Homework Help: Help with coordinate transformations

1. Oct 9, 2011

frogster

1. The problem statement, all variables and given/known data

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 $\vec{A}$, the magnetic field is defined as:
$$\vec{B} = \nabla \times \vec{A}$$
In coordinate system number S we have
$$A_x=x,A_y=0,A_z=0$$
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.

2. Relevant equations
$$\vec{B} = \nabla \times \vec{A}$$
To change coordinate systems by rotating about the z-axis.
$$x' = \cos(\theta) x - \sin(\theta) y$$
$$y' = \sin(\theta) x + \cos(\theta) y$$
And the inverse is just
$$x = \cos(\theta) x' + \sin(\theta) y'$$
$$y = -\sin(\theta) x' + \cos(\theta) y'$$
3. The attempt at a solution

In coordinate system S, we have
$$\vec{B} = \nabla \times \vec{A} = 0$$
rotating this vector just trivially gives in S'
$$\vec{B'} = 0$$
Now to trying to rotate A
$$A'_{x'} = \cos(\theta) x,\ A'_{y'} = - \sin(\theta) x,\ A'_{z'} = 0$$
$$A'_{x'} = \cos(\theta) (\cos(\theta)x' + \sin(\theta)y'),\ A'_{y'} = - \sin(\theta) (\cos(\theta)x' + \sin(\theta)y'),\ A'_{z'} = 0$$
But now we have
$$B'_{x'}=0,\ B'_{y'}=0,\ B'_{z'}=\frac{\partial}{\partial x'} A'_{y'} - \frac{\partial}{\partial y'} A'_{x'} \neq 0$$
$$\vec{B'} \neq 0$$
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.

2. Oct 9, 2011

praharmitra

What you have done is right! You have stumbled upon the fact that the magnetic field B is not really a vector. It is part of a two-tensor called the field strength tensor. Rotation does actually give rise to magnetic field.

3. Oct 10, 2011

Dick

It does most emphatically NOT!! A is a vector. You should have rotated A just like you rotated the coordinates. A_x'=cos(theta)*A_x and A_y'=sin(theta)*A_x. You have a sign mistake. You don't start mixing electric and magnetic fields until you start doing special relativity with boosts. A rotation won't do it.

Last edited: Oct 10, 2011
4. Oct 10, 2011

frogster

Doh! It seems so obvious now.

Thanks. I always have trouble finding factors of two and sign errors. Sometimes if I take a break and then work out the problem from scratch, I can fix the errors, but for some reason I can never find them staring at the math ... like my brain skips over it because it "knew what I meant" or something. I have the same problem with spelling mistakes in essays.

Besides just reworking the problem, do you guys have any suggestions from experience for how to find sign errors? Thanks.