Editing SSC/Structure/BiPolytropes/Analytic51

__FORCETOC__
=BiPolytrope with n<sub>c</sub> = 5 and n<sub>e</sub> = 1 (Pt 1)=
<table border="1" align="center" width="100%" colspan="8">
<tr>
  <td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/Structure/BiPolytropes/Analytic51|Part I:&nbsp; (n<sub>c</sub>,n<sub>e</sub>) = (5,1) BiPolytrope]]
&nbsp;
  </td>
  <td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/Structure/BiPolytropes/Analytic51/Pt2|Part II:&nbsp; Example Models]]
&nbsp;
  </td>
  <td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/Structure/BiPolytropes/Analytic51/Pt3|Part III:&nbsp; Limiting Mass]]
&nbsp;
  </td>
  <td align="center" bgcolor="lightblue"><br />[[SSC/Structure/BiPolytropes/Analytic51/Pt4|Part IV:&nbsp; Free Energy]]
&nbsp;
  </td>
</tr>
</table>


{| class="PGEclass" style="float:left; margin-right: 20px; border-style: solid; border-width: 3px border-color: black"
|- 
! style="height: 125px; width: 125px; background-color:white;" |<font size="-1">[[H_BookTiledMenu#MoreModels|<b>Eggleton, Faulkner<br />&amp; Cannon (1998)<br /><br />Analytic</b>]]<br />(n<sub>c</sub>, n<sub>e</sub>) = (5, 1)</font>
|}
[[File:CommentButton02.png|right|100px|Comment by J. E. Tohline on 30 March 2013:  As far as I have been able to determine, this analytic structural model has not previously been published in a refereed, archival journal.       Subsequent comment by J. E. Tohline on 23 June 2013:  Last night I stumbled upon an article by Eagleton, Faulkner, and Cannon (1998) in which this identical analytically definable bipolytrope has been presented.  Insight drawn from this article is presented in an additional subsection, below.]]
Here we construct a [[SSC/Structure/BiPolytropes#BiPolytropes|bipolytrope]] in which the core has an <math>~n_c=5</math> polytropic index and the envelope has an <math>~n_e=1</math> polytropic index.  This system is particularly interesting because the entire structure can be described by closed-form, analytic expressions.  In deriving the properties of this model, we will follow the [[SSC/Structure/BiPolytropes#Solution_Steps|general solution steps for constructing a bipolytrope]] that we have outlined elsewhere.   
&nbsp;<br />
&nbsp;<br />
&nbsp;<br />
&nbsp;<br />

==Steps 2 &amp; 3==
Based on the discussion [[SSC/Structure/Polytropes#n_.3D_5_Polytrope|presented elsewhere of the structure of an isolated <math>n=5</math> polytrope]], the core of this bipolytrope will have the following properties:
<div align="center">
<math>
\theta(\xi) = \biggl[ 1 + \frac{1}{3}\xi^2 \biggr]^{-1/2} ~~~~\Rightarrow ~~~~  \theta_i = \biggl[ 1 + \frac{1}{3}\xi_i^2 \biggr]^{-1/2} ;
</math>

<math>
\frac{d\theta}{d\xi} = - \frac{\xi}{3}\biggl[ 1 + \frac{1}{3}\xi^2 \biggr]^{-3/2} ~~~~\Rightarrow ~~~~  \biggl(\frac{d\theta}{d\xi}\biggr)_i = - \frac{\xi_i}{3}\biggl[ 1 + \frac{1}{3}\xi_i^2 \biggr]^{-3/2} \, .
</math>
</div>
The first zero of the function <math>~\theta(\xi)</math> and, hence, the surface of the corresponding isolated <math>~n=5</math> polytrope is located at <math>~\xi_s = \infty</math>.  Hence, the interface between the core and the envelope can be positioned anywhere within the range, <math>~0 < \xi_i < \infty</math>.

