Physical background

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

${\displaystyle {\frac {dM_{x}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{x}}$

${\displaystyle {\frac {dM_{y}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{y}}$

${\displaystyle {\frac {dM_{z}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{z}}$

or, in vector notation:

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

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

The relaxation terms,

${\displaystyle \left(-{\frac {M_{x}}{T_{2}}},-{\frac {M_{y}}{T_{2}}},-{\frac {M_{z}-M_{0}}{T_{1}}}\right)}$

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

As macroscopic equations

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:

${\displaystyle {\frac {dM_{x}(t)}{dt}}=\gamma \left(M_{y}(t)B_{z}(t)-M_{z}(t)B_{y}(t)\right)-{\frac {M_{x}(t)}{T_{2}}}}$

${\displaystyle {\frac {dM_{y}(t)}{dt}}=\gamma \left(M_{z}(t)B_{x}(t)-M_{x}(t)B_{z}(t)\right)-{\frac {M_{y}(t)}{T_{2}}}}$

${\displaystyle {\frac {dM_{z}(t)}{dt}}=\gamma \left(M_{x}(t)B_{y}(t)-M_{y}(t)B_{x}(t)\right)-{\frac {M_{z}(t)-M_{0}}{T_{1}}}}$

The above form is further simplified assuming

${\displaystyle M_{xy}=M_{x}+iM_{y}{\text{ and }}B_{xy}=B_{x}+iB_{y}\,}$

where i = √−1. After some algebra one obtains:

${\displaystyle {\frac {dM_{xy}(t)}{dt}}=-i\gamma \left(M_{xy}(t)B_{z}(t)-M_{z}(t)B_{xy}(t)\right)-{\frac {M_{xy}(t)}{T_{2}}}}$.

${\displaystyle {\frac {dM_{z}(t)}{dt}}=i{\frac {\gamma }{2}}\left(M_{xy}(t){\overline {B_{xy}(t)}}-{\overline {M_{xy}}}(t)B_{xy}(t)\right)-{\frac {M_{z}(t)-M_{0}}{T_{1}}}}$

where

${\displaystyle {\overline {M_{xy}}}=M_{x}-iM_{y}}$.

is the complex conjugate of M_xy. The real and imaginary parts of M_xy correspond to M_x and M_y respectively. M_xy is sometimes called transverse nuclear magnetization.

Matrix form

The Bloch equations can be recast in matrix-vector notation:

${\displaystyle {\frac {d}{dt}}\left({\begin{array}{c}M_{x}\\M_{y}\\M_{z}\end{array}}\right)=\left({\begin{array}{ccc}-{\frac {1}{T_{2}}}&\gamma B_{z}&-\gamma B_{y}\\-\gamma B_{z}&-{\frac {1}{T_{2}}}&\gamma B_{x}\\\gamma B_{y}&-\gamma B_{x}&-{\frac {1}{T_{1}}}\end{array}}\right)\left({\begin{array}{c}M_{x}\\M_{y}\\M_{z}\end{array}}\right)+\left({\begin{array}{c}0\\0\\{\frac {M_{0}}{T_{1}}}\end{array}}\right)}$

Solution of Bloch equations with T₁, T₂ → ∞

Assume that:

• at t = 0 the transverse nuclear magnetization M_xy(0) experiences a constant magnetic field B(t) = (0, 0, B₀);
• B₀ is positive;
• there are no longitudinal and transverse relaxations (that is T₁ and T₂ → ∞).

Then the Bloch equations are simplified to:

${\displaystyle {\frac {dM_{xy}(t)}{dt}}=-i\gamma M_{xy}(t)B_{0}}$,

${\displaystyle {\frac {dM_{z}(t)}{dt}}=0}$.

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

${\displaystyle M_{xy}(t)=M_{xy}(0)e^{-i\gamma B_{0}t}}$,

${\displaystyle M_{z}(t)=M_{0}={\text{const}}\,}$.

