Editing SSC/Structure/BiPolytropes/51RenormaizePart2 (section)

==Building Each Model==

===Basic Equilibrium Structure===
Most of the details underpinning the following summary relations can be [[SSC/Structure/BiPolytropes/Analytic51Renormalize#BiPolytrope_with_(nc,_ne)_=_(5,_1)|found here]].

<div align="center"><b>New Normalization</b></div>

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

<tr>
  <td align="right"><math>\tilde\rho</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>\rho \biggl[\biggl( \frac{K_c}{G} \biggr)^{3 / 2} \frac{1}{M_\mathrm{tot}} \biggr]^{-5} \, ;</math></td>
</tr>

<tr>
  <td align="right"><math>\tilde{P}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>P \biggl[K_c^{-10} G^{9} M_\mathrm{tot}^{6}  \biggr] \, ;</math></td>
</tr>

<tr>
  <td align="right"><math>\tilde{r}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>r \biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr]\, ,</math></td>
</tr>

<tr>
  <td align="right"><math>\tilde{M}_r</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>\frac{M_r}{M_\mathrm{tot}} \, ;</math></td>
</tr>

<tr>
  <td align="right"><math>\tilde{H}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>H \biggl[K_c^{-5 / 2} G^{3 / 2} M_\mathrm{tot} \biggr] \, ;</math></td>
</tr>

<tr>
  <td align="right"><math>\tilde{t}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>t \biggl[K_c^{15} G^{-13} M_\mathrm{tot}^{-10} \biggr]^{1 / 4} \, .</math></td>
</tr>
</table>

<span id="VariableProfiles">Note:&nbsp;</span>  For an n = 5 polytrope (like our bipolytrope's core), the units of the polytropic constant, <math>K_c</math>, are <math>\biggl[ \frac{\mathrm{length}^{13}}{\mathrm{mass} \cdot \mathrm{time}^{10}} \biggr]^{1 / 5}</math>.

<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center">
Quantity
  </td>
  <td align="center">
[[SSC/Structure/BiPolytropes/Analytic51#Steps_2_&_3|Core]]<br />
<math>0 \le \xi \le \xi_i</math><br />
----
<math>\theta = \biggl[ 1 + \frac{1}{3}\xi^2 \biggr]^{-1/2}</math><br />
<math>\frac{d\theta}{d\xi} = - \frac{\xi}{3}\biggl[ 1 + \frac{1}{3}\xi^2 \biggr]^{-3/2}</math>
  </td>
  <td align="center">
[[SSC/Structure/BiPolytropes/Analytic51#Step_6:_Envelope_Solution|Envelope]]<br />
<math>\eta_i \le \eta \le \eta_s</math><br />
----
<math>\phi = A \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr]</math><br />
<math>\frac{d\phi}{d\eta} = -\frac{A}{\eta^2} \biggl[ \sin(\eta-B) - \eta\cos(\eta-B)\biggr] </math>
  </td>
</tr>

<tr>
  <td align="center">
<math>\tilde{r}</math>
  </td>
  <td align="center">
<math>\mathcal{m}_\mathrm{surf}^{-2} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{4} \biggl(\frac{3}{2\pi}\biggr)^{1/2} \xi</math>
  </td>
  <td align="center">
<math>\mathcal{m}_\mathrm{surf}^{-2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{3} \theta^{-2}_i (2\pi)^{-1/2}\eta </math>
  </td>
</tr>

<tr>
  <td align="center">
<math>\tilde{\rho}</math>
  </td>
  <td align="center">
<math>\mathcal{m}_\mathrm{surf}^5 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-10} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2}</math>
  </td>
  <td align="center">
<math>\mathcal{m}_\mathrm{surf}^5 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-9} \theta^{5}_i \phi </math>
  </td>
</tr>

<tr>
  <td align="center">
<math>\tilde{P}</math>
  </td>
  <td align="center">
<math>\mathcal{m}_\mathrm{surf}^6 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-12} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3} </math>
  </td>
  <td align="center">
<math>\mathcal{m}_\mathrm{surf}^6 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-12} \theta^{6}_i \phi^{2}</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>\tilde{M}_r</math>
  </td>
  <td align="center">
<math>
\mathcal{m}_\mathrm{surf}^{-1} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{2} 
\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>
  <td align="center">
<math>\mathcal{m}_\mathrm{surf}^{-1}~ \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)</math>
  </td>
</tr>

