Consider a spherical coordinate system. Let a point be represented by $(r, \theta, \phi)$, in that order. Now that the order of the coordinates is established, I can define unit vectors that form a right-handed coordinate system. Suppose the radial unit vector $\vec{e}_r$ points radially outward from the origin to the point, and the polar unit vector $\vec{e}_{\theta}$ points in the direction of increasing $\theta$. From the right hand rule, the azimuthal unit vector $\vec{e}_{\phi}$ points in the direction of increasing $\phi$. Below is a diagram with an arbitrary vector $\vec{r}=(r,\theta,\phi)=r \vec{e}_r$. The Polar Unit Vector in Terms of Spherical Coordinates Another axis called the $w$-axis is included, to help define $\vec{e}_{\theta}$ in terms of symbols. Since $\theta$ increases from 0 on the $+z$-axis to $\pi$ radians on the $-z$-axis, as $\theta$ increases $\vec{r}$ traces a curved path in the $z-w$ plane. Also, $\vec{r}$ undergoes a counterclockwise rotation from the $+z$-axis to the $+w$-axis, if the $+z$-axis is on the right. The length of the radius does not affect the direction of the polar unit vector, so I set $r=1$. $\vec{e}_{\theta} = -\sin\theta \vec{e}_z + \cos\theta \vec{e}_w$. This equation can be obtained with trigonometry. This unit vector has a length of 1. The next step is to solve for $\vec{e}_w$. From vector addition, $\vec{e}_w = r’ \cos\phi \vec{e}_x + r’ \sin\phi \vec{e}_y $. To get the direction for $\vec{e}_w$, I disregard $r=1$ and set $r’=1$. Then $\vec{e}_w = \cos\phi \vec{e}_x + \sin\phi \vec{e}_y $. This unit vector also has a length of 1. Substituting the previous expression for $\vec{e}_w$ into the initial expression for $\vec{e}_{\theta}$, $\vec{e}_{\theta} = -\sin\theta \vec{e}_z + \cos\theta (\cos\phi \vec{e}_x + \sin\phi \vec{e}_y)$. Simplifying, $ \boxed { \vec{e}_{\theta} = \cos\theta \cos\phi \vec{e}_x + \cos\theta \sin\phi \vec{e}_y -\sin\theta \vec{e}_z }$. The Polar Unit Vector in Terms of Cartesian […]
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