Editing ThreeDimensionalConfigurations/HomogeneousEllipsoids (section)

=Work In Progress=
==Derivation of Expression for Gravitational Potential==

In &sect;373 (p. 700) of his book titled, ''Hydrodynamics'', [<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>] states that, "<font color="maroon">The gravitation-potential, at internal points, of a uniform mass enclosed by the surface</font>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>], <font color="#00CC00">&sect;373, Eq. (1)</font></td></tr>
</table>

<font color="maroon">&hellip; may be written</font>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\Phi(\vec{x})}{G}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\pi \rho (\alpha_0 x^2 + \beta_0 y^2 + \gamma_0 z^2 - \chi_0) \, ,
</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>], <font color="#00CC00">&sect;373, Eq. (4)</font></td></tr>
</table>
<font color="maroon">where, as in &sect;114,</font>"
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\alpha_0}{abc} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\int_0^\infty \frac{d\lambda}{(a^2 + \lambda)\Delta} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\beta_0}{abc} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\int_0^\infty \frac{d\lambda}{(b^2 + \lambda)\Delta} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\gamma_0}{abc} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\int_0^\infty \frac{d\lambda}{(c^2 + \lambda)\Delta} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\chi_0}{abc} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\int_0^\infty \frac{d\lambda}{\Delta} \, ,</math>
  </td>
</tr>
<tr><td align="center" colspan="7">[<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>], <font color="#00CC00">&sect;373, Eqs. (5) &amp; (6)</font></td></tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Delta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[(a^2 + \lambda)(b^2 + \lambda)(c^2 + \lambda)]^{1 / 2} \, .</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>], <font color="#00CC00">&sect;373, Eq. (3)</font></td></tr>
</table>
Although different variable names have been used, it is easy to see the correspondence between these expressions and the ''defining integral expressions'' that we have drawn from the more recent publications of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>] and [<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>] and [[#Gravitational_Potential|presented above]].  Here, we are interested in demonstrating how [<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>] derived his expression for the potential inside (and on the surface of) an homogeneous ellipsoid.

==Acceleration at the Pole==
===Prolate Spheroids===

In our above review, for consistency, we assumed that the longest axis of the ellipsoid was aligned with the <math>~x</math>-axis in all cases &#8212; for prolate spheroids as well as for oblate spheroids and for the more generic, triaxial ellipsoids.  In this discussion, in order to better align with the operational features of a standard cylindrical coordinate system, we will orient the prolate-spheroidal configuration such that its major axis and, hence, its axis of symmetry aligns with the <math>~z</math>-axis while the center of the spheroid remains at the center of the (cylindrical) coordinate grid.  In this case, the surface will be defined by the ellipse,
<div align="center">
<math>~\frac{\varpi^2}{a_3^2} + \frac{z^2}{a_1^2} = 1 ~~~~\Rightarrow ~~~~ \varpi = a_3\sqrt{1-z^2/a_1^2} \, ,</math>
</div>
and the gravitational potential will be given by the expression, 

<div align="center">
<math>
~\Phi(\vec{x}) = -\pi G \rho \biggl[ I_\mathrm{BT} a_1^2 - \biggl(A_1 z^2 + A_3 \varpi^2 \biggr) \biggr].
</math>
</div>

<span id="FirstDetermination">
The magnitude of the gravitational acceleration at the pole <math>~(\varpi, z) = (0, a_1)</math> of this prolate spheroid can be obtained from the gravitational potential via the expression,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{A} \equiv \biggl|- \frac{\partial \Phi}{\partial z}\biggr|_{a_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi G \rho A_1 a_1 \, ,</math>
  </td>
</tr>
</table>
</div>

where, [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Prolate_Spheroids_.28a1_.3E_a2_.3D_a3.29|as above]],

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
~A_1
</math>
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>
\ln\biggl[ \frac{1+e}{1-e} \biggr] \frac{(1-e^2)}{e^3} - \frac{2(1-e^2)}{e^2} \, .
</math>
  </td>
</tr>
</table>