<tr>
  <td align="left" colspan="3">
Note that, for a given specification of the molecular-weight ratio, <math>\mu_e/\mu_c</math>, and the interface location, <math>\xi_i</math>, 
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\theta_i</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl(1+\frac{1}{3}\xi_i^2\biggr)^{-1 / 2} \, ,</math></td>
</tr>

<tr>
  <td align="right"><math>\eta_i</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl(\frac{\mu_e}{\mu_c}\biggr)~ \sqrt{3} \theta_i^2 \xi_i \, ,</math></td>
</tr>

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

<tr>
  <td align="right"><math>A</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\eta_i\biggl(1 + \Lambda_i^2\biggr)^{1 / 2} \, ,</math></td>
</tr>

<tr>
  <td align="right"><math>B</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\eta_i - \frac{\pi}{2} + \tan^{-1}(\Lambda_i) \, ,</math></td>
</tr>

<tr>
  <td align="right"><math>\eta_s</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\pi + B = 
\frac{\pi}{2} + \eta_i + \tan^{-1}(\Lambda_i) \, ,</math></td>
</tr>
</table>
in which case,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\mathcal{m}_\mathrm{surf}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl(\frac{2}{\pi}\biggr)^{1 / 2} \frac{A\eta_s}{\theta_i} \, ,</math></td>
</tr>

<tr>
  <td align="right"><math>{\tilde\rho}_c</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf}^5 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-10} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^2 \sqrt{3} ~ \biggl[ \frac{\xi_i^3 \theta_i^4}{A\eta_s}\biggr] \, ,</math></td>
</tr>

<tr>
  <td align="right"><math>q \equiv \frac{r_\mathrm{core}}{R}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl(\frac{\mu_e}{\mu_c}\biggr) \sqrt{3} ~ \biggl[ \frac{\xi_i \theta_i^2}{\eta_s}\biggr] \, .</math></td>
</tr>
</table>

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

===One of the Linearized Equations===
From an [[SSC/Perturbations#Summary_Set_of_Linearized_Equations|accompanying discussion]], the linearized "Euler + Poisson Equations" is,

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

<tr>
  <td align="right"><math>\frac{P_0}{\rho_0} \frac{dp}{dr_0}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(4x + p)g_0 + \omega^2 r_0 x
\, .
</math>
  </td>
</tr>
</table>

If we shift to our [[#Basic_Equilibrium_Structure|above-specified, new normalization]] and insert the relation, 
<math>g_0 = GM_r/r_0^2</math>, we have,

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

<tr>
  <td align="right"><math>\omega^2 \biggl[\frac{r_0}{g_0}\biggr] x + (4x + p)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl\{
\frac{P_0 r_0}{\rho_0 G M_r} 
\biggr\}
\frac{dp}{d\ln r_0}  
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\tilde{P}\biggl[K_c^{-10} G^{9} M_\mathrm{tot}^{6}  \biggr]^{-1} 
\tilde{r}\biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr]^{-1}
\frac{1}{\tilde\rho}\biggl[\biggl( \frac{K_c}{G} \biggr)^{3 / 2} \frac{1}{M_\mathrm{tot}} \biggr]^{-5}
\biggl\{
\frac{1}{G M_\mathrm{tot}\tilde{M}_r} 
\biggr\}  
\frac{dp}{d\ln \xi}  
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\tilde{P}\biggl[K_c^{10} G^{-9} M_\mathrm{tot}^{-6}  \biggr] 
\tilde{r}\biggl[K_c^{-5 / 2}G^{5 / 2} M_\mathrm{tot}^{2}  \biggr]
\frac{1}{\tilde\rho}\biggl[K_c^{- 15 / 2} G^{15 / 2} M_\mathrm{tot}^5 \biggr]
\biggl[ G^{-1} M_\mathrm{tot}^{-1}\biggr] \tilde{M}_r^{-1} 
\frac{dp}{d\ln \xi}  
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl\{
\tilde{P} 
\tilde{r}
\tilde\rho^{-1}
\tilde{M}_r^{-1}  
\biggr\}
\frac{dp}{d\ln \xi}  \, .
</math>
  </td>
</tr>
</table>
Throughout the core, then,

