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 $\dot{\hat{z}}\times\hat{z}.$
Bloch Equations – Bloch Equations in a Rotated Frame
bloch-spheremagnetic fieldsmagnetic-momentquantum-spintorque
Related Solutions
This is a very good question that many people ask when they learn about the Bloch sphere. There are two important reasons
Physical reason
The most famous two state system in quantum mechanics is the spin 1/2 particle. When you draw the spin 1/2 states on the Bloch sphere something very nice happens, the points on the Bloch sphere correspond to the physical directions of the states. For example the $|\uparrow\,\rangle$ state corresponds to a point that points in the $+z$ direction. Similarly the state $|x+\rangle=\tfrac{ 1}{\sqrt{2}}(|\uparrow\,\rangle+|\downarrow\,\rangle)$ points in the $+x$ direction on the Bloch sphere. To be more precise for some state $|\psi\rangle$ the vector of expectation values given by $\langle\psi|\vec S|\psi\rangle=\langle \vec S\rangle =(\langle S_x\rangle,\langle S_y\rangle,\langle S_z\rangle)^T$ points in the same direction as the Bloch vector. See also this picture:
Mathematical reason
Another reason that the Bloch representation is good is that states which are physically the same correspond to the same point on the Bloch sphere. This is also mentioned in the comments. In your diagram $|0\rangle$ and $-|0\rangle$ are antipodal points but they represent the same state up to an arbitrary phase. Remember that $|\psi\rangle$ and $e^{i\alpha}|\psi\rangle$ represent the same state. The Bloch sphere intuitively leaves out any global phase. To represent the full state on a Bloch sphere you would have to specify an additional phase for any point. In Steane's "Introduction to Spinors" this is represented by drawing a state as a little flag. The direction of the flag gives the point on the Bloch sphere and the angle that flag makes gives the global phase. So in this representation the states $|0\rangle$ and $-|0\rangle$ both point in the z-direction but the first one has its flag rotated to $\alpha=0$ and the second one has it rotated to $\alpha=\pi$ radians. Interestingly when you rotate a spin 1/2 particle by $360^\circ$ its state will pick up a minus sign and you would have to rotate it by another $360^\circ$ before it becomes exactly the same state.
The non linearities arise when you consider the feedback loop. The magnetic moment can generate a field of its own. $\mathbf{B}$ will no longer be the externally applied field, but will rather depend on $\mathbf{M}$ hence the nonlinearity. These nonlinearities give rise to new behaviors such as synchronization and chaotic motion.
Check out this article: Abergel D. "Chaotic solutions of the feedback driven Bloch equations." Phys Lett A 2002;302:17–22.
Hope this helps and tell me if you find some mistakes.
Best Answer
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:
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)