📂Mathematical Physics

Unit Vectors of the Spherical Coordinate System Expressed in Terms of Unit Vectors of the Cartesian Coordinate System

Spherical Coordinate System’s Unit Vectors

$$\begin{align*} \hat{\mathbf{r}} &= \cos\phi \sin\theta\hat{\mathbf{x}} + \sin\phi \sin\theta\hat{\mathbf{y}} + \cos\theta\hat{\mathbf{z}} \\ \hat{\boldsymbol{\theta}} &= \cos\phi \cos\theta \hat{\mathbf{x}} + \sin\phi \cos\theta \hat{\mathbf{y}} - \sin\theta\hat{\mathbf{z}} \\ \hat{\boldsymbol{\phi}} &= -\sin\phi \hat{\mathbf{x}} + \cos\phi \hat{\mathbf{y}} \end{align*}$$

Derivation

First, calculate $\hat{\mathbf{r}}$ and then use it to derive the other two.

Radial Direction Unit Vector $\hat{\mathbf{r}}$

$$\hat{\mathbf{r}}=r\hat{\mathbf{r}}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}}$$

Therefore, dividing both sides by $r$ gives:

$$\begin{align*} \hat{\mathbf{r}}&=\frac{x}{r}\hat{\mathbf{x}}+\frac{y}{r}\hat{\mathbf{y}}+\frac{z}{r}\hat{\mathbf{z}} \\ &= \frac{x}{r \sin\theta}\sin\theta\hat{\mathbf{x}}+\frac{y}{r \sin\theta}\sin\theta\hat{\mathbf{y}}+\cos\theta\hat{\mathbf{z}} \\ &= \cos\phi \sin\theta \hat{\mathbf{x}} + \sin\phi \sin\theta\hat{\mathbf{y}} + \cos\theta\hat{\mathbf{z}} =\hat{\mathbf{r}}(\theta,\phi) \end{align*}$$

Polar Angle Direction Unit Vector $\hat{\boldsymbol{\theta}}$

$\hat{\boldsymbol{\theta}}$ is obtained from the $\hat{\mathbf{r}}$ direction, where $\phi$ remains the same, and only $\theta$ increases by $\dfrac{\pi}{2}$, as follows:

$$\begin{align*} \hat{\boldsymbol{\theta}} &= \hat{\mathbf{r}} \left(\theta+\dfrac{\pi}{2}, \phi \right) \\ &= \cos\phi \sin\left(\theta+\dfrac{\pi}{2}\right) \hat{\mathbf{x}} + \sin\phi \sin\left(\theta+\dfrac{\pi}{2}\right)\hat{\mathbf{y}} + \cos\left(\theta+\dfrac{\pi}{2}\right)\hat{\mathbf{z}} \\ &= \cos\phi \cos\theta \hat{\mathbf{x}} + \sin\phi \cos\theta\hat{\mathbf{y}} - \sin\theta\hat{\mathbf{z}} \end{align*}$$

Azimuthal Angle Direction Unit Vector $\hat{\boldsymbol{\phi}}$

Given $\hat{\boldsymbol{\phi}}=\hat{\mathbf{r}} \times \hat{\boldsymbol{\theta}}$, it follows that:

$$\begin{align*} \hat{\boldsymbol{\phi}} &= \begin{vmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \cos\phi \sin\theta & \sin\phi \sin\theta\ & \cos\theta \\ \cos\phi \cos\theta & \sin\phi \cos\theta & -\sin\theta \end{vmatrix} \\ &= (-\sin\phi \sin^2\theta-\sin\phi \cos^2\theta)\hat{\mathbf{x}}+ (\cos\phi \cos^2\theta + \cos\phi \sin^2\theta)\hat{\mathbf{y}} \\ &\quad +(\cos\phi \sin\theta \sin\phi \cos\theta -\cos\phi \sin\theta \sin\phi \cos\theta)\hat{\mathbf{z}} \\ &= -\sin\phi (\sin^2\theta + \cos^2\theta) \hat{\mathbf{x}} + \cos\phi (\cos^2 \theta + \sin^2\theta) \hat{\mathbf{y}} \\ &= -\sin\phi \hat{\mathbf{x}} + \cos\phi \hat{\mathbf{y}} \end{align*}$$

Or, one can consider it in this way: When determining the direction of $\hat{\boldsymbol{\phi}}$, $\theta$ does not influence. The direction is solely determined by the values of $r$ and $\phi$, regardless of the value of $\theta$. Moreover, the direction of $\hat{\boldsymbol{\phi}}$ is that in which $\phi$ has increased by $\dfrac{\pi}{2}$ from the $\hat{\mathbf{r}}$ direction. Therefore, the term $\theta$ disappears from $\hat{\mathbf{r}}$, and instead of $\phi$, $\phi + \dfrac{\pi}{2}$ is substituted.

$$\begin{align*} \hat{\boldsymbol{\phi}} &= \cos{(\phi+\dfrac{\pi}{2} )}\hat{\mathbf{x}} + \sin{(\phi + \dfrac{\pi}{2})}\hat{\mathbf{y}} \\ &= -\sin \phi \hat{\mathbf{x}}+ \cos \phi \hat{\mathbf{y}} \end{align*}$$

■