In this post, I relate coefficients of unit vectors to derivatives and to a Jacobian matrix that was used in a previous post. Unit Vectors 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 $ $\vec{e}_{\phi} = -\sin\phi \vec{e}_x + \cos\phi \vec{e}_{y}$. In this coordinate system, a point is represented by $(r, \theta, \phi)$. Derivatives Recall the following nine derivatives from previous posts. $\frac{dx(r)}{dr}\big|_{r^+}=\sin\theta\cos\phi$ $\frac{dy(r)}{dr}\big|_{r^+} = \sin\theta\sin\phi$ $\frac{dz(r)}{dr}\big|_{r^+} = \cos\theta$ $\frac{dx(\theta)}{d\theta} \big|_{\theta^+} = r\cos\theta \cos\phi$ $\frac{d y(\theta)}{d \theta} \big|_{\theta^+} = r \cos\theta \sin\phi $ $\frac{dz(\theta)}{d\theta} \big|_{\theta^+} = -r \sin\theta$ $\frac{dx(\phi)}{d\phi}\big|_{\phi^+} = -r\sin\theta\sin\phi$ $\frac{dy(\phi)}{d\phi}\big|_{\phi^+} = r \sin\theta \cos\phi$ $\frac{dz(\phi)}{d\phi}\big|_{\phi^+} = 0$ Notice that these derivatives are very similar to the coefficients of the unit vectors. In particular, $ \vec{e}_r = \frac{dx(r)}{dr}\big|_{r^+} \vec{e}_x + \frac{dy(r)}{dr}\big|_{r^+} \vec{e}_y + \frac{dz(r)}{dr}\big|_{r^+} \vec{e}_z $ $ \vec{e}_{\theta} = \frac{1}{r} \frac{dx(\theta)}{d\theta} \big|_{\theta^+} \vec{e}_x + \frac{1}{r} \frac{d y(\theta)}{d \theta} \big|_{\theta^+}  \vec{e}_y  + \frac{1}{r} \frac{dz(\theta)}{d\theta} \big|_{\theta^+}  \vec{e}_z $ $\vec{e}_{\phi} = \frac{1}{r \sin\theta} \frac{dx(\phi)}{d\phi}\big|_{\phi^+}  \vec{e}_x + \frac{1}{r \sin\theta} \frac{dy(\phi)}{d\phi}\big|_{\phi^+}  \vec{e}_{y} + \frac{dz(\phi)}{d\phi}\big|_{\phi^+}  \vec{e}_z$. The extra factors such as $\frac{1}{r}$ and $\frac{1}{r \sin\theta} $ account for change with respect to an arc length $s = r \alpha$ in which $\alpha$ is an arbitrary angle. The above equations can be represented with a matrix equation: Next, recall this Jacobian matrix: This Jacobian matrix is very similar to the first 3×3 matrix in the current post. The two matrices in this post are related by switching the row and columns. Switching the rows and columns of a matrix is also called taking the transpose of a matrix. References [1]