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==Mass Profile==
==Mass Profile==


===The Core===
The core has <math>~n_c = 5 \Rightarrow \gamma_c = 1+1/n_c = 6/5</math>.  Referring to the general relation [[User:Tohline/SSC/BipolytropeGeneralization_Version2#Partitioning_the_Mass|as established in our accompanying overview]], and using <math>~\rho_0</math> to represent the central density, we can write,
The core has <math>~n_c = 5 \Rightarrow \gamma_c = 1+1/n_c = 6/5</math>.  Referring to the general relation [[User:Tohline/SSC/BipolytropeGeneralization_Version2#Partitioning_the_Mass|as established in our accompanying overview]], and using <math>~\rho_0</math> to represent the central density, we can write,
<div align="center">
<div align="center">
Line 256: Line 257:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>(\mathrm{For}~0 \leq x \leq q)</math> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<math>~M_r\biggr|_\mathrm{core} </math>
<math>~M_r </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 287: Line 287:
</table>
</table>
</div>
</div>
Note that, when <math>~x \rightarrow q</math>, <math>~M_r \rightarrow M_\mathrm{core} = \nu M_\mathrm{tot}</math>. Hence, this last expression gives,
Note that, when <math>~x \rightarrow q</math>, <math>~M_r|_\mathrm{core} \rightarrow M_\mathrm{core} = \nu M_\mathrm{tot}</math>. Hence, this last expression gives,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 327: Line 327:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>(\mathrm{For}~0 \leq x \leq q)</math> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<math>~M_r\biggr|_\mathrm{core} </math>
<math>~M_r </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 343: Line 342:
</table>
</table>
</div>
</div>
and the coefficient, <math>~a_\xi</math>, will be determined only after the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, has been determined, via the relation,
and, after the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, has been determined from the free-energy analysis, the coefficient, <math>~a_\xi</math>, can be determined via the relation,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 362: Line 361:
</div>
</div>


==Gravitational Potential Energy==
<table border="1" cellpadding="5" align="center" width="90%">
Here we follow the steps that have been [[User:Tohline/SSC/BipolytropeGeneralization_Version2#Separate_Contributions_to_Gravitational_Potential_Energy|outlined in an accompanying overview]] to determine the separate contributions to the gravitational potential energy.  Let's do the core first.  In this case, <math>~\rho_\mathrm{core}(x) = \rho_c = </math> constant &#8212; hence, also, <math>~[\rho(x)/\bar\rho]_\mathrm{core} = 1</math>.  As has been demonstrated above, the corresponding <math>~M_r</math> function is,
<tr><td align="left">
 
MORE USEFUL:
 
Letting, <math>\ell \equiv \xi/\sqrt{3}</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~a_\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{\ell_i}{q} \biggr)^2 \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\tilde\mathfrak{f}_\mathrm{Mcore}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{\bar\rho}{\rho_c} \biggr)_\mathrm{core}
=
(1 + \ell_i^2)^{-3/2} \, .
</math>
  </td>
</tr>
</table>
</div>
 
</td></tr>
</table>
 
===The Envelope===
The envelope has <math>~n_e = 1 \Rightarrow \gamma_e = 1+1/n_e = 2</math>.  Again, referring to the general relation [[User:Tohline/SSC/BipolytropeGeneralization_Version2#Partitioning_the_Mass|as established in our accompanying overview]], and continuing to use <math>~\rho_0</math> to represent the central density, we can write,
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<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 369: Line 409:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[ \frac{M_r(x)}{M_\mathrm{tot}}\biggr]_\mathrm{core} </math>
<math>(\mathrm{For}~q \leq x \leq 1)</math> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<math>~M_r </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 376: Line 417:
   <td align="left">
   <td align="left">
<math>
<math>
\biggl( \frac{\nu}{q^3} \biggr) x^3 \, .
M_\mathrm{tot} \biggl\{\nu + \biggl( \frac{1-\nu}{1-q^3} \biggr) \int_{x_i}^{x}  3
\biggl[ \frac{\rho(x)}{\bar\rho} \biggr]_\mathrm{env} 
x^2 dx \biggr\} \, .
</math>
</math>
   </td>
   </td>
Line 383: Line 426:
</table>
</table>
</div>
</div>
Hence,  
 
