Parabolic coordinates

Parabolic coordinates are an alternative system of coordinates for three dimensions. They are orthogonal. Conversion from Cartesian to parabolic coordinates is effected by means of the following equations:

<math> \eta = - z + \sqrt{ x^2 + y^2 + z^2 }, <math>

<math> \xi = z + \sqrt{ x^2 + y^2 + z^2 }, <math>

<math> \phi = \arctan {y \over x}. <math>

<math>

\begin{vmatrix}d\eta\\d\xi\\d\phi\end{vmatrix} = \begin{vmatrix}

   \frac{x}{\sqrt{x^2+y^2+z^2}}

& \frac{y}{\sqrt{x^2+y^2+z^2}} &-1+\frac{z}{\sqrt{x^2+y^2+z^2}}\\

   \frac{x}{\sqrt{x^2+y^2+z^2}}

& \frac{y}{\sqrt{x^2+y^2+z^2}} &1 +\frac{z}{\sqrt{x^2+y^2+z^2}}\\ \frac{-y}{x^2+y^2}&\frac{x}{x^2+y^2}&0 \end{vmatrix} \cdot \begin{vmatrix}dx\\dy\\dz\end{vmatrix} <math>

<math>\eta\ge 0,\quad\xi\ge 0<math>

If φ=0 then a cross-section is obtained; the coordinates become confined to the x-z plane:

<math> \eta = -z + \sqrt{ x^2 + z^2}, <math>

<math> \xi = z + \sqrt{ x^2 + z^2}. <math>

If η=c (a constant), then

<math> \left. z \right|_{\eta = c} = {x^2 \over 2 c} - {c \over 2}. <math>

This is a parabola whose focus is at the origin for any value of c. The parabola's axis of symmetry is vertical and the concavity faces upwards.

If ξ=c then

<math> \left. z \right|_{\xi = c} = {c \over 2} - {x^2 \over 2 c}. <math>

This is a parabola whose focus is at the origin for any value of c. Its axis of symmetry is vertical and the concavity faces downwards.

Now consider any upward parabola η=c and any downward parabola ξ=b. It is desired to find their intersection:

<math> {x^2 \over 2 c} - {c \over 2} = {b \over 2} - {x^2 \over 2 b}, <math>

regroup,

<math> {x^2 \over 2 c} + {x^2 \over 2 b} = {b \over 2} + {c \over 2}, <math>

factor out the x,

<math> x^2 \left( {b + c \over 2 b c} \right) = {b + c \over 2}, <math>

cancel out common factors from both sides,

<math> x^2 = b c, \,<math>

take the square root,

<math> x = \sqrt{b c}. <math>

x is the geometric mean of b and c. The abscissa of the intersection has been found. Find the ordinate. Plug in the value of x into the equation of the upward parabola:

<math> z_c = {b c \over 2 c} - {c \over 2} = {b - c \over 2}, <math>

then plug in the value of x into the equation of the downward parabola:

<math> z_b = {b \over 2} - {b c \over 2 b} = {b - c \over 2}. <math>

z_c = z_b, as should be. Therefore the point of intersection is

<math> P : \left( \sqrt{b c}, {b - c \over 2} \right). <math>

Draw a pair of tangents through point P, each one tangent to each parabola. The tangential line through point P to the upward parabola has slope:

<math> {d z_c \over d x} = {x \over c} = { \sqrt{ b c} \over c} = \sqrt{ b \over c} = s_c. <math>

The tangent through point P to the downward parabola has slope:

<math> {d z_b \over d x} = - {x \over b} = { - \sqrt{ b c } \over b} = - \sqrt{ {c \over b} } = s_b. <math>

The products of the two slopes is

<math> s_c s_b = - \sqrt{ {b \over c}} \sqrt{ {c \over b}} = -1. <math>

The product of the slopes is negative one, therefore the slopes are perpendicular. This is true for any pair of parabolas with concavities in opposite directions.

Such a pair of parabolas intersect at two points, but when φ is restricted to zero, it actually confines the other coordinates η and ξ to move in a half-plane with x>0, because x<0 corresponds to φ=π.

Thus a pair of coordinates η and ξ specify a unique point on the half-plane. Then letting φ range from 0 to 2π the half-plane revolves with the point (around the z-axis as its hinge): the parabolas form paraboloids. A pair of opposing paraboloids specifies a circle, and a value of φ specifies a half-plane which cuts the circle of intersection at a unique point. The point's Cartesian coordinates are [Menzel, p. 139]:

<math> x = \sqrt{\xi \eta} \cos \phi, <math>

<math> y = \sqrt{\xi \eta} \sin \phi, <math>

<math> z = \begin{matrix}\frac{1}{2}\end{matrix} ( \xi - \eta ). <math>

<math>

\begin{vmatrix}dx\\dy\\dz\end{vmatrix} = \begin{vmatrix}

\frac{1}{2}\sqrt{\frac{\xi}{\eta}}\cos\phi

&\frac{1}{2}\sqrt{\frac{\eta}{\xi}}\cos\phi &-\sqrt{\xi\eta}\sin\phi\\

\frac{1}{2}\sqrt{\frac{\xi}{\eta}}\sin\phi

&\frac{1}{2}\sqrt{\frac{\eta}{\xi}}\sin\phi &\sqrt{\xi\eta}\cos\phi\\ -\frac{1}{2}&\frac{1}{2}&0 \end{vmatrix} \cdot \begin{vmatrix}d\eta\\d\xi\\d\phi\end{vmatrix} <math>

See also: spherical coordinates, cylindrical coordinates, Cartesian coordinates.

Reference

• Menzel, Donald H., Mathematical Physics, Dover Publications, 1961.