## 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 perpendicular to the line segment from the origin to the point $(x,y,0)$ 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 […]