Editing Apps/MaclaurinSpheroidSequence (section)

==Oblate Spheroidal Coordinates==
Following the lead of {{ Bardeen71 }}, {{ HE83 }}, and {{ HE84 }} &#8212; also see the succinct summary that is provided in Appendix A (pp. ) of {{ HTE87 }} &#8212; let's shift to oblate-spheroidal coordinates <math>(\xi, \eta, \phi)</math> which are related to Cartesian coordinates via the relations,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>x</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_0\biggl[(1+\xi^2)(1 - \eta^2) \biggr]^{1 / 2} \cos\phi \, ,</math></td>
</tr>
<tr>
  <td align="right"><math>y</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_0\biggl[(1+\xi^2)(1 - \eta^2) \biggr]^{1 / 2} \sin\phi \, ,</math></td>
</tr>
<tr>
  <td align="right"><math>z</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_0\xi\eta \, .</math></td>
</tr>
</table>
For axisymmetric configurations, such as Maclaurin spheroids, we also appreciate that,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\varpi \equiv (x^2 + y^2)^{1 / 2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_0 \biggl[(1+\xi^2)(1 - \eta^2) \biggr]^{1 / 2}\, .</math></td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Bardeen71 }}, §IV, p. 429, Eq. (12)<br />
{{ HE83 }}, &sect;A.1, p. 587, Eq. (1)
  </td>
</tr>
</table>
In this coordinate system, the surface of the Maclaurin spheroid is marked by a specific value of the coordinate, <math>\xi</math> &#8212; call it, <math>\xi_s</math> &#8212; and points along the surface (in any meridional plane) are identified by varying <math>\eta</math> from zero (equatorial plane) to unity (the pole).  Given that the eccentricity of the spheroid is <math>e = [1 - c^2/a^2]^{1 / 2}</math>, we understand that,

<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>a</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_0 (1+\xi_s^2)^{1 / 2}\, ,</math></td>
</tr>
<tr>
  <td align="right"><math>c</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_0 \xi_s\, ,</math></td>
</tr>
<tr>
  <td align="right"><math>\Rightarrow ~~~ e^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
1 - (a_0 \xi_s)^2 \biggl[a_0^2 (1+\xi_s^2)\biggr]^{-1} 
=
1 - \frac{\xi_s^2}{ (1+\xi_s^2)}
=
\frac{1}{ (1+\xi_s^2)}
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ Bardeen71 }}, §IV, p. 429, Eq. (14)
  </td>
</tr>
<tr>
  <td align="right"><math>\Rightarrow ~~~ \xi_s^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{ e^2} - 1
\, .</math>
  </td>
</tr>
</table>
Also, in order for the volume of the spheroid to remain constant &#8212; and equal to that of a sphere of the same total mass and density &#8212; along the sequence of spheroids we understand that,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right"><math>\frac{M}{\rho} = \frac{4\pi a^2c}{3}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{4\pi}{3} a^3 \biggl(\frac{c}{a}\biggr)
=
\frac{4\pi}{3} a^3 \biggl[1 - e^2\biggr]^{1 / 2}
</math>
  </td>
</tr>
<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{3M}{4\pi\rho} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
a_0^3 (1+\xi_s^2)^{3 / 2} \biggl\{
\xi_s^2 \biggl[(1+\xi_s^2)\biggr]^{-1}
\biggr\}^{1 / 2}
</math>
  </td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
a_0^3 \xi_s (1+\xi_s^2) 
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HE83 }}, &sect;A.2, p. 588, Eq. (10)<br />
{{ HTE87 }}, p. 610, Eq. (A5)
  </td>
</tr>
<tr>
  <td align="right"><math>\Rightarrow ~~~ a_0^3 </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{3M}{4\pi\rho}\biggr) \biggl[ \xi_s (1+\xi_s^2) \biggr]^{-1}
=
\biggl(\frac{3M}{4\pi\rho}\biggr) \frac{e^3}{(1-e^2)^{1 / 2}} \, .
</math>
  </td>
</tr>
</table>


