In a general orthogonal coordinate system we denote the coordinates by \( y_i \). The position \( \boldsymbol{r} \) is then a function of \( y_i \) and time \( t \) which we denoted by \( \boldsymbol{r} = \boldsymbol{r}(y_i,t) \). A one-to-one correspondence between the general orthogonal coordinates \( y_i \) and the Cartesian coordinates \( x_i \). For convenience we denote the the general orthogonal coordinate system for the \( y \)-system.
The base vectors \( \boldsymbol{g}_i \) (not unit vectors) of the \( y \)-system, are tangent vectors to the coordinate lines of the \( y \)-coordinates: $$ \begin{equation} \boldsymbol{g}_i = \partd{\boldsymbol{r}}{y_i} = \partd{\boldsymbol{r}}{x_k} \partd{x_k}{y_i}= \boldsymbol{e}_k \partd{x_k}{y_i} \quad \Leftrightarrow \quad \boldsymbol{e}_k = \partd{y_i}{x_k} \boldsymbol{g}_i \tag{8.13} \end{equation} $$ with corresponding magnitude: $$ \begin{equation} h_i = \sqrt{\boldsymbol{g}_i \cdot \boldsymbol{g}_i} = \sqrt{\partd{x_k}{y_i}\partd{x_k}{y_i}}, \quad \mathrm{no \; sum \; over \; i} \tag{8.14} \end{equation} $$ Further, due to orthogonality of the coordinate lines: $$ \begin{equation} \boldsymbol{g}_i \cdot \boldsymbol{g}_j = 0 \quad \mathrm{for} \quad i \neq j \quad \Rightarrow \partd{x_k}{y_i} \partd{x_k}{y_j} = 0 \quad \mathrm{for} \quad i \neq j \tag{8.15} \end{equation} $$ From equation (8.14) and (8.15) we get the following relation: $$ \begin{equation} \partd{x_k}{y_i} \partd{x_k}{y_j} = h_i^2 \, \delta_{ij} \tag{8.16} \end{equation} $$ From the fundamental properties of the general orthogonal coordinate system we have: $$ \begin{equation} \partd{y_i}{y_j} = \partd{y_i}{x_k} \partd{x_k}{y_j} = \delta_{ij} \tag{8.17} \end{equation} $$ Thus, the components in the sum in equation (8.17) may be interpreted as the product of two matrices with the components: $$ \begin{equation} \left [ \partd{y_i}{x_k} \right ] = \left [\partd{x_k}{y_i} \right ]^{-1} \tag{8.18} \end{equation} $$ Then from equation (8.16) and (8.18) we get: $$ \begin{equation} \left ( \partd{y_i}{x_k} \partd{y_j}{x_k} \right ) = \left (\partd{x_k}{y_i} \partd{x_k}{y_j} \right )^{-1} = \frac{1}{h_i^2} \, \delta_{ij} \tag{8.19} \end{equation} $$
The del-operator is normally defined as follows in a Cartesian coordinate system: $$ \begin{equation} \nabla () = \boldsymbol{e}_k \partd{()}{x_k} \tag{8.20} \end{equation} $$ A representation of the del-operator in a general orthogonal coordinate system may then be found by substitution of equation (8.13) and (8.19) into equation (8.20): $$ \begin{equation} \nabla () = \left ( \partd{y_i}{x_k} \, \boldsymbol{g}_i\right ) \; \left( \partd{()}{y_j} \partd{y_j}{x_k} \right ) % = \sum_i \left (\frac{1}{h_i^2} \delta_{ij} \right ) \boldsymbol{g}_i \partd{()}{y_j} = \sum_i \frac{1}{h_i^2} \boldsymbol{g}_i \partd{()}{y_i} \tag{8.21} \end{equation} $$ By introducing a unit base vector in the y-system: $$ \begin{equation} \boldsymbol{e}_i^y = \frac{\boldsymbol{g}_i}{h_i} \tag{8.22} \end{equation} $$ the expression for the del-operator take the form: $$ \begin{equation} \nabla () = \sum_i \frac{1}{h_i} \boldsymbol{e}_i^y \partd{()}{y_i} \tag{8.23} \end{equation} $$
The physical components of a vector \( \boldsymbol{a} \) and a second order tensor \( \boldsymbol{A} \) are defined in the following manner: $$ \begin{equation} \boldsymbol{a} = a^y_k \boldsymbol{e}^y_k, \qquad \boldsymbol{A} = A^y_{kl} \, \boldsymbol{e}^y_k \otimes \boldsymbol{e}^y_l \tag{8.24} \end{equation} $$
From the expression in equation (8.21) we get for a scalar \( \alpha \): $$ \begin{equation} \nabla \alpha = \sum_i \frac{1}{h_i} \boldsymbol{e}_i^y \partd{\alpha}{y_i} \tag{8.25} \end{equation} $$
and for the divergence of a vector \( \boldsymbol{a} \): $$ \begin{equation} \nabla \cdot \boldsymbol{a} = \sum_i \frac{1}{h_i} \boldsymbol{e}_i^y \partd{(a^y_k \boldsymbol{e}^y_k)}{y_i} \tag{8.26} \end{equation} $$ rotation of a vector \( \boldsymbol{a} \): $$ \begin{equation} \nabla \times \boldsymbol{a} = \sum_i \frac{1}{h_i} \boldsymbol{e}_i^y \times \partd{(a^y_k \boldsymbol{e}^y_k)}{y_i} \tag{8.27} \end{equation} $$ The divergence of a second order tensor \( \boldsymbol{A} \) is: $$ \begin{equation} \mathrm{div} \boldsymbol{A} = \boldsymbol{A} \cdot \stackrel{\leftarrow}{\nabla} = \sum_i \frac{1}{h_i}\partd{\left (A^y_{kl} \boldsymbol{e}^y_k \otimes \boldsymbol{e}^y_l \right )}{y_i} \cdot \boldsymbol{e}_i^y \tag{8.28} \end{equation} $$
This expression for the \( \mathrm{div} \boldsymbol{A} \) in equation (8.28) may be expanded by using the chain rule and taking into account \( (\boldsymbol{a} \otimes \boldsymbol{b}) \cdot \boldsymbol{c} = \boldsymbol{a} \, (\boldsymbol{b} \cdot \boldsymbol{c}) \) and the orthogonality of the y-system (i.e., \( \boldsymbol{e}_l^y \cdot \boldsymbol{e}_i^y = \delta_{li} \)): $$ \begin{equation} \mathrm{div} \boldsymbol{A} = \sum_i \frac{1}{h_i}\partd{A^y_{ki}}{y_i} \boldsymbol{e}_k^y + % \frac{1}{h_i} A^y_{ki} \partd{\boldsymbol{e}_k^y}{y_i} + % \frac{1}{h_i} A^y_{ki} \boldsymbol{e}_k \otimes \partd{\boldsymbol{e}_l^y}{y_i} \cdot \boldsymbol{e}^y_i \tag{8.29} \end{equation} $$