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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Conversions between spherical and cylindrical coordinates

Our textbook discusses conversions between cartesian and other systems, but we don't have a lot of inforation to convert between spherical and cylindrical, for example.

Coverting coordiantes can be simple, but converting a field will take a few extra steps. Not only do we have to convert the field's functions, but also the unit vectors we use to express the field

In this brief entry, we will derive expressions to convert a spherical field (expressed using $r,\theta ,\varphi$ with unit vectors $\hat{r},\hat{\theta },\hat{\varphi }$) into a cylindrical field (epxressed using $s,\varphi ,z$ and the unite vectors $\hat{s},\hat{\varphi },\hat{z}$) Let's start by recalling some unit vector conversions (as seen on our textbook): $$\begin{array}{rl}\hat{r}& =\sin \left(\theta \right)\cos \left(\varphi \right)\hat{x}+\sin \left(\theta \right)\sin \left(\varphi \right)\hat{y}+\cos \left(\theta \right)\hat{z}\\ \hat{\theta }& =\cos \left(\theta \right)\cos \left(\varphi \right)\hat{x}+\cos \left(\theta \right)\sin \left(\varphi \right)\hat{y}-\sin \left(\theta \right)\hat{z}\\ \hat{\varphi }& =-\sin \left(\varphi \right)\hat{x}+\cos \left(\varphi \right)\hat{y}\end{array}$$

And

$$\begin{array}{rl}\hat{x}& =\cos \left(\varphi \right)\hat{s}-\sin \left(\varphi \right)\hat{\varphi }\\ \hat{y}& =\sin \left(\varphi \right)\hat{s}+\cos \left(\varphi \right)\hat{\varphi }\\ \hat{z}& =\hat{z}\end{array}$$

With those two conversions in mind, we can find expressions for the spherical unit vectors in terms of cylindrical unit vectors:

$$\begin{array}{rl}\text{spherical}& \to \text{cylindrical}\\ \hat{r}& =\sin \left(\theta \right)\hat{s}+\cos \left(\theta \right)\hat{z}\\ \hat{\theta }& =\cos \left(\theta \right)\hat{s}-\sin \left(\theta \right)\hat{z}\\ \hat{\varphi }& =\hat{\varphi }\end{array}$$ The conversion between components is easy to derive geometrically: $$\begin{array}{rl}r& =\sqrt{s^2+z^2}\\ \theta & =\arccos \left(\frac{z}{\sqrt{s^2+z^2}}\right)\\ \varphi & =\varphi \end{array}$$

We can use these to write the spherical unit vectors using only cylindrical variables. It is relevant to recall that

$$\sin (\arccos (\omega ))=\sqrt{1-{\omega }^2}$$

So

$$\begin{array}{rl}\hat{r}& =\sqrt{1-\frac{z^2}{s^2+z^2}}\hat{s}+\frac{z}{\sqrt{s^2+z^2}}\hat{z}\\ \hat{\theta }& =\frac{z}{\sqrt{s^2+z^2}}\hat{s}-\sqrt{1-\frac{z^2}{s^2+z^2}}\hat{z}\\ \hat{\varphi }& =\hat{\varphi }\end{array}$$

With the last two sets of equations we will be able to convert any spherical vector field

$$\mathbf{v}=f_r(r,\theta ,\varphi )\hat{r}+f_{\theta }(r,\theta ,\varphi )\hat{\theta }+f_{\varphi }(r,\theta ,\varphi )\hat{\varphi }$$

Into a cylindrical vector field

$$\mathbf{v}=f_s(s,\varphi ,z)\hat{s}+f_{\varphi }(s,\varphi ,z)\hat{\varphi }+f_z(s,\varphi ,z)\hat{z}$$

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