Editing SSC/Structure/BiPolytropes/AnalyzeStepFunction (section)

=Try Again, With Detailed Example=

==Model Amodel2==

Under the "AModel2FD" tab of an Excel spreadsheet titled, "qAndNuMaxAug21", we have constructed a discrete model of the core of a bipolytrope that has the following properties:

<table border="1" align="center" cellpadding="5">
<tr>
  <td align="center" colspan="2">Model A</td>
</tr>

<tr>
  <td align="center" colspan="1"><math>\mu_e/\mu_c</math></td>
  <td align="right">0.31</td>
</tr>

<tr>
  <td align="center" colspan="1"><math>\nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}}</math></td>
  <td align="right">0.337217</td>
</tr>

<tr>
  <td align="center" colspan="1"><math>q \equiv \frac{r_\mathrm{core}}{R_\mathrm{tot}}</math></td>
  <td align="right">0.0755023</td>
</tr>

<tr>
  <td align="center" colspan="2">Core Properties (at interface)</td>
</tr>

<tr>
  <td align="center" colspan="1"><math>\xi_i</math></td>
  <td align="right">9.0149598</td>
</tr>

<tr>
  <td align="center" colspan="1"><math>\theta_i</math></td>
  <td align="right">0.1886798</td>
</tr>

<tr>
  <td align="center" colspan="1"><math>\frac{d\theta}{d\xi}\biggr|_i</math></td>
  <td align="right">-0.0201845</td>
</tr>

<tr>
  <td align="center" colspan="1"><math>M_\mathrm{core} = 
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr]_i</math></td>
  <td align="right">6.8009303</td>
</tr>

<tr>
  <td align="center" colspan="1"><math>r_\mathrm{core} = \biggl(\frac{3}{2\pi}\biggr)^{1 / 2} \xi_i</math></td>
  <td align="right">6.2292317</td>
</tr>
</table>
We have divided this model into 100 equally spaced radial zones (center and surface included), that is, each with the following radial-shell thickness:  &nbsp; <math>\delta \xi \equiv \xi_i/99 = 0.0910602</math>, and <math>\delta r = r_\mathrm{core}/99 = 0.0629215</math>.  Analytic expressions providing the physical properties of our <math>(n_c, n_e) = (5, 1)</math> bipolytropes at each radial shell have been drawn from [[SSC/Structure/BiPolytropes/Analytic51#BiPolytrope_with_nc_=_5_and_ne_=_1|an accompanying chapter]] &#8212; for example &hellip;
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right"><math>r^*</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\biggl(\frac{3}{2\pi}\biggr)^{1 / 2}\xi \, ,</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\rho^*</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\biggl(1 + \frac{\xi^2}{3}\biggr)^{-5 / 2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right"><math>P^*</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right"><math>m</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr] \, .</math>
  </td>
</tr>
</table>

We note as well that, in equilibrium,
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right"><math>g_0 \equiv \frac{m}{(r^*)^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- 4\pi (r^*)^2 \cdot \frac{dP^*}{dm}
=
- \frac{1}{\rho^*} \cdot \frac{dP^*}{dr^*}
\, .
</math>
  </td>
</tr>
</table>
Hence, in equilibrium we have,
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right"><math>\frac{dP^*}{dm}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{4\pi (r^*)^2\rho^*} \biggl( \frac{\xi}{r^*} \biggr) \frac{dP^*}{d\xi}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\xi}{4\pi (r^*)^3\rho^*} \cdot \biggl[-3\biggl(1 + \frac{\xi^2}{3}\biggr)^{-4} \frac{2\xi}{3}\biggr]
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-\biggl(\frac{2 \pi}{3^3} \biggr)^{1 / 2} 
\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}  
\xi^{-1} \, ;
</math>
  </td>
</tr>
</table>
and,

<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right"><math>4\pi (r^*)^2 \cdot \frac{dP^*}{dm}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- 2^2\pi \biggl[ \biggl(\frac{3}{2\pi}\biggr)^{1 / 2}\xi \biggr]^2  
\biggl(\frac{2 \pi}{3^3} \biggr)^{1 / 2} 
\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}  
\xi^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- 2   
\biggl(\frac{2 \pi}{3} \biggr)^{1 / 2} 
\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}  
\xi
\, ;
</math>
  </td>
</tr>
</table>
and,

<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right"><math>4\pi (r^*)^4 \cdot \frac{dP^*}{dm}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \biggl[ \biggl(\frac{3}{2\pi}\biggr)^{1 / 2}\xi \biggr]^2 
2 \biggl(\frac{2 \pi}{3} \biggr)^{1 / 2} 
\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}  
\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \biggl(\frac{2 \cdot 3}{\pi}\biggr)^{1 / 2} 
\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}  
\xi^3
\, .
</math>
  </td>
</tr>
</table>

The following table provides a sample of variable values in the central region (shells 0 - 4) and near the interface (shells 95 - 99) of the equilibrium configuration.