==Step 4:  Throughout the core (0 &le; &xi; &le; &xi;<sub>i</sub>)==
<div align="center">
<table border="0" cellpadding="3">
<tr>
  <td align="center" colspan="3">
Specify:  <math>K_c</math> and <math>\rho_0 ~\Rightarrow</math>
  </td>
  <td colspan="2">
&nbsp;
  </td>
</tr>
<tr>
  <td align="right">
<math>\rho</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\rho_0 \theta^{n_c}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\rho_0 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>P</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>K_c \rho_0^{1+1/n_c} \theta^{n_c + 1}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>K_c \rho_0^{6/5} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>r</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{(n_c + 1)K_c}{4\pi G} \biggr]^{1/2} \rho_0^{(1-n_c)/(2n_c)} \xi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{K_c}{G\rho_0^{4/5}} \biggr]^{1/2} \biggl(\frac{3}{2\pi}\biggr)^{1/2} \xi</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>M_r</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>4\pi \biggl[ \frac{(n_c + 1)K_c}{4\pi G} \biggr]^{3/2} \rho_0^{(3-n_c)/(2n_c)} \biggl(-\xi^2 \frac{d\theta}{d\xi} \biggr)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{K_c^3}{G^3 \rho_0^{2/5} } \biggr]^{1/2} \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>
</tr>
</table>

</div>

<table border="1" align="center" cellpadding="5" width="80%"><tr><td align="left">

Equations (A2) from {{ EFC98full }} &#8212; hereafter, {{ EFC98hereafter }} &#8212; present the same relations but adopt the following notations:<font color="red" size="+2"><b>*</b></font>

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

<tr>
  <td align="right"><math>P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>p_0 \theta^6</math></td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp; <math>\Rightarrow~~~</math></td>
  <td align="right"><math>K_c</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>p_0 \rho_0^{-6/5} \, ;</math></td>
</tr>

<tr>
  <td align="right"><math>\rho</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\rho_0 \theta^5</math></td>
<td align="center">&nbsp; &nbsp; &nbsp; where: </td>
  <td align="right"><math>\theta</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>a_\mathrm{efc} \biggl[a_\mathrm{efc}^2 + r^2 \biggr]^{-1 / 2} 
= \biggl[1 + \biggl(\frac{r}{a_\mathrm{efc}}\biggr)^2 \biggr]^{-1 / 2}</math></td>
</tr>

<tr>
  <td align="right"><math>a_\mathrm{efc}^2 </math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>\frac{18p_0}{4\pi G \rho_0^2}</math></td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp; <math>\Rightarrow~~~</math></td>
  <td align="right"><math>a_\mathrm{efc}^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{18K_c \rho_0^{6/5}}{4\pi G \rho_0^2}
=
3 \biggl[ \frac{K_c }{G \rho_0^{4/5}} \cdot \biggl( \frac{3}{2\pi}\biggr) \biggr]</math></td>
</tr>

<tr>
  <td align="center" colspan="3">&nbsp;</td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp; <math>\Rightarrow~~~</math></td>
  <td align="right"><math>a_\mathrm{efc}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\sqrt{3} \biggl[ \frac{K_c }{G \rho_0^{4/5}} \biggr]^{1 / 2} \biggl( \frac{3}{2\pi}\biggr)^{1 / 2}
=
\sqrt{3}\biggl(\frac{r}{\xi}\biggr)
</math></td>
</tr>

<tr>
  <td align="center" colspan="3">&nbsp;</td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp; <math>\Rightarrow~~~</math></td>
  <td align="right"><math>\frac{r}{a_\mathrm{efc}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{\xi}{\sqrt{3}} \, ;</math></td>
</tr>
</table>
Hence,

<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>
</table>
which matches our expression for the core's polytrope function, <math>\theta</math>.

