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I'm going to find a relation between Cartesian and elliptic unit vectors.

For ellipse with major and minor semiaxes $a$ and $b$ Cartesian coordinates are:
$$\begin{align*} x&=f \cosh(\xi)\cos(\eta)\\ y&=f \sinh(\xi)\sin(\eta) \end{align*}$$ where $$f = \sqrt{a^2-b^2}.$$

Elliptic coordinates are: $$\begin{align*} \xi &= \Re\left(\mathrm{arcosh}\left( {{x + iy} \over {f}}\right)\right)\\ \eta &= \Im\left(\mathrm{arcosh}\left( {{x + iy} \over {f}}\right)\right) \end{align*}$$ (these equations are taken from [1, 9.11 p.170]).

According to [2, (6) p.481] elliptic unit vectors are related to Cartesian by this matrix equation: $${\begin{pmatrix} \mathbf{\vec{e_x}} \\ \mathbf{\vec{e_y}} \end{pmatrix}} = \frac1{\sqrt{D}} {\begin{pmatrix} \sinh(\xi)\cos(\eta) & -\cosh(\xi)\sin(\eta) \\ \cosh(\xi)\sin(\eta) & \sinh(\xi)\sin(\eta) \end{pmatrix}} {\begin{pmatrix} \mathbf{\vec{e_\xi}} \\ \mathbf{\vec{e_\eta}} \end{pmatrix}}, $$ where $$ D = { 1 \over 2 } (\cosh(2\xi) - \cos(2\eta)) $$ is a determinant of a matrix (see above). This matrix is build from a partial derivatives $$ \begin{pmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial x}{\partial \eta} \\ \frac{\partial y}{\partial \xi} & \frac{\partial y}{\partial \eta} \end{pmatrix} $$ (AFAIU, Jacobian, see [3]).

I need to find the matrix $M$ of inverse transformation: $${\begin{pmatrix} \mathbf{\vec{e_\xi}} \\ \mathbf{\vec{e_\eta}} \end{pmatrix}} = M {\begin{pmatrix} \mathbf{\vec{e_x}} \\ \mathbf{\vec{e_y}} \end{pmatrix}} $$ Any ideas?

References:
1. J. C. Gutiérrez-Vega and S. Chávez-Cerda, "Probability distributions in classical and quantum elliptic billiards," Revista Mexicana de Física, Vol. 47, no. 5, pp. 480-488, Oct. 2001.
2. N. W. McLachlan, "Theory and applications of Mathieu functions," Oxford University Press, 1937.
3. http://en.wikipedia.org/wiki/Curvilinear_coordinates

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This question is also on mathoverflow.net/questions/82371/…. –  N0rbert Dec 1 '11 at 14:22
    
You can't invert the $2\times 2$ matrix that you have? –  J. M. Dec 1 '11 at 14:53
    
2J.M. Yes I can. It is straightforward. But is it correct? –  N0rbert Dec 1 '11 at 20:04
    
It's certainly easy to test: generate known points, apply the known transformation, then the purported inverse, then check if what you now have look like what you started with. –  J. M. Dec 1 '11 at 23:47

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