<table border="1" align="center" cellpadding="5">
<tr>
  <td align="center" colspan="8">Analytically Determined Physical Properties at Various Radial Shells</td>
</tr>

<tr>
  <td align="center">Shell</td>
  <td align="center"><math>\xi</math></td>
  <td align="center"><math>m = -4\pi (r^*)^4\frac{dP^*}{dm} </math></td>
  <td align="center"><math>r^*</math></td>
  <td align="center"><math>\rho^*</math></td>
  <td align="center"><math>P^*</math></td>
  <td align="center"><math>-~\frac{dP^*}{dm}</math></td>
  <td align="center"><math>-4\pi (r^*)^2\frac{dP^*}{dm} = g_0</math></td>
</tr>

<tr>
  <td align="center">0</td>
  <td align="right">0.000000</td>
  <td align="right">0.000000</td>
  <td align="right">0.000000</td>
  <td align="right">1.000000</td>
  <td align="right">1.000000</td>
  <td align="center"><math>\infty</math></td>
  <td align="right">0.000000</td>
</tr>

<tr>
  <td align="center">1</td>
  <td align="right">0.091060</td>
  <td align="right">0.001039</td>
  <td align="right">0.062922</td>
  <td align="right">0.993123</td>
  <td align="right">0.991754</td>
  <td align="right">5.275715</td>
  <td align="right">0.262476</td>
</tr>

<tr>
  <td align="center">2</td>
  <td align="right">0.182120</td>
  <td align="right">0.008211</td>
  <td align="right">0.125843</td>
  <td align="right">0.972886</td>
  <td align="right">0.967552</td>
  <td align="right">2.605474</td>
  <td align="right">0.518508</td>
</tr>

<tr>
  <td align="center">3</td>
  <td align="right">0.273181</td>
  <td align="right">0.027155</td>
  <td align="right">0.188765</td>
  <td align="right">0.940420</td>
  <td align="right">0.928937</td>
  <td align="right">1.701968</td>
  <td align="right">0.762083</td>
</tr>

<tr>
  <td align="center">4</td>
  <td align="right">0.364241</td>
  <td align="right">0.062586</td>
  <td align="right">0.251686</td>
  <td align="right">0.897462</td>
  <td align="right">0.878252</td>
  <td align="right">1.241164</td>
  <td align="right">0.988001</td>
</tr>

<tr>
  <td align="center" colspan="8"><font size="+1"><b>&vellip;</b></td>
</tr>

<tr>
  <td align="center">95</td>
  <td align="right">8.650719</td>
  <td align="right">6.769823</td>
  <td align="right">5.977546</td>
  <td align="right">0.000292</td>
  <td align="right">0.000057</td>
  <td align="right">0.000422</td>
  <td align="right">0.189466</td>
</tr>

<tr>
  <td align="center">96</td>
  <td align="right">8.741779</td>
  <td align="right">6.777942</td>
  <td align="right">6.040467</td>
  <td align="right">0.000277</td>
  <td align="right">0.000054</td>
  <td align="right">0.000405</td>
  <td align="right">0.185762</td>
</tr>

<tr>
  <td align="center">97</td>
  <td align="right">8.832839</td>
  <td align="right">6.785828</td>
  <td align="right">6.103389</td>
  <td align="right">0.000264</td>
  <td align="right">0.000051</td>
  <td align="right">0.000389</td>
  <td align="right">0.182163</td>
</tr>

<tr>
  <td align="center">98</td>
  <td align="right">8.923900</td>
  <td align="right">6.793488</td>
  <td align="right">6.166310</td>
  <td align="right">0.000251</td>
  <td align="right">0.000048</td>
  <td align="right">0.000374</td>
  <td align="right">0.178666</td>
</tr>

<tr>
  <td align="center">99</td>
  <td align="right">9.014960</td>
  <td align="right">6.800930</td>
  <td align="right">6.229232</td>
  <td align="right">0.000239</td>
  <td align="right">0.000045</td>
  <td align="right">0.000359</td>
  <td align="right">0.175267</td>
</tr>
</table>

This last expression is precisely the same (in magnitude, but opposite in sign) as the expression that we have presented for <math>m</math>.  If we therefore return to <font color="red>STEP 6</font>, we appreciate that the right-hand side of the "eigenvector" expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>+\biggl[ \frac{\omega^2}{G\rho_c} \biggr] \biggl(\frac{\delta r^*}{r^*}\biggr) (r^*)</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
\biggl[ 4\pi (r^*)^2 \cdot \frac{dP^*}{dm} + \frac{m}{(r^*)^2}\biggr] \, ,
</math>
  </td>
</tr>
</table>
goes to zero at every radial shell if the configuration is in equilibrium.