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

<tr>
  <td align="right"><math>\omega^2 \biggl[\frac{r_0}{g_0}\biggr] x + (4x + p)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf}^6 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-12} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3} 
\biggl\{
\mathcal{m}_\mathrm{surf}^5 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-10} 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} 
\mathcal{m}_\mathrm{surf}^{-1} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{2} 
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr] 
\biggr\}^{-1}
\biggl[
\mathcal{m}_\mathrm{surf}^{-2} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{4} 
\biggl(\frac{3}{2\pi}\biggr)^{1/2} \xi 
\biggr]\frac{dp}{d\ln \xi}  
</math>
  </td>
</tr>

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

\biggl\{
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{5/2} 
\biggl[ \xi^{-3} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} \biggr] 
\biggr\}

\frac{dp}{d\ln \xi}  
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{2\xi^2}\biggl( 1 + \frac{1}{3}\xi^2 \biggr) 
\frac{dp}{d\ln \xi}  \, .
</math>
  </td>
</tr>
</table>
Let's relate this to the [[SSC/Stability/InstabilityOnsetOverview#Displacement_Functions_Summary|displacement functions summary]], namely,

<table border="1" align="center" width="60%" cellpadding="8"><tr><td align="left">

<div align="center"><b>Summary &hellip;</b></div>

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

<tr>
  <td align="right"><math>x_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\xi \theta^{n}}\biggr) 
\frac{d\theta}{d\xi}\biggr] \, ,</math>
  </td>
</tr>

<tr>
  <td align="right"><math>d_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
-3 \biggl\{1   
- 
\frac{(n-3)}{2}\biggl[ \frac{1}{\theta^{n+1}} \biggl( \frac{d\theta}{d\xi}\biggr)^2 \biggr] \biggr\}
\, ,
</math></td>
</tr>

<tr>
  <td align="right"><math>p_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\biggl( \frac{n+1}{n} \biggr) d_P
\, .
</math></td>
</tr>
</table>

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

From the structural solution for [[SSC/Structure/Polytropes#Primary_E-Type_Solution_2|equilibrium,]] <math>n=5</math> polytropes, we know that,

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

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

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{d\theta}{d\xi}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{\xi}{3}\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}
\, .
</math>
  </td>
</tr>
</table>
Therefore, for <math>n=5</math> structures,

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

<tr>
  <td align="right"><math>x_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\frac{6}{5}\biggl[1 - \frac{1}{2}\biggl( \frac{1}{\xi \theta^{5}}\biggr) 
\frac{\xi}{3}\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}\biggr] </math>
  </td>
</tr>

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

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
1 -  \frac{\xi^2}{15} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>d_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math> 
-3 + 3 \biggl(1 + \frac{\xi^2}{3}\biggr)^3 \frac{\xi^2}{9}\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
-3 + \frac{\xi^2}{3}
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>p_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
-\frac{18}{5}\biggl[1 - \frac{\xi^2}{9}\biggr]
\, .
</math>
  </td>
</tr>
</table>
Hence, we find that,

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

<tr>
  <td align="right"><math>\omega^2 \biggl[\frac{r_0}{g_0}\biggr] x </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{2\xi}\biggl( 1 + \frac{1}{3}\xi^2 \biggr) 
\frac{d(p_P)}{d\xi} - (4x_P + p_P) 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{2\xi}\biggl( 1 + \frac{1}{3}\xi^2 \biggr) 
\frac{4\xi}{5} 
- 4\biggl[ 1 -  \frac{\xi^2}{15} \biggr] 
+ \biggl[ \frac{18}{5}\biggl(1 - \frac{\xi^2}{9}\biggr)  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl( 1 + \frac{1}{3}\xi^2 \biggr) 
\frac{2}{5} - 4 + \frac{4\xi^2}{15}
+ \frac{18}{5} - \frac{2\xi^2}{5}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl( 1 + \frac{1}{3}\xi^2 \biggr) 
\frac{2}{5} 
- \frac{2}{5} - \frac{2\xi^2}{15}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
0 \, .
</math>
  </td>
</tr>
</table>
For completeness, note that the LHS can be rewritten as,

