# transforming vector potential with a coordinate rotation

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?

• This looks more like something for physics.SE... – J. M. is a poor mathematician Oct 8 '11 at 13:45
• it seems you want to know why $\nabla\times (RA)\ne 0$, where $R$ is a rotation matrix. – Shiyu Oct 8 '11 at 14:41

A magnetic field shoudn't appear from no-where if I merely rotate my coordinate system

That's right, it should not :-)

What am I doing wrong?

It seems to be a simple sign error (or you may have to look up the contravariant transformation law for vectors, e.g. on Wikipedia here). We would like to compute $A(x', y', z')$, so remember that we have to compute, for the first component, $$A_1(x', y', z') = \frac{\partial x'}{\partial x} * A_1(x, y, z) + \frac{\partial x'}{\partial y} * A_2(x, y, z) + \frac{\partial x'}{\partial z} * A_3(x, y, z) = \frac{\partial x'}{\partial x} * x = x \cos(\theta)$$

and inserting $x = \cos(θ) x'+ \sin(θ) y'$ we get $$A_1(x', y', z') = \cos(\theta) ( \cos(θ) x'+ \sin(θ) y')$$

and for the second: $$A_2(x', y', z') = \frac{\partial y'}{\partial x} * A_1(x, y, z) + \frac{\partial y'}{\partial y} * A_2(x, y, z) + \frac{\partial y'}{\partial z} * A_3(x, y, z) = \frac{\partial x'}{\partial x} * x = x \sin(\theta)$$ Inserting again $x = \cos(θ) x'+ \sin(θ) y'$ we get

$$A_2(x', y', z') = \sin(\theta) ( \cos(θ) x'+ \sin(θ) y')$$ For the z-component of $B$ we now get $$B_{z'} = \partial_{x'} A_2(x', y' ,z') - \partial_{y'} A_1(x', y' ,z') = \sin(\theta) \cos(\theta) - \sin(\theta) \cos(\theta) = 0$$ I hope I did not screw up :-)