<table border="1" align="center" cellpadding="10" width="80%"><tr><td align="left">
<font color="red">ASIDE:</font>&nbsp; Next, we will focus on building a discrete, finite-difference representation of each equilibrium model, as well as of this "eigenvector" expression.  In doing so, we will immediately find that the two terms on the right-hand-side ''do not'' exactly sum to zero, even for an equilibrium configuration. We should nevertheless try to construct a finite-difference representation for which the two terms cancel to a relatively high degree of precision.   We have considered rewriting this "eigenvector" expression in the form,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>+\biggl[ \frac{\omega^2}{G\rho_c} \biggr] \biggl(\frac{\delta r^*}{r^*}\biggr) (r^*)^3</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
\biggl[ 4\pi (r^*)^4 \cdot \frac{dP^*}{dm} + m \biggr] \, ,
</math>
  </td>
</tr>
</table>

because this form isolates the Lagrangian mass, <math>m</math>, which is time-invariant.  But as the numbers in the third column of the above table illustrate, the terms on the right-hand-side vary by several orders of magnitude as we move from shell 1 to shell 100.  If it is written instead in the form that we have initially suggested, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>+\biggl[ \frac{\omega^2}{G\rho_c} \biggr] \biggl(\frac{\delta r^*}{r^*}\biggr) (r^*)</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
\biggl[ 4\pi (r^*)^2 \cdot \frac{dP^*}{dm} + \frac{m}{(r^*)^2}\biggr] \, ,
</math>
  </td>
</tr>
</table>
then, as is illustrated by the numbers in the last column of the table and by the following figure, the terms on the right-hand-side vary by no more than one order of magnitude as we move from the center to the surface of the configuration.
<div align="center">[[File:G0variationBipolytrope.png|450px|Variation of <math>g_0</math> from center to interface]]</div>  This choice should facilitate cancellation to a higher degree of precision in our finite-difference-based model.
</td></tr></table>

==Even Simpler Core==
<font color="green"><b>STEP 0:</b></font> &nbsp; We construct a finite-difference representation of the initial (unperturbed) equilibrium configuration by dividing the model into 99 radial zones that are equally spaced in <math>0 \le \xi \le \xi_i</math>.  The initial radial coordinate of each zone and the corresponding initial enclosed mass are given, respectively, by the expressions &hellip; 

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

<tr>
  <td align="right">
<math>[r_j]_0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{3}{2\pi}\biggr)^{1 / 2} \xi_j \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; 
  <td align="right">
<math>m_j</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr] \, .
</math>
  </td>
</tr>
</table>

The mass, <math>m_j</math>, will serve as our Lagrangian coordinate; that is, we will perturb the model by modifying the radial location of each shell while fixing the enclosed mass.

<font color="green"><b>STEP 1:</b></font> &nbsp; Guess the eigenvector, <math>{\delta r}_i</math>, remembering that a reasonably good trial eigenfunction for the core is one that has a "parabolic dependence on the radius," namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{x}{\alpha_\mathrm{scale}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 - \frac{\xi^2}{15} \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; where, &nbsp; &nbsp; &nbsp; 
  <td align="right">
<math>\xi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{2\pi}{3}\biggr)^{1 / 2} r_0 \, .
</math>
  </td>
</tr>
</table>

This means,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>x(r_0)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\alpha_\mathrm{scale}\biggl[1 - \biggl(\frac{2\pi}{45}\biggr)r_0^2  \biggr]
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{dx}{dr_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\alpha_\mathrm{scale}\biggl(\frac{4\pi}{45}\biggr)r_0 
\, .
</math>
  </td>
</tr>
</table>

Once a "guess" for the fractional displacement vector, <math>(\delta r)_j = [x \cdot r_0]_j\, ,</math> has been specified, we recognize that the perturbed location of each radial shell is given by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>r_j</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(r_0)_j [ 1 + x_j] \, .
</math>
  </td>
</tr>
</table>

<font color="green"><b>STEP 2:</b></font> &nbsp; Our finite-difference representation of the mass-density at each radial shell in the equilibrium configuration (subscript "0") is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[{\bar\rho}_{j-1/2} \biggr]_0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[m_j - m_{j-1}\biggr]_0\biggl[\frac{4\pi}{3} (r_j^3 - r_{j-1}^3)  \biggr]_0^{-1} \, .
</math>
  </td>
</tr>
</table>
After perturbing the radial location of each shell &#8212; that is, after setting <math>r_j = (r_0)_j + [x \cdot r_0]_j</math> &#8212; the resulting finite-difference representation of the perturbed mass density of each shell is given by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>{\bar\rho}_{j-1/2} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[m_j - m_{j-1}\biggr]_0\biggl[\frac{4\pi}{3} (r_j^3 - r_{j-1}^3)  \biggr]^{-1} \, .
</math>
  </td>
</tr>
</table>
(Note that we retain a subscript "0" on the mass, <math>m_j</math>, because it serves as our Lagrangian identifier for each shell.)

