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 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:
[1] https://en.wikipedia.org/wiki/Unit_vector
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