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(Concentric Ellipsoidal (T6) Coordinates)
(Concentric Ellipsoidal (T6) Coordinates)
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When <math>~\lambda_1 = a</math>, we obtain the standard definition of an ellipsoidal surface, it being understood that, <math>~q^2 = a^2/b^2</math> and <math>~p^2 = a^2/c^2</math>.  (We will assume that <math>~a > b > c</math>, that is, <math>~p^2 > q^2 > 1</math>.)  What is the expression for the unit vector normal to the surface at <math>~(x, y, z)</math> when written in terms of Cartesian unit vectors?
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When <math>~\lambda_1 = a</math>, we obtain the standard definition of an ellipsoidal surface, it being understood that, <math>~q^2 = a^2/b^2</math> and <math>~p^2 = a^2/c^2</math>.  (We will assume that <math>~a > b > c</math>, that is, <math>~p^2 > q^2 > 1</math>.)   
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A vector, <math>~\bold{\hat{n}}</math>, that is normal to the <math>~\lambda_1</math> = constant surface is given by the gradient of the function,
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<table border="0" cellpadding="5" align="center">
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<tr>
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  <td align="right">
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<math>~F(x, y, z)</math>
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  </td>
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  <td align="center">
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<math>~\equiv</math>
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  </td>
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  <td align="left">
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<math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} - \lambda_1 \, .</math>
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  </td>
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</tr>
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</table>
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In Cartesian coordinates, this means,
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<table border="0" cellpadding="5" align="center">
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<tr>
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  <td align="right">
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<math>~\bold{\hat{n}}(x, y, z)</math>
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  </td>
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  <td align="center">
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<math>~=</math>
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  </td>
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  <td align="left">
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<math>~
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\hat\imath \biggl( \frac{\partial F}{\partial x} \biggr)
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+ \hat\jmath \biggl( \frac{\partial F}{\partial y} \biggr)
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+ \hat{k} \biggl( \frac{\partial F}{\partial z} \biggr)
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</math>
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  </td>
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</tr>
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<tr>
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  <td align="right">
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&nbsp;
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  </td>
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  <td align="center">
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<math>~=</math>
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  </td>
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  <td align="left">
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<math>~
 +
\hat\imath \biggl[ x(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr]
 +
+ \hat\jmath \biggl[ q^2y(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr]
 +
+ \hat\jmath \biggl[ p^2 z(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr]
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</math>
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  </td>
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</tr>
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<tr>
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  <td align="right">
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&nbsp;
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  </td>
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  <td align="center">
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<math>~=</math>
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  </td>
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  <td align="left">
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<math>~
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\hat\imath \biggl( \frac{x}{\lambda_1} \biggr)
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+ \hat\jmath \biggl( \frac{q^2y}{\lambda_1} \biggr)
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+ \hat\jmath \biggl(\frac{p^2 z}{\lambda_1} \biggr) \, ,
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</math>
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  </td>
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</tr>
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</table>
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where it is understood that this expression is only to be evaluated at points, <math>~(x, y, z)</math>, that lie on the selected <math>~\lambda_1</math> surface &#8212; that is, at points for which the function, <math>~F(x,y,z) = 0</math>.  The length of this normal vector is given by the expression,
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<table border="0" cellpadding="5" align="center">
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<tr>
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  <td align="right">
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<math>~[ \bold{\hat{n}} \cdot \bold{\hat{n}} ]^{1 / 2}</math>
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  </td>
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  <td align="center">
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<math>~=</math>
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  </td>
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  <td align="left">
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<math>~
 +
\biggl[ \biggl( \frac{\partial F}{\partial x} \biggr)^2 + \biggl( \frac{\partial F}{\partial y} \biggr)^2 + \biggl( \frac{\partial F}{\partial z} \biggr)^2 \biggr]^{1 / 2}
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</math>
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  </td>
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</tr>
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</table>
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 +
 
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 +
The properly normalized
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Next, we appreciate that the vector that is normal to theWhat is the expression for the unit vector normal to the surface at <math>~(x, y, z)</math> when written in terms of Cartesian unit vectors?
Well, to start with we know that <math>~\lambda_1^2</math> is constant across the entire surface, so at any point on this specified surface we must find,
Well, to start with we know that <math>~\lambda_1^2</math> is constant across the entire surface, so at any point on this specified surface we must find,

Revision as of 10:09, 27 October 2020

Contents

Concentric Ellipsoidal (T6) Coordinates

Whitworth's (1981) Isothermal Free-Energy Surface
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Background

Building on our general introduction to Direction Cosines in the context of orthogonal curvilinear coordinate systems, and on our previous development of T3 (concentric oblate-spheroidal) and T5 (concentric elliptic) coordinate systems, here we explore the creation of a concentric ellipsoidal (T6) coordinate system. This is motivated by our desire to construct a fully analytically prescribable model of a nonuniform-density ellipsoidal configuration that is an analog to Riemann S-Type ellipsoids.

Orthogonal Coordinates

We start by defining a "radial" coordinate whose values identify various concentric ellipsoidal shells,

~\lambda_1

~\equiv

~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} \, .

When ~\lambda_1 = a, we obtain the standard definition of an ellipsoidal surface, it being understood that, ~q^2 = a^2/b^2 and ~p^2 = a^2/c^2. (We will assume that ~a > b > c, that is, ~p^2 > q^2 > 1.)

A vector, ~\bold{\hat{n}}, that is normal to the ~\lambda_1 = constant surface is given by the gradient of the function,

~F(x, y, z)

~\equiv

~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} - \lambda_1 \, .

In Cartesian coordinates, this means,

~\bold{\hat{n}}(x, y, z)

~=

~
\hat\imath \biggl( \frac{\partial F}{\partial x} \biggr)
+ \hat\jmath \biggl( \frac{\partial F}{\partial y} \biggr)
+ \hat{k} \biggl( \frac{\partial F}{\partial z} \biggr)

 

~=

~
\hat\imath \biggl[ x(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr]
+ \hat\jmath \biggl[ q^2y(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr]
+ \hat\jmath \biggl[ p^2 z(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr]

 

~=

~
\hat\imath \biggl( \frac{x}{\lambda_1} \biggr)
+ \hat\jmath \biggl( \frac{q^2y}{\lambda_1} \biggr)
+ \hat\jmath \biggl(\frac{p^2 z}{\lambda_1} \biggr) \, ,

where it is understood that this expression is only to be evaluated at points, ~(x, y, z), that lie on the selected ~\lambda_1 surface — that is, at points for which the function, ~F(x,y,z) = 0. The length of this normal vector is given by the expression,

~[ \bold{\hat{n}} \cdot \bold{\hat{n}} ]^{1 / 2}

~=

~
\biggl[ \biggl( \frac{\partial F}{\partial x} \biggr)^2 + \biggl( \frac{\partial F}{\partial y} \biggr)^2 + \biggl( \frac{\partial F}{\partial z} \biggr)^2 \biggr]^{1 / 2}


The properly normalized

Next, we appreciate that the vector that is normal to theWhat is the expression for the unit vector normal to the surface at ~(x, y, z) when written in terms of Cartesian unit vectors?

Well, to start with we know that ~\lambda_1^2 is constant across the entire surface, so at any point on this specified surface we must find,

~0

~=

~2x dx + 2q^2y dy + 2p^2z dz \, .

See Also

Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2020 by Joel E. Tohline
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