<font color="green"><b>STEP 3:</b></font> &nbsp; The pressure can be determined in the equilibrium configuration (subscript "0") and after the perturbation from knowledge of the density and the chosen adiabatic index, <math>\gamma_c = 6/5</math>, via knowledge of the (fixed) specific entropy, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[ P_{j-1/2} \biggr]_0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ {\bar\rho}_{j-1/2}\biggr]_0^{\gamma_c} \cdot \exp\biggl[ \frac{\mu (\gamma-1)s}{\mathfrak{R}} \biggr]
\, ,
</math>
  </td>
<td align="center">&nbsp;  &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>P_{j-1/2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ {\bar\rho}_{j-1/2}\biggr]^{\gamma_c} \cdot \exp\biggl[ \frac{\mu (\gamma-1)s}{\mathfrak{R}} \biggr]
\, .
</math>
  </td>
</tr>
</table>
The pressure perturbation can therefore be obtained from the simple difference,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[ \delta P \biggr]_{j-1/2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
P_{j - 1 / 2} - \biggl[ P_{j - 1 / 2}\biggr]_0 \, .
</math>
  </td>
</tr>
</table>

[[File:DisplacementFunctionAModel2.png|350px|right|Displacement Function]]<font color="green"><b>STEP 4:</b></font> &nbsp; If, in the context of our [[#Step_6|above discussion of the perturbed "Euler + Poisson Equations"]], we set LHS = RHS, we obtain,

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

<tr>
  <td align="right">
<math>\biggl[\frac{\omega_i^2}{G\rho_c}\biggr] x r_0</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>\biggl\{
4\pi r_0^2\biggl[ \frac{d(P_0 + \delta P)}{dm}\biggr]
+ g_0 \biggr\}
- 4xg_0
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl\{
4\pi r_0^2\biggl[ \frac{d(\delta P)}{dm}\biggr]\biggr\}
- 4xg_0
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ x</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>\biggl\{
4\pi r_0^2\biggl[ \frac{d(\delta P)}{dm}\biggr]\biggr\}\biggl\{ \biggl[\frac{\omega_i^2}{G\rho_c}\biggr] r_0 + 4g_0 \biggr\}^{-1} \, .
</math>
  </td>
</tr>
</table>
So for a specified value of the square of the oscillation frequency, <math>[\omega^2/(G\rho_c)]</math> &#8212; same value for all shells &#8212; the strategy should be to (a) guess <math>x_j</math>; (b) evaluate the right-hand-side of this last expression; (c) if the RHS does not equal the "guessed" eigenvector, <math>x_j</math>, then you need to guess a new eigenvector; (d) repeat!

In the figure shown here on the right, the black curve displays the variation with mass, <math>m</math>, of the (parabolic-shaped) displacement function, <math>x/\alpha_\mathrm{scale}</math>, that served as our initial "guess;" while the red dots show how the right-hand side of this last expression varies with <math>m</math> for the case, <math>\omega_i^2 = 0</math>.  They lie almost exactly on top of one another, as hoped/expected.

==Transition==

In transitioning from the core to the envelope, all of the above (green) <font color="green">STEPS</font> will remain the same except for the 1<sup>st</sup> Law's treatment of the pressure.  Following along the lines of our [[Appendix/Ramblings/PatrickMotl#May_5_(following_a_phone_conversation_with_Patrick)|''Ramblings'' idea exchange with Patrick Motl]], the mass-density is generically related to the pressure via the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{s}{\Re/\bar{\mu}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{(\gamma_g-1)}\ln \biggl[ \frac{P}{(\gamma_g-1)\rho^{\gamma_g}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{s \bar\mu (\gamma_g-1)}{\Re}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\ln \biggl[ \frac{P}{(\gamma_g-1)\rho^{\gamma_g}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~P 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(\gamma_g - 1) \rho^{\gamma_g}\exp \biggl[\frac{s \bar\mu (\gamma_g-1)}{\Re}\biggr]\, .</math>
  </td>
</tr>
</table>
Given that, in polytropic configurations for which we make the association, <math>\gamma_g = (n+1)/n</math>, the pressure is related to the density via the expression, <math>P = K\rho^{\gamma_g}</math>, we appreciate that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>K</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(\gamma_g - 1) \exp \biggl[\frac{s \bar\mu (\gamma_g-1)}{\Re}\biggr] \, .
</math>
  </td>
</tr>
</table>

===Core===
For the core, we set <math>\gamma_g = 6/5</math>.  Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
P 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>K_c \rho^{6/5} \, .</math>
  </td>
</tr>
</table>

Normalizing the pressure and the density as we have in [[SSC/Structure/BiPolytropes/Analytic51#Normalization|a closely related discussion of the structure of bipolytropes]], we have throughout the core,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
P^* = \frac{P}{K_c\rho_0^{6/5}}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(\rho^*)^{6/5} \, .</math>
  </td>
</tr>
</table>

Now, in this normalized expression we see that the polytropic constant for the core is, <math>K^*_c = 1</math>.  This means that the specific entropy of all the fluid in the core is given by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>1</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{5} \exp \biggl[\frac{s^*_c {\bar\mu}_c }{5\Re}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{s^*_c {\bar\mu}_c }{\Re}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
5\ln(5 ) \, .
</math>
  </td>
</tr>
</table>

