Coordinates rotation and function change

444 Views Asked by At

In the Cartesian coordinates $(x,y)$, I have a vector function $\bar{f}(x)=\hat{x}A\cos(yk)$, where $A$ and $k$ are constants. I make now a 45 degrees rotation (in the same plane) to the new set of Cartesian coordinates $(u,v)$. Since $\sin 45=\cos 45=1/\sqrt{2}$, I know that we will have:

$x=(u-v)/\sqrt{2}$

$y=(u+v)/\sqrt{2}$

Which will give us the function in the new coordinates systems as: $\bar{f}(u,v)=\frac{\hat{u}-\hat{v}}{\sqrt{2}}A\cos[(u+v)k/\sqrt{2}]$.

Is this the correct way to express the function in the new coordinates? Is there any missing scaling factor of any kind?

Thanks.

1

There are 1 best solutions below

0
On BEST ANSWER

That looks exactly correct to me. You can check by plugging in $(x, y) = (1, 0)$ or $(x, y = (1,1)$ and comparing to what you get when you plug in $(u, v) = (1, -1)/\sqrt{2}$ (for the first) or $(u, v) = (1, 0)$ (for the second). Since the results are the same, your expression in $uv$-coords is fine.