# Del operator in Cylindrical coordinates (problem in partial differentiation)

I am currently reviewing basic vector analysis and trying to understand every single detail, however, I got stuck in some derivation.

What I want to show is the following:

Given the del operator (i.e., vector differential operator) in Cartesian coordinates $(x,y,z)$

$$\nabla=\frac{\partial }{\partial x}\mathbf{a}_x+\frac{\partial }{\partial y}\mathbf{a}_y+\frac{\partial }{\partial z}\mathbf{a}_z$$

show that the corrseponding operator in Cylindrical coordinates $(\rho, \phi ,z)$ is given by$$\nabla=\frac{\partial }{\partial\rho}\mathbf{a}_\rho+\frac{1}{\rho}\frac{\partial }{\partial \phi}\mathbf{a}_\phi+\frac{\partial }{\partial z}\mathbf{a}_z$$

I tried one approach. However, for curiosity I tried a different method but I couldn't get it right.

# Approach #1:

From the point-to-point transformation $$\rho=\sqrt{x^2+y^2}, \; \phi=\text{tan}\frac{y}{x}$$

partial differentiation with respect to $x$ and $y$ yields \begin{align} \frac{\partial \rho}{\partial x} &=\frac{x}{\sqrt{x^2+y^2}}=\frac{\rho \, \text{cos}\phi}{\rho}=\text{cos}\phi \\ \frac{\partial \rho}{\partial y}&=\frac{y}{\sqrt{x^2+y^2}}=\frac{\rho \, \text{sin}\phi}{\rho}=\text{sin}\phi \end{align} and \begin{align} \frac{\partial \phi}{\partial x}&=\frac{-y}{x^2}\frac{1}{1+(\frac{y}{x})^2}=\frac{-y}{x^2+y^2}=\frac{-\rho \, \text{sin}\phi}{\rho^2}=\frac{-\text{sin}\phi}{\rho} \\ \frac{\partial \phi}{\partial y}&=\frac{1}{x}\frac{1}{1+(\frac{y}{x})^2}=\frac{x}{x^2+y^2}=\frac{\rho \, \text{cos}\phi}{\rho^2}=\frac{\text{cos}\phi}{\rho} \end{align}

Now, plugging these in the chain rule differentiation formulas \begin{align} \frac{\partial }{\partial x}&=\frac{\partial }{\partial \rho}\;\frac{\partial \rho}{\partial x}+\frac{\partial }{\partial \phi}\;\frac{\partial \phi}{\partial x} \\ \frac{\partial }{\partial y}&=\frac{\partial }{\partial \rho}\;\frac{\partial \rho}{\partial y}+\frac{\partial }{\partial \phi}\;\frac{\partial \phi}{\partial y} \end{align}

and making use of the unit vector transformation from Cartesian to Cylindrical \begin{align} \mathbf{a}_x&=\text{cos}\phi\;\mathbf{a}_\rho-\text{sin}\phi\;\mathbf{a}_\phi\\ \mathbf{a}_y&=\text{sin}\phi\;\mathbf{a}_\rho+\text{cos}\phi\;\mathbf{a}_\phi \end{align}

We get \begin{align} \nabla&=\frac{\partial }{\partial x}\mathbf{a}_x+\frac{\partial }{\partial y}\mathbf{a}_y+\frac{\partial }{\partial z}\mathbf{a}_z \\ &=\left (\frac{\partial }{\partial \rho}\;\frac{\partial \rho}{\partial x}+\frac{\partial }{\partial \phi}\;\frac{\partial \phi}{\partial x} \right )\left ( \text{cos}\phi\;\mathbf{a}_\rho-\text{sin}\phi\;\mathbf{a}_\phi \right )\\ &+\left ( \frac{\partial }{\partial \rho}\;\frac{\partial \rho}{\partial y}+\frac{\partial }{\partial \phi}\;\frac{\partial \phi}{\partial y} \right )\left ( \text{sin}\phi\;\mathbf{a}_\rho+\text{cos}\phi\;\mathbf{a}_\phi \right )+\frac{\partial }{\partial z}\mathbf{a}_z \\ &=\left (\frac{\partial }{\partial \rho}\;\text{cos}\phi+\frac{\partial }{\partial \phi}\;\frac{-\text{sin}\phi}{\rho} \right )\left ( \text{cos}\phi\;\mathbf{a}_\rho-\text{sin}\phi\;\mathbf{a}_\phi \right )\\ &+\left ( \frac{\partial }{\partial \rho}\;\text{sin}\phi+\frac{\partial }{\partial \phi}\;\frac{\text{cos}\phi}{\rho} \right )\left ( \text{sin}\phi\;\mathbf{a}_\rho+\text{cos}\phi\;\mathbf{a}_\phi \right )+\frac{\partial }{\partial z}\mathbf{a}_z\\ &=\left ( \text{sin}^2\phi+\text{cos}^2\phi \right )\frac{\partial }{\partial \rho}\mathbf{a}_\rho+\frac{1}{\rho}\left ( \text{sin}^2\phi+\text{cos}^2\phi \right )\frac{\partial }{\partial \phi}\mathbf{a}_\phi+\frac{\partial }{\partial z}\mathbf{a}_z\\ &=\frac{\partial }{\partial\rho}\mathbf{a}_\rho+\frac{1}{\rho}\frac{\partial }{\partial \phi}\mathbf{a}_\phi+\frac{\partial }{\partial z}\mathbf{a}_z \end{align}