===Envelope===
For the envelope, we set <math>\gamma_g = 2</math>.  Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
P 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>K_e \rho^{2} \, .</math>
  </td>
</tr>
</table>

Adopting the same pressure and density normalizations as used in the core, we have throughout the envelope,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
P^* = \frac{P}{K_c\rho_0^{6/5}}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl(\frac{K_e}{K_c} \biggr) \rho_0^{4/5} (\rho^*)^2 \, .</math>
  </td>
</tr>
</table>
Now, at the [[SSC/Structure/BiPolytropes/Analytic51#Step_5:_Interface_Conditions|interface of the equilibrium model]], we know that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\frac{K_e}{K_c}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\rho_0^{-4/5}\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-4}_i
=
\rho_0^{-4/5}\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl[1 + \frac{\xi_i^2}{3}\biggr]^2 \, .
</math>
  </td>
</tr>
</table>
Hence, throughout the envelope,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
P^* 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl[1 + \frac{\xi_i^2}{3}\biggr]^2 (\rho^*)^2 
\, ,
</math>
  </td>
</tr>
</table>

in which case we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>K^*_e</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl[1 + \frac{\xi_i^2}{3}\biggr]^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \frac{s^*_e \bar\mu_e}{\Re}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\ln\biggl\{ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl[1 + \frac{\xi_i^2}{3}\biggr]^2 \biggr\} \, .
</math>
  </td>
</tr>
</table>

===<font color="green">STEP 3</font> Clarification===

It is critically important to appreciate that the manner in which the pressure is determined at discrete locations in our finite-difference model must be different in the envelope and the core.  Keeping in mind that, in general,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
P 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(\gamma_g - 1) \rho^{\gamma_g}\exp \biggl[\frac{s \bar\mu (\gamma_g-1)}{\Re}\biggr]\, ,</math>
  </td>
</tr>
</table>
for <math>(n_c, n_e) = (5, 1)</math> bipolytropes, we have &hellip;

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

<tr>
  <td align="right">
Core <math>(\gamma_g=6/5)</math>: &nbsp; &nbsp; &nbsp; <math>P_{j-1/2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{5}\biggl[ {\bar\rho}_{j-1/2}\biggr]^{6/5} \cdot \exp\biggl[ \frac{\mu_c s^*_c}{5\mathfrak{R}} \biggr]
=
\biggl[ {\bar\rho}_{j-1/2}\biggr]^{6/5} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
Envelope <math>(\gamma_g=2)</math>: &nbsp; &nbsp; &nbsp; <math>P_{j-1/2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ {\bar\rho}_{j-1/2}\biggr]^{2} \cdot \exp\biggl[ \frac{\mu_e s^*_e}{\mathfrak{R}} \biggr]
=
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl[1 + \frac{\xi_i^2}{3}\biggr]^2 \biggl[ {\bar\rho}_{j-1/2}\biggr]^{2} 
\, .
</math>
  </td>
</tr>
</table>

===Interface===
Drawing from our [[SSC/Structure/BiPolytropes/Analytic51|chapter in which the bipolytrope is constructed]], we note the following determination of the density and pressure at the interface, as viewed from the perspective of the core and, separately, from the perspective of the envelope.

<table border="1" align="center">
<tr>
  <td align="center" colspan="2">
Properties at the (Unperturbed) Interface
  </td>
</tr>
<tr>
  <td align="center" width="50%">
<b>Core</b><br />
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\frac{s^*_c \bar\mu_c}{\Re}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
5\ln(5) 
</math>
  </td>
</tr>
</table>

  </td>
  <td align="center">
<b>Envelope</b><br />
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\frac{s^*_e \bar\mu_e}{\Re}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\ln\biggl\{ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta_i^{-4} \biggr\} 
</math>
  </td>
</tr>
</table>
  </td>
</tr>

<tr>
<td align="left">

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>(\rho^*)_i </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\theta_i^5</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
(P^*)_i 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{5} (\rho^*)_i^{6/5}\exp \biggl[\frac{s_c \bar\mu_c }{5\Re}\biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(\rho^*)_i^{6/5}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\theta_i^{6}</math>
  </td>
</tr>
</table>

</td>
<td align="left">

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>(\rho^*)_i </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
(P^*)_i 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(\rho^*)_i^{2}\exp \biggl[\frac{s_e \bar\mu_e }{\Re}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(\rho^*)_i^{2}\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta_i^{-4} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\theta_i^6</math>
  </td>
</tr>
</table>

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


----

In what follows, we will continue to focus on the interface but, for simplicity, will drop the "i" subscript; instead, we will use a "0" subscript to indicate a property at the ''unperturbed'' interface.


