Editing Aps/MaclaurinSpheroidFreeFall (section)

===Case of the Oblate Spheroid===
====Governing, Dimensionless Differential Equations====
We are primarily interested in determining, for various initial values of the parameter pair <math>(e_0, \Omega)</math>, how rapidly an oblate spheroid collapses to an infinitesimally thin disk <math>(e \rightarrow 1)</math>, and what the radius of this disk is at the instant it forms.  As has been detailed above, according to {{ LB62hereafter }} the time-dependent behavior of <math>R</math> and <math>Z</math> &#8212; and, hence, of <math>e</math> &#8212; can be obtained by integrating the following pair of governing relations:
<table border="0" cellpadding="3" align="center">

<tr>
  <td align="right">
<math>\ddot{R}  - \frac{\Omega^2}{R^3}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>- 2A R \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\ddot{Z} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- 2C Z \, .</math>
  </td>
</tr>
</table>

Turning both of these 2<sup>nd</sup>-order ODEs into a pair of 1<sup>st</sup>-order ODEs, then multiplying through by <math>\tau_\mathrm{ff} \equiv [3\pi/(32G\rho_0)]^{1 / 2}</math> or, as appropriate, by <math>\tau_\mathrm{ff}^2</math>, we have,
<table border="0" cellpadding="3" align="center">

<tr>
  <td align="right">
<math>\tau_\mathrm{ff} \cdot \frac{d(\tau_\mathrm{ff}\dot{R})}{dt}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>\frac{3\pi}{32G\rho_0} \biggl[ \frac{\Omega^2}{R^3}- 2A R \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>\frac{3\pi^2}{16} \biggl\{ 
\frac{\Omega^2/(2\pi G \rho_0)}{R^3} - \biggl[\frac{\rho(t)}{\rho_0}  \biggr]\frac{A(t)}{\pi G \rho(t)} \cdot R 
\biggr\} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\tau_\mathrm{ff} \cdot \frac{dR}{dt}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>\tau_\mathrm{ff}\dot{R} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\tau_\mathrm{ff} \cdot \frac{d(\tau_\mathrm{ff}\dot{Z})}{dt}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{3\pi}{32G\rho_0} \biggl[ 2C Z \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{3\pi^2}{16} \biggl\{ \biggl[\frac{\rho(t)}{\rho_0}  \biggr] \frac{C(t) }{\pi G \rho(t)}\cdot Z \biggr\} 
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\tau_\mathrm{ff} \cdot \frac{dZ}{dt}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>\tau_\mathrm{ff}\dot{Z} \, .</math>
  </td>
</tr>
</table>
Adopting the shorthand notation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\sigma_0^2</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\frac{\Omega^2}{2\pi G\rho_0}   \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\tau</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>\frac{t}{\tau_\mathrm{ff}}   \, ,</math>
  </td>
</tr>
</table>
and pulling [[#Table1|expressions for the oblate spheroid from Table 1]] above, we have,
<table border="0" cellpadding="3" align="center">

<tr>
  <td align="right">
<math>\frac{d(\tau_\mathrm{ff}\dot{R})}{d\tau}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>\frac{3\pi^2}{16} \biggl\{ 
\frac{\sigma_0^2}{R^3} - \frac{1}{R Z} \biggl[ 
-\frac{(1-e^2)}{e^2} + 
~\frac{(1-e^2)^{1 / 2}}{ e^3 } \cdot
\sin^{-1}e
\biggr] \biggr\} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{dR}{d\tau}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>\tau_\mathrm{ff}\dot{R} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{d(\tau_\mathrm{ff}\dot{Z})}{d\tau}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{3\pi^2}{16} \biggl\{ \frac{1}{R^2} \biggl[ 
\frac{2}{e^2} - 
~\frac{2(1-e^2)^{1 / 2}}{ e^3 } \cdot \sin^{-1}e \biggr] \biggr\} 
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{dZ}{d\tau}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>\tau_\mathrm{ff}\dot{Z} \, .</math>
  </td>
</tr>
</table>

====Discrete Representations====
Along a uniformly segmented, discrete time grid &#8212; time step, <math>\Delta = (\tau_{n+1}-\tau_{n})</math> &#8212; let's define the following discretized variables:
<ul>
  <li>At integer values of <math>\Delta</math>: &nbsp; <math>R</math> and <math>Z</math></li>
  <li>At half-integer values of <math>\Delta</math>: &nbsp; <math>F \equiv (\tau_\mathrm{ff} \dot{R}) \, ; K \equiv (\tau_\mathrm{ff} \dot{Z})</math></li>
</ul>
The four finite-difference relations are, then:
<table border="0" cellpadding="3" align="center">

