# Example of a Problem Made Easier with Skew Coordinates

Skew or oblique coordinate systems are coordinate systems where the angle between the axes is not 90 degrees. The second answer to this question has formulas to convert between these systems with an arbitrary angle, as well as a helpful diagram to illustrate the situation.

Are there any examples of problems which would be hard (or at least harder) to solve in orthogonal coordinate systems, or at least the Cartesian coordinate system, but is reduced to an easy/easier problem when taken in skew coordinates? More generally, are there any applications of skew coordinate systems?
Since choosing a certain value of the skew angle would transform suitable parallelograms into rectangles, that could be a certain simplification to start from. The less trivial, the better.

• I assume you are thinking of the finite dimensional case. It may be interesting to note that, if one is given any basis, there is an inner product with respect to which that basis is orthonormal. – Jonas Dahlbæk Dec 30 '14 at 18:30
• Here's an answer I wrote where I effectively defined a non-orthogonal coordinate system in which the solutions were easy to characterize. The problem was to find an ellipse passing through two given points with specified tangents. – Rahul Dec 30 '14 at 18:32

The main result is repeated here for convenience: $$\left[ \begin{array}{c} x - x_1 \\ y - y_1 \end{array} \right] = \left[ \begin{array}{cc} (x_2 - x_1) & (x_3 - x_1) \\ (y_2 - y_1) & (y_3 - y_1) \end{array} \right] \left[ \begin{array}{c} \xi \\ \eta \end{array} \right]$$ Note that the triangle is half a parallelogram. And instead of transforming a parallelogram into a rectangle, an arbitrary triangle is transformed into a rectangular isosceles triangle. Therefore the above is in fact equivalent with any skew linear transformation (+ translation) : $$\left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{cc} \alpha & \beta \\ \gamma & \delta \end{array} \right] \left[ \begin{array}{c} \xi \\ \eta \end{array} \right] + \left[ \begin{array}{c} p \\ q \end{array} \right]$$ If we define the coordinates of our transformed triangle as:
$\,(x_1,y_1) = (p,q)\,$ , $\,x_2 = \alpha + p\,$ , $\,x_3 = \beta + p\,$ , $\,y_2 = \gamma + q\,$ , $\,y_3 = \delta + q\,$.
$(1)=(-1,-1)\, ,\, (2)=(+1,-1)\, ,\, (3)=(-1,+1)\, ,\, (4)=(+1,+1)\,$ has bilinear interpolation: $$f = \frac{1}{4}(1-\xi)(1-\eta)f_1 + \frac{1}{4}(1+\xi)(1-\eta)f_2 + \frac{1}{4}(1-\xi)(1+\eta)f_3 + \frac{1}{4}(1+\xi)(1+\eta)f_4 \\ = \sum_{k=1}^4 N_k(\xi,\eta)\,f_k \qquad \mbox{with} \qquad N_k(\xi,\eta) = \frac{1}{4}(1\pm\xi)(1\pm\eta)$$ The accompanying (isoparametric) transformation is found by replacing $f$ with $x$ and $y$ : $$x = \sum_{k=1}^4 N_k(\xi,\eta)\,x_k \qquad ; \qquad y = \sum_{k=1}^4 N_k(\xi,\eta)\,y_k$$ The bilinear transformation becomes linear again if the quadrilateral is a parallelogram , because then $\,x_1+x_4=x_2+x_3\,$ and $\,y_1+y_4=y_2+y_3\,$ (: diagonals property) , as substituted in: $$f = (f_1+f_2+f_3+f_4)/4 + \xi\,(-f_1+f_2-f_3+f_4)/4 + \eta \, (-f_1-f_2+f_3+f_4)/4 \\ + \xi\eta\,(f_1-f_2-f_3+f_4)$$ and hence the last term becomes zero.