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

<tr>
  <td align="right"><math>\omega^2 \biggl\{\frac{r_0}{g_0}\biggr\} x </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\omega^2 x \biggl\{r_0^3 G^{-1} M_r^{-1}\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\omega^2 x \biggl\{\tilde{r}^3 \biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr]^{-3}G^{-1}
 \biggl[M_\mathrm{tot}\tilde{M}_r\biggr]^{-1}\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\omega^2 x \biggl[ \frac{\tilde{r}^3}{\tilde{M}_r} \biggr]  \biggl[ K_c^{-15 / 2} G^{13 / 2}
M_\mathrm{tot}^5 \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\omega^2 x \biggl[ \frac{\tilde{r}^3}{\tilde{M}_r} \biggr]  \biggl[ K_c^{-15} G^{13} M_\mathrm{tot}^{10} \biggr]^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\omega^2 x \biggl[ K_c^{-15} G^{13} M_\mathrm{tot}^{10} \biggr]^{1 / 2} 
\biggl[\mathcal{m}_\mathrm{surf}^{-2} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{4} \biggl(\frac{3}{2\pi}\biggr)^{1/2} \xi\biggr]^3 
\biggl\{ \mathcal{m}_\mathrm{surf}^{-1} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{2} 
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr]\biggr\}^{-1} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\omega^2 x \biggl[ K_c^{-15} G^{13} M_\mathrm{tot}^{10} \biggr]^{1 / 2} 
\biggl[\mathcal{m}_\mathrm{surf}^{-6} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{12} \biggl(\frac{3}{2\pi}\biggr)^{3/2} \xi^3\biggr] 
\biggl\{ \mathcal{m}_\mathrm{surf} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2} 
\biggl( \frac{\pi }{2\cdot 3} \biggr)^{1/2} \biggl[ \xi^{-3} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} \biggr]\biggr\} 
</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)
\omega^2 x \biggl[ K_c^{-15} G^{13} M_\mathrm{tot}^{10} \biggr]^{1 / 2} 
\biggl[\mathcal{m}_\mathrm{surf}^{-5} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{10}  \biggr] 
\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{3/2} \, .
</math>
  </td>
</tr>
</table>

===Rescale for Bonnor-Ebert-Type Analysis===
Let's rescale all these relations in such a way that the mass in the core remains constant along the sequence.

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

<tr>
  <td align="right"><math>\rho^* \equiv \tilde\rho \nu^5</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\rho \biggl[\biggl( \frac{K_c}{G} \biggr)^{3 / 2} \frac{1}{M_\mathrm{tot}} \biggr]^{-5} \biggl(\frac{M_\mathrm{core}}{M_\mathrm{tot}}\biggr)^5
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>P^* \equiv \tilde{P} \nu^6</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
P \biggl[K_c^{-10} G^{9} M_\mathrm{tot}^{6}  \biggr] 
\biggl(\frac{M_\mathrm{core}}{M_\mathrm{tot}}\biggr)^6
\, ;</math>
  </td>
</tr>

<tr>
  <td align="right"><math>r^* \equiv \tilde{r} \nu^{-2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
r \biggl[\biggl( \frac{K_c}{G} \biggr)^{5 / 2} M_\mathrm{tot}^{-2}  \biggr]\biggl(\frac{M_\mathrm{core}}{M_\mathrm{tot}}\biggr)^{-2}
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>M^* \equiv \tilde{M}_r \nu^{-1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{M_r}{M_\mathrm{tot}} \biggl(\frac{M_\mathrm{core}}{M_\mathrm{tot}}\biggr)^{-1}\, ;</math></td>
</tr>

<tr>
  <td align="right"><math>H^* \equiv \tilde{H} \nu</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
H \biggl[K_c^{-5 / 2} G^{3 / 2} M_\mathrm{tot} \biggr] \biggl(\frac{M_\mathrm{core}}{M_\mathrm{tot}}\biggr)
\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>t^* \equiv \tilde{t} \nu^{-5/2}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>t \biggl[K_c^{15} G^{-13} M_\mathrm{tot}^{-10} \biggr]^{1 / 4} \biggl(\frac{M_\mathrm{core}}{M_\mathrm{tot}}\biggr)^{-5/2}\, .</math></td>
</tr>
</table>

===Additional Relations===

====Core====
The analytically prescribed radial pressure gradient in the core can be obtained as follows.

