Let $(u_1,u_2,u_3)$ represent the three coordinates in a general curvilinear system in $\mathbb{R}^3$, and let $\mathbf{e}_i$ be the unit vector that points the direction of increasing $u_i$. Let $\mathbf{e}_i \cdot \mathbf{e}_j=\delta_{ij}$, and $\mathbf{e}_i\times\mathbf{e}_j=\varepsilon_{ijk}\mathbf{e}_k$ for $\{i,j,k\}=\{1,2,3\}$. For any $\mathbf{r}$ in $\mathbb{R}^3$, let $\frac{\partial \mathbf{r}}{\partial u_i}=h_i\mathbf{e}_i$, where $h_i$ is the scale vector. We then have $dr^2=(\sum_{i=1}^3 h_idu_i)^2$. This is called orthogonal curvilinear coordinate system. Moreover, by making use of $\nabla u_i=h_i^{-1}\mathbf{e}_i$, we can write the operator $\nabla$ as $\nabla=\sum_{i=1}^3 \mathbf{e}_i h_i^{-1}\frac{\partial}{\partial u_i}$. Then, the gradient formula for the orthogonal curvilinear system is obvious.

We have some tricks to obtain the divergence and curl formulas in the orthognal curvilinear system. For example, we may use the relation $\nabla\cdot\left(\frac{\mathbf{e}_1}{h_2h_3}\right)=\nabla\cdot(\nabla \mathbf{e}_2\times\nabla\mathbf{e}_3)=0$ to obtain the divergence formula. However, we may consider a direct approach: Compute the derivative of a unit vector against a coordinate, i.e., compute the value of $\frac{\partial \mathbf{e}_j}{\partial u_i}$. If $i=j$, by noticing $\frac{\partial \mathbf{e}_i}{\partial u_i}=\varepsilon_{ijk}\frac{\partial(\mathbf{e}_j\times\mathbf{e}_k)}{\partial u_i}$, it suffices to compute $\frac{\partial \mathbf{e}_j}{\partial u_i}$ for $j\neq i$. Actually, there is a simple formula about it: For $j\neq i$, $\frac{\partial \mathbf{e}_j}{\partial u_i}=h_j^{-1}\frac{\partial h_i}{\partial u_j}\mathbf{e}_i$. My question is how to derive this formula.


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