Rotating Reference Frame

• Motivated to look at a reference frame where $\vec{B}_{0}$, $\vec{B}_{1}$ are static and $\vec{\mu}$ will also be static when $\vec{B}_{1}$ is off, and relaxation is neglected.

• Go over some basic conventions of rotating reference frames, and reminders of time-dependent frames.

• Remember when $\hat{x}=C \rightarrow \hat{x}(t)$ going to need to pull out chain rule to find time derivatives.

  
  
  
  
  
  
  
  

• Use convention that rotating frame will be primed.

• Look at time derivative of an arbitrary vector $\vec{V}(t)$
  

  $$\, \hspace{-0.5in}\frac{d V_{x}}{d t}\hat{x} + \frac{d V_{y}}{d t} \hat{y} + \frac{d V_{z}}{d t} \hat{z}$$

  $$= \frac{d V_{x'}}{d t} \hat{x}'(t) + \frac{d V_{y'}}{d t} \hat{y}'(... ...'(t)}{d t} + V_{y'}\frac{d \hat{y}'(t)}{d t} + V_{z'}\frac{d \hat{z}'(t)} {d t}$$ (3.3)

  
  

• Frame rotates with angular velocity $\Omega$ about a vector $\hat{\Omega}$. In MRI we choose $\hat{\Omega}$ to be parallel to $\vec{B}_{0}$. Also convention is that $\hat{B}_{0} = \hat{z}$

• $\hat{x'}$ is being rotated with respect to the lab. A vector being rotated with respect to the laboratory frame is going to have a time derivative that can be related to the axis and rate of rotation according to
  

  $$\frac{d \vec{C}}{dt} = \vec{\Omega} \times \vec{C}$$ (3.4)

  
  
  therefore
  

  $$\frac {d\hat{x}'} {dt} = \vec{\Omega} \times \hat{x'} \\$$ (3.5)

  
  

  
  
  
  
  
  
  
  

• Can re-write 3.3
  

  $${\frac {d\vec{V}} {dt}} = \left(\frac {d \vec{V}} {dt}\right)' + \vec{\Omega} \times \vec{V}\\$$ (3.6)

  
  
  where $\left(\frac {d \vec{V}} {dt}\right)' = dV'/dt \hat{x'}$