Editing SSC/Virial/PolytropesSummary (section)

===Relating and Reconciling Two Mass-Radius Relationships for n = 5 Polytropes===

Now, let's examine the case of pressure-truncated, <math>~n=5</math> polytropes.  As we have discussed in the context of [[User:Tohline/SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|detailed force-balanced models]], [http://adsabs.harvard.edu/abs/1983ApJ...268..165S Stahler (1983)] has deduced that all <math>~n=5</math> equilibrium configurations obey the mass-radius relationship,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{M}{M_\mathrm{SWS}} \biggr)^2 - 5 \biggl( \frac{M}{M_\mathrm{SWS}} \biggr)\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr)
+ \frac{2^2 \cdot 5 \pi}{3} \biggl( \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr)^4 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, ,</math>
  </td>
</tr>
</table>
</div>
where, as [[User:Tohline/SSC/Virial/PolytropesSummary#Detailed_Force-Balanced_Solution_2|reviewed above]], the mass and radius normalizations, <math>~M_\mathrm{SWS}</math> and <math>~R_\mathrm{SWS}</math>, may be treated as constants once the parameters <math>~K</math> and <math>~P_e</math> are specified.  In contrast to this, the mass-radius relationship that we have just derived ''from the virial theorem'' for pressure-truncated, <math>~n=5</math> polytropes is,
<div align="center">
<math>
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{mod}} \biggr)^2
- \biggl( \frac{R_\mathrm{eq}}{R_\mathrm{mod}} \biggr)^{2/5} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{mod}} \biggr)^{6/5} 
+ \biggl( \frac{R_\mathrm{eq}}{R_\mathrm{mod}} \biggr)^4 = 0 \, ,
</math>
</div>
where the mass and radius normalizations,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_\mathrm{mod}\biggr|_{n=5}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~M_\mathrm{SWS} \biggl( \frac{3\mathcal{B}}{4\pi} \biggr)^{5/3} \biggl[ \frac{2\cdot 5\pi}{3^2 \mathcal{A}} \biggr]^{3/2} 
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~R_\mathrm{mod}\biggr|_{n=5}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~R_\mathrm{SWS} \biggl( \frac{3\mathcal{B}}{4\pi}\biggr)^{5/6} \biggl[ \frac{2\cdot 5\pi}{3^2\mathcal{A}} \biggr]^{1/2} \, ,</math>
  </td>
</tr>
</table>

depend, not only on <math>~K</math> and <math>~P_e</math> via the definitions of <math>~M_\mathrm{SWS}</math> and <math>~R_\mathrm{SWS}</math>, but also on the structural form factors via the free-energy coefficients, <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math>.  While these two separate mass-radius relationships are similar, they are not identical.  In particular, the middle term involving the cross-product of the mass and radius contains different exponents in the two expressions.  It is not immediately obvious how the two different polynomial expressions can be used to describe the same physical sequence.

This apparent discrepancy is reconciled as follows:  The structural form factors &#8212; and, hence, the free-energy coefficients &#8212; vary from equilibrium configuration to equilibrium configuration.  So it does not make sense to discuss ''evolution along the sequence'' that is defined by the second of the two polynomial expressions.  If you want to know how a given system's equilibrium radius will change ''as its mass changes'', the first of the two polynomials will do the trick.  However, the equilibrium radius of ''a given system'' can be found by looking for extrema in the free-energy function while holding the free-energy coefficients, <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math>, constant; more importantly, the relative stability ''of a given equilibrium system'' can be determined by analyzing the behavior of the system's free energy ''while holding the free-energy coefficients constant''.  Dynamically stable versus dynamically unstable configurations can be readily distinguished from one another along the sequence that is defined by the second polynomial expression; they cannot be readily distinguished from one another along the sequence that is defined by the first polynomial expression.  It is useful, therefore, to determine how to map a configuration's position on one of the sequences to the other. 

