The simplest 3D case of a skew coordinate system modifies a
Cartesian coordinate system into an
affine coordinate system by rotating one of the axes (say the
x axis) by some angle \phi, while staying orthogonal to one of the remaining two axes. For this example, the
x axis of a Cartesian coordinate has been bent toward the
z axis by \phi, remaining orthogonal to the
y axis.
Algebra and useful quantities Let \mathbf e_1, \mathbf e_2, and \mathbf e_3 respectively be unit vectors along the x, y, and z axes. These represent the
covariant basis; computing their dot products gives the
metric tensor: : [g_{ij}] = \begin{pmatrix} 1&0&\sin(\phi)\\ 0&1&0\\ \sin(\phi)&0&1 \end{pmatrix} ,\qquad [g^{ij}] = \frac{1}{\cos^2(\phi)} \begin{pmatrix} 1&0&-\sin(\phi)\\ 0&\cos^2(\phi)&0\\ -\sin(\phi)&0&1 \end{pmatrix} where :\quad g_{13} = \cos\left(\frac \pi 2 - \phi\right) = \sin(\phi) and :\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 the
gradient of a scalar function
f is :\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 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: :\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 \mathbf a is :\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 \mathbf A :\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 :\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: :(\mathbf a \cdot \nabla) = \biggl(\sum_i a^i e_i\biggr) \cdot \biggl(\sum_i \frac{\partial}{\partial q^i} \mathbf e^i\biggr) = \biggl(\sum_i a^i \frac{\partial}{\partial q^i}\biggr) 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 :\nabla \times \mathbf a = \sum_{i, j, k} \mathbf e_k \epsilon^{ijk} \frac{\partial a_j}{\partial q^i} = ::\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). ==See also==