User:Tohline/Appendix/Ramblings/T4Integrals

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Whitworth's (1981) Isothermal Free-Energy Surface
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Integrals of Motion in T4 Coordinates

In an accompanying Wiki document, we have derived the properties of an orthogonal, axisymmetric, T3 coordinate system in which the first coordinate, <math>\lambda_1</math>, defines a family of concentric oblate-spheroidal surfaces whose (uniform) flattening is defined by a parameter <math>q \equiv R_\mathrm{eq}/Z_\mathrm{pole}</math>. In a separate, but related, Wiki document, we attempt to derive the <math>3^\mathrm{rd}</math> isolating integral of motion for massless particles that move inside a flattened, axisymmetric potential whose equipotential surfaces align with <math>\lambda_1 = \mathrm{constant}</math> surfaces in the special (quadratic) case when <math>q^2 = 2</math>. While examining this special case, we noticed that, in T3 Coordinates, the <math>h_1</math> and <math>h_2</math> scale factors are only a function of the coordinate ratio <math>\lambda_1/\lambda_2</math>. This has led us to wonder whether it might be more fruitful to search for the <math>3^\mathrm{rd}</math> isolating integral using a coordinate system in which one of the coordinates is defined by this T3-coordinate ratio.

It is with this in mind that we explore the development of a new T4 coordinate system. From the very beginning we will restrict the T4-coordinate definition to the special case of <math>q^2 = 2</math> because, at present, we think that the coordinate T3-coordinate ratio <math>\lambda_1/\lambda_2</math> is only interesting in the quadratic case. (See, for example, the polynomial root derived to complete the T1-coordinate inversion for the cubic case <math>q^2=3</math>; it is another combination of the T3 coordinates that appears to be relevant in the cubic case.)

STOP!

(7/06/2010)

As defined, below, this is not an orthogonal coordinate system.

Definition

In what follows, the coordinates <math>(\lambda_1,\lambda_2,\lambda_3)</math> refer to T3 Coordinates. Let's define a set of orthogonal T4 Coordinates for the special (quadratic) case <math>q^2=2</math> such that,

<math> \xi_1 </math>

<math> \equiv </math>

<math> (\lambda_1^2 + \lambda_2^2)^{1/2} </math>

<math> = </math>

<math> \varpi\biggl[1 + \sinh^2\Zeta + (\sinh\Zeta)^{2/(1-q^2)} \biggr]^{1/2} </math>

<math> \xrightarrow{~~(q^2=2)~~} </math>

<math> \varpi\biggl[1 + \sinh^2\Zeta + \frac{1}{\sinh^2\Zeta} \biggr]^{1/2} ; </math>

<math> \xi_2 </math>

<math> \equiv </math>

<math> \frac{\lambda_2}{\lambda_1} </math>

<math> = </math>

<math> \biggl[ \frac{(\sinh\Zeta)^{2/(1-q^2)}}{1+\sinh^2\Zeta} \biggr]^{1/2} </math>

<math> \xrightarrow{~~(q^2=2)~~} </math>

<math> \biggl[ \frac{1}{\sinh^2\Zeta(1+\sinh^2\Zeta)} \biggr]^{1/2} ; </math>

<math> \tan\xi_3 </math>

<math> \equiv </math>

<math> \frac{y}{x} , </math>

 

 

 

 

where,

<math> \sinh^2\Zeta \equiv \biggl(\frac{qz}{\varpi}\biggr)^2 ~~~~\xrightarrow{~~(q^2=2)~~} ~~~~ \frac{2z^2}{\varpi^2} . </math>

The coordinate inversion — from <math>(\xi_1,\xi_2,\xi_3)</math> back to <math>(\lambda_1,\lambda_2,\lambda_3)</math> — is straightforward. Specifically,

<math> \lambda_1 </math>

<math> = </math>

<math> \xi_1 \cos\biggl[ \tan^{-1}\xi_2 \biggr] ; </math>

<math> \lambda_2 </math>

<math> = </math>

<math> \xi_1 \sin\biggl[ \tan^{-1}\xi_2 \biggr] ; </math>

<math> \lambda_3 </math>

<math> = </math>

<math> \xi_3 . </math>

Here are some relevant partial derivatives:

<math> \frac{\partial\sinh^2\Zeta}{\partial\varpi} = -\frac{4z^2}{\varpi^3} ; </math>

<math> \frac{\partial\sinh^2\Zeta}{\partial z} = + \frac{4z}{\varpi^2} . </math>

Partial derivatives with respect to cylindrical coordinates are,

 