<tr>
  <td align="right">
<math>\frac{F_{n+1/2} - F_{n-1/2}}{\Delta}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>\frac{3\pi^2}{16} \biggl\{ 
\frac{\sigma_0^2}{R^3} - \frac{1}{R Z} \biggl[ 
-\frac{(1-e^2)}{e^2} + 
~\frac{(1-e^2)^{1 / 2}}{ e^3 } \cdot
\sin^{-1}e
\biggr] \biggr\}_n \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{R_{n+1} - R_n}{\Delta}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>F_{n+1/2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{K_{n+1/2} - K_{n-1/2}}{\Delta}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{3\pi^2}{16} \biggl\{ \frac{1}{R^2} \biggl[ 
\frac{2}{e^2} - 
~\frac{2(1-e^2)^{1 / 2}}{ e^3 } \cdot \sin^{-1}e \biggr] \biggr\}_n 
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{Z_{n+1} - Z_{n}}{\Delta}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>K_{n+1/2} \, .</math>
  </td>
</tr>
</table>

<font color="red">STEP #0:</font> 
<ul>
  <li>Choose time-independent values of the parameters, <math>(c_0/a_0)</math> and <math>\beta_0</math>, which means that, <math>\sigma_0^2 = 3\pi^2\beta_0/8</math> and <math>e_0 = [1 - (c_0/a_0)^2]^{1 / 2}</math>.  Also set the (uniform) integration time step, <math>\Delta</math>; for 201 time steps, for example, set <math>\Delta \approx 0.005</math>.<br />
The example values shown below (inside parentheses) assume that <math>(c_0/a_0, \beta_0) = (0.90, 0.0) ~~\Rightarrow (e_0, \sigma_0^2) = (0.43588989, 0)</math>, which corresponds to one of the model-evolutions presented by {{ LMS65 }}.</li>
</ul>

<font color="red">STEP #1:</font> Initially, that is, at integration time step, <math>n = 0</math>
<ul>
  <li>Set <math>R_0 = 1</math>, <math>Z_0=1</math>, and <math>e = e_0</math>; this means that the RHS of the first and third discrete evolution equations is known.</li>
  <li>Given that the configuration is collapsing ''from rest'', we want to set <math>\dot{R}_0</math> and <math>\dot{Z}_0</math> to zero. This is accomplished by establishing reflection symmetry through the (time) origin, that is, by setting <math>F_{+1/2}= - F_{-1/2}</math> and <math>K_{+1/2}= - K_{-1/2}</math>.  Initially, then, the LHS of the first and third discrete equations are, respectively, <math>2F_{+1/2}/\Delta</math> and <math>2K_{+1/2}/\Delta</math>.</li>
  <li>Use the first and third discrete equations to solve for the "R" and "Z" velocities at time step <math>n=0</math>, namely,</li>
</ul>
<table border="0" width="80%" cellpadding="3" align="center">

<tr>
  <td align="right">
<math>F_{+1/2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>\frac{\Delta}{2} \cdot \frac{3\pi^2}{16} \biggl\{ 
\sigma_0^2 - \biggl[ 
-\frac{(1-e_0^2)}{e_0^2} + 
~\frac{(1-e_0^2)^{1 / 2}}{ e_0^3 } \cdot
\sin^{-1}e_0
\biggr] \biggr\} \, ,</math>
  </td>
  <td align="right" width="25%">&nbsp;&nbsp;&nbsp; <math>(-2.952452 \times 10^{-3})</math></td>
</tr>

<tr>
  <td align="right">
<math>K_{+1/2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \frac{\Delta}{2} \cdot \frac{3\pi^2}{16} \biggl\{ \biggl[ 
\frac{2}{e_0^2} - 
~\frac{2(1-e_0^2)^{1 / 2}}{ e_0^3 } \cdot \sin^{-1}e_0 \biggr] \biggr\} 
\, .</math>
  </td>
  <td align="right" width="25%">&nbsp;&nbsp;&nbsp; <math>(-3.347849\times 10^{-3})</math></td>
</tr>
</table>
<ul>
  <li>From the second and fourth discrete relations, determine the advanced coordinate positions.</li>
</ul>

<table border="0" width="80%" cellpadding="3" align="center">

<tr>
  <td align="right">
<math>R_{+1} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>1 + \Delta \cdot F_{+1/2} \, ,</math>
  </td>
  <td align="right" width="25%">&nbsp;&nbsp;&nbsp; <math>(1.0-1.476226 \times 10^{-5})</math></td>
</tr>

