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In electrodynamics, given the vector potential $\vec{A}$, the magnetic field is defined as:
$\vec{B} = \nabla \times \vec{A}$

I'm having trouble figuring out how a coordinate transformation (a rotation) affects these vectors.

If in coordinate system number 1 we have $A_x=x,A_y=0,A_z=0$. Then $\vec{B}=0$.
Now change coordinate systems by rotating.
$x' = \cos(\theta) x - \sin(\theta) y$
$y' = \sin(\theta) x + \cos(\theta) y$
And the inverse is just
$x = \cos(\theta) x' + \sin(\theta) y'$
$y = -\sin(\theta) x' + \cos(\theta) y'$

So $A'_{x'} = \cos(\theta) x,\ A'_{y'} = - \sin(\theta) x,\ A'_{z'} = 0$
$A'_{x'} = \cos(\theta) (\cos(\theta)x' + \sin(\theta)y'),\ A'_{y'} = - \sin(\theta) (\cos(\theta)x' + \sin(\theta)y'),\ A'_{z'} = 0$

But now we have
$B'_{x'}=0,\ B'_{y'}=0,\ B'_{z'}=\frac{\partial}{\partial x'} A'_{y'} - \frac{\partial}{\partial y'} A'_{x'} \neq 0$

A magnetic field shoudn't appear from no-where if I merely rotate my coordinate system. What am I doing wrong? What is the correct way to do these transformations?

  • 0
    This looks more like something for [physics.SE](http://physics.stackexchange.com/)...2011-10-08
  • 0
    it seems you want to know why $\nabla\times (RA)\ne 0$, where $R$ is a rotation matrix.2011-10-08

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