Thus the transverse magnetization, M_xy, rotates around the z axis with angular frequency ω₀ = γB₀ in clockwise direction (this is 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

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

then

${\displaystyle M_{x}(t)={\text{Re}}\left(M_{xy}(t)\right)=M_{xy}(0)\cos(\omega _{0}t)}$,

${\displaystyle M_{y}(t)={\text{Im}}\left(M_{xy}(t)\right)=-M_{xy}(0)\sin(\omega _{0}t)}$,

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 field B₀ along z axis the transverse magnetization M_xy rotates around this axis in clockwise direction with angular frequency ω₀. If the observer were rotating around the same axis in clockwise direction with angular frequency Ω, M_xy it would appear to her or him rotating with angular frequency ω₀ - Ω. Specifically, if the observer were rotating around the same axis in clockwise direction with angular frequency ω₀, the transverse magnetization M_xy would appear to her or him stationary.

This can be expressed mathematically in the following way:

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

Obviously:

${\displaystyle M_{z}'(t)=M_{z}(t)\,}$.

What is M_xy′(t)? Expressing the argument at the beginning of this section in a mathematical way:

${\displaystyle M_{xy}'(t)=e^{+i\Omega t}M_{xy}(t)\,}$.

Equation of motion of transverse magnetization in rotating frame of reference

What is the equation of motion of M_xy′(t)?

${\displaystyle {\frac {dM_{xy}'(t)}{dt}}={\frac {d\left(M_{xy}(t)e^{+i\Omega t}\right)}{dt}}=e^{+i\Omega t}{\frac {dM_{xy}(t)}{dt}}+i\Omega e^{+i\Omega t}M_{xy}(t)=e^{+i\Omega t}{\frac {dM_{xy}(t)}{dt}}+i\Omega M_{xy}'(t)}$

Substitute from the Bloch equation in laboratory frame of reference:

${\displaystyle {\begin{aligned}{\frac {dM_{xy}'(t)}{dt}}&=e^{+i\Omega t}\left[-i\gamma \left(M_{xy}(t)B_{z}(t)-M_{z}(t)B_{xy}(t)\right)-{\frac {M_{xy}(t)}{T_{2}}}\right]+i\Omega M_{xy}'(t)\\&=\left[-i\gamma \left(M_{xy}(t)e^{+i\Omega t}B_{z}(t)-M_{z}(t)B_{xy}(t)e^{+i\Omega t}\right)-{\frac {M_{xy}(t)e^{+i\Omega t}}{T_{2}}}\right]+i\Omega M_{xy}'(t)\\&=-i\gamma \left(M_{xy}'(t)B_{z}(t)-M_{z}(t)B_{xy}'(t)\right)+i\Omega M_{xy}'(t)-{\frac {M_{xy}'(t)}{T_{2}}}\\\end{aligned}}}$

But by assumption in the previous section: B_z′(t) = B_z(t) = B₀ + ΔB_z(t) and M_z(t) = M_z′(t). Substituting into the equation above:

${\displaystyle {\begin{aligned}{\frac {dM_{xy}'(t)}{dt}}&=-i\gamma \left(M_{xy}'(t)(B_{0}+\Delta B_{z}(t))-M_{z}'(t)B_{xy}'(t)\right)+i\Omega M_{xy}'(t)-{\frac {M_{xy}'(t)}{T_{2}}}\\&=-i\gamma B_{0}M_{xy}'(t)-i\gamma \Delta B_{z}(t)M_{xy}'(t)+i\gamma B_{xy}'(t)M_{z}'(t)+i\Omega M_{xy}'(t)-{\frac {M_{xy}'(t)}{T_{2}}}\\&=i(\Omega -\omega _{0})M_{xy}'(t)-i\gamma \Delta B_{z}(t)M_{xy}'(t)+i\gamma B_{xy}'(t)M_{z}'(t)-{\frac {M_{xy}'(t)}{T_{2}}}\\\end{aligned}}}$

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

• i (Ω - ω₀) M_xy′(t) is the Larmor term in the frame of reference rotating with angular frequency Ω. Note that it becomes zero when Ω = ω₀.
• The -i γ ΔB_z(t) M_xy′(t) term describes the effect of magnetic field inhomogeneity (as expressed by ΔB_z(t)) 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 i γ B_xy′(t) M_z(t) describes the effect of RF field (the B_xy′(t) factor) on nuclear magnetization. For an example see below.
• - M_xy′(t) / T₂ describes the loss of coherency of transverse magnetization.