<table border="1" align="center">
<tr>
  <td align="center" colspan="2">
Properties at the ''Perturbed Interface
  </td>
</tr>
<tr>
  <td align="center" width="50%">
<b>Core</b><br />
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\frac{s^*_c \bar\mu_c}{\Re}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
5\ln(5) 
</math>
  </td>
</tr>
</table>

  </td>
  <td align="center">
<b>Envelope</b><br />
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\frac{s^*_e \bar\mu_e}{\Re}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\ln\biggl\{ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta_i^{-4} \biggr\} 
</math>
  </td>
</tr>
</table>
  </td>
</tr>

<tr>
<td align="left">

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>\rho^*_c = (\rho^*_c)_0 + (\delta \rho)_c </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\theta_i^5 + (\delta \rho)_c</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
P^*_c  = (P^*_c)_0 + (\delta P)_c 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \rho^*_c \biggr]^{6/5}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[\theta_i^5 + (\delta \rho)_c\biggr]^{6/5}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>\theta_i^6\biggl[1 + \frac{6(\delta \rho)_c}{5\theta_i^5}\biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Rightarrow ~~~ (\delta P)_c</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>\frac{6\theta_i(\delta \rho)_c}{5}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Rightarrow ~~~ p_c \equiv \frac{(\delta P)_c}{(P^*_c)_0}</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>\frac{6(\delta \rho)_c}{5\theta_i^5}</math>
  </td>
</tr>
</table>

</td>
<td align="left">

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>\rho^*_e = (\rho^*_e)_0 + (\delta\rho)_e</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5+ (\delta\rho)_e</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
P^*_e = (P^*_e)_0 + (\delta P)_e
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \rho^*_e \biggr]^{2}\exp \biggl[\frac{s_e \bar\mu_e }{\Re}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5+ (\delta\rho)_e\biggr]^{2}
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta_i^{-4} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>\theta_i^6\biggl[ 1 + 2(\delta\rho)_e \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i^{-5}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ (\delta P)_e</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>2(\delta\rho)_e \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\Rightarrow ~~~ p_e \equiv \frac{(\delta P)_e}{(P^*_e)_0}</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>\frac{2(\delta\rho)_e}{\theta_i^5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} </math>
  </td>
</tr>
</table>

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

Now, in order for <math>(\delta P)_c = (\delta P)_e</math> at the interface, we must have,

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

<tr>
  <td align="right">
<math>
2(\delta\rho)_e \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{6\theta_i(\delta \rho)_c}{5}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~(\delta\rho)_e   
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{3}{5}\biggl( \frac{\mu_e}{\mu_c} \biggr) (\delta \rho)_c
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \frac{(\delta\rho)_e }{\theta_i^5}  
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{3}{5}\biggr) \frac{(\delta \rho)_c}{\theta_i^5}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~\biggl[ \frac{(\delta\rho)}{\rho^*_0} \biggr]_e 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{3}{5}\biggr) \biggl[ \frac{(\delta\rho)}{\rho^*_0} \biggr]_c
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~d_e 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{\gamma_c}{\gamma_e}\biggr) d_c \, .
</math>
  </td>
</tr>
</table>

We also know that, according to the linearized equation of continuity, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\biggl[r_0 \frac{dx}{dr_0} + 3x + d \biggr]_e
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[r_0 \frac{dx}{dr_0} + 3x + d \biggr]_c \, .
</math>
  </td>
</tr>
</table>
Given that, at the interface, <math>x_e = x_c</math>, this means that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\biggl[r_0 \frac{dx}{dr_0}\biggr]_e   
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[r_0 \frac{dx}{dr_0}\biggr]_c - \biggl[\biggl(\frac{\gamma_c}{\gamma_e}\biggr)- 1\biggr] d_c 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[r_0 \frac{dx}{dr_0}\biggr]_c 
+ \biggl[\biggl(\frac{\gamma_c}{\gamma_e}\biggr)- 1\biggr] \biggl\{
\biggl[r_0 \frac{dx}{dr_0}\biggr]_c + 3x 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
3x\biggl[\biggl(\frac{\gamma_c}{\gamma_e}\biggr)- 1\biggr] 
+ \biggl(\frac{\gamma_c}{\gamma_e}\biggr) 
\biggl[r_0 \frac{dx}{dr_0}\biggr]_c  \, ;
</math>
  </td>
</tr>
</table>
or, dividing through by <math>x</math>,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\biggl[\frac{d\ln x}{d\ln r_0}\biggr]_e   
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
3\biggl[\biggl(\frac{\gamma_c}{\gamma_e}\biggr)- 1\biggr] 
+ \biggl(\frac{\gamma_c}{\gamma_e}\biggr) 
\biggl[\frac{d\ln x}{d\ln r_0}\biggr]_c  \, .
</math>
  </td>
</tr>
</table>

This exactly matches the interface constraint presented in [[SSC/Stability/BiPolytropes#Interface_Conditions|a related discussion in a separate chapter]].