----

Now, look at the {{ EFC98hereafter }} expression for the core's integrated mass.
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>M_r</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{4\pi \rho_0 a_\mathrm{efc}^3}{3} \cdot \frac{r^3}{(a_\mathrm{efc}^2 + r^2)^{3/ 2}}</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{4\pi \rho_0 }{3} \biggl[ a_\mathrm{efc}^3 \biggr] 
(r/a_\mathrm{efc})^3 \biggl[1 + (r/a_\mathrm{efc})^2 \biggr]^{-3/ 2}</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{4\pi \rho_0 }{3} \biggl[ \frac{K_c }{G \rho_0^{4/5}} \cdot \biggl( \frac{3^2}{2\pi}\biggr)  \biggr]^{3 / 2} 
\frac{\xi^3}{3^{3 / 2}} \biggl[1 + \frac{\xi^2}{3} \biggr]^{-3/ 2}</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{4\pi }{3} \biggl( \frac{3}{2\pi}\biggr)^{3 / 2} \biggl[ \frac{K_c^3}{G^3}\cdot \frac{\rho_0^2}{ \rho_0^{12/5}}  \biggr]^{1 / 2} 
\xi^3 \biggl[1 + \frac{\xi^2}{3} \biggr]^{-3/ 2}</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl[ \frac{K_c^3}{G^3\rho_0^{2/5}} \biggr]^{1 / 2}\biggl( \frac{2\cdot 3}{\pi} \biggr)^{1 / 2} 
\xi^3 \biggl(1 + \frac{\xi^2}{3} \biggr)^{-3/ 2}</math></td>
</tr>
</table>
This expression matches ours.
</td></tr></table>

==Step 5: Interface Conditions==

<div align="center">
<table border="0" cellpadding="3">
<tr>
  <td colspan="3">
&nbsp;
  </td>
  <td align="left" colspan="2">
Setting <math>n_c=5</math>, <math>n_e=1</math>, and <math>\phi_i = 1 ~~~~\Rightarrow</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{\rho_e}{\rho_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{n_c}_i \phi_i^{-n_e}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{5}_i </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl( \frac{K_e}{K_c} \biggr) </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\rho_0^{1/n_c - 1/n_e}\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-(1+1/n_e)} \theta^{1 - n_c/n_e}_i</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</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{\eta_i}{\xi_i}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{n_c + 1}{n_e+1} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c}\biggr) \theta_i^{(n_c-1)/2} \phi_i^{(1-n_e)/2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>3^{1/2} \biggl( \frac{\mu_e}{\mu_c}\biggr) \theta_i^{2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\biggl( \frac{d\phi}{d\eta} \biggr)_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{n_c + 1}{n_e + 1} \biggr]^{1/2} \theta_i^{- (n_c + 1)/2} \phi_i^{(n_e+1)/2} \biggl( \frac{d\theta}{d\xi} \biggr)_i</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>3^{1/2} \theta_i^{- 3} \biggl( \frac{d\theta}{d\xi} \biggr)_i</math>
  </td>
</tr>
</table>
</div>

==Step 6:  Envelope Solution==

Adopting equation (8) of {{ Beech88full }}, the most general solution to the <math>n=1</math> Lane-Emden equation can be written in the form,
<div align="center">
<math>
\phi = A \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr] \, ,
</math>
</div>
where <math>A</math> and <math>B</math> are constants.  The first derivative of this function is,
<div align="center">
<math>
\frac{d\phi}{d\eta} = \frac{A}{\eta^2} \biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr] \, .
</math>
</div>


<table border="1" align="center" width="80%"  cellpadding="5"><tr><td align="left" bgcolor="lightgreen">
For purposes of determining the envelope mass &#8212; see the light-green text box in [[#Step_8:_Throughout_the_envelope_(ηi_≤_η_≤_ηs)|Step #8]] below &#8212; note that,

<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right"><math>-\eta^2 \frac{d\phi}{d\eta}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>A\biggl[\sin(\eta-B) - \eta\cos(\eta - B)  \biggr]\, .</math></td>
</tr>
</table>

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

<!--
<table border="1" align="center" cellpadding="5" width="80%"><tr><td align="left">

Equations (A2) from {{ EFC98 }} present the same relations but adopt the following notations:<font color="red" size="+2"><b>*</b></font>