Similarly, the equation of motion of M_z in the rotating frame of reference is:

${\displaystyle {\frac {dM_{z}'(t)}{dt}}=i{\frac {\gamma }{2}}\left(M'_{xy}(t){\overline {B'_{xy}(t)}}-{\overline {M'_{xy}}}(t)B'_{xy}(t)\right)-{\frac {M_{z}-M_{0}}{T_{1}}}}$

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

When the external field has the form:

${\displaystyle B_{x}(t)=B_{1}\cos \omega t}$

${\displaystyle B_{y}(t)=-B_{1}\sin \omega t}$

${\displaystyle B_{z}(t)=B_{0}}$,

We define:

${\displaystyle \epsilon$ :=\gamma B_{1}} and ${\displaystyle \Delta$ :=\gamma B_{0}-\omega } ,

and get (in the matrix-vector notation):

${\displaystyle {\frac {d}{dt}}\left({\begin{array}{c}M'_{x}\\M'_{y}\\M'_{z}\end{array}}\right)=\left({\begin{array}{ccc}-{\frac {1}{T_{2}}}&\Delta &-\epsilon \\-\Delta &-{\frac {1}{T_{2}}}&\epsilon \\-\epsilon &-\epsilon &-{\frac {1}{T_{1}}}\end{array}}\right)\left({\begin{array}{c}M'_{x}\\M'_{y}\\M'_{z}\end{array}}\right)+\left({\begin{array}{c}0\\0\\{\frac {M_{0}}{T_{1}}}\end{array}}\right)}$

Relaxation of transverse nuclear magnetization M_xy

Assume that:

• The nuclear magnetization is exposed to constant external magnetic field in the z direction B_z′(t) = B_z(t) = B₀. Thus ω₀ = γB₀ and ΔB_z(t) = 0.
• There is no RF, that is B_xy' = 0.
• 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 transverse nuclear magnetization, M_xy'(t) simplifies to:

${\displaystyle {\frac {dM_{xy}'(t)}{dt}}=-{\frac {M_{xy}'}{T_{2}}}}$

This is a linear ordinary differential equation and its solution is

${\displaystyle M_{xy}'(t)=M_{xy}'(0)e^{-t/T_{2}}}$.

where M_xy'(0) is the transverse nuclear magnetization in the rotating frame at time t = 0. 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 Ω = ω₀), the vector of transverse nuclear magnetization, M_xy(t) 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 B_z′(t) = B_z(t) = B₀. Thus ω₀ = γB₀ and ΔB_z(t) = 0.
• There is no RF, that is B_xy' = 0.
• 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, M_z(t) simplifies to:

${\displaystyle {\frac {dM_{z}(t)}{dt}}=-{\frac {M_{z}(t)-M_{z,\mathrm {eq} }}{T_{1}}}}$

This is a linear ordinary differential equation and its solution is

${\displaystyle M_{z}(t)=M_{z,\mathrm {eq} }-[M_{z,\mathrm {eq} }-M_{z}(0)]e^{-t/T_{1}}}$

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 B_z′(t) = B_z(t) = B₀. Thus ω₀ = γB₀ and ΔB_z(t) = 0.
• At t = 0 an RF pulse of constant amplitude and frequency ω₀ is applied. That is B'_xy(t) = B'_xy is constant. Duration of this pulse is τ.
• The rotating frame of reference rotates with an angular frequency Ω = ω₀.
• T₁ and T₂ → ∞. Practically this means that τ ≪ T₁ and T₂.

Then for 0 ≤ t ≤ τ:

${\displaystyle {\begin{aligned}{\frac {dM_{xy}'(t)}{dt}}=i\gamma B_{xy}'M_{z}(t)\end{aligned}}}$

${\displaystyle {\frac {dM_{z}(t)}{dt}}=i{\frac {\gamma }{2}}\left(M'_{xy}(t){\overline {B'_{xy}}}-{\overline {M'_{xy}}}(t)B'_{xy}\right)}$