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 …
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 …
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 …
Reasoning about Left and Right Handed Coordinate Systems Read More »
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 …
Del is More than an Upside Down Triangle, Part 1 Read More »
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 …
How to Integrate in a Spherical Coordinate System Read More »
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 …
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 …
