# derive-it.com

## 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 looking for a unit vector that is perpendicular to this line segment. A unit vector has a magnitude and a direction, and these properties of the unit vector do not change if I translate the unit vector to the origin. So, I am looking for a unit vector that is

1. perpendicular to the line segment from the origin to the point $(x,y,0)$
2. points in the direction of increasing $\phi$

This is a geometrical problem, so I include a diagram. The arrow for the angle $\phi$ indicates the direction of increasing $\phi$. This diagram shows the unit vector $\vec{e}_{\phi}$ for the point $(x,y,0)=(r,\phi,90)$. The radius is $r\sin\theta$ because this diagram is for spherical coordinates instead of polar coordinates.

## Determining $\beta$

The next step is to express $\beta$ in terms of $\phi$. First,

$\alpha = 90 – \phi$.

Using the grey line,

$90 + \beta + \alpha = 180$.

Substituting $90 – \phi$ for $\alpha$,

$90 + \beta + (90 – \phi) = 180$.

Simplifying,

$\beta + 90 – \phi = 90$, or

$\beta = \phi$.

## An Expression for $\vec{e}_{\phi}$

Now that $\beta$ has been determined, it is possible to find the side lengths of the triangle formed by $\beta$ and the $+y$-axis. Since a unit vector has a magnitude of one, the hypotenuse of this triangle is 1. Using the side lengths written in the diagram, and noting that the $x$-coordinate for the unit vector is negative in the diagram,

$\boxed{\vec{e}_{\phi} = -\sin\phi \vec{e}_x + \cos\phi \vec{e}_{y}}$.

At this point, it is necessary to extend the domain of the sine and cosine functions such that the expression for $\vec{e}_{\phi}$ is valid for any azimuthal unit vector in the $x-y$ plane. Assuming that the previously derived trigonometric identities for sine and cosine are still valid for the updated sine and cosine functions, derivatives involving variation with respect to $\phi$ are:

$\frac{dx(\phi)}{d\phi}\big|_{\phi^+} = -r\sin\theta\sin\phi$

$\frac{dy(\phi)}{d\phi}\big|_{\phi^+} = r \sin\theta \cos\phi$

$\frac{dz(\phi)}{d\phi}\big|_{\phi^+} = 0$

These derivatives were not needed to find an expression for $\vec{e}_{\phi}$, but it is possible to see the relation between these derivatives and $\vec{e}_{\phi}$. Just consider $r\sin\theta$ as part of the arc length corresponding to $\phi$. Then the coefficients of the Cartesian unit vectors are in the expressions for the derivatives.

## Converting to Cartesian Coordinates

The next step is to determine $\sin\phi$ and $\cos\phi$ in terms of Cartesian coordinates. This can be done with the triangle with hypotenuse $r\sin\theta$, base $x$, and height $y$.

Using the trigonometric definitions of sine and cosine,

$\sin\phi = \frac{y}{r\sin\theta}$

$\cos\phi = \frac{x}{r\sin\theta}$.

This is not the final result because $r\sin\theta$ is still in spherical coordinates.

From the Pythagorean theorem and using spherical coordinates,

$r\sin\theta = \sqrt{x^2 + y^2}$.

Therefore,

$\boxed{\vec{e}_{\phi} = -\frac{y}{\sqrt{x^2 + y^2}} \vec{e}_x + \frac{x}{\sqrt{x^2 + y^2}} \vec{e}_{y}}$.

References:

 https://en.wikipedia.org/wiki/Unit_vector