Editing SSC/Virial/PolytropesEmbedded/SecondEffortAgain/Pt2 (section)

====Plotting Stahler's Relation====

<span id="MRplot">[[File:CorrectedStahlerN5.png|thumb|300px|Mass (Y) vs. radius (X) plot]]Switching, again, to the shorthand notation,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{X}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{Y}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{M_\mathrm{tot}}{M_\mathrm{SWS}} \, ,</math>
  </td>
</tr>
</table>
</div>
the equilibrium mass-radius relation defined by the first of the two polynomial expressions can be plotted straightforwardly in either of two ways.  

=====Quadratic Equation=====
One way is to recognize that the polynomial is a quadratic equation whose solution is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{Y}_\pm</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{5}{2} \mathcal{X} \biggl\{ 1 \pm \biggl[ 1 - \biggl( \frac{2^4\cdot \pi}{3\cdot 5} \biggr) \mathcal{X}^2 \biggr]^{1/2} \biggr\} \, .</math>
  </td>
</tr>
</table>
</div>
In the figure shown here on the right &#8212; see also the bottom panel of [[SSC/Structure/PolytropesEmbedded#Stahler1983Fig17|Figure 2 in our accompanying discussion of detailed force-balance models]] &#8212; Stahler's mass-radius relation has been plotted using the solution to this quadratic equation; the green segment of the displayed curve was derived from the ''positive'' root while the segment derived from the ''negative'' root is shown in orange.  The two curve segments meet at the maximum value of the normalized equilibrium radius, namely, at
<div align="center">
<math>\mathcal{X}_\mathrm{max} \equiv \biggl[ \frac{3\cdot 5}{2^4 \pi} \biggr]^{1/2} \approx 0.54627 \, .</math>
</div>
We note that, when <math>~\mathcal{X} = \mathcal{X}_\mathrm{max}</math>, <math>~\mathcal{Y} = (5\mathcal{X}_\mathrm{max}/2) \approx 1.36569</math>.  Along the entire sequence, the maximum value of <math>~\mathcal{Y}</math> occurs at the location where <math>~d\mathcal{Y}/d\mathcal{X} = 0</math> along the segment of the curve corresponding to the ''positive'' root.  This occurs along the upper segment of the curve where <math>~\mathcal{X}/\mathcal{X}_\mathrm{max} = \sqrt{3}/2</math>, at the location,
<div align="center">
<math>\mathcal{Y}_\mathrm{max} \equiv \biggl[ \frac{3^3 \cdot 5^2}{2^6 } \biggr]^{1/2} \mathcal{X}_\mathrm{max} 
= \biggl[ \frac{3^4 \cdot 5^3}{2^{10} \pi } \biggr]^{1/2}  \approx 1.77408 \, .</math>
</div>

=====Parametric Relations=====
The other way is to determine the normalized mass and normalized radius individually through Stahler's pair of parametric relations.  Drawing partly from our [[SSC/Virial/PolytropesSummary#Detailed_Force-Balanced_Solution_2|above discussion]] and partly from a separate discussion where we provide a [[SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|tabular summary of the properties of pressure-truncated <math>n=5</math> polytropes]], these are,
<div align="center">
<table border="0" cellpadding="3">

<tr>
  <td align="right">
<math>
~\mathcal{X}\biggr|_{n=5}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{5}{4\pi} \biggr)^{1/2} \tilde\xi \tilde\theta^{2} =
\biggl\{ \frac{3\cdot 5}{2^2 \pi} \biggl[ \frac{\tilde\xi^2/3}{(1+\tilde\xi^2/3)^{2}} \biggr] \biggr\}^{1/2} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
~\mathcal{Y}\biggr|_{n=5}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{5^3}{4\pi} \biggr)^{1/2} \tilde\theta (- \tilde\xi^2 \tilde\theta^') =
\biggl[  \biggl( \frac{3 \cdot 5^3}{2^2\pi} \biggr) \frac{(\tilde\xi^2/3)^3}{(1+\tilde\xi^2/3)^{4}} \biggr]^{1/2} \, .
</math>
  </td>
</tr>
</table>
</div>
The entire sequence will be traversed by varying the Lane-Emden parameter, <math>~\tilde\xi</math>, from zero to infinity.  Using the first of these two expressions, we have determined, for example, that the point along the sequence corresponding to the maximum normalized equilibrium radius, <math>~\mathcal{X}_\mathrm{max}</math>, is associated with an embedded <math>~n=5</math> polytrope whose truncated, dimensionless Lane-Emden radius is,
<div align="center">
<math>
~\tilde\xi \biggr|_{\mathcal{X}_\mathrm{max}} = 3^{1/2} \, .
</math>
</div>
Similarly, we have determined that the point along the sequence that corresponds to the maximum dimensionless mass, <math>~\mathcal{Y}_\mathrm{max}</math>, is associated with an embedded <math>~n=5</math> polytrope whose truncated, dimensionless Lane-Emden radius is, precisely,
<div align="center">
<math>
~\tilde\xi \biggr|_{\mathcal{Y}_\mathrm{max}} = 3 \, .
</math>
</div>