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

[[File:CorrectedStahlerN5.png|thumb|300px|Pressure vs. pressure plot]]Switching, again, to the shorthand notation,
<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.  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}</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 [[User:Tohline/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>


The other way is to determine the normalized mass and normalized radius individually through Stahler's pair of parametric relations.  Drawing partly from our [[User:Tohline/SSC/Virial/PolytropesSummary#Detailed_Force-Balanced_Solution_2|above discussion]] and partly from a separate discussion where we provide a [[User:Tohline/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}} = \frac{1}{5^{1/2}} \biggl[ 2^5\pi - 15 + 2^3\pi^{1/2}(2^4\pi-15)^{1/2} \biggr]^{1/2} 
\approx 5.8264 \, .
</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>

====Plotting the Virial Theorem Relation====

The relevant relation is obtained by plugging <math>~n = 5</math> into the [[User:Tohline/SSC/Virial/PolytropesSummary#ConciseVirialXY|general mass-radius relation derived above]], repeated here for clarity:

<table border="1" cellpadding="8" align="center">
<tr><td>

<div align="center">
<math>
\mathcal{X}^4 \biggl[\frac{4\pi (5-n)}{3} \biggr] - \mathcal{X}^{(n-3)/n} \mathcal{Y}^{(n+1)/n}  (4\pi)^{-1/n} \mathfrak{b}_n + \mathcal{Y}^2 \biggl(\frac{n+1}{n}\biggr) 
= 0 
</math>
</div>
where,
<div align="center">
<math>\mathfrak{b}_n = \biggl[ (n+1) (-\tilde\theta^')^2 + \biggl( \frac{5-n}{3} \biggr)\tilde\theta^{n+1} \biggr]
\biggl( \frac{\tilde\xi}{-\tilde\theta^'} \biggr)^{(n+1)/n} 
</math>
</div>

</td></tr>
</table>
We will begin by plugging <math>~n = 5</math> into these expressions everywhere except for the coefficient <math>~(5-n)</math>, which we will leave unresolved, for the time being, in order to better appreciate the interplay of various terms.  We obtain,
<div align="center">
<table border="0" cellpadding="5">

<tr>
  <td align="right">
<math>~
\frac{6}{5} \mathcal{Y}^2 - \biggl( \frac{\mathfrak{b}_I^5 }{4\pi} \cdot \mathcal{X}^{2} \mathcal{Y}^{6}\biggr)^{1/5}  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(n-5)\biggl[ \biggl(\frac{4\pi}{3} \biggr) \mathcal{X}^4 -  \biggl( \frac{\mathfrak{b}_{II}^5}{4\pi} \cdot \mathcal{X}^{2} \mathcal{Y}^{6}\biggr)^{1/5} \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
where, if they are to be assigned values that are actually associated with a particular detailed force-balance model having truncation radius, <math>~\tilde\xi</math>,
<div align="center">
<table border="0" cellpadding="5">

<tr>
  <td align="right">
<math>~
\mathfrak{b}_I^5  
</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl[ (n+1) (-\tilde\theta^')^2 \biggl( \frac{\tilde\xi}{-\tilde\theta^'} \biggr)^{(n+1)/n} \biggr]^5_{n=5} 
= 2^5\cdot 3^5 \biggl[ (-\tilde\theta^')^4 {\tilde\xi}^6 \biggr]  
= 2^5 \cdot 3 \cdot \tilde\xi^{10} (1+\tilde\xi^2/3)^{-6}\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~
\mathfrak{b}_{II}^5
</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{1}{3} ~ \tilde\theta^{n+1}  \biggl( \frac{\tilde\xi}{-\tilde\theta^'} \biggr)^{(n+1)/n} \biggr]^5_{n=5}
= \frac{1}{3^5} ~ \tilde\theta^{30}  \biggl( \frac{\tilde\xi}{-\tilde\theta^'} \biggr)^{6} 
= 3(1+\tilde\xi^2/3)^{-6} \, .
</math>
  </td>
</tr>
</table>
</div>