Drawing on the derivation of [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#BiPolytrope_with_nc_.3D_5_and_ne_.3D_1|detailed force-balance models of <math>~(n_c, n_e) = (5, 1)</math> bipolytropes]], the density profile [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Step_8:__Throughout_the_envelope_.28.2|throughout the envelope]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl[ \frac{\rho(\eta)}{\rho_0} \biggr]_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5 \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
where definitions of the constants <math>~A</math> and <math>~B</math> are given in an [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Parameter_Values|accompanying table of parameter values]], and the dimensionless radial coordinate is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<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_i^2 (2\pi)^{1/2} r \, .</math>
  </td>
</tr>
</table>
</div>
Using the same radial and mass-density normalizations as defined, above, for the core, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~r^* (\rho_0^*)^{2/5}\biggl[ \frac{G}{K_c} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2}
R_\mathrm{norm} \rho_\mathrm{norm}^{2/5}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~r^* (\rho_0^*)^{2/5} \biggl( \frac{3}{4\pi} \biggr)^{2/5}  \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2} \, .
</math>
  </td>
</tr>
</table>
</div>
Next, we set,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 389: Line 497:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~W_\mathrm{grav}\biggr|_\mathrm{core}</math>
<math>~r^*</math>
  </td>
  <td align="center">
<math>~\rightarrow~</math>
  </td>
  <td align="left">
<math>
~ x \chi_\mathrm{eq} \, ;
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\rho_0^*</math>
  </td>
  <td align="center">
<math>~\rightarrow~</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{env} \biggl( \frac{{\bar\rho}_\mathrm{env}}{\rho_\mathrm{norm}} \biggr)
= \biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{env} \frac{(1-\nu) M_\mathrm{tot}/[(1-q^3) R_\mathrm{edge}^3]_\mathrm{eq}}{M_\mathrm{tot}/R_\mathrm{norm}^3}
= \frac{1-\nu}{1-q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{env}  \chi_\mathrm{eq}^{-3} \, .
</math>
  </td>
</tr>
 
</table>
</div>
Hence, we can write,
<div align="center">
<math>~\eta = b_\eta x \, ,</math>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~b_\eta</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
~\chi_\mathrm{eq}^{-1/5} \biggl[ \frac{1-\nu}{1-q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{env}  \biggr]^{2/5}
\biggl( \frac{3}{4\pi} \biggr)^{2/5}  \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2}  \, .
</math>
  </td>
</tr>
</table>
</div>
In which case,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5 \biggl[ \frac{\sin(b_\eta x - B)}{b_\eta x} \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
so,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~M_r \biggr|_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
<td align="left">
  <td align="left">
<math>
<math>
- E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{\nu}{q^3} \biggr)
M_\mathrm{tot} \biggl\{\nu + \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq} \int_{q}^{x}  3
\int_0^{q} 3\biggl( \frac{\nu}{q^3} \biggr)x^4 dx  
A \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5 \biggl[ \frac{\sin(b_\eta x - B)}{b_\eta x} \biggr] 
x^2 dx \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\nu M_\mathrm{tot} +
M_\mathrm{tot} \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq}  \biggl( \frac{\mu_e}{\mu_c} \biggr)
\frac{3A\theta_i^5 }{b_\eta}
\int_{q}^{x} \sin(b_\eta x - B) x dx  
</math>
</math>
   </td>
   </td>
Line 409: Line 610:
<math>~=</math>
<math>~=</math>
   </td>
   </td>
  <td align="left">
  <td align="left">
<math>\nu M_\mathrm{tot} +
M_\mathrm{tot} \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq}  \biggl( \frac{\mu_e}{\mu_c} \biggr)
\frac{3A\theta_i^5 }{b_\eta}
\biggl[ - \frac{\sin(B-b_\eta x) + b_\eta x \cos(B - b_\eta x)}{b_\eta^2} \biggr]_q^x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\nu M_\mathrm{tot} +
M_\mathrm{tot}  \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq}  \biggl( \frac{\mu_e}{\mu_c} \biggr)
\frac{3A\theta_i^5 }{b_\eta^3}
\biggl[ C_1-\sin(B-b_\eta x) -xb_\eta \cos(B - b_\eta x) \biggr] \, ,
</math>
  </td>
</tr>
 
</table>
</div>
where, <math>~C_1</math> is a constant obtained by evaluating the integral at the interface <math>~(x = x_i = q)</math>, specifically,
<div align="center">
<math>
C_1 \equiv \sin(B-b_\eta q)+ b_\eta q \cos(B - b_\eta q) \, .
</math>
</div>
 