<math> \frac{\partial}{\partial \varpi} </math>

<math> \frac{\partial}{\partial z} </math>

<math> \frac{\partial}{\partial \phi} </math>

<math>{\xi_1}</math>

<math> \frac{\varpi}{\xi_1 z^2}\biggl(\varpi^2 + z^2 \biggr) </math>

<math> \frac{1}{2\xi_1 z^3}\biggl(4z^4 - \varpi^4 \biggr) </math>

<math> 0 </math>

<math>\xi_2</math>

<math> \frac{2\xi_2^3 z^2}{\varpi^5}(\varpi^2 + 4z^2) </math>

<math> - \frac{2\xi_2^3 z}{\varpi^4}(\varpi^2 + 4z^2) </math>

<math> 0 </math>

<math>\xi_3</math>

<math> 0 </math>

<math> 0 </math>

<math> 1 </math>

Hence, the partials with respect to Cartesian coordinates are,

 

<math> \frac{\partial}{\partial x} </math>

<math> \frac{\partial}{\partial y} </math>

<math> \frac{\partial}{\partial z} </math>

<math>\xi_1</math>

<math> \frac{x}{(1-q^2)\xi_1} \biggl[ 1 + \frac{q^4 z^2}{\varpi^2} - \frac{q^2 \xi_1^2}{\varpi^2} \biggr] </math>

<math>\xrightarrow{(q^2=2)}~~~ \frac{x}{\xi_1 z^2} (\varpi^2 + z^2)</math>

<math> \frac{y}{(1-q^2)\xi_1} \biggl[ 1 + \frac{q^4 z^2}{\varpi^2} - \frac{q^2 \xi_1^2}{\varpi^2} \biggr] </math>

<math>\xrightarrow{(q^2=2)}~~~ \frac{y}{\xi_1 z^2} (\varpi^2 + z^2) </math>

<math> - \frac{\varpi^2}{(1-q^2)\xi_1 z} \biggl[ 1 + \frac{q^4 z^2}{\varpi^2} - \frac{\xi_1^2}{\varpi^2} \biggr] </math>

<math>\xrightarrow{(q^2=2)}~~~ + \frac{1}{2\xi_1 z^3}\biggl(4z^4 - \varpi^4 \biggr) </math>

<math>\xi_2</math>

<math> ~~ </math>

<math> ~~ </math>

<math> ~~ </math>

<math>\xi_3</math>

<math> -\frac{y}{\varpi^2} </math>

<math> +\frac{x}{\varpi^2} </math>

<math> 0 </math>


The scale factors are,

<math>h_1^2</math>

<math>=</math>

<math> \biggl[ \biggl( \frac{\partial\xi_1}{\partial x} \biggr)^2 + \biggl( \frac{\partial\xi_1}{\partial y} \biggr)^2 + \biggl( \frac{\partial\xi_1}{\partial z} \biggr)^2 \biggr]^{-1} </math>

<math>=</math>

<math> \biggl[ \biggl( \frac{\partial\xi_1}{\partial \varpi} \biggr)^2 + \biggl( \frac{\partial\xi_1}{\partial z} \biggr)^2 \biggr]^{-1} </math>

<math> = </math>

<math> \biggl[ \frac{4\xi_1^2 z^6 }{(\varpi^2 + 4z^2)(\varpi^6 + 4z^6)} \biggr] </math>

<math>h_2^2</math>

<math>=</math>

<math> \biggl[ \biggl( \frac{\partial\xi_2}{\partial x} \biggr)^2 + \biggl( \frac{\partial\xi_2}{\partial y} \biggr)^2 + \biggl( \frac{\partial\xi_2}{\partial z} \biggr)^2 \biggr]^{-1} </math>

<math>=</math>

<math> \biggl[ \biggl( \frac{\partial\xi_2}{\partial \varpi} \biggr)^2 + \biggl( \frac{\partial\xi_2}{\partial z} \biggr)^2 \biggr]^{-1} </math>

<math> = </math>

<math> \frac{\varpi^{10}}{4\xi_2^6 z^2} \biggl[ \frac{1}{(\varpi^2 + 4z^2)^2(\varpi^2 + z^2)} \biggr] </math>

<math>h_3^2</math>

<math>=</math>

<math> \biggl[ \biggl( \frac{\partial\xi_3}{\partial x} \biggr)^2 + \biggl( \frac{\partial\xi_3}{\partial y} \biggr)^2 + \biggl( \frac{\partial\xi_3}{\partial z} \biggr)^2 \biggr]^{-1} </math>

<math>=</math>

<math> \varpi^2 </math>

 

 

where,        <math>~~</math>.

The position vector is,

<math>\vec{x}</math>

<math>=</math>

<math> \hat\imath x + \hat\jmath y + \hat{k}z </math>

<math>=</math>

<math> ~~ </math>

See Also

 

Whitworth's (1981) Isothermal Free-Energy Surface

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