<table border="1" align="center" width="90%" cellpadding="8">
<tr><td align="left">

From Appendix A of {{ HTE87 }} &#8212; hereafter {{ HTE87hereafter }} &#8212; we also appreciate that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\Omega^2 \equiv \frac{\omega_0^2}{4\pi G \rho}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\xi q_2(\xi) \, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HE83 }}, &sect;A.2, p. 588, Eq. (9)<br />
{{ HTE87hereafter }}, p. 610, Eq. (A4)
  </td>
</tr>

<tr>
  <td align="right"><math>L</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{8\pi}{15}\biggr) \rho \omega_0 a_0^5 \xi(1+\xi^2)^2 \, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HE83 }}, &sect;A.2, p. 588, Eq. (11)<br />
{{ HTE87hereafter }}, p. 610, Eq. (A6)
  </td>
</tr>

<tr>
  <td align="right"><math>T_\mathrm{rot}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{4\pi}{15}\biggr) \rho \omega_0^2 a_0^5 \xi(1+\xi^2)^2 \, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HTE87hereafter }}, p. 610, Eq. (A7)
  </td>
</tr>

<tr>
  <td align="right"><math>W_\mathrm{grav}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \biggl(\frac{16\pi^2}{15}\biggr) G \rho^2 a_0^5 \xi^2 (1+\xi^2)^2 q_0(\xi)\, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HTE87hereafter }}, p. 610, Eq. (A8)
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{T_\mathrm{rot}}{|W_\mathrm{grav}|}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{4\pi}{15}\biggr) \rho \omega_0^2 a_0^5 \xi(1+\xi^2)^2
\biggl[ \biggl(\frac{16\pi^2}{15}\biggr) G \rho^2 a_0^5 \xi^2 (1+\xi^2)^2 q_0(\xi) \biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[ \frac{\omega_0^2}{4\pi G \rho}\biggr] \frac{1}{\xi q_0(\xi)}
=
\frac{q_2(\xi)}{q_0(\xi)} \, ,
</math>
  </td>
</tr>
</table>
where &#8212; see Eqs. (A15) - (A17) of {{ HTE87hereafter }} and Appendix A (p. 443) of {{ Bardeen71hereafter }}&#8212; the first three spheroidal wave functions of the second kind are,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>q_0(\xi)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\tan^{-1}(1/\xi) \, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HTE87hereafter }}, p. 610, Eq. (A15)
  </td>
</tr>

<tr>
  <td align="right"><math>q_1(\xi)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-\xi \tan^{-1}(1/\xi) + 1\, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HTE87hereafter }}, p. 610, Eq. (A16)
  </td>
</tr>

<tr>
  <td align="right"><math>q_2(\xi)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{2}\biggl[(3\xi^2 + 1)\tan^{-1}(1/\xi) - 3\xi\biggr] 
\, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ HTE87hereafter }}, p. 610, Eq. (A17)
  </td>
</tr>
</table>

<hr />

'''Check:'''  &nbsp;Given that, <math>\xi^2 = (1-e^2)/e^2</math>, we have,

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\tan^{-1}(1/\xi)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\sin^{-1}\biggl[ \frac{1}{\sqrt{\xi^2+1}} \biggr] = \sin^{-1}e\, ,
</math>
  </td>
</tr>
</table>
in which case:

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\frac{\omega_0^2}{2\pi G\rho}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{(1 - e^2)^{1 / 2}}{e}\biggl\{ \biggl[\frac{3(1-e^2)}{e^2} + 1\biggr]\sin^{-1}e - \frac{3(1-e^2)^{1 / 2}}{e} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[(3-2e^2)(1 - e^2)^{1 / 2} \biggr]\cdot \frac{\sin^{-1}e}{e^3} - \frac{3(1-e^2)}{e^2} \, ;
</math>
&nbsp; &nbsp; &nbsp; &nbsp; [[Apps/MaclaurinSpheroidSequence#MaclaurinFrequency|(matches here)]]
  </td>
</tr>

