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