Parabolic coordinates

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Parabolic coordinate system

Parabolic coordinates are a two-dimensional orthogonal coordinate system in which the coordinate lines are confocal parabolas. A three-dimensional version of parabolic coordinates is obtained by rotating the two-dimensional system about the symmetry axis of the parabolas.

Parabolic coordinates have found many applications, e.g., the treatment of the Stark effect and the potential theory of the edges.

1 Two-dimensional parabolic coordinates
2 Two-dimensional scale factors
3 Three-dimensional parabolic coordinates
4 Three-dimensional scale factors
5 An alternative formulation
6 See also
7 References

[edit] Two-dimensional parabolic coordinates

Two-dimensional parabolic coordinates $(σ,τ)$ are defined by the equations

$x = \sigma \tau\,$

$y = \frac{1}{2} \left( \tau^{2} - \sigma^{2} \right)$

The curves of constant $σ$ form confocal parabolae

$2y = \frac{x^{2}}{\sigma^{2}} - \sigma^{2}$

that open upwards (i.e., towards $+ y$ ), whereas the curves of constant $τ$ form confocal parabolae

$2y = -\frac{x^{2}}{\tau^{2}} + \tau^{2}$

that open downwards (i.e., towards $- y$ ). The foci of all these parabolae are located at the origin.

[edit] Two-dimensional scale factors

The scale factors for the parabolic coordinates $(σ,τ)$ are equal

$h_{\sigma} = h_{\tau} = \sqrt{\sigma^{2} + \tau^{2}}$

Hence, the infinitesimal element of area is

$dA = \left( \sigma^{2} + \tau^{2} \right) d\sigma d\tau$

and the Laplacian equals

$\nabla^{2} \Phi = \frac{1}{\sigma^{2} + \tau^{2}} \left( \frac{\partial^{2} \Phi}{\partial \sigma^{2}} + \frac{\partial^{2} \Phi}{\partial \tau^{2}} \right)$

Other differential operators such as $\nabla \cdot \mathbf{F}$ and $\nabla \times \mathbf{F}$ can be expressed in the coordinates $(σ,τ)$ by substituting the scale factors into the general formulae found in orthogonal coordinates.

[edit] Three-dimensional parabolic coordinates

The two-dimensional parabolic coordinates form the basis for two sets of three-dimensional orthogonal coordinates. The parabolic cylindrical coordinates are produced by projecting in the $z$ -direction. Rotation about the symmetry axis of the parabolae produces a set of confocal paraboloids, forming a coordinate system that is also known as "parabolic coordinates"

x = στcosφ

y = στsinφ

$z = \frac{1}{2} \left(\tau^{2} - \sigma^{2} \right)$

where the parabolae are now aligned with the $z$ -axis, about which the rotation was carried out. Hence, the azimuthal angle $φ$ is defined

$\tan \phi = \frac{y}{x}$

The surfaces of constant $σ$ form confocal paraboloids

$2z = \frac{x^{2} + y^{2}}{\sigma^{2}} - \sigma^{2}$

that open upwards (i.e., towards $+ z$ ) whereas the surfaces of constant $τ$ form confocal paraboloids

$2z = -\frac{x^{2} + y^{2}}{\tau^{2}} - \tau^{2}$

that open downwards (i.e., towards $- z$ ). The foci of all these paraboloids are located at the origin.