which is the desired result.

# Approach #2:

How can I get the same result starting from the point-to-point transformation $$x=\rho \, \text{cos}\phi,\; y=\rho \, \text{sin}\phi$$ by using partial differentiation? Maybe implicit differentiation?

• Hint: $\mathbf a_\rho=cos\phi\;\mathbf a_x+sin\phi\;\mathbf a_y$ and $\mathbf a_\phi=cos\phi\;\mathbf a_y-sin\phi\;\mathbf a_x$. – amd Sep 21 '15 at 17:47
• As Dr. MV's answer shows, you do not really need the full equations of coordinate change to perform differential computations. You only need to know their derivatives. – Giuseppe Negro Sep 21 '15 at 21:13

The coordinate transformation from polar to rectangular coordinates is given by

\begin{align} x&=\rho \cos \phi \tag 1\\\\ y&=\rho \sin \phi \tag 2 \end{align}

Now, suppose that the coordinate transformation from Cartesian to polar coordinates as given by

\begin{align} \rho&=\rho (x,y)=\sqrt{x^2+y^2} \\\\ \phi&=\phi(x,y) = \begin{cases} \arctan(y/x)&,x>0\\\\ \arctan(y/x)+\pi&,x<0,y\ge 0\\\\ \arctan(y/x)-\pi&,x<0,y<0\\\\ \pi/2&,x=0,y>0\\\\ \pi/2&,x=0,y<0\\\\ \end{cases} \end{align}

were unavailable in closed form. We can still proceed to develop a transformation of the gradient operator from Cartesian coordinates to polar.

To do so, we use the differential relationships

\begin{align} dx&=\frac{\partial x}{\partial \rho}d\rho+\frac{\partial x}{\partial \phi}d\phi\\\\ dy&=\frac{\partial y}{\partial \rho}d\rho+\frac{\partial y}{\partial \phi}d\phi\\\\ d\rho&=\frac{\partial \rho}{\partial x}dx+\frac{\partial \rho}{\partial y}dy\\\\ d\phi&=\frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy \end{align}

Defining the Wronskian $W$ as

\begin{align} W&=\frac{\partial \rho \cos \phi}{\partial \rho}\frac{\partial \rho \sin \phi}{\partial \phi}-\frac{\partial \rho \cos \phi}{\partial \phi}\frac{\partial \rho \sin \phi}{\partial \rho}\\\\ &=\rho \end{align}

we find that

\begin{align} \frac{\partial \rho }{\partial x}&=\frac{1}{W}\frac{\partial \rho \sin \phi}{\partial \phi}\\\\&= \cos \phi\\\\ \frac{\partial \rho }{\partial y}&=-\frac{1}{W}\frac{\partial \rho \cos \phi}{\partial \phi}\\\\&= \sin \phi\\\\ \frac{\partial \phi }{\partial x}&=-\frac{1}{W}\frac{\partial \rho \sin \phi}{\partial \rho}\\\\&=- \frac{\sin \phi}{\rho}\\\\ \frac{\partial \phi }{\partial y}&=\frac{1}{W}\frac{\partial \rho \sin \phi}{\partial \phi}\\\\&=\frac{ \cos \phi}{\rho} \end{align}

Therefore, we have

\begin{align} \hat x \frac{\partial }{\partial x}+\hat y \frac{\partial }{\partial y}&=(\hat \rho \cos \phi -\hat \phi \sin \phi)\left(\frac{\partial \rho}{\partial x}\frac{\partial }{\partial \rho}+\frac{\partial \phi}{\partial x}\frac{\partial }{\partial \phi}\right)+(\hat \rho \sin \phi +\hat \phi \cos \phi)\left(\frac{\partial \rho}{\partial y}\frac{\partial }{\partial \rho}+\frac{\partial \phi}{\partial y}\frac{\partial }{\partial \phi}\right)\\\\ &=\hat \rho \frac{\partial }{\partial \rho}+\hat \phi \frac{1}{\rho}\frac{\partial }{\partial \phi} \end{align}

as was to be shown!

