.. _nabla: ############## Nabla operator ############## *********** Description *********** I consider the following operator in the Cartesian coordinate: .. math:: \vec{\nabla} \equiv \sum_i \vec{e}_i \pder{}{x^i}, which frequently appears and can operate upon arbitrary order of tensors. Now I aim at describing this relation on the general orthogonal coordinate systems. Using :ref:`the basis vector transform ` .. math:: \fromgtoc and the chain rule, I notice .. math:: \sum_i \vec{e}_i \pder{}{x^i} = \sum_{ijk} \vec{E}_j \pder{X^j}{x^i} \pder{X^k}{x^i} \pder{}{X^k}. By using :ref:`the relation of the transformation matrices `: .. math:: \jacobiconv, I have .. math:: \sum_i \pder{X^j}{x^i} \pder{X^k}{x^i} = \frac{1}{H_j H_j} \frac{1}{H_k H_k} \sum_i \pder{x^i}{X^j} \pder{x^i}{X^k}, and by adopting :ref:`the relation of the metric tensor `: .. math:: \metrictensor, this yields .. math:: \frac{1}{H_j H_j} \frac{1}{H_k H_k} H_j H_k \delta_{jk} = \frac{1}{H_j} \frac{1}{H_k} \delta_{jk}. Thus .. math:: \sum_{jk} \vec{E}_j \frac{1}{H_j} \frac{1}{H_k} \delta_{jk} \pder{}{X^k} & = \sum_j \vec{E}_j \frac{1}{H_j H_j} \pder{}{X^j} \\ & = \sum_j \vec{E}^j \pder{}{X^j}. In summary, .. math:: \vec{\nabla} & \equiv \sum_i \vec{e}_i \pder{}{x^i} \\ & = \sum_i \vec{E}^i \pder{}{X^i} \\ & = \sum_i \vec{\hat{E}}_i \frac{1}{H_i} \pder{}{X^i}.