<div id="KimuraApplication">
<table border="1" width="90%" align="center" cellpadding="8">
<tr><td align="left">
Referring back to [[SSC/Structure/PolytropesEmbedded#Turning_Points|our review of turning points]] along equilibrium sequences and, especially, the work of {{ Kimura81bfull }}, we appreciate that the point that corresponds to the maximum mass, <math>~\mathcal{Y}_\mathrm{max}</math>, is the turning point that Kimura refers to as the "extremum in M<sub>1</sub>" along a p<sub>1</sub> sequence.  [[SSC/Structure/PolytropesEmbedded#Location_of_Pressure_Limit|As we have highlighted]], according to Kimura, this point should occur along the sequence where <math>~h_G=0</math>, that is, where the following condition applies:
<div align="center">
<math>~\frac{\tilde\theta^{n+1}}{(\tilde\theta^')^2} = \frac{(n-3)}{2}  \, .</math>
</div>
For the specific case being studied here, namely, <math>~n = 5</math> polytropic configurations, we therefore expect from Kimura's work that <math>~[\tilde\theta^6/(\tilde\theta^')^2] = 1</math> at the "maximum mass" turning point.  Given that,
<div align="center">
<math>~\tilde\xi \biggr|_{\mathcal{Y}_\mathrm{max}} = 3</math>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<math>~\Rightarrow</math>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<math>\tilde\theta_{n=5} = \frac{1}{2}</math>&nbsp;&nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp; <math>\tilde\theta_{n=5}^' = -\frac{1}{8} \, ,</math>
</div>
we see that Kimura's condition holds and, hence, that our identification of the location along the sequence of the maximum mass matches Kimura's identification of the location of that turning point.

We appreciate, as well, that the point corresponding to the maximum normalized equilibrium radius, <math>\mathcal{X}_\mathrm{max}</math>, is the turning point that Kimura would reference as the "extremum in r<sub>1</sub>" along a p<sub>1</sub> sequence.  Following Kimura's analysis [[SSC/Structure/PolytropesEmbedded#TurningPointXmax|we have shown that this point occurs along the sequence where the following condition applies]]:
<div align="center">
<math>~\frac{\xi (-\theta^')}{\tilde\theta} = \frac{2}{(n-1)} \, ,</math>
</div>
that is, for the specific case being studied here, we should expect <math>~[\tilde\xi (-\tilde\theta^')/\tilde\theta] = 1/2</math> at the "maximum radius" turning point.  Given that,
<div align="center">
<math>~\tilde\xi \biggr|_{\mathcal{X}_\mathrm{max}} = 3^{1/2}</math>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<math>~\Rightarrow</math>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<math>\tilde\theta_{n=5} = 2^{-1/2}</math>&nbsp;&nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp; 
<math>\tilde\theta_{n=5}^' = -(2^3 \cdot 3)^{-1/2} \, ,</math>
</div>
we see that Kimura's condition holds and, hence, that our identification of the location along the sequence of the maximum radius matches Kimura's identification of the location of that turning point.
</td></tr>
</table>
</div>

=====Discussion=====
Notice that if the quadratic equation is used to map out the mass-radius relationship, the parameter, <math>~\tilde\xi</math>, never explicitly enters the discussion.  Instead, a radius <math>~0 \le \mathcal{X} \le \mathcal{X}_\mathrm{max}</math> is specified and the ''two'' equilibrium masses associated with <math>~\mathcal{X}</math> &#8212; call them, <math>~\mathcal{Y}_+</math> and <math>~\mathcal{Y}_-</math> &#8212; are determined.  (The values of the two masses are degenerate at both limiting values of <math>~\mathcal{X}</math>.)  If the pair of parametric relations is used, instead, only ''one'' value of the mass is obtained for each specified value of <math>~\tilde\xi</math>.  As <math>~\tilde\xi</math> is increased from <math>~0</math> to <math>~\sqrt{3}</math>, <math>~\mathcal{X}</math> increases monotonically from <math>~0</math> to <math>~\mathcal{X}_\mathrm{max}</math> and the corresponding mass is (only) <math>~\mathcal{Y}_-</math>; that is, as <math>~\tilde\xi</math> is increased from <math>~0</math> to <math>~\sqrt{3}</math>, we move away from the origin in a counter-clockwise direction along the lower segment (colored orange in the [[#MRplot|above figure]]) of the plotted equlibrium sequence.  Then, as <math>~\tilde\xi</math> is increased from <math>~\sqrt{3}</math> to <math>~\infty</math>, we continue to move in a counter-clockwise direction along the equilibrium sequence, but now along the upper segment (colored green in the [[#MRplot|above figure]]) of the sequence, back to the origin; that is to say, <math>~\mathcal{X}</math> steadily decreases from <math>~\mathcal{X}_\mathrm{max}</math> back to <math>~0</math> and this time the relevant associated mass is the positive root of the quadratic relation, <math>~\mathcal{Y}_+</math>. 

Clearly, then, each value of <math>~\mathcal{X}</math> is associated with two different values of the parametric parameter, <math>~\tilde\xi</math>.  By inverting the <math>~\mathcal{X}(\tilde\xi)</math> parametric expression we see that, the two values of <math>~\tilde\xi</math> associated with a given equilibrium radius are,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tilde\xi_\pm</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{\frac{3}{\alpha} \biggl[ 1 \pm \sqrt{1 - \alpha^2} \biggr]  \biggr\}^{1/2} \, ,</math>
  </td>
</tr>
</table>
</div>

where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\alpha</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{(\mathcal{X}/\mathcal{X}_\mathrm{xmax})^2}{2-(\mathcal{X}/\mathcal{X}_\mathrm{xmax})^2} \, .</math>
  </td>
</tr>
</table>
</div>
We note as well that, for a given equilibrium radius, <math>~\mathcal{X}</math>, the ''ratio'' of the two mass solutions is given by a very simple expression, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\mathcal{Y}_-}{\mathcal{Y}_+} = \frac{\tilde\xi_-^2}{3}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; or &nbsp; &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{\mathcal{Y}_+}{\mathcal{Y}_-} = \frac{\tilde\xi_+^2}{3} \, .</math>
  </td>
</tr>
</table>
</div>
This implies, as well, that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tilde\xi_+ \cdot \tilde\xi_-</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\, .</math>
  </td>
</tr>
</table>
</div>