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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?

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    it seems you want to know why $\nabla\times (RA)\ne 0$, where $R$ is a rotation matrix.2011-10-08

1 Answers 1

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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 :-)