Math. If the observer were rotating around the same axis in clockwise direction with angular frequency Template:Math, Template:Math it would appear to them rotating with angular frequency Template:Math. Specifically, if the observer were rotating around the same axis in clockwise direction with angular frequency ω₀, the transverse magnetization Template:Math would appear to her or him stationary.

This can be expressed mathematically in the following way:

• Let Template:Math the Cartesian coordinate system of the laboratory (or stationary) frame of reference, and
• Template:Math be a Cartesian coordinate system that is rotating around the z axis of the laboratory frame of reference with angular frequency Template:Math. This is called the rotating frame of reference. Physical variables in this frame of reference will be denoted by a prime.

Obviously:

$${M_z}^{\prime }(t)=M_z(t).$$

What is Template:Math? Expressing the argument at the beginning of this section in a mathematical way:

$$M_{x^{\prime }y}(t)=e^{+i\mathrm{\Omega }t}{M}_{xy}(t).$$

Equation of motion of transverse magnetization in rotating frame of reference

What is the equation of motion of Template:Math?

$$\begin{array}{cc}\frac{d{M}_{x^{\prime }y}(t)}{dt}& =\frac{d}{dt}(M_{xy}(t)e^{+i\mathrm{\Omega }t})\\ & =e^{+i\mathrm{\Omega }t}\frac{d{M}_{xy}(t)}{dt}+i\mathrm{\Omega }{e}^{+i\mathrm{\Omega }t}{M}_{xy}(t)\\ & =e^{+i\mathrm{\Omega }t}\frac{d{M}_{xy}(t)}{dt}+i\mathrm{\Omega }{M}_{x^{\prime }y}(t)\end{array}$$

Substitute from the Bloch equation in laboratory frame of reference:

$$\begin{array}{cc}\frac{d{M}_{x^{\prime }y}(t)}{dt}& =e^{i\mathrm{\Omega }t}[-i\gamma (M_{xy}{B}_z-M_z{B}_{xy})-\frac{M_{xy}}{T_2}]+i\mathrm{\Omega }{M}_{x^{\prime }y}\\ & =[-i\gamma (M_{xy}{e}^{i\mathrm{\Omega }t}{B}_z-M_z{B}_{xy}{e}^{i\mathrm{\Omega }t})-\frac{M_{xy}{e}^{i\mathrm{\Omega }t}}{T_2}]+i\mathrm{\Omega }{M}_{x^{\prime }y}\\ & =-i\gamma (M_{x^{\prime }y}{B}_z-M_z{B}_{x^{\prime }y})+i\mathrm{\Omega }{M}_{x^{\prime }y}-\frac{M_{x^{\prime }y}}{T_2}\end{array}$$

But by assumption in the previous section: Template:Math and Template:Math. Substituting into the equation above:

$$\begin{array}{cc}\frac{d{M}_{x^{\prime }y}(t)}{dt}& =-i\gamma (M_{x^{\prime }y}(B_0+\mathrm{\Delta }{B}_z)-{M_z}^{\prime }{B}_{x^{\prime }y})+i\mathrm{\Omega }{M}_{x^{\prime }y}-\frac{M_{x^{\prime }y}}{T_2}\\ & =-i\gamma B_0{M}_{x^{\prime }y}-i\gamma \mathrm{\Delta }{B}_z{M}_{x^{\prime }y}+i\gamma B_{x^{\prime }y}{M_z}^{\prime }+i\mathrm{\Omega }{M}_{x^{\prime }y}-\frac{M_{x^{\prime }y}}{T_2}\\ & =i(\mathrm{\Omega }-{\omega }_0)M_{x^{\prime }y}-i\gamma \mathrm{\Delta }{B}_z{M}_{x^{\prime }y}+i\gamma B_{x^{\prime }y}{M_z}^{\prime }-\frac{M_{x^{\prime }y}}{T_2}\end{array}$$

This is the meaning of terms on the right hand side of this equation:

• Template:Math is the Larmor term in the frame of reference rotating with angular frequency Ω. Note that it becomes zero when Template:Math.
• The Template:Math term describes the effect of magnetic field inhomogeneity (as expressed by Template:Math) on the transverse nuclear magnetization; it is used to explain T₂^*. It is also the term that is behind MRI: it is generated by the gradient coil system.
• The Template:Math describes the effect of RF field (the Template:Math factor) on nuclear magnetization. For an example see below.
• Template:Math describes the loss of coherency of transverse magnetization.