Now, this expression can be significantly simplified by drawing on earlier results of this section as well as on attributes of the corresponding detailed force-balanced model.  First, independent of the specific density profiles that define the structure of a bipolytrope, the ratio of the ''mean'' densities of the two structural regions is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\bar\rho_e}{\bar\rho_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{q^3(1-\nu)}{\nu(1-q^3)} \, .</math>
  </td>
</tr>
</table>
</div>
Hence the bracketed pre-factor of the second term of the expression for <math>~M_r|_\mathrm{env}</math> may be rewritten as,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\nu}{q^3}  \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{core} \biggr]_\mathrm{eq} \, .</math>
  </td>
</tr>
</table>
</div>
But, from the [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy5_1#The_Core|above derivation of the mass profile in the core]], we know that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>\biggl[\biggl( \frac{\rho_0} {{\bar\rho}}\biggr)_\mathrm{core} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( 1 + a_\xi q^2 \biggr)^{3/2} = \biggl( 1 + \frac{1}{3} \xi_i^2 \biggr)^{3/2} = \theta_i^{-3} \, ,
</math>
  </td>
</tr>
</table>
</div>
where the final step comes from knowledge of the expression for <math>~\theta_i</math> drawn from the detailed force-balanced model (see, for example, the associated [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Parameter_Values|''Parameter Values'' table]]).  Hence, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~M_r \biggr|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\nu M_\mathrm{tot} +
M_\mathrm{tot}  \biggl( \frac{\mu_e}{\mu_c} \biggr)
\frac{3\nu A\theta_i^2 }{(b_\eta q)^3}
\biggl[ C_1-\sin(B-b_\eta x) -xb_\eta \cos(B - b_\eta x) \biggr] \, ,
</math>
  </td>
</tr>
 
</table>
</div>
and note that the expression for the coefficient, <math>~b_\eta</math>, becomes simpler as well, specifically,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~b_\eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
<math>
- E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{\nu}{q^3} \biggr)^2 \biggl( \frac{3}{5} q^5 \biggr)  
~\chi_\mathrm{eq}^{-1/5} \biggl( \frac{3}{4\pi} \biggr)^{2/5} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2\biggl( \frac{\nu}{q^3 \theta_i^3} \biggr)^{2/5}
</math>
</math>
   </td>
   </td>
</tr>
</tr>


<tr>
  <td align="right">
<math>~\Rightarrow ~~~~\chi_\mathrm{eq} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~b_\eta^{-5} \biggl( \frac{3^4  \pi}{2^3} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^5 \frac{\nu^2 \theta_i^4}{q^6}  \, .
</math>
  </td>
</tr>
</table>
</table>
</div>
</div>


Now, let's do the envelope.  In this case, <math>~\rho_\mathrm{env}(x) = \rho_e = </math> constant; hence, also, <math>~[\rho(x)/\bar\rho]_\mathrm{env} = 1</math>.  As [[User:Tohline/SSC/BipolytropeGeneralization#ExampleMass|shown elsewhere]], the corresponding <math>~M_r</math> function is,
Next &#8212; and, again, drawing from knowledge of the [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Step_6:__Envelope_Solution|internal structure of the detailed force-balanced model]], in particular, realizing that,
<div align="center">
<math>b_\eta q = \eta_i = 3^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 \xi_i \, ,</math>
</div>
&#8212; note that the constant, <math>~C_1</math>, can be rewritten as,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 425: Line 762:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[ \frac{M_r(x)}{M_\mathrm{tot}} \biggr]_\mathrm{env} </math>
<math>~C_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~b_\eta q \cos(b_\eta q - B) - \sin(b_\eta q- B) </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\eta_i \cos(\eta_i  - B) - \sin(\eta_i - B) </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\eta_i^2}{A}  \biggl( \frac{d\phi}{d\eta} \biggr)_i  = - \frac{1}{A} \biggl( \frac{\mu_e}{\mu_c} \biggr)^2 3^{1/2} \theta_i^4 \xi_i^3</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 432: Line 805:
   <td align="left">
   <td align="left">
<math>
<math>
\nu  + \biggl(\frac{1-\nu}{1-q^3} \biggr) (x^3 - q^3) \, .
~- \frac{1}{A} \biggl( \frac{\mu_e}{\mu_c} \biggr)^2 3^{1/2} \theta_i^4 \biggl[ 3^{-1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i^{-2} b_\eta q\biggr]^3  
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- \frac{1}{3A\theta_i^2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} (b_\eta q)^3 \, ,
</math>
</math>
   </td>
   </td>
Line 438: Line 825:
</table>
</table>
</div>
</div>
Hence,
 
which means,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~M_r \biggr|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\nu M_\mathrm{tot} -
\nu M_\mathrm{tot}  \biggl\{ 1 - \frac{1}{C_1} \biggl[ \sin(B-b_\eta x) + xb_\eta \cos(B - b_\eta x) \biggr] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{\nu M_\mathrm{tot}}{C_1} \biggl[ \sin(B-b_\eta x) + xb_\eta \cos(B - b_\eta x) \biggr] \, .
</math>
  </td>
</tr>
 
