Let $\;U \subset \mathbb R^2\;$ be a smooth open domain. On $\;\partial U\;$ consider the positively oriented orthonormal basis $\;(\mathcal v,\mathcal τ)\;$ where $\;\mathcal v\;$ is the outer normal and $\;\mathcal τ\;$ is the tangential vector.
Fix $\;x\in \partial U\;$ and take $\;\{\mathcal v_x,\mathcal τ_x \}\;$. Now fix this system and drop the subscript $\;x\;$.
Let $\;x'\;$ be the coordinates of an arbitary vector in $\;\mathbb R^2\;$ with respect to $\;\{\mathcal v,\mathcal τ\}\;$ and $\;A\;$ the rotation: $\;x'=Ax\;$ where $\;A\;$ is orthogonal.
I'm interested in finding:
1. The coordinates of $\;x'\;$ with respect to $\;\{\mathcal v,\mathcal τ\}\;$.
2. The coordinates of $A\mathcal v\;$
and understand the geometry behind these changes.
My attempt:
Below I tried to picture the above as I understand them:
The coordinates of $\;x'\;$ with respcect to $\;\{x_1,x_2\}\;$ are : $\;x'=(x_1,x_2)\;$ while the coordinates of $\;x'\;$ with respect to $\;\{\mathcal v,\mathcal τ\}\;$ should be: $\;x'=({x_1}',{x_2}')=(\langle x',\mathcal τ \rangle\;\langle x',\mathcal v \rangle)\;$ where $\;\langle \cdot, \cdot \rangle \;$ stands for the Euclidean inner product.
Questions:
- Is the above visualization correct?
- Does the orthogonal transformation $\;x'=Ax\;$ represent the rotation of the axes $\;\{x_1,x_2\}\;$ to $\;\{\mathcal v,\mathcal τ\}\;$?
- What will $\;A\mathcal v\;$ be like?
- Except the rotaion of the axes, there is also a translation. Where did it show up?
All these coordinates and transformations confuse me a lot. I would really appreciate if someone could make all that clear to me.I apologize in advance if the above are quite elementary but I focused mostly on just solving equations in the past rather than understanding their geometric meaning (which is completely wrong).
Thanks!