[edit] Three-dimensional scale factors

The scale factors $h σ$ and $h τ$ are the same as in the two-dimensional case. The scale factor for the azimuthal angle $φ$ equals

h φ = στ

Hence, the infinitesimal volume element is

$dV = \sigma\tau \left( \sigma^{2} + \tau^{2} \right) d\sigma d\tau d\phi$

and the Laplacian is given by

$\nabla^{2} \Phi = \frac{1}{\sigma^{2} + \tau^{2}} \left[ \frac{1}{\sigma} \frac{\partial}{\partial \sigma} \left( \sigma \frac{\partial \Phi}{\partial \sigma} \right) + \frac{1}{\tau} \frac{\partial}{\partial \tau} \left( \tau \frac{\partial \Phi}{\partial \tau} \right) + \left( \frac{1}{\sigma^{2}} + \frac{1}{\tau^{2}} \right) \frac{\partial^{2} \Phi}{\partial \phi^{2}} \right]$

Other differential operators such as $\nabla \cdot \mathbf{F}$ and $\nabla \times \mathbf{F}$ can be expressed in the coordinates $(σ,τ,φ)$ by substituting the scale factors into the general formulae found in orthogonal coordinates.

[edit] An alternative formulation

Conversion from Cartesian to parabolic coordinates is effected by means of the following equations:

$\eta = - z + \sqrt{ x^2 + y^2 + z^2 },$

$\xi = z + \sqrt{ x^2 + y^2 + z^2 },$

$\phi = \arctan {y \over x}.$

$\begin{vmatrix}d\eta\\d\xi\\d\phi\end{vmatrix} = \begin{vmatrix} \frac{x}{\sqrt{x^2+y^2+z^2}} & \frac{y}{\sqrt{x^2+y^2+z^2}} &-1+\frac{z}{\sqrt{x^2+y^2+z^2}}\\ \frac{x}{\sqrt{x^2+y^2+z^2}} & \frac{y}{\sqrt{x^2+y^2+z^2}} &1 +\frac{z}{\sqrt{x^2+y^2+z^2}}\\ \frac{-y}{x^2+y^2}&\frac{x}{x^2+y^2}&0 \end{vmatrix} \cdot \begin{vmatrix}dx\\dy\\dz\end{vmatrix}$

$\eta\ge 0,\quad\xi\ge 0$

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

$\eta = -z + \sqrt{ x^2 + z^2},$

$\xi = z + \sqrt{ x^2 + z^2}.$

If η=c (a constant), then

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

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

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

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:

${x^2 \over 2 c} - {c \over 2} = {b \over 2} - {x^2 \over 2 b},$

regroup,

${x^2 \over 2 c} + {x^2 \over 2 b} = {b \over 2} + {c \over 2},$

factor out the x,

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

cancel out common factors from both sides,

$x^2 = b c, \,$

take the square root,

$x = \sqrt{b c}.$

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:

$z_c = {b c \over 2 c} - {c \over 2} = {b - c \over 2},$

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

$z_b = {b \over 2} - {b c \over 2 b} = {b - c \over 2}.$

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

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

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:

${d z_c \over d x} = {x \over c} = { \sqrt{ b c} \over c} = \sqrt{ b \over c} = s_c.$

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

${d z_b \over d x} = - {x \over b} = { - \sqrt{ b c } \over b} = - \sqrt{ {c \over b} } = s_b.$

The products of the two slopes is

$s_c s_b = - \sqrt{ {b \over c}} \sqrt{ {c \over b}} = -1.$

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]:

$x = \sqrt{\xi \eta} \cos \phi,$

$y = \sqrt{\xi \eta} \sin \phi,$

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

$\begin{vmatrix}dx\\dy\\dz\end{vmatrix} = \begin{vmatrix} \frac{1}{2}\sqrt{\frac{\xi}{\eta}}\cos\phi &\frac{1}{2}\sqrt{\frac{\eta}{\xi}}\cos\phi &-\sqrt{\xi\eta}\sin\phi\\ \frac{1}{2}\sqrt{\frac{\xi}{\eta}}\sin\phi &\frac{1}{2}\sqrt{\frac{\eta}{\xi}}\sin\phi &\sqrt{\xi\eta}\cos\phi\\ -\frac{1}{2}&\frac{1}{2}&0 \end{vmatrix} \cdot \begin{vmatrix}d\eta\\d\xi\\d\phi\end{vmatrix}$