Objective of this Post
The objective of this post is to form (4) in this reference.
Partial Derivatives Relating Cartesian Coordinates to Spherical Coordinates
Suppose $F(x,y,z) \equiv f(x) g(y) h(z)$. In the previous post, the following three equations were written.
$ \frac{\partial F(x,y,z)}{\partial x} = \cos\phi \sin\theta \frac{\partial F}{\partial r} + \frac{\cos \phi \cos\theta}{r} \frac{\partial F}{\partial \theta} – \frac{\sin\phi}{r \sin\theta}\frac{\partial F}{\partial \phi} $.
$ \frac{\partial F(x,y,z)}{\partial y} = \sin\phi \sin\theta \frac{\partial F}{\partial r} + \frac{\sin\phi \cos\theta}{r} \frac{\partial F}{\partial \theta} + \frac{\cos \phi}{r \sin\theta} \frac{\partial F}{\partial \phi} $.
$ \frac{\partial F(x,y,z)}{\partial z} = \cos\theta \frac{\partial F}{\partial r} – \frac{\sin\theta}{r} \frac{\partial F}{\partial \theta} $.
These three equations correspond to (2) in this reference.
Unit Vectors for a Spherical Coordinate System
Next, recall from this post about unit vectors that three unit vectors for a right-handed spherical coordinate system are
$ \vec{e}_r = \sin \theta \cos\phi \vec{e}_x + \sin \theta \sin\phi \vec{e}_y + \cos \theta \vec{e}_z $
$ \vec{e}_{\theta} = \cos\theta \cos\phi \vec{e}_x + \cos\theta \sin\phi \vec{e}_y -\sin\theta \vec{e}_z $
$\vec{e}_{\phi} = -\sin\phi \vec{e}_x + \cos\phi \vec{e}_{y}$.
These three equations correspond to (3) in this reference. In this coordinate system, a point is represented by $(r, \theta, \phi)$.
Equations with Unit Vectors
Reference [2] suggests that the following three equations are valid.
$\vec{e}_x \stackrel{\text{?}}{=} \sin\theta \cos\phi \vec{e}_r + \cos\theta \cos\phi \vec{e}_\theta – \sin\phi \vec{e}_\phi $
$\vec{e}_y \stackrel{\text{?}}{=} \sin\theta\sin\phi \vec{e}_r + \cos\theta\sin\phi \vec{e}_\theta + \cos\phi \vec{e}_\phi $
$\vec{e}_z \stackrel{\text{?}}{=} \cos\theta \vec{e}_r – \sin\theta \vec{e}_\theta $
These three equations are now checked, using known information
Checking the Equation for $\vec{e}_x$
Checking the Equation for $\vec{e}_y$
Checking the Equation for $\vec{e}_z$
Conclusion
Therefore,
$\vec{e}_x = \sin\theta \cos\phi \vec{e}_r + \cos\theta \cos\phi \vec{e}_\theta – \sin\phi \vec{e}_\phi $
$\vec{e}_y = \sin\theta\sin\phi \vec{e}_r + \cos\theta\sin\phi \vec{e}_\theta + \cos\phi \vec{e}_\phi $
$\vec{e}_z = \cos\theta \vec{e}_r – \sin\theta \vec{e}_\theta $
These three equations correspond to (4) in Reference 2.