## The Azimuthal Unit Vector

In this post, I write the azimuthal unit vector $\vec{e}_{\phi}$ in terms of Cartesian coordinates. Here, $\phi$ is the azimuthal angle in the $x-y$ plane. As noted in this post, $\vec{e}_{\phi}$ points in the direction of increasing $\phi$. Geometrical Setup Since $\vec{e}_{\phi}$ is perpendicular to the line segment from the origin to the point $(x,y,0)$, I am …

## A Right Handed Spherical Coordinate System

This post introduced the following questions. What direction does $\vec{e}_{\phi}$ point? Modern convention dictates that $\vec{e}_{\phi}$ should point in the direction of increasing $\phi$, but what is the reason for this convention? If there are only two unique configurations for three mutually perpendicular unit vectors in light of the ambiguous orientation of each unit vector, it seems likely …

## Reasoning about Left and Right Handed Coordinate Systems

A coordinate system can be defined by three perpendicular unit vectors. If the coordinate system is Cartesian, which direction does the $+x$ axis point? To resolve this problem, I define an orientation–a coordinate system that is oriented in a certain direction in three-dimensional space. How many unique orientations are there? Here, a unique orientation is an …

## Del is More than an Upside Down Triangle, Part 1

What is Del? In math, the symbol $\vec{\nabla}$ is called “del.” This symbol is defined in terms of Cartesian coordinates. $\vec{\nabla} \equiv \frac{d}{dx}\vec{e}_x + \frac{d}{dy}\vec{e}_y + \frac{d}{dz}\vec{e}_z$ The right side is a sum of unit vectors. So $\vec{\nabla}$ is a vector. This is why I write $\vec{\nabla}$ instead of just $\nabla$. Is it possible to …

## How to Integrate in a Spherical Coordinate System

Review of Integration Integration with Cartesian coordinates is simple. The general form is $\int\int\int f(x,y,z)dxdydz$ in which $f(x,y,z)$ is an arbitrary function of the Cartesian coordinates. However, there may be cases in which integrating with spherical coordinates is more convenient. Given the above, general form for integration with Cartesian coordinates, how can one integrate in a spherical …

## The Jacobian Matrix

Using conclusions from previous posts, the following nine derivatives have been determined. $\frac{dx(r)}{dr}\big|_{r^+}=\sin\theta\cos\phi$ $\frac{dy(r)}{dr}\big|_{r^+} = \sin\theta\sin\phi$ $\frac{dz(r)}{dr}\big|_{r^+} = \cos\theta$ $\frac{dx(\phi)}{d\phi}\big|_{\phi^+} = -r\sin\theta\sin\phi$ $\frac{dy(\phi)}{d\phi}\big|_{\phi^+} = r \sin\theta \cos\phi$ $\frac{dz(\phi)}{d\phi}\big|_{\phi^+} = 0$ $\frac{dx(\theta)}{d\theta} \big|_{\theta^+} = r\cos\theta \cos\phi$ $\frac{d y(\theta)}{d \theta} \big|_{\theta^+} = r \cos\theta \sin\phi$ $\frac{dz(\theta)}{d\theta} \big|_{\theta^+} = -r \sin\theta$ Next, recall the following result from this …

## How to Use the Product Rule

I would like to evaluate two more derivatives. They are $\frac{dx}{d\phi}\big|_{\phi^+}$ and $\frac{dz}{d\theta}\big|_{\theta^+}$ given $x = r \sin\theta\cos\phi$ and $z = r\cos\theta$.   Start with $\frac{dx}{d\phi}\big|_{\phi^+}$. The first step is to substitute $x$ with $r \sin\theta \cos\phi$. $\frac{dx}{d\phi}\big|_{\phi^+} = \frac{d r \sin\theta \cos \phi}{d\phi}\big|_{\phi^+}$. The next step is …