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

<tr>
  <td align="right"><math>P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>p_0 \theta^6</math></td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp; <math>\Rightarrow~~~</math></td>
  <td align="right"><math>K_c</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>p_0 \rho_0^{-6/5} \, ;</math></td>
</tr>
</table>

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

From Step 5, above, we know the value of the function, <math>\phi</math> and its first derivative at the interface; specifically,
<div align="center">
<math>
\phi_i = 1~~~~\mathrm{and}
~~~~\biggl( \frac{d\phi}{d\eta}\biggr)_i =3^{1/2} \theta_i^{- 3} \biggl( \frac{d\theta}{d\xi} \biggr)_i~~~~
\mathrm{at}~~~~\eta_i =3^{1/2} \xi_i \biggl( \frac{\mu_e}{\mu_c}\biggr) \theta_i^{2} \, .</math>
</div>
From this information we can determine the constants <math>A</math> and <math>B</math>; specifically,
<div align="center">
<math>
\eta_i - B = \tan^{-1}(\Lambda_i^{-1}) = \frac{\pi}{2}- \tan^{-1}(\Lambda_i) \, ,
</math>

<math>
A = \frac{\phi_i \eta_i}{\sin(\eta_i - B)} = \phi_i \eta_i (1 + \Lambda_i^2)^{1/2} \, ,
</math>
</div>
where,
<div align="center">
<math>
\Lambda_i = \frac{1}{\eta_i} + \frac{1}{\phi_i} \biggl(\frac{d\phi}{d\eta}\biggr)_i \, .
</math>
</div>

==Step 7==
The surface will be defined by the location, <math>\eta_s</math>, at which the function <math>\phi(\eta)</math> first goes to zero, that is, 
<div align="center">
<math>
\eta_s = \pi + B =  \frac{\pi}{2} + \eta_i + \tan^{-1}(\Lambda_i) \, .
</math>
</div>

<table border="1" align="center" cellpadding="5" width="80%"><tr><td align="left">

Equations (A2) from {{ EFC98 }} present the same relations but adopt the following notations:<font color="red" size="+2"><b>*</b></font>

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

<tr>
  <td align="right"><math>P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>p_0 \theta^6</math></td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp; <math>\Rightarrow~~~</math></td>
  <td align="right"><math>K_c</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>p_0 \rho_0^{-6/5} \, ;</math></td>
</tr>
</table>

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

==Step 8:  Throughout the envelope (&eta;<sub>i</sub> &le; &eta; &le; &eta;<sub>s</sub>)==
<div align="center">
<table border="0" cellpadding="3">
<tr>
  <td align="center" colspan="5">
&nbsp;
  </td>
  <td align="left" colspan="2">
Knowing:  <math>K_e/K_c</math> and <math>\rho_e/\rho_0</math> from Step 5 &nbsp; <math>\Rightarrow</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>\rho</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\rho_e \phi^{n_e}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\rho_0 \biggl(\frac{\rho_e}{\rho_0}\biggr) \phi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\rho_0 \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{5}_i \phi</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>P</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>K_e \rho_e^{1+1/n_e} \phi^{n_e + 1}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>K_c \rho_0^{6/5} \biggl(\frac{K_e \rho_0^{4/5}}{K_c}\biggr) \biggl(\frac{\rho_e}{\rho_0}\biggr)^{2} \phi^{2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>K_c \rho_0^{6/5}  \theta^{6}_i \phi^{2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>r</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{(n_e + 1)K_e}{4\pi G} \biggr]^{1/2} \rho_e^{(1-n_e)/(2n_e)} \eta</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{K_c}{G \rho_0^{4/5}} \biggr]^{1/2} \biggl( \frac{K_e \rho_0^{4/5}}{K_c} \biggr)^{1/2} (2\pi)^{-1/2}\eta</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{K_c}{G \rho_0^{4/5}} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>M_r</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>4\pi \biggl[ \frac{(n_e + 1)K_e}{4\pi G} \biggr]^{3/2} \rho_e^{(3-n_e)/(2n_e)} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{K_c^3}{G^3 \rho_0^{2/5}} \biggr]^{1/2} \biggl( \frac{K_e \rho_0^{4/5}}{K_c} \biggr)^{3/2} \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl( \frac{2}{\pi} \biggr)^{1/2} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{K_c^3}{G^3 \rho_0^{2/5}} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)</math>