We should also be able to derive this expression for <math>~\mathcal{A}</math> by integrating the <math>~z</math>-component of the differential acceleration over the mass distribution, that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\int \biggl[ \frac{G }{r^2} \cdot \frac{(a_1-z)}{r} \biggr] dm = \int \biggl[ \frac{(a_1-z)G }{r^3} \biggr] 2\pi \varpi d\varpi dz</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi G\rho \int^{a_1}_{-a_1} (a_1-z)dz \int_0^{a_3\sqrt{1-z^2/a_1^2}} [\varpi^2+(z-a_1)^2]^{-3/2}\varpi d\varpi \, ,</math>
  </td>
</tr>
</table>
</div>
where the distance, <math>~r</math>, has been measured from the pole, that is,
<div align="center">
<math>~r^2 = \varpi^2 + (z-a_1)^2 \, .</math>
</div>

Performing the integral over <math>~\varpi</math> gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi G\rho \int^{a_1}_{-a_1} (a_1-z)dz \biggl\{ -[\varpi^2+(z-a_1)^2]^{-1/2} \biggr\}_0^{a_3\sqrt{1-z^2/a_1^2}} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi G\rho \int^{a_1}_{-a_1} (a_1-z)dz \biggl\{ \frac{1}{z - a_1} -\biggl[ a_3^2 \biggl(1-\frac{z^2}{a_1^2} \biggr)
+ a_1^2\biggl(1-\frac{z}{a_1}\biggr)^2 \biggr]^{-1/2} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - 2\pi G\rho a_1 \int^{1}_{-1}  d\zeta \biggl\{ \frac{1-\zeta}{1-\zeta } - (1-\zeta)\biggl[ \biggl(\frac{a_3}{a_1}\biggr)^2 \biggl(1-\zeta^2 \biggr)
+ \biggl(1-\zeta\biggr)^2 \biggr]^{-1/2} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2\pi G\rho a_1 \int^{1}_{-1}  d\zeta \biggl\{ 
(1-\zeta) [ (2-e^2) - 2\zeta + e^2\zeta^2 ]^{-1/2} -1
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
where, <math>~\zeta\equiv z/a_1</math>.  For later reference, we will identify the expression inside the curly braces as the function, <math>~\mathcal{Z}</math>; specifically,
<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>~(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>~- 1 - \frac{\zeta}{\sqrt{X}} + \frac{1}{\sqrt{X}}  \, ,</math>
  </td>
</tr>
</table>
</div>
where,  in an effort to line up with notation found in integral tables, in this last expression we have used the notation, 
<math>~X \equiv a + b\zeta + c\zeta^2</math> and, in our case,
<div align="center">
<math>a \equiv (2-e^2)\, ,</math> &nbsp; &nbsp; &nbsp;
<math>b \equiv -2\, ,</math> &nbsp; &nbsp; &nbsp; and  &nbsp; &nbsp; &nbsp;
<math>c \equiv e^2\, .</math>
</div>

We find that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\int_{-1}^1 \mathcal{Z} d\zeta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \zeta\biggr|_{-1}^{1} - \biggl\{ \frac{\sqrt{X}}{c} \biggr\}_{-1}^1
+\biggl[1 + \frac{b}{2c} \biggr]\int_{-1}^1 \frac{d\zeta}{\sqrt{X}}
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- 2 + \frac{2}{e^2} 
+\biggl[\frac{e^2-1}{e^3} \biggr] \biggl\{ \ln \biggl[2e^2 - 2  \biggr] 
- \ln \biggl[4e - 2e^2 - 2  \biggr] \biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{1-e^2}{e^3} \biggr] \ln \biggl[\frac{1+e}{1-e} \biggr] -2\biggl[\frac{1-e^2 }{e^2}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A_1 \, .
</math>
  </td>
</tr>
</table>
</div>

Hence, we have,
<div align="center">
<math>~\mathcal{A} = 2\pi G\rho a_1  \biggl[ \int_{-1}^1 \mathcal{Z} d\zeta\biggr]= 2\pi G \rho A_1 a_1 \, ,</math>
</div>
which exactly matches the result [[#FirstDetermination|obtained, above]], by taking the derivative of the potential.