Do coordinate transformation equations imply constant θ?

[stevendaryl]

Suppose that I have a two-dimensional coordinate system $(x,y)$ and I change to a new coordinate system $(u,v)$. What I know is that there is some function $\theta(u,v)$ such that:

1. $\dfrac{\partial x}{\partial u} = cos(\theta)$
2. $\dfrac{\partial x}{\partial v} = -sin(\theta)$
3. $\dfrac{\partial y}{\partial u} = sin(\theta)$
4. $\dfrac{\partial y}{\partial v} = cos(\theta)$

My question is: do these 4 equations imply that $\theta =$ constant? (so that the relationship between the coordinate systems is linear)

 

[wabbit]

Looks like it yes, if you compute the cross derivatives
$\frac{\partial^2x}{\partial u\partial v} =-\sin\theta\frac{\partial\theta}{\partial v}=-\cos\theta\frac{\partial\theta}{\partial u}\\\frac{\partial^2y}{\partial u\partial v} =\cos\theta\frac{\partial\theta}{\partial v}=-\sin\theta\frac{\partial\theta}{\partial u}$
you get a set of equations that require $\frac{\partial\theta}{ \partial u}=\frac{\partial\theta}{ \partial v}=0$

 

[stevendaryl]

Looks like it yes, if you compute the cross derivatives
$\frac{\partial^2x}{\partial u\partial v} =-\sin\theta\frac{\partial\theta}{\partial v}=-\cos\theta\frac{\partial\theta}{\partial u}\\\frac{\partial^2y}{\partial u\partial v} =\cos\theta\frac{\partial\theta}{\partial v}=-\sin\theta\frac{\partial\theta}{\partial u}$
you get a set of equations that require $\frac{\partial\theta}{ \partial u}=\frac{\partial\theta}{ \partial v}=0$

Thank you! I tried exactly that, but I made a stupid sign error, and found them consistent.

 

[wabbit]

Ah yes, signs are the spawn of the devil :)