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

<tr>
  <td align="right"><math>\frac{d\tilde{M}_r}{d\xi}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf}^{-1} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{2} 
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} 
\biggl\{ 
3\xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} 
- 
\xi^4 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf}^{-1} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{2} 
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} 
\biggl\{ 
3\xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr) 
- 
\xi^4  
\biggr\}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf}^{-1} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{2} 
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} 
\biggl\{ 
3\xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{d\xi}{d\tilde{M}_r}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\mathcal{m}_\mathrm{surf} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2} 
\biggl( \frac{\pi }{2\cdot 3} \biggr)^{1/2} 
\biggl\{ 
\frac{1}{3\xi^2 }\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{5/2} 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
Also,

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

<tr>
  <td align="right"><math>\frac{d\tilde{P}}{d\xi}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \mathcal{m}_\mathrm{surf}^6 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-12} \biggl\{
2\xi\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-4} 
\biggr\}
</math>
  </td>
</tr>
</table>

Hence,

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

<tr>
  <td align="right"><math>\frac{d\tilde{P}}{d\tilde{M}_r}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \mathcal{m}_\mathrm{surf}^6 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-12} \biggl\{
2\xi\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-4} 
\biggr\}
\cdot
\mathcal{m}_\mathrm{surf} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2} 
\biggl( \frac{\pi }{2\cdot 3} \biggr)^{1/2} 
\biggl\{ 
\frac{1}{3\xi^2 }\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{5/2} 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \mathcal{m}_\mathrm{surf}^7 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-14} \biggl( \frac{2\pi }{3^3} \biggr)^{1/2} 
\frac{1}{\xi}\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \, .
</math>
  </td>
</tr>
</table>

For comparison, in [[SSCpt2/SolutionStrategies#Solution_Strategies|hydrostatic balance]] we expect &hellip;

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

<tr>
  <td align="right">
<math>\frac{dP}{dM_r} = \frac{dP}{dr} \cdot \frac{dr}{dM_r}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{GM_r \rho}{r^2} \cdot \frac{1}{4\pi r^2\rho}
=
- \frac{GM_r }{4\pi r^4} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \frac{d\tilde{P}}{d\tilde{M}_r} = \biggl[ \frac{dP}{dM_r}\biggr] \cdot
\biggl[ K_c^{-10} G^9 M_\mathrm{tot}^6 \biggr]M_\mathrm{tot}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{G M_r }{4\pi r^4} 
\biggl[ K_c^{-10} G^9 M_\mathrm{tot}^7 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{\tilde{M}_r }{4\pi r^4} 
\biggl[ K_c^{-10} G^{10} M_\mathrm{tot}^8 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{\tilde{M}_r }{4\pi \tilde{r}^4} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \biggl\{
\mathcal{m}_\mathrm{surf}^{7} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-14} 
\biggl(\frac{2^2\pi^2}{3^2}\biggr) 
\biggr\} 
\frac{1}{4\pi} 
\biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \frac{1}{\xi} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \biggl\{
\mathcal{m}_\mathrm{surf}^{7} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-14} 
\biggr\} 
\biggl( \frac{2\cdot \pi}{ 3^3 } \biggr)^{1/2} \biggl[ \frac{1}{\xi} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr] \, .
</math>
  </td>
</tr>
</table>
<span id="Takeaway">This matches our earlier expression, as it should. </span> 

<table border="1" align="center" cellpadding="8" width="60%"><tr><td align="left">
<div align="center">'''Takeaway Expression'''</div>

<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right">
<math>
\frac{d\tilde{P}}{d\tilde{M}_r} 
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{\tilde{M}_r }{4\pi \tilde{r}^4} 
</math>
  </td>
</tr>
</table>

