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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.

Ninad Munshi
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ueir
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  • Your second set of equations doesn't quite make sense. Why do you have $\partial_r$ in terms of $\partial_x$ and $\partial_\phi$ ? – Ninad Munshi Oct 17 '19 at 09:36
  • I'm very sorry for making confusion. It's typo. I fixed them. – ueir Oct 17 '19 at 09:41
  • Unfortunately that one line wasn't what I was hinting at you to fix. You shouldn't have any "half and half" coordinate systems (i.e. writing a variable in one coordinate system in terms of variables from both its own coordinate system and variables from another coordinate system). Your entire line of reasoning for the first half is not correct because of this error. And $\frac{\partial y}{\partial r} \neq -r\sin\phi$ – Ninad Munshi Oct 17 '19 at 09:44
  • I fix one more time. I believe it's corrected. – ueir Oct 17 '19 at 09:57
  • I apologize for potentially coming off as curt. My original plan was to plug back in what you found into the original equations and see what happened from there, and wasn't able to do so. – Ninad Munshi Oct 17 '19 at 10:02
  • I appreciate your any reply. – ueir Oct 17 '19 at 10:09
  • I have found the logical error. I appreciate any feedback you can give me. – Ninad Munshi Oct 17 '19 at 10:14

1 Answers1

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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.

Ninad Munshi
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