Coordinate Systems

04 Nov 2024

We need to take a quick detour. In order to calculate the moment of inertia of an object, we need to be able to algebraically describe its geometry. Choosing the right coordinate system can vastly simplify the analysis. In this post, I will briefly go over the standard coordinate systems used in physics - Cartesian, cylindrical, and spherical.

Cartesian Coordinates

Conceptually, this is the most intuitive and obvious coordinate system. Given any point in $3\text{D}$ space, pick any three orthogonal directions and measure the point’s distance from a fixed origin in those directions. This coordinate system most naturally describes cuboid geometries and also typically used in irregular geometries.

$$\b{r} = x \; \u{x} + y \; \u{y} + z \; \u{z}$$

The benefit of this coordinate system is that each unit vector is constant. No matter what is going on inside the coordinate system, the $\u{x}$ unit vector always points along the $x$-axis (a property novice calculus students take for granted). Therefore, doing calculus with vectors in this coordinate system is very easy. The cost of this is that parameterizing complicated geometries can become unwieldy. This is why we have alternative coordinate systems.

Cylindrical Coordinates

Now we get a little more complicated (but not too much). We are using the polar coordiantes $(s, \phi)$ to describe the point’s projection onto the $xy$-plane, then we simply go straight in the $z$ direction. As the name suggests, this coordinate system is best suited for cylindrically-shaped objects.

$$\b{r} = s \; \u{s} + z \; \u{z}$$

We have to be careful with the $\u{s}$ as it is not constant. Its direction changes depending on the coordinate we are describing, which becomes relevant when taking derivatives. For example, $\frac{\partial}{\partial \phi} \u{s} = \u{\phi}$ (we prove this fact when finding the moment of inertia of a circle). The benefit of this small annoyance is (usually) a reduction in complexity in parameterization.

Below is the conversion between Cartesian and cylindrical coordinates. We will particularly use the conversion between the unit vectors. Deriving these formulas is a good exercise to make sure you understand the coordinate system.

$$\begin{align} &\begin{cases} x = s \ \cos \phi \\ y = s \ \sin \phi \\ z = z \end{cases} \qquad&& \begin{cases} \u{x} = \cos \phi \ \u{s} - \sin \phi \ \u{\phi} \\ \u{y} = \sin \phi \ \u{s} + \cos \phi \ \u{\phi} \\ \u{z} = \u{z} \end{cases} \\[15pt] &\begin{cases} s = \sqrt{x^2 + y^2} \\ \phi = \arctan(y / x) \\ z = z \end{cases} \qquad&& \begin{cases} \u{s} = \cos \phi \ \u{x} + \sin \phi \ \u{y} \\ \u{\phi} = - \sin \phi \ \u{x} + \cos \phi \ \u{y} \\ \u{z} = \u{z} \end{cases} \end{align}$$

Spherical Coordinates

Finally, we have the coordinate system which is conceptually the most difficult since none of the unit vectors are constant. Notice the direction we measure $\theta$ (it’s probably the opposite of what you would expect). Also notice that even though the unit vectors change depending on the point of interest, they always remain perpendicular to each other.

$$\b{r} = r \; \u{r}$$

This coordinate system finishes what cylindrical coordinates started. In this coordinate system, none of the unit vectors are constant. However, due to the nature $\u{\theta}$ and $\u{\phi}$ there is usually a symmetry argument to simplify them. Again, it is trading one type of complexity for another. In certain geometries, it greatly simplifies the analysis.

Finally, we give the conversion between spherical coordinates and the two previous coordinate systems. If you’ve never worked with spherical coordinates, I recommend you try to derive these formulas. It’s much more simple than it looks.

$$\begin{align} &\begin{cases} x = r \ \sin \theta \ \cos \phi \\ y = r \ \sin \theta \ \sin \phi \\ z = r \ \cos \theta \end{cases} \qquad&& \begin{cases} \u{x} = \sin \theta \ \cos \phi \ \u{r} + \cos \theta \cos \phi \ \u{\theta} - \sin \phi \ \u{\phi} \\ \u{y} = \sin \theta \ \sin \phi \ \u{r} + \cos \theta \sin \phi \ \u{\theta} + \cos \phi \ \u{\phi} \\ \u{z} = \cos \theta \ \u{r} - \sin \theta \ \u{\theta} \end{cases} \\[15pt] &\begin{cases} r = \sqrt{x^2 + y^2 + z^2} \\ \theta = \arctan(\sqrt{x^2 + y^2} / z) \\ \phi = \arctan(y / x) \end{cases} \qquad&& \begin{cases} \u{r} = \sin \theta \ \cos \phi \ \u{x} + \sin \theta \sin \phi \ \u{y} + \cos \theta \ \u{z} \\ \u{\theta} = \cos \theta \ \cos \phi \ \u{x} + \cos \theta \sin \phi \ \u{y} - \sin \theta \ \u{z} \\ \u{\phi} = - \sin \phi \ \u{x} + \cos \phi \ \u{y} \end{cases} \end{align}$$

$$\begin{align} &\begin{cases} s = r \ \sin \theta \\ \phi = \phi \\ z = r \cos \theta \end{cases} \qquad&& \begin{cases} \u{s} = \sin \theta \ \u{r} + \cos \theta \ \u{\theta} \\ \u{\phi} = \u{\phi} \\ \u{z} = \cos \theta \ \u{r} - \sin \theta \ \u{\theta} \end{cases} \\[15pt] &\begin{cases} r = \sqrt{s^2 + z^2} \\ \theta = \arctan(s / z) \\ \phi = \phi \end{cases} \qquad&& \begin{cases} \u{r} = \sin \theta \ \u{s} + \cos \theta \ \u{z} \\ \u{\theta} = \cos \theta \ \u{s} - \sin \theta \ \u{z} \\ \u{\phi} = \u{\phi} \end{cases} \end{align}$$