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

====Envelope====
Given that, for the envelope,

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

<tr>
  <td align="right">
<math>
\tilde{M}_r
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\mathcal{m}_\mathrm{surf}^{-1}~ \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} 
A\biggl[ \sin(\eta-B) - \eta\cos(\eta-B) \biggr] \, ,</math>&nbsp; &nbsp; &nbsp; and,
  </td>
</tr>

<tr>
  <td align="right">
<math>
\tilde{r}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\mathcal{m}_\mathrm{surf}^{-2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{3} \theta^{-2}_i (2\pi)^{-1/2}\eta \, ,</math>
  </td>
</tr>
</table>
we deduce that,

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

<tr>
  <td align="right">
<math>
\frac{d\tilde{P}}{d\tilde{M}_r} = - \frac{\tilde{M}_r}{4\pi \tilde{r}^4}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{1}{4\pi}\biggr)
\mathcal{m}_\mathrm{surf}^{-1}~ \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} 
A\biggl[ \eta\cos(\eta-B) -\sin(\eta-B) \biggr] \cdot
\biggl[
\mathcal{m}_\mathrm{surf}^{-2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{3} \theta^{-2}_i (2\pi)^{-1/2}\eta
\biggr]^{-4}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{1}{2^4\pi^2} \cdot \frac{2}{\pi} \cdot 2^4 \pi^4\biggr)^{1 / 2} 
\mathcal{m}_\mathrm{surf}^{7}~ \theta^{7}_i  
A\biggl[ \eta\cos(\eta-B) -\sin(\eta-B) \biggr] \cdot
\biggl[
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-12} \eta^{-4}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl( 2\pi\biggr)^{1 / 2} 
\mathcal{m}_\mathrm{surf}^{7}~ \theta^{7}_i  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-12}  \cdot
\frac{A}{\eta^4}\biggl[ \eta\cos(\eta-B) -\sin(\eta-B) \biggr] \cdot
</math>
  </td>
</tr>
</table>

As a cross-check &hellip;

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

<tr>
  <td align="right">
<math>
\frac{d\tilde{P}}{d\eta}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\mathcal{m}_\mathrm{surf}^6 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-12} \theta^{6}_i 
\biggl[2\phi \cdot \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>2\mathcal{m}_\mathrm{surf}^6 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-12} \theta^{6}_i 
\cdot \frac{A^2}{\eta^3}
\cdot
\biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr] \sin(\eta - B) \, ,</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>
\frac{d\tilde{M}_r}{d\eta}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
A \mathcal{m}_\mathrm{surf}^{-1}~ \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} 
\frac{d}{d\eta}\biggl[ \sin(\eta-B) - \eta\cos(\eta-B) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
A \mathcal{m}_\mathrm{surf}^{-1}~ \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} 
\biggl\{ 
\eta\sin(\eta-B)
\biggr\} 
\, .
</math>
  </td>
</tr>
</table>
That is,

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

<tr>
  <td align="right">
<math>
\frac{d\tilde{P}}{d\tilde{M}_r}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>2\mathcal{m}_\mathrm{surf}^6 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-12} \theta^{6}_i 
\cdot \frac{A^2}{\eta^3}
\cdot
\biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr] \sin(\eta - B) 
\biggl\{
A \mathcal{m}_\mathrm{surf}^{-1}~ \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} 
\biggl[ 
\eta\sin(\eta-B)
\biggr] 
\biggr\}^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>(2\pi)^{1/2}  \mathcal{m}_\mathrm{surf}^7 \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-12} \theta^{7}_i 
\cdot \frac{A}{\eta^4}
\cdot
\biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr]  \, .
</math>
  </td>
</tr>
</table>
<font color="red">Correct!</font>

====Time-Dependent Euler Equation====

We begin with the form of the,

<div align="center">
<span id="PGE:Euler"><font color="#770000">'''Euler Equation'''</font></span><br />