</table>
</div>
 
<table border="1" cellpadding="5" align="center" width="90%">
<tr><td align="left">
 
MORE USEFUL:
 
Letting, <math>\ell \equiv \xi/\sqrt{3}</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~b_\eta = \eta_s</math>
  </td>
  <td align="center">
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
and
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
</td>
  <td align="left">
<math>~b_\eta q = \eta_i = 3\biggl( \frac{\mu_e}{\mu_c} \biggr) \ell_i (1 + \ell_i^2)^{-1}  \, ,</math>
  </td>
</tr>
</table>
</div>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\tilde\mathfrak{f}_\mathrm{Menv} \equiv \biggl( \frac{\bar\rho}{\rho_c}\biggr)_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{q^3(1-\nu)}{\nu(1-q^3)} \cdot \tilde\mathfrak{f}_\mathrm{Mcore}
</math>
  </td>
</tr>
</table>
</div>
 
</td></tr>
</table>
 
==Gravitational Potential Energy==
 
===The Core===
 
Borrowing from our derivation, above, of the mass distribution in this type of bipolytrope, the expression for the gravitational potential energy in the core that has been [[User:Tohline/SSC/BipolytropeGeneralization_Version2#Separate_Contributions_to_Gravitational_Potential_Energy|outlined in our accompanying overview]] may be written as,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 444: Line 916:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~W_\mathrm{grav}\biggr|_\mathrm{env}</math>
<math>~W_\mathrm{grav}\biggr|_\mathrm{core}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 451: Line 923:
  <td align="left">
  <td align="left">
<math>
<math>
- E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{1-\nu}{1-q^3} \biggr)
- E_\mathrm{norm} \cdot \chi^{-1} \biggl[ \frac{\nu}{q^3} \biggl(\frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq}
\biggl\{ \int_{q}^{1}   
\int_0^{q} 3x \biggl[\frac{M_r(x)}{M_\mathrm{tot}} \biggr]_\mathrm{core}  \biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{core}  dx
\biggl[ \nu -q^3 \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr]3x dx + \int_{q}^{1}   \biggl[ \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] 3x^4 dx
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
- E_\mathrm{norm} \cdot \chi^{-1} \biggl[ \frac{\nu}{q^3} \biggl( 1 + a_\xi q^2 \biggr)^{3/2}  \biggr]_\mathrm{eq}
\int_0^{q} 3x \biggl\{
\nu \biggl( \frac{x^3}{q^3} \biggr) \biggl[ \frac{ 1 + a_\xi x^2 }{ 1 + a_\xi q^2 } \biggr]^{-3/2}
\biggr\}  
\biggr\}  
\biggl( 1 + a_\xi x^2 \biggr)^{-5/2}  dx
</math>
</math>
   </td>
   </td>
Line 468: Line 956:
  <td align="left">
  <td align="left">
<math>
<math>
- E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{1-\nu}{1-q^3} \biggr)
- E_\mathrm{norm} \cdot \chi^{-1} \biggl[ 3\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3}  \biggr]_\mathrm{eq}
\biggl\{ \frac{3}{2}
\int_0^{q} x^4 \biggl( 1 + a_\xi x^2 \biggr)^{-4} dx
\biggl[ \nu  -q^3 \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] (1-q^2) + \frac{3}{5} \biggl[ \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] (1-q^5)
\biggr\}  
</math>
</math>
   </td>
   </td>
Line 485: Line 971:
  <td align="left">
  <td align="left">
<math>
<math>
- \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1}  
- E_\mathrm{norm} \cdot \chi^{-1} \biggl[ 3\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggr]_\mathrm{eq}
\biggl[ \frac{1}{\nu} \biggl( \frac{1-\nu}{1-q^3} \biggr)\biggr] \biggl\{ \frac{5}{2}
\biggl\{
\biggl[ 1  - \frac{q^3}{\nu} \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] (q-q^3) + \biggl[ \frac{q}{\nu} \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] (1-q^5)
\frac{a_\xi^{1/2} q(3a_\xi^2 q^4 - 8a_\xi q^2 - 3) + 3(a_\xi q^2 +1)^3 \tan^{-1}(a_\xi^{1/2} q)}{48 a_\xi^{5/2}(a_\xi q^2 + 1)^3}
\biggr\}  
\biggr\}  
</math>
</math>
Line 502: Line 988:
  <td align="left">
  <td align="left">
<math>
<math>
- \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1}  
- E_\mathrm{norm} \cdot \chi^{-1} \biggl[\biggl(\frac{3}{2^4}\biggr) a_\xi^{-5/2}\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggr]_\mathrm{eq}
\biggl[ \frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3} \biggr)\biggr] \biggl\{ \frac{5}{2}
\biggl[
\biggl[ 1  - \frac{q^3}{\nu} \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] \biggl(\frac{1}{q^2}-1 \biggr) + \biggl[ \frac{q^3}{\nu} \biggl(\frac{1-\nu}{1-q^3} \biggr)
a_\xi^{1/2} q(a_\xi^2 q^4 - \frac{8}{3}a_\xi q^2 - 1) (a_\xi q^2 +1)^{-3} + \tan^{-1}(a_\xi^{1/2} q)
\biggr] \biggl( \frac{1}{q^5}-1\biggr) \biggr\} \, .
\biggr] \, .
</math>
</math>
   </td>
   </td>
Line 512: Line 998:
</table>
</table>
</div>
</div>
Realizing from the above mass segregation derivation that,
 
