Express $\partial_x$ by $\partial_r$ and $\partial_{\phi}$

Question

I want to express $\partial_x$ and $\partial_y$ by $\partial_r$ and $\partial_{\phi}$.

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

$$ \frac{\partial }{\partial \phi} = \frac{\partial x}{\partial \phi}\frac{\partial }{\partial x} + \frac{\partial y}{\partial \phi}\frac{\partial }{\partial y}$$

, substituting $x=r\cos\phi$ and $y=r\sin\phi$, we get follows.

$$ \frac{\partial }{\partial r} = \cos{\phi}\frac{\partial }{\partial x} +\sin\phi\frac{\partial }{\partial y}$$

$$ \frac{\partial }{\partial \phi} = -r\sin{\phi}\frac{\partial }{\partial x} + r\cos\phi\frac{\partial }{\partial y}$$ Rewrite this by matrix as below. $$ \begin{pmatrix} \frac{\partial }{\partial r} \\ \frac{1}{r}\frac{\partial }{\partial \phi} \\ \end{pmatrix}= \begin{pmatrix} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \\ \end{pmatrix} \begin{pmatrix} \frac{\partial }{\partial x} \\ \frac{\partial }{\partial y} \\ \end{pmatrix}$$ $$\Leftrightarrow \begin{pmatrix} \frac{\partial }{\partial x} \\ \frac{\partial }{\partial y} \\ \end{pmatrix}= \begin{pmatrix} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \\ \end{pmatrix} \begin{pmatrix} \frac{\partial }{\partial r} \\ \frac{1}{r}\frac{\partial }{\partial \phi} \\ \end{pmatrix}$$ Then, we can express $\partial_x$ and $\partial_y$ by $\partial_r$ and $\partial_\phi$, we get follows.

$$ \frac{\partial }{\partial x} = \cos\phi\frac{\partial }{\partial r} - \frac{\sin \phi}{r}\frac{\partial }{\partial \phi}$$

$$ \frac{\partial }{\partial y} = \sin\phi\frac{\partial }{\partial r} + \frac{\cos\phi}{r}\frac{\partial }{\partial \phi}$$

However, if we directly calculate $\partial_x$ and $\partial_y$ from $ \frac{\partial }{\partial x} = \frac{\partial r}{\partial x}\frac{\partial }{\partial r} + \frac{\partial \phi}{\partial x}\frac{\partial }{\partial \phi}$ and $ \frac{\partial }{\partial y} = \frac{\partial r}{\partial y}\frac{\partial }{\partial r} + \frac{\partial \phi}{\partial y}\frac{\partial }{\partial \phi}$, result does not meet with each other. For example, \begin{align} &\frac{\partial \phi}{\partial x} \\ =&\frac{1}{r}\frac{\partial \phi}{\partial \cos\phi} \\ =&\frac{1}{r\sin\phi} \end{align} Here, I use the relationship, $\frac{\partial f(x)}{\partial x} = \left(\frac{\partial x}{\partial f(x)}\right)^{-1}$. However, from the first calculation, this $\frac{\partial \phi}{\partial x}$ should be equal to $-\frac{\sin\phi}{r}$. What is the origin of this contradiction? I find this error (many times) when I'm scoring freshman's physics class test as a TA, however I cannot nicely explain why such latter calculation fails.

Answer 1

The first line of reasoning is correct and the second is not. Anyone who teaches multivariable calculus for a living would immediately be suspicious of the statement $\frac{\partial x}{\partial f} = \left(\frac{\partial f}{\partial x}\right)^{-1}$ since that isn't true for partial derivatives in general (the famous triple product $\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x} = -1$, for example), but that actually is not the problem here, and your use of that statement is indeed correct.

The reason is subtle and has to do with the definition of partial differentiation. Whenever we partial differentiate, we are also implicitly demanding that we hold a specific set of variables constant. If you have any familiarity with statistical mechanics/thermodynamics, then you know that Maxwell's relations depend heavily on which variables are being held constant during a specific partial differentiation.

But when you evaluate

$$\frac{\partial \phi}{\partial x} = \frac{\partial \phi}{\partial (r\cos\phi)}$$

...what exactly is being held constant there? In this case, the quantity being held constant is $r\sin\phi$, because this partial derivative is being taken w.r.t. $x$ with the implicit assumption that $y$ is being held constant. So just pulling $r$ out of the partial like you did in saying $\partial (r\cos\phi) = r\partial \cos\phi$ wouldn't be correct as it is not the variable being held constant.

What happens if we do pull out the correct constant?

$$\frac{\partial \phi}{\partial (r\cos\phi)} = \frac{1}{r\sin\phi}\frac{\partial \phi}{\partial (\cot\phi)} = \frac{-\sin^2\phi}{r\sin\phi} = -\frac{\sin\phi}{r}$$

using the reciprocal derivative trick, which does indeed give us the correct answer.