<math>\frac{dv_r}{dt} = - \frac{1}{\rho}\frac{dP}{dr} - \frac{d\Phi}{dr} </math><br />
</div>
that is broadly relevant to studies of radial oscillations in [[SSCpt1/PGE#PGE_for_Spherically_Symmetric_Configurations|spherically symmetric configurations]].  Recognizing from, for example, a [[SSC/Dynamics/FreeFall#Assembling_the_Key_Relations|related discussion]] that, <math>v_r = dr/dt</math>, and that,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\frac{d\Phi}{dr}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{GM_r}{r^2}</math></td>
</tr>
</table>

we obtain our
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="center" colspan="3">
<font color="#770000">'''Desired Form of the Euler Equation'''</font>
  </td>
</tr>
<tr>
  <td align="right"><math>\frac{d^2r}{dt^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>- \frac{1}{\rho} \frac{dP}{dr} -\frac{GM_r}{r^2} \, .</math></td>
</tr>
</table>

Given as well that,

<div align="center">
{{Math/EQ_SSmassConservation01}}
</div>
we see that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\frac{dM_r}{4\pi r^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\rho dr</math></td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{d^2r}{dt^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>- 4\pi r^2 \frac{dP}{dM_r} -\frac{GM_r}{r^2} </math></td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{1}{4\pi r^2}\cdot \frac{d^2r}{dt^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>- \frac{dP}{dM_r} -\frac{GM_r}{4\pi r^4} \, .</math></td>
</tr>
</table>

<span id="NormalizedEuler">Next,</span> if as [[#Core|above]], we multiply through by <math>\biggl[ K_c^{-10} G^9 M_\mathrm{tot}^7 \biggr]</math>, we obtain the relevant,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="center" colspan="3">
<font color="#770000">'''Normalized Euler Equation'''</font>
  </td>
</tr>
<tr>
  <td align="right"><math>\frac{1}{4\pi \tilde{r}^2}\cdot \frac{d^2\tilde{r}}{d\tilde{t}^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>- \frac{d\tilde{P}}{d\tilde{M}_r} -\frac{\tilde{M}_r}{4\pi \tilde{r}^4} \, ,</math></td>
</tr>
</table>
where, as a reminder, the dimensionless time is,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\tilde{t}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>t \biggl[K_c^{15} G^{-13} M_\mathrm{tot}^{-10} \biggr]^{1 / 4} \, .</math></td>
</tr>
</table>


<table border="1" align="center" width=80%" cellpadding="8">
<tr><td align="left">
<div align="center"><font color="red"><b>CAUTION!</b></font> &nbsp; Regarding Our Chosen Lagrangian Fluid Marker</div>

If we were to use <math>\tilde{r}</math> as our primary Lagrangian fluid marker, we would be in a position to analytically specify the function, <math>\tilde{M}_r(\tilde{r})</math>.  Here, however, we will call upon <math>\tilde{M}_r</math> rather than <math>\tilde{r}</math> to serve as the primary Lagrangian fluid marker because mass facilitates our efforts to highlight a variety of important physical properties of bipolytropic configurations. We will therefore need to specify the function, <math>\tilde{r}(\tilde{M}_r)</math> instead of <math>\tilde{M}_r(\tilde{r})</math>.  For the core, this choice does not introduce any particularly difficult computational challenges because we can invert the <math>\tilde{M}_r(\tilde{r})</math> relationship analytically to obtain &hellip;  

<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right"><math>\xi^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
3\biggl[ 3\biggl(\frac{c_m}{\tilde{M}_r}\biggr)^{2/3} - 1\biggr]^{-1} \, ,
</math>
  </td>
</tr>
</table>
where,

<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right"><math>c_m</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
m_\mathrm{surf}^{-1} \biggl(\frac{\mu_e}{\mu_c}\biggr)^2 \biggl(\frac{2\cdot 3}{\pi}\biggr)^{1 / 2} \, .
</math>
  </td>
</tr>
</table>

This is not the case for the envelope, however; we will not be able to analytically specify <math>\tilde{r}(\tilde{M}_r)</math>.  This is unfortunate, as a ''numerical'' (rather than analytic) specification will necessarily introduce additional errors into our solution of the displacement function &#8212; which already is a small and error-prone quantity.  We will nevertheless proceed along this line.
</td></tr>
</table>