Induction equation in a rotating frame In his book, Tom Cravens refers to the induction equation (perfectly conducting limit) in a rotating frame, the Sun. See link below:
https://books.google.com/books?id=xw50iZep-McC&pg=PA222&lpg=PA222&dq=induction+equation+rotating+frame&source=bl&ots=rZyW98JmqJ&sig=uGL3JWPYtAJAGQUA26KKmx5gCe8&hl=en&sa=X&ved=0CDEQ6AEwA2oVChMInpTW2fWuxwIVCM2ACh3EcgkO#v=onepage&q=induction%20equation%20rotating%20frame&f=false
My question is how to arrive from the static frame equation (6.87):
$\frac{\partial B}{\partial t}$ = $\nabla\times (u' \times B$)  +$\nabla \times( (\Omega \times  r)\times B)$
$u'$ denotes the velocity in the rotating frame, rotating with angular velocity $\Omega$, $r$ denotes the position in spherical coordinates.
to the rotating frame equation (6.88):
$\frac{\partial B}{\partial t}|_R = \nabla \times (u'\times B$) 
Where $|_R$ denotes the derivative in the rotating frame, so it should only add on $\Omega\times B$ which he says cancel out by expending the second term of the RHS of (6.87)
I tried expanding the double cross product using vector identities, where we have one term canceling from Maxwell's equation ($\nabla \cdot B=0$) but the three other terms remain still.
Can someone help me figure it out ?
Thank you
 A: I think I found the answer, it was a bit tricky so I am posting it -
Using vector calculus identity we can start by expanding $\frac{\partial B}{\partial t}$:
$\frac{\partial B}{\partial t}$=$\nabla \times (u' \times B)  +\nabla \times ((\Omega \times r)\times B$)
Here we need to use $\nabla\times\left(\mathbf{A}\times\mathbf{B}\right)=\mathbf{A}\left(\nabla\cdot\mathbf{B}\right)-\mathbf{B}\left(\nabla\cdot\mathbf{A}\right)+\left(\mathbf{B}\cdot\nabla\right)\mathbf{A}-\left(\mathbf{A}\cdot\nabla\right)\mathbf{B}$
So that $\nabla$x (($\Omega$ x $r$)x$B$) = $(\Omega \times r)(\nabla \cdot B)-B(\nabla \cdot (\Omega \times r))+(B \cdot \nabla)(\Omega \times r)-((\Omega \times r) \cdot \nabla)(B)$
Using the fact that $\nabla \cdot B=0$ and $\nabla \cdot (\Omega \times r)=0$ The two first terms are equal to zero
Another very useful vector trick is:
$(B \cdot \nabla)(\Omega \times r)=\Omega \times (B \cdot \nabla)r = \Omega \times B$ 
We are left with: 
$\frac{\partial B}{\partial t}$=$\nabla$x ($u' \times B) +\Omega \times B -((\Omega \times r) \cdot \nabla)(B)$ 
Here $\frac{\partial B}{\partial t}$ is taken in the static frame, we need to convert that to the rotating frame:
$\frac{\partial B}{\partial t}|_{static}= \frac{DB}{Dt}|_{static} - u_\Omega \cdot \nabla(B)$ 
to take in account advection as the velocity of the field with the frame is $u_\Omega=(\Omega \times r)$
$\frac{DB}{Dt}|_{static}=\frac{DB}{Dt}|_{rotating}+\Omega \times B$
But $\frac{DB}{Dt}|_{rotating}$ is identical to $\frac{\partial B}{\partial t}|_{rotating}$ because the frame is moving with $u_{\Omega}$ already.
Combining everything yields to:
$\nabla \times (u' \times B) +\Omega \times B -((\Omega \times r) \cdot \nabla)(B)=\frac{\partial B}{\partial t}|_{rotating} + \Omega \times B - ((\Omega \times r) \cdot \nabla)(B)$
After simplifying - $\frac{\partial B}{\partial t}|_{rotating}=\nabla \times (u' \times B)$ ....
Now there is probably a physical argument that can spare me all the troubles but I really needed to see it mathematically. I think what was really tricky was to use the total derivative to be able to use the formula relating the derivatives in the different frames (/!\ that relation does not hold for partial derivatives)
Check that link for more info: http://physics.princeton.edu/~mcdonald/examples/rotatingEM.pdf