Now, if we plug <math>~n=5</math> into the remaining unresolved <math>~(n-5)</math> coefficient, the righthand side goes to zero and the mass-radius relationship provided by the virial theorem becomes,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~\frac{6}{5}  \mathcal{Y}^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\mathfrak{b}_I^5 }{4\pi} \cdot \mathcal{X}^{2} \mathcal{Y}^{6}\biggr)^{1/5}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \mathcal{Y}^4 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\mathcal{X}^{2} \biggl[ \biggl( \frac{5^5}{2^5\cdot 3^5}\biggr) \frac{\mathfrak{b}^5_{I}}{4\pi} \biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \mathcal{Y} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\mathcal{X}^{1/2} \biggl( \frac{5^5 \mathfrak{b}^5_{I}}{2^7\cdot 3^5 \pi}\biggr)^{1/4} \, .  </math>
  </td>
</tr>
</table>
</div>
Hence, for a given value of the structural form factor(s) &#8212; which implies a specific value of the constant coefficient, <math>~\mathfrak{b}_I</math> &#8212; the scalar virial theorem defines a relationship where the normalized mass <math>~(\mathcal{Y})</math> varies as the square root of the normalized radius <math>~(\mathcal{X})</math>.  On the other hand, if we demand that the expression inside the square brackets on the righthand side of the virial theorem relation go to zero on its own &#8212; without relying on the leading coefficient to knock it zero &#8212; the mass-radius relationship provided by the virial theorem becomes,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~\frac{4\pi}{3} ~ \mathcal{X}^4  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\mathfrak{b}_{II}^5}{4\pi} \cdot \mathcal{X}^{2} \mathcal{Y}^{6}\biggr)^{1/5} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \mathcal{Y}^6 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\mathcal{X}^{18} \biggl[ \biggl( \frac{4\pi}{3}\biggr)^5 \frac{4\pi}{\mathfrak{b}^5_{II}} \biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \mathcal{Y} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>4\pi \mathcal{X}^{3} ( 3^5\mathfrak{b}^5_{II} )^{-1/6} \, .  </math>
  </td>
</tr>
</table>
</div>
<!-- COMMENT 1 .......

In order to graphically display the mass-radius relation derived from the virial theorem in the <math>~\mathcal{X}-\mathcal{Y}</math> plane, as desired, we must first write out the expressions for the free-energy coefficients.  After setting <math>~M_\mathrm{limit}/M_\mathrm{tot} = 1</math> in the [[User:Tohline/SSC/Virial/PolytropesSummary#Structural_Form_Factors|above summary expressions]], we obtain for all polytropic indexes,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\frac{1}{5} \cdot \frac{\tilde\mathfrak{f}_W}{\tilde\mathfrak{f}_M^2}  = \frac{1}{5-n} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde\mathfrak{f}_A}{\tilde\mathfrak{f}_M^{(n+1)/n}} =
\frac{1}{3(5-n) ( 4\pi )^{1/n}} \biggl[ 3(n+1) (\tilde\theta^')^2 + (5-n)\tilde\theta^{n+1} \biggr]
\biggl( \frac{\tilde\xi}{\tilde\theta^'} \biggr)^{(n+1)/n} \, .
</math>
  </td>
</tr>
</table>
</div>
In an effort not to be caught dividing by zero while investigating the specific case of <math>~n=5</math> polytropes, we will use as shorthand notation,
<div align="center">
<math>\mathfrak{b}_5 \equiv \biggl[ (5-n)\mathcal{B}\biggr]_{n=5} 
= \biggl[ \frac{2^3\cdot 3^5}{ \pi} \cdot (-\tilde\theta^')^{4} \tilde\xi^6 \biggr]^{1/5}
= \biggl[ \frac{2^3 \cdot 3}{\pi} \biggl(1 + \frac{\tilde\xi^2}{3} \biggr)^{-6} \tilde\xi^{10} \biggr]^{1/5}
\, ,</math>
</div>
where we have inserted the definition of <math>~\tilde\theta^'</math> as provided for <math>~n=5</math> polytropic structures in another section's [[User:Tohline/SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|summary table]].  For pressure-truncated <math>~n=5</math> polytropes, we therefore have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{R_\mathrm{eq}}{R_\mathrm{mod}} = \mathcal{X} \cdot \frac{R_\mathrm{SWS}}{R_\mathrm{mod}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\mathcal{X} \cdot \biggl( \frac{4\pi}{3\mathcal{B}}\biggr)^{5/6} \biggl[ \frac{3^2\mathcal{A}}{2\cdot 5\pi} \biggr]^{1/2}
= (5-n)^{1/3} \mathcal{X} \cdot \biggl( \frac{4\pi}{3\mathfrak{b}_5}\biggr)^{5/6} \biggl[ \frac{3^2}{2\cdot 5\pi} \biggr]^{1/2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
(5-n)^{1/3} \mathcal{X} \cdot \biggl( \frac{2^{7}\cdot 3\pi^2}{5^3} \biggr)^{1/6} \mathfrak{b}_5^{-5/6} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{M_\mathrm{tot}}{M_\mathrm{mod}} = \mathcal{Y} \cdot \frac{M_\mathrm{SWS}}{M_\mathrm{mod}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~ \mathcal{Y} \cdot \biggl( \frac{4\pi}{3\mathcal{B}} \biggr)^{5/3} \biggl[ \frac{3^2 \mathcal{A}}{2\cdot 5\pi} \biggr]^{3/2}
= (5-n)^{1/6}\mathcal{Y} \cdot \biggl( \frac{4\pi}{3\mathfrak{b}_5} \biggr)^{5/3} \biggl[ \frac{3^2}{2\cdot 5\pi} \biggr]^{3/2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
(5-n)^{1/6}\mathcal{Y} \cdot \biggl( \frac{2^{11}\cdot 3^8 \pi}{5^9} \biggr)^{1/6}  \mathfrak{b}_5^{-5/3} \, .
</math>
  </td>
</tr>
</table>
</div>

