Nabla operator

Description

I consider the following operator in the Cartesian coordinate:

$$\vec{\nabla} \equiv \sum_i \vec{e}_i \pder{}{x^i},$$

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 the basis vector transform

$$\fromgtoc$$

and the chain rule, I notice

$$\sum_i \vec{e}_i \pder{}{x^i} = \sum_{ijk} \vec{E}_j \pder{X^j}{x^i} \pder{X^k}{x^i} \pder{}{X^k}.$$

By using the relation of the transformation matrices:

$$\jacobiconv,$$

I have

$$\sum_i \pder{X^j}{x^i} \pder{X^k}{x^i} = \frac{1}{H_j H_j} \frac{1}{H_k H_k} \sum_i \pder{x^i}{X^j} \pder{x^i}{X^k},$$

and by adopting the relation of the metric tensor:

$$\metrictensor,$$

this yields

$$\frac{1}{H_j H_j} \frac{1}{H_k H_k} H_j H_k \delta_{jk} = \frac{1}{H_j} \frac{1}{H_k} \delta_{jk}.$$

Thus

$$\begin{split}\sum_{jk} \vec{E}_j \frac{1}{H_j} \frac{1}{H_k} \delta_{jk} \pder{}{X^k} & = \sum_j \vec{E}_j \frac{1}{H_j H_j} \pder{}{X^j} \\ & = \sum_j \vec{E}^j \pder{}{X^j}.\end{split}$$

In summary,

$$\begin{split}\vec{\nabla} & \equiv \sum_i \vec{e}_i \pder{}{x^i} \\ & = \sum_i \vec{E}^i \pder{}{X^i} \\ & = \sum_i \vec{\hat{E}}_i \frac{1}{H_i} \pder{}{X^i}.\end{split}$$