Now, as has already been stated in [[#Step_5|Step 5, above]], for our analytic guess for the core's displacement function,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[\frac{d\ln x}{d\ln r_0}\biggr]_c</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
\biggl(\frac{2}{15}\biggr) \xi^2 \biggl[\frac{\xi^2}{15}-1\biggr]^{-1}
=
2.4526969
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl[\frac{d\ln x}{d\ln r_0}\biggr]_e</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
0.2716182
\, .
</math>
  </td>
</tr>
</table>
where the numerical evaluation has been presented specifically for the core/envelope interface of the critical "Model A" configuration in which, <math>\xi_i = 9.0149598</math>.  Also, given that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\frac{x}{r_0}   
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\alpha_\mathrm{scale}}{\xi} \biggl(\frac{2\pi}{3}\biggr)^{1 / 2} \biggl[1 - \frac{\xi^2}{15}\biggr]
=
(-\alpha_\mathrm{scale}) \cdot 0.7092314 \, ,
</math>
  </td>
</tr>
</table>
we find that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[\frac{dx}{dr_0}\biggr]_c</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
(-\alpha_\mathrm{scale}) \cdot 1.7395297
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl[\frac{d x}{d r_0}\biggr]_e</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
(-\alpha_\mathrm{scale}) \cdot 0.1926402
\, .
</math>
  </td>
</tr>
</table>

==Mapping Between Lagrangian Coordinates (Mass and Radius)==
===Core===
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>r^*</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
c_r \xi \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>M^*</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
c_M \xi^3\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}
\, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>c_r</math>
  </td>
  <td align="center">
<math>\equiv</math>
  <td align="left">
<math>
\biggl( \frac{3}{2\pi} \biggr)^{1 / 2}
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>c_M</math>
  </td>
  <td align="center">
<math>\equiv</math>
  <td align="left">
<math>
\biggl( \frac{2\cdot 3}{\pi} \biggr)^{1 / 2}
\, .
</math>
  </td>
</tr>
</table>
Establish the derivative of mass with respect to radius:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>dr^*</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
c_r d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>dM^*</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
c_M \biggl\{
3\xi^2\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}
- \biggl(\frac{3}{2}\biggr)
\xi^3\biggl(1 + \frac{\xi^2}{3}\biggr)^{-5 / 2} \cdot \frac{2\xi}{3}
\biggr\}d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
c_M \biggl(1 + \frac{\xi^2}{3}\biggr)^{-5 / 2}\biggl\{
3\xi^2\biggl(1 + \frac{\xi^2}{3}\biggr)
- \xi^4
\biggr\}d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
c_M \biggl(1 + \frac{\xi^2}{3}\biggr)^{-5 / 2}\biggl\{
3\xi^2
\biggr\}d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
\frac{3c_M}{c_r} \biggl(1 + \frac{\xi^2}{3}\biggr)^{-5 / 2} \, .
\xi^2 dr^*
</math>
  </td>
</tr>
</table>
<span id="FirstDerivation">Hence,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\frac{dr^*}{dM^*}
</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
\frac{c_r}{3c_M} \biggl(1 + \frac{\xi^2}{3}\biggr)^{5 / 2}
\xi^{-2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
\frac{1}{6} 
\biggl(1 + \frac{\xi^2}{3}\biggr)^{5 / 2}
\xi^{-2} 
=
8.5761836
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \frac{dx}{dM^*} = \frac{dr^*}{dM^*} \cdot \frac{dx}{dr_0}
</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
(-\alpha_\mathrm{scale}) \cdot 1.7395297 \times 8.5761836 
=(-\alpha_\mathrm{scale}) \cdot
14.9185261 \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="8"><tr><td align="left">
<font color="red"><b>ASIDE:</b></font>&nbsp; &nbsp; Following along the lines of a [[Appendix/Ramblings/NonlinarOscillation#Foundation|separate but similar discussion]], let's invert the <math>M^*(\xi)</math> function to obtain a desired <math>r^*(M^*)</math> function.
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>M^*</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
c_M \xi^3\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl(1 + \frac{\xi^2}{3}\biggr)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{c_M}{M^*}\biggr)^{2 / 3} \xi^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 3 + \xi^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
3\biggl(\frac{c_M}{M^*}\biggr)^{2 / 3} \xi^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ 
\biggl[ 3\biggl(\frac{c_M}{M^*}\biggr)^{2 / 3} -1 \biggr]\xi^2
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
3 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ 
\xi
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl(\frac{c_M}{M^*}\biggr)^{2 / 3} - \frac{1}{3} \biggr]^{-1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ 
r^*
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
c_r\biggl[ \biggl(\frac{c_M}{M^*}\biggr)^{2 / 3} - \frac{1}{3} \biggr]^{-1 / 2}
\, .
</math>
  </td>
</tr>
</table>