So, the cross term that appears in the mass-radius relation obtained from the virial theorem may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl(\frac{R_\mathrm{eq}}{R_\mathrm{mod}}\biggr)^{2/5}\biggl(\frac{M_\mathrm{tot}}{M_\mathrm{mod}}\biggr)^{6/5} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{\biggl[  
(5-n)^{1/3} \mathcal{X} \cdot \biggl( \frac{2^{7}\cdot 3\pi^2}{5^3} \biggr)^{1/6} \mathfrak{b}_5^{-5/6} 
\biggr]^2 \times \biggl[
(5-n)^{1/6}\mathcal{Y} \cdot \biggl( \frac{2^{11}\cdot 3^8 \pi}{5^9} \biggr)^{1/6}  \mathfrak{b}_5^{-5/3} 
\biggr]^6 \biggr\}^{1/5}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(5-n)^{1/3} \mathcal{X}^{2/5} \mathcal{Y}^{6/5}
\biggl[ \biggl( \frac{2^{7}\cdot 3\pi^2}{5^3} \biggr)^{1/3} \mathfrak{b}_5^{-5/3} \cdot
\biggl( \frac{2^{11}\cdot 3^8 \pi}{5^9} \biggr)  \mathfrak{b}_5^{-10}
\biggr]^{1/5}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(5-n)^{1/3} \mathcal{X}^{2/5} \mathcal{Y}^{6/5}
\biggl( \frac{2^{8}\cdot 3^{5} \pi}{5^{6}} \biggr)^{1/3} 
\mathfrak{b}_5^{-7/3} \, ,
</math>
  </td>
</tr>
</table>
</div>
and, in entirety, the virial theorem relation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(5-n)^{1/3}\biggl\{ \mathcal{Y}^2 \cdot \biggl( \frac{2^{11}\cdot 3^8 \pi}{5^9} \biggr)^{1/3}  \mathfrak{b}_5^{-10/3}
- \mathcal{X}^{2/5} \mathcal{Y}^{6/5} \biggl( \frac{2^{8}\cdot 3^{5} \pi}{5^{6}} \biggr)^{1/3} \mathfrak{b}_5^{-7/3}
+ (5-n) \mathcal{X}^4 \cdot \biggl( \frac{2^{7}\cdot 3\pi^2}{5^3} \biggr)^{2/3} \mathfrak{b}_5^{-10/3} \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

