Transforming partial derivatives to polar coordinates

Question

I have to convert the following expression $V(x,y) = x\dfrac{\partial f}{\partial y} - y\dfrac{\partial f}{\partial x}$ to polar coordinates. How do i express the partial derivatives in terms of $r$ and $\theta$? The answer should be something like $V(r,\theta) = \dfrac{\partial F}{\partial \theta}$, where $F(r,\theta) = f(r\cos\theta, r\sin\theta)$.

Answer 1

OP is rightfully using a different notation for the function depending on which coordinates it depends on: $f(x,y)$ vs. $F(r,\theta)\,.$

By the chain rule,

\begin{align} \partial_rF&=(\cos\theta)\,\partial_xf+(\sin\theta)\,\partial_yf\,,\\[2mm] \tfrac1r\partial_\theta F&=(-\sin\theta)\,\partial_xf+(\cos\theta)\,\partial_yf\,. \end{align} Therefore, \begin{align} \partial_xf&=(\cos\theta)\,\partial_rF+(-\sin\theta)\,\tfrac1r\partial_\theta F\,,\\[2mm] \partial_yf&=(\sin\theta)\,\partial_rF+(\cos\theta)\,\tfrac1r\partial_\theta F\,. \end{align} Then, \begin{align}\require{cancel} &x\,\partial_yf-y\,\partial_xf\\[2mm] &=\cancel{(r\cos\theta\sin\theta)\,\partial_rF}+(\cos^2\theta)\,\partial_\theta F-\cancel{(r\sin\theta\cos\theta)\,\partial_rF}+(\sin^2\theta)\,\partial_\theta F\\[2mm] &=\partial_\theta F\,.\tag{1} \end{align}

Another approach is to consider the one-form $$ x\,dy-y\,dx $$ that is the dual of the vector field $x\,\partial_y-y\,\partial_x\,.$ Using the chain rule in the form $$ dx=(\cos\theta)\,dr-(r\sin\theta)\,d\theta\,,\quad dy=(\sin\theta)\,dr+(r\cos\theta)\,d\theta\tag{2} $$ gives \begin{align}\require{cancel} &x\,dy-y\,dx\\[2mm] &=\cancel{(r\cos\theta)(\sin\theta)\,dr}+(r^2\cos^2\theta)\,d\theta-\cancel{(r\sin\theta)(\cos\theta)\,dr}+(r^2\sin^2\theta)\,d\theta\\[2mm] &=r^2\,d\theta\,.\tag{3} \end{align} The factor $r^2$ seems odd but it is correct because the vector $x\,\partial_y-y\,\partial_x=\partial_\theta$ has length $\sqrt{x^2+y^2}=r\,.$ From (2) we see that the covector $r\,d\theta$ has length one. Therefore $r^2\,d\theta$ has length $r$ as it must.

• It follows that the vector that corresponds to $r^2\,d\theta$ is $\partial_\theta\,.$

This is compatible with the metric in polar coordinates being $$ \boldsymbol{g}=dr\otimes dr+r^2\,d\theta\otimes d\theta $$ and the musical isomorphism by which $$ \partial_\theta{}^\flat=\boldsymbol{g}(\partial_\theta\,,\,.\,)=r^2\,d\theta\,. $$

Answer 2

Write $x=r\cos\theta$ and $y=r\sin\theta$ and apply the chain rule \begin{equation}\frac{\partial f}{\partial \theta}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}\end{equation}

We can substitute $\frac{\partial x}{\partial \theta}=-r\sin\theta= -y$ and $\frac{\partial y}{\partial \theta}=r\cos\theta=x$

Therefore \begin{equation}\frac{\partial f}{\partial \theta}=-y\frac{\partial f}{\partial x}+x\frac{\partial f}{\partial y}=V(x,y)\end{equation}

Answer 3

By the chain rule:

$\frac{\partial f}{\partial \theta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}$

$\frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}$

Which can be expressed as a matrix.

$\begin{pmatrix} \frac{\partial f}{\partial \theta} \\ \frac{\partial f}{\partial r} \end{pmatrix} = \begin{pmatrix} \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta} \\ \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \end{pmatrix} \begin{pmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y}\end{pmatrix}$

The matrix's inverse can be used to do a reverse coordinate transform without having to reuse the chain rule.

You can effectively bypass the coordinate transform or much calculation at all.

$V(x,y)=x \frac{\partial f}{\partial y}-y\frac{\partial f}{\partial x }=\vec{r} \cdot (\nabla f \times \hat{k})$

The term in parentheses is obviously perpendicular to $\hat{k}$. That means it only has $\hat{\theta} $ and $\hat{r}$ components. $\vec{r}$ has no $\hat{\theta}$ component in polar coordinates. It follows that only the $\hat{r}$ component of the term in parentheses plays a role. $\hat{k}$ is the normal to the plane, so up to a sign it just swaps x for y coordinates, or r for theta coordinates. This means the only useful term in parentheses is $\frac{1}{r}\frac{\partial f}{\partial \theta}\hat{r}$.Since $\vec{r} = r \hat{r}$ then the dot product is $\frac{\partial f}{\partial \theta}$ establishing the desired result.