  </td>
</tr>
</table>

</div>


<table border="1" align="center" width="80%"  cellpadding="5"><tr><td align="left" bgcolor="lightgreen">
<div align="center">'''n = 1 Polytropic Envelope'''</div>
Let's verify the expression for the pressure by integrating the hydrostatic-balance equation, 
<div align="center">
{{Math/EQ_SShydrostaticBalance01}} 
</div>
From our [[SSC/Structure/Polytropes|introductory discussion]] of the 
<div align="center">
<span id="LaneEmdenEquation"><font color="#770000">'''Lane-Emden Equation'''</font></span>
<br />

<math>
\frac{1}{\eta^2} \frac{d}{d\eta}\biggl(\eta^2 \frac{d\phi}{d\eta} \biggr) = \phi \, .
</math>
</div>
we appreciate that, for this <math>n=1</math> envelope, <math>\phi = A[\sin(\eta-B)/\eta]</math>, in which case,

<div align="center">
<math>\rho = \biggl[\rho_0 \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{5}_i \biggr] A \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr]</math>
</div>
and,
<div align="center">
<math>
r = \biggl[ \frac{K_c}{G \rho_0^{4/5}} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta \, .
</math>
</div>
Combining these expressions with the differential expression for <math>M_r</math>, namely, <math>dM_r/dr = 4\pi r^2 \rho</math>, we find that,
<table align="center" border="0" cellpadding="5">

<tr>
  <td align="right">
<math>
~M_r(\eta) - M_\mathrm{core}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_{r_i}^r 4\pi r^2 \rho~ dr  
</math>
  </td>  
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~4\pi \biggl\{ \biggl[ \frac{K_c}{G \rho_0^{4/5}} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2} \biggr\}^3
\biggl[\rho_0 \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{5}_i \biggr] A 
\int_{\eta_i}^{\eta} \eta\biggl[ \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>
\biggl(\frac{2}{\pi}\biggr)^{1 / 2}\biggl\{ \biggl( \frac{K_c}{G } \biggr)^{3/2}\rho_0^{-1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggr\}\theta^{-1}_i A 
\int_{(\upsilon_i + B)}^{(\upsilon+B)} (\upsilon + B)\sin(\upsilon)~ d\upsilon
</math>
  </td>  
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{2}{\pi}\biggr)^{1 / 2}\biggl\{ \biggl( \frac{K_c}{G } \biggr)^{3/2}\rho_0^{-1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggr\}\theta^{-1}_i A 
\biggl[\sin(\eta - B) - (\eta - B)\cos(\eta - B) - ( B)\cos(\eta - B)\biggr]_{\eta_i}^{\eta}
</math>
  </td>  
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{2}{\pi}\biggr)^{1 / 2}\biggl\{ \biggl( \frac{K_c}{G } \biggr)^{3/2}\rho_0^{-1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggr\}\theta^{-1}_i A 
\biggl[\sin(\eta - B) - \eta \cos(\eta - B) \biggr]_{\eta_i}^{\eta}
</math>
  </td>  
</tr>
</table>

where we have temporarily utilized the variable shift, <math>\upsilon \equiv (\eta - B)</math>.
</td></tr></table>


<table border="1" align="center" cellpadding="5" width="80%"><tr><td align="left">

An examination of their equations (A3) reveals that {{ EFC98hereafter }} continue to use <math>\theta</math> to represent the polytropic function throughout the envelope &#8212; for clarity, we will attach the subscript <math>\theta_\mathrm{efc} </math> &#8212; whereas we use <math>\phi</math>. Henceforth we will assume that these functions are interchangeable, that is, <math>\theta_\mathrm{efc} \leftrightarrow \phi</math>, and examine whether or not their various physical parameter expressions match ours. 