Now, differentiate this expression &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\frac{dr^*}{dM^*}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{c_r}{2}\biggl[ \biggl(\frac{M^*}{c_M}\biggr)^{-2 / 3} - \frac{1}{3} \biggr]^{-3 / 2}
\biggl(\frac{2c_M^{2 / 3}}{3}\biggr)\biggl(M^*\biggr)^{-5 / 3}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{c_r}{3c_M}\biggl[ \biggl(\frac{M^*}{c_M}\biggr)^{-2 / 3} - \frac{1}{3} \biggr]^{-3 / 2}
\biggl(\frac{M^*}{c_M}\biggr)^{-5 / 3} \, .
</math>
  </td>
</tr>
</table>
Or, given that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl(\frac{M^*}{c_M}\biggr)^{-1 / 3}</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
\frac{1}{\xi}\biggl(1 + \frac{\xi^2}{3}\biggr)^{1 / 2}
\, ,
</math>
  </td>
</tr>
</table>
we can also write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\frac{dr^*}{dM^*}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{c_r}{3c_M}\biggl[ \frac{1}{\xi^2}\biggl(1 + \frac{\xi^2}{3}\biggr) - \frac{1}{3} \biggr]^{-3 / 2}
\frac{1}{\xi^5}\biggl(1 + \frac{\xi^2}{3}\biggr)^{5 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{c_r}{3c_M}
\frac{1}{\xi^2}\biggl(1 + \frac{\xi^2}{3}\biggr)^{5 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{1}{6} \biggr)
\frac{1}{\xi^2}\biggl(1 + \frac{\xi^2}{3}\biggr)^{5 / 2} \, .
</math>
  </td>
</tr>
</table>
This exactly matches the expression for this derivative that we derived [[#FirstDerivation|immediately above]].
</td></tr></table>

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

<tr>
  <td align="right">
<math>r^*</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
e_r \eta
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>M^*</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
e_M\biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
e_M \cdot A \biggl[ \sin(\eta-B) - \eta\cos(\eta-B) \biggr] \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>e_r</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>e_M</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{e_r}{e_M}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}
\biggl[
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2}
\biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i \biggl(\frac{1}{2\pi}\biggr)^{1 / 2}
\biggl[
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{2} \theta_i \biggl( \frac{\pi}{2} \biggr)^{1/2}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2}\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \, ,
</math>
  </td>
</tr>
</table>
and from the [[SSC/Structure/BiPolytropes/Analytic51#Parameter_Values|parameter values]] associated with the equilibrium configuration,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\Lambda_i</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\frac{1}{\eta_i} + \biggl( \frac{d\phi}{d\eta}\biggr)_i
=
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\frac{1}{3^{1 / 2} \theta_i^2 \xi_i} - \frac{\xi_i}{3^{1 / 2}}
=
\frac{1}{3^{1 / 2}\xi_i} \biggl[
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\biggl(1 + \frac{\xi_i^2}{3}\biggr) - \xi_i^2
\biggr]
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>A</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\eta_i(1 + \Lambda_i^2)^{1 / 2}
=
\biggl(\frac{\mu_e}{\mu_c}\biggr) 3^{1 / 2}\xi_i \biggl(1 + \frac{\xi_i^2}{3}\biggr)^{-1} \biggl(1 + \Lambda_i^2 \biggr)^{1 / 2}
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>B</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\eta_i - \frac{\pi}{2} + \tan^{-1}(\Lambda_i)
=
\biggl(\frac{\mu_e}{\mu_c}\biggr) 3^{1 / 2}\xi_i \biggl(1 + \frac{\xi_i^2}{3}\biggr)^{-1}- \frac{\pi}{2} + \tan^{-1}(\Lambda_i)
\, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="8"><tr><td align="left">
Note, as well, that
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>A</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\frac{\eta_i}{\sin(\eta_i - B)} \, .
</math>
  </td>
</tr>
</table>

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

Establish the derivative of mass with respect to radius:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>dr^*</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
e_r d\eta
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>dM^*</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
e_M\cdot A \biggl[ \eta\sin(\eta-B) \biggr]d\eta 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{e_M\cdot A}{e_r} \biggl[ \eta\sin(\eta-B) \biggr]dr^* 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ dM^*</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{e_M\cdot A}{e_r} \biggl[ \eta\sin(\eta-B) \biggr]dr^*
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \frac{dr^*}{dM^*}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{e_r}{e_M\cdot A} \biggl[ \eta\sin(\eta-B) \biggr]^{-1} \, .
</math>
  </td>
</tr>
</table>

Hence, precisely at the interface,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\biggl[ \frac{dr^*}{dM^*} \biggr]_i
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{e_r}{e_M\cdot \eta_i^2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{6}\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \biggl[
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2} \theta_i^{-4} \xi_i^{-2}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{6}\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-5}_i \xi_i^{-2}
\, .
</math>
  </td>
</tr>
</table>
Given that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\biggl[\frac{d x}{d r_0}\biggr]_e</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
(-\alpha_\mathrm{scale}) \cdot 0.1926402
\, ,
</math>
  </td>
</tr>
</table>
we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\biggl[\frac{d x}{d M^*}\biggr]_e  
=
\biggl[\frac{d r^*}{d M^*}\biggr]_e \cdot 
\biggl[\frac{d x}{d r_0}\biggr]_e</math>
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
(-\alpha_\mathrm{scale}) \cdot 0.1926402 \cdot \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} 
\biggl[ 8.5761836 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  <td align="left">
<math>
(-\alpha_\mathrm{scale}) \cdot 5.2394120 \, .
</math>
  </td>
</tr>
</table>