There is a great way to rotate a Cartesian vector field about the origin described in Rotating vector functions.
Instead, let us suppose that we have a velocity vector field in polar coordinates i.e., $\vec{V}(r,\theta)\equiv V_r ~\hat{r} + V_\theta ~\hat{\theta}$, where $V_r$ is the radial velocity component and $V_\theta$ is the azimuthal velocity component.
How would one rotate this polar velocity vector field by angle $\alpha$ clockwise about the origin $(x,y)=(0,0)$?
I suppose one could convert $V_r$ and $V_\theta$ into their corresponding Cartesian components $V_x$ and $V_y$, rotate those fields via the method described in Rotating vector functions, and then convert it back.
Is there a more elegant way?
Best Answer
It will turn out that the rotation in polar coordinates is nothing else than a shift of coordinate functions by an angle $\alpha$. Note that we usually shift a function $g$ on $\mathbb R$ by $\alpha$ by going to the function $x\mapsto g(x\color{red}{-}\alpha)$.
A very convenient method to formalize vector fields in different coordinates is to treat them as derivations. This is very common in differential geometry. In $2d$ Cartesian coordinates a vector field is then $$ X=X_1\frac{\partial}{\partial x_1}+X_2\frac{\partial}{\partial x_2} $$ where $X_1,X_2$ are scalar functions on $\mathbb R^2$.
From the Cartesian-polar transformation \begin{align} r&=\sqrt{x_1^2+x_2^2}\,,& \theta&=\arctan\frac{x_2}{x_1}\,\\ x_1&=r\cos\theta\,,&x_2&=r\sin\theta \end{align} we get \begin{align} \frac{\partial x_1}{\partial r}&=\cos\theta\,,&\frac{\partial x_1}{\partial \theta}=-r\sin\theta\,,\\ \frac{\partial x_2}{\partial r}&=\sin\theta\,,&\frac{\partial x_2}{\partial \theta}=r\cos\theta\,. \end{align} By the chain rule we have for any differentiable function $f$ on $\mathbb R^2$ \begin{align} \frac{\partial f}{\partial r}&=\frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial r}+\frac{\partial f}{\partial x_2}\frac{\partial x_2}{\partial r} =\frac{\partial f}{\partial x_1}\cos\theta+\frac{\partial f}{\partial x_2}\sin\theta\,,\\[3mm] \frac{\partial f}{\partial \theta}&=\frac{\partial f}{\partial x_1}\frac{\partial x_1}{\partial \theta}+\frac{\partial f}{\partial x_2}\frac{\partial x_2}{\partial \theta} =-\frac{\partial f}{\partial x_1}r\sin\theta+r\frac{\partial f}{\partial x_2}\cos\theta\,. \end{align} In matrix notation and dropping the test function this is the transformation rule for the basis vector fields: $$ \begin{pmatrix}\frac{\partial}{\partial r}\\\frac{\partial}{\partial \theta} \end{pmatrix}=\begin{pmatrix}\cos\theta &\sin\theta\\-r\sin\theta&r\cos\theta\end{pmatrix}\begin{pmatrix}\frac{\partial}{\partial x_1}\\\frac{\partial}{\partial x_2} \end{pmatrix}. $$ Clearly, $$ \begin{pmatrix}\frac{\partial}{\partial x_1}\\\frac{\partial}{\partial x_2} \end{pmatrix}=\begin{pmatrix}\cos\theta &-\frac{1}{r}\sin\theta\\\sin\theta&\frac{1}{r}\cos\theta\end{pmatrix}\begin{pmatrix}\frac{\partial}{\partial r}\\\frac{\partial}{\partial \theta} \end{pmatrix}. $$ The vector field $X$ in polar coordinates is therefore, \begin{align} X'&=X_1'\Big(\cos\theta\frac{\partial}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial}{\partial \theta}\Big)+X_2'\Big(\sin\theta\frac{\partial}{\partial r}+\frac{1}{r}\cos\theta\frac{\partial}{\partial \theta}\Big)\\[3mm] &=\Big(X_1'\cos\theta+X_2'\sin\theta\Big)\frac{\partial}{\partial r}+ \Big(-X_1'\frac{1}{r}\sin\theta+X_2'\frac{1}{r}\cos\theta\Big)\frac{\partial}{\partial \theta} \end{align} where $X'_1(r,\theta)=X_1(x_2,x_2)$ and $X'_2(r,\theta)=X_2(x_2,x_2)\,.$
The whole point is now that nothing is simpler than rotating the vector field $X'$ by an angle $\alpha$ around the origin.
This rotation is a shift which gives \begin{align} X'' &=\Big(X_1''\cos(\theta-\alpha)+X_2'\sin(\theta-\alpha)\Big)\frac{\partial}{\partial r}+ \Big(-X_1''\frac{1}{r}\sin(\theta-\alpha)+X_2''\frac{1}{r}\cos(\theta-\alpha)\Big)\frac{\partial}{\partial \theta} \end{align} where $X''_1(r,\theta)=X'_1(r,\theta-\alpha)$ and $X''_2(r,\theta)=X_2'(r,\theta-\alpha)\,.$