[[File:CommentButton02.png|right|100px|Comment by J. E. Tohline:  As detailed in the text, there appears to be a type-setting error in both of these expressions; as published by EFC98, the exponent on the coefficient of theta_i should be 6 and 5, respectively, whereas it appears as 4.]]In their Eqs. (A3), {{ EFC98hereafter }} state that, throughout the envelope,

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

<tr>
  <td align="right"><math>P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>p_0 \theta_c^4 \theta_\mathrm{efc}^2 = K_c \rho_0^{6 / 5} \theta_i^4 \phi^2</math></td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp; and,</td>
  <td align="right"><math>\rho</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{\rho_0}{\alpha} \cdot \theta_c^4 \theta_\mathrm{efc} 
= \rho_0\biggl(\frac{\mu_e}{\mu_c}\biggr)  \theta_i^4 \phi \, ,</math></td>
</tr>
</table>

where, in both expressions, we have replaced their label for the value of the polytropic function at the core-envelope interface, <math>\theta_c</math>, with our label, <math>\theta_i</math>. Both of their expressions match ours <font color="red"><b>EXCEPT &hellip; NOTE:</b></font> &nbsp; in both of their expressions, <math>\theta_i</math> is raised to the 4<sup>th</sup> power whereas, according to our derivation, this interface value should be raised to the 6<sup>th</sup> power in the expression for pressure and it should be raised to the 5<sup>th</sup> power in the expression for the density.  We are confident that our expressions are the correct ones and therefore presume that type-setting errors are present in both of the {{ EFC98hereafter }} expressions.

----

We state that the envelope's polytropic function has the form,

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

<tr>
  <td align="right"><math>\phi</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>-\frac{A}{\eta} \cdot \sin(B_\mathrm{ours}-\eta) \, ,</math></td>
</tr>
</table>
where,

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

<tr>
  <td align="right"><math>\eta</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\biggl[ \frac{G \rho_0^{4/5}}{K_c} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{2}_i (2\pi)^{1 / 2}~r \, .</math>
  </td>
</tr>
</table>

Again, as part of their set of (A3) equations, {{ EFC98hereafter }} define the parameter,

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

<tr>
  <td align="right"><math>\beta </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{3\theta_c^2}{\alpha } \biggl[a_\mathrm{efc}\biggr]^{-1} </math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>3 \theta_c^2 \biggl(\frac{\mu_e}{\mu_c}\biggr) 
\biggl\{\sqrt{3} \biggl[ \frac{K_c }{G \rho_0^{4/5}} \biggr]^{1 / 2} \biggl( \frac{3}{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{G \rho_0^{4/5}}{K_c} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{2}_i (2\pi)^{1 / 2} \, .</math>
  </td>
</tr>
</table>

This is the same expression that relates <math>r</math> to <math>\eta</math> in our derivation of the envelope's properties; specifically, for example, we can write,

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

<tr>
  <td align="right" bgcolor="pink"><math>\eta = \beta r \, .</math></td>
</tr>
</table>
And, rewriting the {{ EFC98hereafter }}  expression for the envelope's polytropic function gives,

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

<tr>
  <td align="right"><math>\theta_\mathrm{efc}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{B_\mathrm{efc}}{r} \sin\biggl[ \beta(r_s - r) \biggr]
=
\frac{\beta B_\mathrm{efc}}{\eta}\sin(\eta_s - \eta) \, .
</math>
  </td>
</tr>
</table>
This matches our expression for the envelope's polytropic function after making the substitutions:
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="center" bgcolor="pink"><math>\theta_\mathrm{efc} ~\rightarrow~ \phi \, ;</math></td>
  <td align="center">&nbsp; &nbsp; &nbsp; &nbsp;</td>
  <td align="center" bgcolor="pink"><math>\eta_s ~\rightarrow~ B_\mathrm{ours} \, ;</math></td>
  <td align="center">&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;</td>
  <td align="center" bgcolor="pink"><math>B_\mathrm{efc} ~\rightarrow~ -\frac{A}{\beta } ~~\Rightarrow 
~~ \frac{B_\mathrm{efc}}{r} = -\frac{A}{\eta }\, .</math></td>
</tr>
</table>

----

From above, our expression for the integrated mass throughout the envelope is,

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

<tr>
  <td align="right" bgcolor="lightgreen">
<math>M_r (\eta) - M_\mathrm{core} = -A \biggl[ \frac{K_c^3}{G^3 \rho_0^{2/5}} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} \biggl[\sin(B - \eta) + \eta\cos(B - \eta) \biggr]_{\eta_i}^{\eta} \, .</math>
  </td>
</tr>
</table>

According to {{ EFC98hereafter }},
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>M_r</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{4\pi \rho_0 \alpha}{9} \biggl[ B_\mathrm{efc} a_\mathrm{efc}^2 \biggr] 
\biggl\{
\sin[\beta(r_s - r)] + \beta r \cos[\beta(r_s - r)]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{4\pi \rho_0 }{9}\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\biggl[ -\frac{A a_\mathrm{efc}^2}{\beta} \biggr] 
\biggl[
\sin(B-\eta) + \eta \cos(B-\eta)
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-A \cdot
\frac{4\pi \rho_0 }{9}\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1} 
\biggl[
\sin(B-\eta) + \eta \cos(B-\eta)
\biggr]
\biggl\{ \biggl[ \frac{G \rho_0^{4/5}}{K_c} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{2}_i (2\pi)^{1 / 2}  \biggr\}^{-1}
\biggl\{ 3 \biggl[ \frac{K_c }{G \rho_0^{4/5}} \cdot \biggl( \frac{3}{2\pi}\biggr) \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-A \cdot
\biggl(\frac{2 }{\pi}\biggr)^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2} \theta^{-2}_i 
\biggl[
\sin(B-\eta) + \eta \cos(B-\eta)
\biggr]
\biggl[ \frac{K_c^3}{G^3 \rho_0^{2/5}} \biggr]^{1/2}
</math>
  </td>
</tr>
</table>

We see that our expression for the envelope mass matches the one presented in equation (A3) of {{ EFC98hereafter }} except for two things: 
<ol><li>The exponent of <math>\theta_i</math> is "-2" in their publication whereas the exponent is "-1" in our derived expression; <font color="red">Why is this?</font> 
</li>
<li>Our expression explicitly indicates that the final bracketed term should be evaluated at two separate radial locations (&eta;<sub>i</sub> and &eta;).</li></ol>
</td></tr></table>

=Related Discussions=
* [[SSC/Structure/PolytropesEmbedded#Embedded_Polytropic_Spheres|Polytropes emdeded in an external medium]]
* [[SSC/Structure/BiPolytropes#BiPolytropes|Constructing BiPolytropes]]
* [[SSC/Structure/BonnorEbert#Pressure-Bounded_Isothermal_Sphere|Bonnor-Ebert spheres]]
** [http://en.wikipedia.org/wiki/Bonnor-Ebert_mass Bonnor-Ebert Mass] according to Wikipedia
** <font color="red">Link has disappeared:</font>  [http://www.astro.umd.edu/~cychen/MATLAB/ASTR320/matlabFrom320spring2011/Bonnor-EbertSphere/html/BonnorEbert.html A MATLAB script to determine the Bonnor-Ebert Mass coefficient] developed by [http://www.astro.umd.edu/people/cychen.html Che-Yu Chen] as a graduate student in the University of Maryland Department of Astronomy
* [[SSC/Structure/LimitingMasses#Sch.C3.B6nberg-Chandrasekhar_Mass|Sch&ouml;nberg-Chandrasekhar limiting mass]]
* [[SSC/Structure/LimitingMasses#Relationship_Between_the_Bonnor-Ebert_and_Sch.C3.B6nberg-Chandrasekhar_Critical_Masses|Relationship between Bonnor-Ebert and Sch&ouml;nberg-Chandrasekhar limiting masses]]

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