A nontrivial solution is obtained by requiring that the terms inside the curly braces sum to zero.  Noting that the third term must be set to zero, on its own, because it retains a leading factor of <math>~(5-n)</math>, the virial theorem relation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{Y}^2 \cdot \biggl( \frac{2^{11}\cdot 3^8 \pi}{5^9} \biggr)^{1/3}  \mathfrak{b}_5^{-10/3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathcal{X}^{2/5} \mathcal{Y}^{6/5} \biggl( \frac{2^{8}\cdot 3^{5} \pi}{5^{6}} \biggr)^{1/3} \mathfrak{b}_5^{-7/3}
</math>
  </td>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Rightarrow~~~~
\mathcal{Y}^{2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathcal{X} \biggl( \frac{5\mathfrak{b}_5}{2 \cdot 3}  \biggr)^{5/2} 
= \mathcal{X} \biggl( \frac{5^5}{2^2 \cdot 3^4 \pi} \biggr)^{1/2} \biggl(1 + \frac{\tilde\xi^2}{3} \biggr)^{-3} \tilde\xi^{5}  \, .
</math>
  </td>
</tr>
</table>
</div>

.......   END COMMENT -->
<!-- COMMENT 2 ........
Let's go back to the [[User:Tohline/SSC/Virial/PolytropesSummary#CompactStahlerVirial|earlier virial expression]] that still contains the <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math> coefficients and multiply through by <math>~5</math> so that the coefficient of the <math>~\mathcal{X}^4</math> term matches the coefficient found in Stahler's relation.  We have,

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

<tr>
  <td align="right">
<math>~
\biggl(\frac{2^2\cdot 5\pi}{3} \biggr) \mathcal{X}^4  - 
5\mathcal{B} \mathcal{X}^{2/5} \mathcal{Y}^{6/5}  
+ 6\mathcal{A} \mathcal{Y}^{2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .
</math>
  </td>
</tr>
</table>
</div>
Next, let's rewrite the other two terms so that they look more like the terms found in Stahler's expression.

