Editing Apps/MaclaurinSpheroids/GoogleBooks (section)

===Comparison With More Traditional Derivation===

In our [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Acceleration_at_the_Pole|separate derivation of the gravitational acceleration at the pole of a prolate spheroid]] that uses a more traditional approach, we obtained this key coefficient by integrating over a different function, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{Z}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{(1-\zeta)}{ [ (2-e^2) - 2\zeta + e^2\zeta^2 ]^{1/2}} -1 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(1-\zeta)^{1/2}}{ [ 2-e^2 - e^2\zeta  ]^{1/2}} -1 \, .</math>
  </td>
</tr>
</table>
</div>

By comparing Maclaurin's approach to the more traditional one, we conclude that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\int_{-1}^0 \biggl[\frac{\mathrm{AQ}}{a_1}\biggr]dz_\mathrm{RN} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\int_{-1}^1 \mathcal{Z} d\zeta \, .</math>
  </td>
</tr>
</table>
</div>

We should be able to demonstrate explicitly that these two integrands are the same.  We begin by recognizing that the relationship between the dimensionless coordinate, <math>~z/a_1</math>, that we have used to decipher Maclaurin's presentation and the dimensionless coordinate, <math>~\zeta</math>, that was [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Acceleration_at_the_Pole|used in our separate analysis]], is,
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<math>\frac{z}{a_1} = -\zeta \, .</math>
</div>

Returning to the first expression in the above subsection that details the [[Apps/MaclaurinSpheroids/GoogleBooks#Intersection_With_Ellipse|intersection of the red line-segment with the blue ellipse]], we can therefore write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1-e^2) (1-\zeta^2)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1-\zeta)^2\tan^2\alpha </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~ \tan^2\alpha </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(1-e^2) (1+\zeta)}{(1-\zeta)} \, .</math>
  </td>
</tr>
</table>
</div>

Recalling that,
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<math>u_\mathrm{RN} = \cos\alpha = [1+\tan^2\alpha]^{-1/2} \, ,</math>
</div>
we see that,
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<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\mathrm{AQ}}{a_1} = \frac{2(1-e^2)u_\mathrm{RN}^2}{1-e^2 u_\mathrm{RN}^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2(1-e^2)[1+\tan^2\alpha]^{-1}}{1-e^2[1+\tan^2\alpha]^{-1}}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2(1-e^2)}{1-e^2 +\tan^2\alpha}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2(1-e^2)(1-\zeta)}{(1-e^2)(1-\zeta) +(1-e^2)(1+\zeta)}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1-\zeta) \, .</math>
  </td>
</tr>
</table>
</div>

Furthermore, we see that the transformation from <math>~u_\mathrm{RN}</math> to <math>~\zeta</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~u_\mathrm{RN} = [1+\tan^2\alpha ]^{-1/2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[1 + \frac{(1-e^2) (1+\zeta)}{(1-\zeta)}  \biggr]^{-1/2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{(1-\zeta) + (1-e^2) (1+\zeta)}{(1-\zeta)}  \biggr]^{-1/2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1-\zeta)^{1/2} (2-e^2 -e^2\zeta)^{-1/2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ du_\mathrm{RN}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[
-2^{-1}(1-\zeta)^{-1/2} (2-e^2 -e^2\zeta)^{-1/2}
-2^{-1}(1-\zeta)^{1/2} (2-e^2 -e^2\zeta)^{-3/2}(-e^2)]d\zeta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2^{-1}(1-\zeta)^{-1/2}  (2-e^2 -e^2\zeta)^{-3/2}[
(2-e^2 -e^2\zeta)-e^2(1-\zeta) ]d\zeta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-(1-e^2)(1-\zeta)^{-1/2}  (2-e^2 -e^2\zeta)^{-3/2} d\zeta \, .</math>
  </td>
</tr>
</table>
</div>

Putting these two expressions together, we conclude that the integrand is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[\frac{\mathrm{AQ}}{a_1}\biggr] du_\mathrm{RN}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\biggl[\frac{(1-e^2)(1-\zeta)^{1/2} }{ (2-e^2 -e^2\zeta)^{3/2}}\biggr] d\zeta </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\biggl[\frac{(1-e^2)}{(2-e^2 -e^2\zeta)}  \biggr]\biggl[\frac{(1-\zeta)^{1/2} }{ (2-e^2 -e^2\zeta)^{1/2}}\biggr] d\zeta \, .</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\biggl[\frac{(1-e^2)}{(2-e^2 -e^2\zeta)}  \biggr]\biggl[1 + \mathcal{Z}\biggr] d\zeta \, .</math>
  </td>
</tr>
</table>
</div>

This does not appear to match the other integrand.  I must be misinterpreting something!

To be done:  I suspect that Maclaurin's approach will look more familiar if I redo the "traditional" derivation from the point of view of a spherical coordinate system (positioning pole A at the origin), in which an axisymmetric volume element takes the form,
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<math>~dV = 2\pi r^2 \sin\alpha dr d\alpha = 2\pi r^2 dr du_\mathrm{RN} \, ,</math>
</div>
where, from above, <math>u_\mathrm{RN} \equiv \cos\alpha \, .</math>