• A way to look at this calculation is that you’re computing the inverse of $\mathrm d\alpha$, where $\alpha:(\rho,\phi)\mapsto(x,y)$ is the known polar to Cartesian coordinate map, from which you can read off the partial derivatives of $\rho$ and $\phi$ with respect to $x$ and $y$. – amd Sep 21 '15 at 22:00
• Related. In this page, the partial derivatives in the spherical coordinate system are computed without computing the inverse to the transformation, by using exactly the same technique as this answer. – Giuseppe Negro Sep 28 '15 at 21:58
• @GiuseppeNegro Thank you for this comment. +1 – Mark Viola Sep 28 '15 at 22:29

Here’s a more-or-less direct way to do this:

The $z$-dimension term will clearly be unchanged, so we need only focus on $x$ and $y$. We have $$\frac{\partial x}{\partial\rho}=\cos\phi, \frac{\partial y}{\partial\rho}=\sin\phi$$ and $$\frac{\partial x}{\partial\phi}=-\rho\sin\phi,\frac{\partial y}{\partial\phi}=\rho\cos\phi$$ Also,\begin{align}\mathbf a_\rho&=\cos\phi\;\mathbf a_x+\sin\phi\;\mathbf a_y \\ \mathbf a_\phi&=\cos\phi\;\mathbf a_y-\sin\phi\;\mathbf a_x \end{align} Since we know the partial derivatives of $x$ and $y$ w.r. to $\rho$ and $\phi$, a natural way to start is to expand $\partial/\partial\rho$ and $\partial/\partial\phi$ using the chain rule. Doing so, and also transforming the unit vectors, we get \begin{align} \frac\partial{\partial\rho}\mathbf a_\rho&=\left(\frac{\partial x}{\partial\rho}\frac\partial{\partial x}+\frac{\partial y}{\partial\rho}\frac\partial{\partial y}\right)\left(\cos\phi\;\mathbf a_x+\sin\phi\;\mathbf a_y\right) \\ &=\left(\cos\phi\frac\partial{\partial x}+\sin\phi\frac\partial{\partial y}\right)\left(\cos\phi\;\mathbf a_x+\sin\phi\;\mathbf a_y\right) \\ &=\cos^2\phi\frac\partial{\partial x}\mathbf a_x+\cos\phi\sin\phi\frac\partial{\partial x}\mathbf a_y+\cos\phi\sin\phi\frac\partial{\partial y}\mathbf a_x+\sin^2\phi\frac\partial{\partial y}\mathbf a_y \end{align} and (rearranged so that the terms line up) \begin{align} \frac\partial{\partial\phi}\mathbf a_\phi&=\left(\frac{\partial x}{\partial\phi}\frac\partial{\partial x}+\frac{\partial y}{\partial\phi}\frac\partial{\partial y}\right)\left(\cos\phi\;\mathbf a_y-\sin\phi\;\mathbf a_x\right) \\ &=\left(-\rho\sin\phi\frac\partial{\partial x}+\rho\cos\phi\frac\partial{\partial y}\right)\left(\cos\phi\;\mathbf a_y-\sin\phi\;\mathbf a_x\right) \\ &=\rho\sin^2\phi\frac\partial{\partial x}\mathbf a_x-\rho\cos\phi\sin\phi\frac\partial{\partial x}\mathbf a_y-\rho\cos\phi\sin\phi\frac\partial{\partial y}\mathbf a_x+\rho\cos^2\phi\frac\partial{\partial y}\mathbf a_y \end{align} That looks promising. Divide the second equation by $\rho$ and add it to the first: $$\frac\partial{\partial\rho}\mathbf a_\rho+\frac 1\rho\frac\partial{\partial\phi}\mathbf a_\phi=\left(\cos^2\phi+\sin^2\phi\right)\frac\partial{\partial x}\mathbf a_x+\left(\cos^2\phi+\sin^2\phi\right)\frac\partial{\partial y}\mathbf a_y=\frac\partial{\partial x}\mathbf a_x+\frac\partial{\partial y}\mathbf a_y$$

Note that in this case, the inverse coordinate map was easy to calculate. Dr. MV’s solution shows that you don’t really need the inverse coordinate map to compute its partial derivatives.