Bloch Equations in a Rotated Frame

Question

Consider the following Bloch equations in the absence of relaxation $$ \frac{dM_x}{dt}=(\mathbf{M}(t)\times\mathbf{B}(t))_x, \ \ \ \ \ \ \frac{dM_y}{dt}=(\mathbf{M}(t)\times\mathbf{B}(t))_y,\ \ \ \ \ \ \ \frac{dM_z}{dt}=(\mathbf{M}(t)\times\mathbf{B}(t))_z $$ where $\mathbf{M}=(M_x,M_y,M_z)$ is a magnetization and $\mathbf{B}$ is an external magnetic field. In a paper due to Aharonov and Stern, https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.69.3593, it is necessary to rotate the coordinates so that $$ \frac{d\mathbf{M}'}{dt}=\mathbf{ M}'\times\left(\mathbf{B}\hat{z}+\dot{\hat{z}}\times\hat{z}\right), $$ where the prime notation denotes the rotated variable, the overhead dot means time derivative, and $\hat{z}$ is the typical Cartesian unit vector $\hat{x}\times\hat{y}$ but with respect to the rotated frame. My question is, how do we write this in component form? I mean explicitly, where in the unrotated frame it's easy to understand $$ \frac{dM_x}{dt}=M_yB_z-M_zB_y, $$ and so on. My main difficulty is in visualizing the term $\dot{\hat{z}}\times\hat{z}.$

Answer 1

Preliminary comments.

• Vector product of the time derivative of a vector of a rotating basis and the vector itself can be related to the angular velocity of the rotating reference frame, see below.

• Your Bloch's equation either is dimensionally wrong, or it's written not using the International System of Units, SI. Using SI, Bloch's equation without relaxation reads

  $\dfrac{d \mathbf{M}}{dt} = \gamma (\mathbf{M} \times \mathbf{B})$.

A more complete answer. Using two different Cartesian reference frames:

• $\{ \mathbf{\hat{e}}^0_i\}_i$ "fixed" reference frame;
• $\{ \mathbf{\hat{e}}^1_i\}_i$ rotation reference frame w.r.t. the "fixed" one, where
  • the relative orientation is given by the rotation tensor $\mathbb{R}$, so that $\mathbf{\hat{e}}^1_i = \mathbb{R} \cdot \mathbf{\hat{e}}^0_i$,
  • the relative angular velocity $\boldsymbol{\omega}$, s.t. $\boldsymbol{\omega}_{\times} := \mathbb{\Omega} \!\!\!\! \mathbb{\Omega} = \dot{\mathbb{R}} \cdot\mathbb{R}^T$, so that $\dot{\mathbf{\hat{e}}}^1_i = \boldsymbol{\omega} \times \mathbf{\hat{e}}^1_i$.

We can write all the vectors in both Cartesian frames, as an example, $\mathbf{M} = M^0_i \mathbf{\hat{e}}^0_i = M^1_i \mathbf{\hat{e}}^1_i$.

Taking time derivative of $\mathbf{M}$, we get

$\dfrac{d\mathbf{M}}{dt} = \dfrac{d}{dt} \left( M^0_i \mathbf{\hat{e}}^0_i \right) = \dot{M}^0_i \mathbf{\hat{e}}^0_i \\ \qquad = \dfrac{d}{dt} \left( M^1_i \mathbf{\hat{e}}^1_i \right) = \dot{M}^1_i \mathbf{\hat{e}}^1_i + M^1_i \dot{\mathbf{\hat{e}}^1_i} = \dot{M}^1_i \mathbf{\hat{e}}^1_i + M^1_i \boldsymbol{\omega} \times \mathbf{\hat{e}}^1_i = \dfrac{^1 d\mathbf{M}}{dt} + \boldsymbol{\omega} \times \mathbf{M}$,

having introduced the operator $\frac{^1 d}{dt}$ that represents the time variation of a quantity, as seen by the moving reference frame, seeing no variation of the vectors of the rotating basis. Using your notation, this corresponds to the time derivative of the components in the rotating frame $\frac{d \mathbf{M}'}{dt}$.

Bloch's equation reads

$\dfrac{d \mathbf{M}}{dt} = \gamma (\mathbf{M} \times \mathbf{B})$,

and using the point of view of the rotating frame

$\dfrac{^1d \mathbf{M}}{dt} = \gamma (\mathbf{M} \times \mathbf{B}) + \mathbf{M} \times \boldsymbol{\omega} = \mathbf{M} \times \left( \gamma \mathbf{B} + \boldsymbol{\omega} \right)$.

Now, the last step. The angular velocity can be written in terms of the time derivatives of the vector basis of the rotating reference frame, rearranging the relation $\dot{\mathbf{\hat{e}}}^1_i = \boldsymbol{\omega} \times \mathbf{\hat{e}}^1_i$, to get

$\boldsymbol{\omega} = -\dfrac{1}{2} \left( \dot{\mathbf{\hat{e}}}^1_i \times \mathbf{\hat{e}}^1_i\right) \qquad \text{(sum over i)}$

and thus

$\dfrac{^1d \mathbf{M}}{dt} = \mathbf{M} \times \left[ \gamma \mathbf{B} - \dfrac{1}{2}\left( \dot{\mathbf{\hat{e}}}^1_i \times \mathbf{\hat{e}}^1_i\right) \right]$.

(If you provide some more details about the rotation of the rotating frame, we could try to rearrange the last term involving time derivative of the vectors of the basis)