<tr>
  <td align="right">
<math>
L_*^2  \equiv \biggl(\frac{4\pi}{3}\biggr)^{1 / 3}\frac{L^2}{G M^{10/3} \rho^{-1 / 3}}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{4\pi}{3}\biggr)^{1 / 3} G^{-1} M^{-10/3}\rho^{1 / 3}
\biggl(\frac{2^3\pi}{3 \cdot 5}\biggr)^2 \rho^2 a_0^{10} \omega_0^2 
\biggl[ \frac{(1-e^2)}{e^{10}}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{4\pi}{3}\biggr)^{1 / 3} G^{-1} M^{-10/3}\rho^{1 / 3}
\biggl(\frac{2^3\pi}{3 \cdot 5}\biggr)^2 \rho^2 
\biggl[ \frac{(1-e^2)}{e^{10}}\biggr]
\biggl[ \biggl(\frac{3M}{4\pi\rho}\biggr)\frac{e^3}{(1-e^2)^{1 / 2}}\biggr]^{10/3} \omega_0^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{6}{25}  
\biggl[\frac{\omega_0^2}{2\pi G\rho}\biggr] (1-e^2)^{- 2/3}\, ;
</math>
&nbsp; &nbsp; &nbsp; &nbsp; [[Apps/MaclaurinToroid#Maclaurin_Spheroid_Reminder|(matches here)]]
  </td>
</tr>

<tr>
  <td align="right"><math>T_\mathrm{rot}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{4\pi}{15}\biggr) 2\pi G \rho^2 a_0^5 \biggl[ \frac{\omega_0^2}{2\pi G \rho} \biggr] \xi(1+\xi^2)^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{2^3\pi^2}{3 \cdot 5}\biggr) G \rho^2 
(a \cdot e)^5
\biggl[ \frac{\omega_0^2}{2\pi G \rho} \biggr] 
\biggl[\frac{(1-e^2)^{1 / 2}}{e^5}\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{2^3\pi^2}{3 \cdot 5}\biggr) G \rho^2 a^5
\biggl[ \frac{\omega_0^2}{2\pi G \rho} \biggr] 
(1-e^2)^{1 / 2} \, ;  
</math>
&nbsp; &nbsp; &nbsp; &nbsp; [[Apps/MaclaurinSpheroidSequence#Alternate_Sequence_Diagrams|(matches here)]]
  </td>
</tr>

<tr>
  <td align="right"><math>W_\mathrm{grav}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \biggl(\frac{16\pi^2}{15}\biggr) G \rho^2 (a \cdot e)^5 \biggl[ \frac{(1-e^2)}{e^6} \cdot \sin^{-1}e\biggr]
</math>
  </td>
</tr>
<tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \biggl(\frac{16\pi^2}{15}\biggr) G \rho^2 a^5 (1-e^2)\cdot \frac{\sin^{-1}e}{e} \, ;
</math>
&nbsp; &nbsp; &nbsp; &nbsp; [[Apps/MaclaurinSpheroidSequence#Alternate_Sequence_Diagrams|(matches here)]]
  </td>
</tr>
<tr>

<tr>
  <td align="right"><math>\frac{T_\mathrm{rot}}{|W_\mathrm{grav}|} = \frac{q_2(\xi)}{q_0(\xi)}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{2\tan^{-1}(1/\xi)}\biggl[(3\xi^2 + 1)\tan^{-1}(1/\xi) - 3\xi\biggr] 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{2e^2\sin^{-1}e}\biggl[(3-2e^2)\sin^{-1}e - 3e(1-e^2)^{1 / 2}\biggr] \, .</math>
&nbsp; &nbsp; &nbsp; &nbsp; [[Apps/MaclaurinSpheroidSequence#tau|(matches here)]]
  </td>
</tr>
</table>

</td></tr>
</table>