<tr>
  <td align="right">
<math>Z_{+1}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>1 + \Delta \cdot K_{+1/2} \, .</math>
  </td>
  <td align="right" width="25%">&nbsp;&nbsp;&nbsp; <math>(1.0-1.673924\times 10^{-5})</math></td>
</tr>
</table>
<ul>
  <li>Determine the eccentricity at this advanced time.</li>
</ul>
<table border="0" width="80%" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>e_{+1}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 1 - ( 1 - e_0^2 )\biggl( \frac{Z_{+1}}{R_{+1}} \biggr)^2 \biggr]^{1 / 2} \, .
</math>
  </td>
  <td align="right" width="25%">&nbsp;&nbsp;&nbsp; <math>(1 - 0.809996797)^{1 / 2} = (0.435893568)</math></td>
</tr>
</table>

<font color="red">STEP #2:</font> &nbsp; Repeat, in sequence for all values of <math>n > 1</math> until <math>Z</math> passes through zero.
<ul>
  <li>Set <math>n \rightarrow n+1</math>.</li>
  <li>Given knowledge of the various variable values at time-step, <math>(n-1/2)</math> and <math>n</math>, use the first and third discrete evolution relations to determine <math>F_{n+1/2}</math> and  <math>K_{n+1/2}</math>; specifically,</li>
</ul>
<table border="0" width="80%" cellpadding="3" align="center">

<tr>
  <td align="right">
<math>F_{n+1/2} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>F_{n-1/2} + \Delta \cdot \frac{3\pi^2}{16} \biggl\{ 
\frac{\sigma_0^2}{R^3} - \frac{1}{R Z} \biggl[ 
-\frac{(1-e^2)}{e^2} + 
~\frac{(1-e^2)^{1 / 2}}{ e^3 } \cdot
\sin^{-1}e
\biggr] \biggr\}_n \, ,</math>
  </td>
  <td align="right" width="25%">&nbsp;&nbsp;&nbsp; <math>(F_{+1/2} -5.905085\times 10^{-3})</math></td>
</tr>

<tr>
  <td align="right">
<math>K_{n+1/2} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>K_{n-1/2} 
- \Delta \cdot \frac{3\pi^2}{16} \biggl\{ \frac{1}{R^2} \biggl[ 
\frac{2}{e^2} - 
~\frac{2(1-e^2)^{1 / 2}}{ e^3 } \cdot \sin^{-1}e \biggr] \biggr\}_n 
\, .</math>
  </td>
  <td align="right" width="25%">&nbsp;&nbsp;&nbsp; <math>(K_{+1/2} - 6.695907\times 10^{-3})</math></td>
</tr>
</table>
<ul>
  <li>Similarly, from the second and fourth discrete relations, determine the advanced coordinate positions.</li>
</ul>
<table border="0" width="80%" cellpadding="3" align="center">

<tr>
  <td align="right">
<math>R_{n+1}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>R_n + \Delta\cdot F_{n+1/2} \, ,</math>
  </td>
  <td align="right" width="25%">&nbsp;&nbsp;&nbsp; <math>(0.99994095)</math></td>
</tr>

<tr>
  <td align="right">
<math>Z_{n+1}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align=left">
<math>Z_n + \Delta \cdot K_{n+1/2} \, .</math>
  </td>
  <td align="right" width="25%">&nbsp;&nbsp;&nbsp; <math>(0.999933042)</math></td>
</tr>
</table>
<ul>
  <li>Determine the eccentricity at this advanced time.</li>
</ul>
<table border="0" width="80%" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>e_{n+1}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 1 - ( 1 - e_0^2 )\biggl( \frac{Z_{n+1}}{R_{n+1}} \biggr)^2 \biggr]^{1 / 2} \, .
</math>
  </td>
  <td align="right" width="25%">&nbsp;&nbsp;&nbsp; <math>(1 - 0.809987188)^{1 / 2} = (0.43590459)</math></td>
</tr>
</table>

====Results====

<table border="1" align="center" cellpadding="3">
<tr>
  <td align="center" colspan="6">
Table 2:  Collapse of a Nonrotating, Pressure-Free Oblate Spheroid
  </td>
</tr>
<tr>
  <td align="center"><math>\frac{c_0}{a_0}</math></td>
  <td align="center"><math>e_0</math></td>
  <td align="center"><math>\tau_c</math></td>
  <td align="center"><math>R_c</math></td>
  <td align="center"><math>\dot{R}_c</math></td>
  <td align="center"><math>\dot{Z}_c</math></td>
</tr>
<tr>
  <td align="center"><math>0.99</math></td>
  <td align="center"><math>0.141</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="center"><math>0.95</math></td>
  <td align="center"><math>0.312</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="center"><math>0.90</math></td>
  <td align="center"><math>0.436</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="center"><math>0.80</math></td>
  <td align="center"><math>0.600</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="center"><math>0.70</math></td>
  <td align="center"><math>0.714</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="center"><math>0.60</math></td>
  <td align="center"><math>0.800</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="center"><math>0.10</math></td>
  <td align="center"><math>0.995</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
</table>