Similarly, the equation of motion of M_z in the rotating frame of reference is: $$\frac{d{M_z}^{\prime }(t)}{dt}=i\frac{\gamma }{2}(M{\text{'}}_{xy}\overline{B{\text{'}}_{xy}}-\overline{M{\text{'}}_{xy}}B{\text{'}}_{xy})-\frac{M_z-M_0}{T_1}$$

Time independent form of the equations in the rotating frame of reference

When the external field has the form:

$$\begin{array}{cc}{B}_x(t)& =B_1\cos \omega t\\ B_y(t)& =-B_1\sin \omega t\\ B_z(t)& =B_0\end{array}$$

We define: $$ϵ:=\gamma B_1\text{and}\mathrm{\Delta }:=\gamma B_0-\omega$$ and get (in the matrix-vector notation):

$$\frac{d}{dt}\left(\begin{array}{c}M{\text{'}}_x\\ M{\text{'}}_y\\ M{\text{'}}_z\end{array}\right)=\left(\begin{array}{ccc}-\frac{1}{T_2}& \mathrm{\Delta }& 0\\ -\mathrm{\Delta }& -\frac{1}{T_2}& ϵ\\ 0& -ϵ& -\frac{1}{T_1}\end{array}\right)\left(\begin{array}{c}M{\text{'}}_x\\ M{\text{'}}_y\\ M{\text{'}}_z\end{array}\right)+\left(\begin{array}{c}0\\ 0\\ \frac{M_0}{T_1}\end{array}\right)$$

Simple solutions

Relaxation of transverse nuclear magnetization M_xy

Assume that:

• The nuclear magnetization is exposed to constant external magnetic field in the z direction Template:Math. Thus Template:Math and Template:Math.
• There is no RF, that is Template:Math.
• The rotating frame of reference rotates with an angular frequency Template:Math.

Then in the rotating frame of reference, the equation of motion for the transverse nuclear magnetization, Template:Math simplifies to:

$$\frac{d{M}_{x^{\prime }y}(t)}{dt}=-\frac{M_{x^{\prime }y}}{T_2}$$

This is a linear ordinary differential equation and its solution is

$$M_{x^{\prime }y}(t)=M_{x^{\prime }y}(0)e^{-t/T_2}.$$

where Template:Math is the transverse nuclear magnetization in the rotating frame at time Template:Math. This is the initial condition for the differential equation.

Note that when the rotating frame of reference rotates exactly at the Larmor frequency (this is the physical meaning of the above assumption Template:Math), the vector of transverse nuclear magnetization, Template:Math appears to be stationary.

Relaxation of longitudinal nuclear magnetization M_z

Assume that:

• The nuclear magnetization is exposed to constant external magnetic field in the z direction Template:Math. Thus Template:Math and Template:Math.
• There is no RF, that is Template:Math.
• The rotating frame of reference rotates with an angular frequency Ω = ω₀.

Then in the rotating frame of reference, the equation of motion for the longitudinal nuclear magnetization, Template:Math simplifies to:

$$\frac{d{M}_z(t)}{dt}=-\frac{M_z(t)-M_{z,\mathrm{e}\mathrm{q}}}{T_1}$$

This is a linear ordinary differential equation and its solution is

$$M_z(t)=M_{z,\mathrm{e}\mathrm{q}}-[M_{z,\mathrm{e}\mathrm{q}}-M_z(0)]e^{-t/T_1}$$

where M_z(0) is the longitudinal nuclear magnetization in the rotating frame at time t = 0. This is the initial condition for the differential equation.

90 and 180° RF pulses

Assume that:

• Nuclear magnetization is exposed to constant external magnetic field in z direction Template:Math. Thus Template:Math and Template:Math.
• At Template:Math an RF pulse of constant amplitude and frequency ω₀ is applied. That is Template:Math is constant. Duration of this pulse is τ.
• The rotating frame of reference rotates with an angular frequency Ω = ω₀.
• Template:Math and Template:Math. Practically this means that Template:Math and Template:Math.

Then for Template:Math:

$$\begin{array}{cc}\frac{d{M}_{x^{\prime }y}(t)}{dt}& =i\gamma B_{x^{\prime }y}{M}_z(t)\\ \frac{d{M}_z(t)}{dt}& =i\frac{\gamma }{2}(M{\text{'}}_{xy}(t)\overline{B{\text{'}}_{xy}}-\overline{M{\text{'}}_{xy}}(t)B{\text{'}}_{xy})\end{array}$$

See also

• The Bloch–Torrey equation is a generalization of the Bloch equations, which includes added terms due to the transfer of magnetization by diffusion.
• Landau-Lifshitz-Gilbert_equation

Further reading

• Charles Kittel, Introduction to Solid State Physics, John Wiley & Sons, 8th edition (2004), Template:ISBN. Chapter 13 is on Magnetic Resonance.