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

===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>