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

<tr>
  <td align="right">
<math>~
\biggl(\frac{2^2\cdot 5\pi}{3} \biggr) \mathcal{X}^4  - 
5\mathcal{X}\mathcal{Y} \biggl( \frac{\mathcal{B}^5 \mathcal{Y}}{\mathcal{X}^3} \biggr)^{1/5}
+ 6\mathcal{A} \mathcal{Y}^{2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\biggl(\frac{2^2\cdot 5\pi}{3} \biggr) \mathcal{X}^4  - 
5\mathcal{X}\mathcal{Y} 
+ \mathcal{Y}^{2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
5\mathcal{X}\mathcal{Y} \biggl[\biggl( \frac{\mathcal{B}^5 \mathcal{Y}}{\mathcal{X}^3} \biggr)^{1/5}-1\biggr]
+(1- 6\mathcal{A} )\mathcal{Y}^{2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \mathcal{Y} \biggl\{
5\mathcal{X}\biggl[\biggl( \frac{\mathcal{B}^5 \mathcal{Y}}{\mathcal{X}^3} \biggr)^{1/5}-1\biggr]
+(1- 6\mathcal{A} )\mathcal{Y} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \mathcal{Y} \biggl\{
5 \mathcal{B} (\mathcal{Y} \mathcal{X}^2 )^{1/5} - 5 \mathcal{X} 
+(1- 6\mathcal{A} )\mathcal{Y} \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
Now, according to Stahler's relation, the lefthand side of our derived expression should be zero when the chosen <math>~(\mathcal{X}, \mathcal{Y})</math> pair identifies an equilibrium configuration.  Therefore, the terms inside the curly brackets on the righthand side of our derived expression should also sum to zero in equilibrium.  Let's see if, indeed, this is the case; as shorthand, we will use,
<div align="center">
<math>\Lambda \equiv \frac{\tilde{\xi}^2}{3} \, .</math>
</div>
From the [[User:Tohline/SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|tabular summary of detailed force-balanced models]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tilde\theta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{(1+\Lambda)^{1/2}} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~-~\tilde\theta'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{3^{1/2}} \cdot
\frac{\Lambda^{1/2}}{(1+\Lambda)^{3/2}}  
~~~~\Rightarrow~~~~ \biggl( \frac{\tilde{\xi}}{-\tilde\theta'} \biggr) = 3(1+\Lambda)^{3/2} \, ,
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~\mathcal{Y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ \frac{3\cdot 5^3}{2^2\pi} \biggl[ 
\frac{\Lambda^3}{(1+\Lambda)^4} \biggr] \biggr\}^{1/2}
=
5\biggl(\frac{3\cdot 5}{2^2\pi}\biggr)^{1/2} 
\frac{\Lambda^{3/2}}{(1+\Lambda)^2}  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \mathcal{Y}\mathcal{X}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
5\biggl(\frac{3\cdot 5}{2^2\pi}\biggr)^{3/2} 
\frac{\Lambda^{5/2}}{(1+\Lambda)^4} \, ;
</math>
  </td>
</tr>
</table>
</div>
and, from the general definitions given, above,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\frac{1}{5-n} \, ,</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
( 4\pi )^{-1/5} \biggl[ \frac{2\cdot 3}{(5-n)} (\tilde\theta^')^2 + \frac{1}{3}\cdot \tilde\theta^{6} \biggr]
\biggl( \frac{\tilde\xi}{\tilde\theta^'} \biggr)^{6/5} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
( 4\pi )^{-1/5} \biggl[ \frac{2}{(5-n)} \frac{\Lambda}{(1+\Lambda)^3} + \frac{1}{3}\cdot \frac{1}{(1+\Lambda)^3} \biggr]
3^{6/5} (1+\Lambda)^{9/5} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{3}{4\pi }\biggr)^{1/5} \biggl[ \frac{2\cdot 3}{(5-n)} \Lambda + 1 \biggr]
(1+\Lambda)^{-6/5} \, .
</math>
  </td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~5\mathcal{B} (\mathcal{Y}\mathcal{X}^2)^{1/5}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~5
\biggl(\frac{3}{4\pi }\biggr)^{1/5} \biggl[ \frac{2\cdot 3}{(5-n)} \Lambda + 1 \biggr](1+\Lambda)^{-6/5}
\cdot
\biggl\{ 5\biggl(\frac{3\cdot 5}{2^2\pi}\biggr)^{3/2} 
\frac{\Lambda^{5/2}}{(1+\Lambda)^4} \biggr\}^{1/5}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3\cdot 5^3}{4\pi }\biggr)^{1/2}  \biggl[ \frac{2\cdot 3}{(5-n)} \Lambda + 1 \biggr]\frac{\Lambda^{1/2}}{(1+\Lambda)^2} \, ;
</math>
  </td>
</tr>
</table>
</div>
while,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1-6\mathcal{A})\mathcal{Y} - 5\mathcal{X}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 1- \frac{2\cdot 3}{(5-n)} \biggr]\biggl(\frac{3\cdot 5^3}{2^2\pi}\biggr)^{1/2} 
\frac{\Lambda^{3/2}}{(1+\Lambda)^2} 
- \biggl( \frac{3\cdot 5^3}{2^2\pi} \biggr)^{1/2} 
\frac{\Lambda^{1/2}}{(1+\Lambda)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{3\cdot 5^3}{2^2\pi} \biggr)^{1/2} \biggl[ \Lambda - \frac{2\cdot 3 }{(5-n)} \Lambda - (1+\Lambda)\biggr] 
\frac{\Lambda^{1/2}}{(1+\Lambda)^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl(\frac{3\cdot 5^3}{4\pi }\biggr)^{1/2}  \biggl[ \frac{2\cdot 3}{(5-n)} \Lambda + 1 \biggr]\frac{\Lambda^{1/2}}{(1+\Lambda)^2} \, .
</math>
  </td>
</tr>
</table>
</div>
So we have the desired result, namely, that these last two expressions have opposite signs but are otherwise identical and, hence, they sum to zero.
........  END COMMENT 2 -->