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frogster
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Homework Statement
I'm having trouble understanding coordinate transformations for vector fields. There are two 'coordinate pieces', the coordinates pieces of the vector at a point changes, and the function describing the field can also be rewritten in terms of the new coordinates. I'm having trouble being precise enough in my math to understand this. Precision is the key. So please be as pedantic as you want.
Here's an example for us to work on --
In electrodynamics, given the vector potential [itex]\vec{A}[/itex], the magnetic field is defined as:
[tex]\vec{B} = \nabla \times \vec{A}[/tex]
In coordinate system number S we have
[tex]A_x=x,A_y=0,A_z=0[/tex]
In coordinate system S', related to S by a rotation about the Z-axis. What is B? Also calculate A, and determine B this way as well.
Homework Equations
[tex]\vec{B} = \nabla \times \vec{A}[/tex]
To change coordinate systems by rotating about the z-axis.
[tex]x' = \cos(\theta) x - \sin(\theta) y[/tex]
[tex]y' = \sin(\theta) x + \cos(\theta) y[/tex]
And the inverse is just
[tex]x = \cos(\theta) x' + \sin(\theta) y'[/tex]
[tex]y = -\sin(\theta) x' + \cos(\theta) y'[/tex]
The Attempt at a Solution
In coordinate system S, we have
[tex] \vec{B} = \nabla \times \vec{A} = 0[/tex]
rotating this vector just trivially gives in S'
[tex] \vec{B'} = 0[/tex]
Now to trying to rotate A
[tex]A'_{x'} = \cos(\theta) x,\ A'_{y'} = - \sin(\theta) x,\ A'_{z'} = 0[/tex]
[tex]A'_{x'} = \cos(\theta) (\cos(\theta)x' + \sin(\theta)y'),\ A'_{y'} = - \sin(\theta) (\cos(\theta)x' + \sin(\theta)y'),\ A'_{z'} = 0[/tex]
But now we have
[tex]B'_{x'}=0,\ B'_{y'}=0,\ B'_{z'}=\frac{\partial}{\partial x'} A'_{y'} - \frac{\partial}{\partial y'} A'_{x'} \neq 0[/tex]
[tex]\vec{B'} \neq 0[/tex]
This disagrees with above, and clearly a magnetic field shouldn't suddenly appear just because I rotated my coordinate system, so I must be doing something wrong.