Skew coordinates

Let ${\displaystyle \mathbf {e} _{1}}$, ${\displaystyle \mathbf {e} _{2}}$, and ${\displaystyle \mathbf {e} _{3}}$ respectively be unit vectors along the ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$ axes. These represent the covariant basis; computing their dot products gives the following components of the metric tensor:

${\displaystyle g_{11}=g_{22}=g_{33}=1\quad$ ;\quad g_{12}=g_{23}=0\quad ;\quad g_{13}=\cos \left({\frac {\pi }{2}}-\phi \right)=\sin(\phi )}

${\displaystyle {\sqrt {g}}=\mathbf {e} _{1}\cdot (\mathbf {e} _{2}\times \mathbf {e} _{3})=\cos(\phi )}$

which are quantities that will be useful later on.

The contravariant basis is given by[2]

${\displaystyle \mathbf {e} ^{1}={\frac {\mathbf {e} _{2}\times \mathbf {e} _{3}}{\sqrt {g}}}={\frac {\mathbf {e} _{2}\times \mathbf {e} _{3}}{\cos(\phi )}}}$

${\displaystyle \mathbf {e} ^{2}={\frac {\mathbf {e} _{3}\times \mathbf {e} _{1}}{\sqrt {g}}}=\mathbf {e} _{2}}$

${\displaystyle \mathbf {e} ^{3}={\frac {\mathbf {e} _{1}\times \mathbf {e} _{2}}{\sqrt {g}}}={\frac {\mathbf {e} _{1}\times \mathbf {e} _{2}}{\cos(\phi )}}}$

The contravariant basis isn't a very convenient one to use, however it shows up in definitions so must be considered. We'll favor writing quantities with respect to the covariant basis.

Since the basis vectors are all constant, vector addition and subtraction will simply be familiar component-wise adding and subtraction. Now, let

${\displaystyle \mathbf {a} =\sum _{i}a^{i}\mathbf {e} _{i}\quad {\mbox{and}}\quad \mathbf {b} =\sum _{i}b^{i}\mathbf {e} _{i}}$

where the sums indicate summation over all values of the index (in this case, i = 1, 2, 3). The contravariant and covariant components of these vectors may be related by

${\displaystyle a^{i}=\sum _{j}a_{j}g^{ij}}$

so that, explicitly,

${\displaystyle a^{1}={\frac {a_{1}-\sin(\phi )a_{3}}{\cos ^{2}(\phi )}},}$

${\displaystyle a^{2}=a_{2},}$

${\displaystyle a^{3}={\frac {-\sin(\phi )a_{1}+a_{3}}{\cos ^{2}(\phi )}}.}$

The dot product in terms of contravariant components is then

${\displaystyle \mathbf {a} \cdot \mathbf {b} =\sum _{i}a^{i}b_{i}=a^{1}b^{1}+a^{2}b^{2}+a^{3}b^{3}+\sin(\phi )(a^{1}b^{3}+a^{3}b^{1})}$

and in terms of covariant components

${\displaystyle \mathbf {a} \cdot \mathbf {b} =\cos ^{2}(\phi )[a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}-\sin(\phi )(a_{1}b_{3}+a_{3}b_{1})].}$

By definition,[3] the gradient of a scalar function f is

${\displaystyle \nabla f=\sum _{i}\mathbf {e} ^{i}{\frac {\partial f}{\partial q^{i}}}={\frac {\partial f}{\partial x}}\mathbf {e} ^{1}+{\frac {\partial f}{\partial y}}\mathbf {e} ^{2}+{\frac {\partial f}{\partial z}}\mathbf {e} ^{3}}$

where ${\displaystyle q_{i}}$ are the coordinates x, y, z indexed. Recognizing this as a vector written in terms of the contravariant basis, it may be rewritten:

${\displaystyle \nabla f={\frac {{\frac {\partial f}{\partial x}}-\sin(\phi ){\frac {\partial f}{\partial z}}}{\cos(\phi )^{2}}}\mathbf {e} _{1}+{\frac {\partial f}{\partial y}}\mathbf {e} _{2}+{\frac {-\sin(\phi ){\frac {\partial f}{\partial x}}+{\frac {\partial f}{\partial z}}}{\cos(\phi )^{2}}}\mathbf {e} _{3}.}$

The divergence of a vector ${\displaystyle \mathbf {a} }$ is

${\displaystyle \nabla \cdot \mathbf {a} ={\frac {1}{\sqrt {g}}}\sum _{i}{\frac {\partial }{\partial q^{i}}}\left({\sqrt {g}}a^{i}\right)={\frac {\partial a^{1}}{\partial x}}+{\frac {\partial a^{2}}{\partial y}}+{\frac {\partial a^{3}}{\partial z}}.}$

and of a tensor ${\displaystyle \mathbf {A} }$

${\displaystyle \nabla \cdot \mathbf {A} ={\frac {1}{\sqrt {g}}}\sum _{i,j}{\frac {\partial }{\partial q^{i}}}\left({\sqrt {g}}a^{ij}\mathbf {e} _{j}\right)=\sum _{i,j}\mathbf {e} _{j}{\frac {\partial a^{ij}}{\partial q^{i}}}.}$

The Laplacian of f is

${\displaystyle \nabla ^{2}f=\nabla \cdot \nabla f={\frac {1}{\cos(\phi )^{2}}}\left({\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}}-2\sin(\phi ){\frac {\partial ^{2}f}{\partial x\partial z}}\right)+{\frac {\partial ^{2}f}{\partial y^{2}}}}$

and, since the covariant basis is normal and constant, the vector Laplacian is the same as the componentwise Laplacian of a vector written in terms of the covariant basis.

While both the dot product and gradient are somewhat messy in that they have extra terms (compared to a Cartesian system) the advection operator which combines a dot product with a gradient turns out very simple:

${\displaystyle (\mathbf {a} \cdot \nabla )=\left(\sum _{i}a^{i}e_{i}\right)\cdot \left(\sum _{i}{\frac {\partial }{\partial q^{i}}}\mathbf {e} ^{i}\right)=\left(\sum _{i}a^{i}{\frac {\partial }{\partial q^{i}}}\right)}$

which may be applied to both scalar functions and vector functions, componentwise when expressed in the covariant basis.

Finally, the curl of a vector is

${\displaystyle \nabla \times \mathbf {a} =\sum _{i,j,k}\mathbf {e} _{k}\epsilon ^{ijk}{\frac {\partial a_{j}}{\partial q^{i}}}=}$

${\displaystyle {\frac {1}{\cos(\phi )}}\left(\left(\sin(\phi ){\frac {\partial a^{1}}{\partial y}}+{\frac {\partial a^{3}}{\partial y}}-{\frac {\partial a^{2}}{\partial z}}\right)\mathbf {e} _{1}+\left({\frac {\partial a^{1}}{\partial z}}+\sin(\phi )\left({\frac {\partial a^{3}}{\partial z}}-{\frac {\partial a^{1}}{\partial x}}\right)-{\frac {\partial a^{3}}{\partial x}}\right)\mathbf {e} _{2}+\left({\frac {\partial a^{2}}{\partial x}}-{\frac {\partial a^{1}}{\partial y}}-\sin(\phi ){\frac {\partial a^{3}}{\partial y}}\right)\mathbf {e} _{3}\right).}$