<table border="1" cellpadding="5" align="center" width="90%">
<tr><td align="left">
 
MORE USEFUL:
 
<div align="center">
<div align="center">
<math>~\frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3} \biggr) = \frac{\rho_e}{\rho_c} \, ,</math>
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~- \chi \biggl[ \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr]_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
<td align="left">
<math>
\biggl[\biggl(\frac{3}{2^4}\biggr) \biggl( \frac{q}{\ell_i}\biggr)^{5}\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + \ell_i^2 \biggr)^{3}  \biggr]_\mathrm{eq}
\biggl[
\ell_i (\ell_i^4 - \frac{8}{3}\ell_i^2 - 1) (\ell_i^2 +1)^{-3} + \tan^{-1}\ell_i
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
<td align="left">
<math>\frac{3}{5}
\biggl[\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + \ell_i^2 \biggr)^{3}  \biggr]_\mathrm{eq}
\biggl(\frac{5}{2^4}\biggr) \biggl( \frac{q}{\ell_i}\biggr)^{5} \biggl[
\ell_i (\ell_i^4 - \frac{8}{3}\ell_i^2 - 1) (\ell_i^2 +1)^{-3} + \tan^{-1}\ell_i
\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
</td></tr>
</table>
 
But, also from our above discussion of the mass profile, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~a_\xi^{-5/2} \biggl( \frac{\nu}{q^3} \biggr)^2 (1 + a_\xi q^2)^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\chi_\mathrm{eq} \biggl( \frac{2^3 \cdot 3^6}{\pi} \biggr)^{1/2} \, .</math>
  </td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{core} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
<td align="left">
<math>
- \frac{\chi_\mathrm{eq}}{\chi} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} 
\biggl[
a_\xi^{1/2} q(a_\xi^2 q^4 - \frac{8}{3}a_\xi q^2 - 1) (a_\xi q^2 +1)^{-3} + \tan^{-1}(a_\xi^{1/2} q)
\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
</div>
this last expression can be rewritten as,
After making the substitution, <math>~(a_\xi^{1/2} q) \rightarrow x_i</math>, this expression agrees with a result for the dimensionless energy, <math>~W^*_\mathrm{core}</math>, [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|derived by Tohline in the context of detailed force-balanced bipolytropes]].
 
===The Envelope===
 
Again, borrowing from our derivation, above, of the mass distribution in this type of bipolytrope, the expression for the gravitational potential energy in the envelope that has been [[User:Tohline/SSC/BipolytropeGeneralization_Version2#Separate_Contributions_to_Gravitational_Potential_Energy|outlined in our accompanying overview]] may be written as,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 522: Line 1,094:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~W_\mathrm{grav}\biggr|_\mathrm{env}</math>
<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
<td align="left">
<math>
- \chi^{-1} \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq}
\int_{q}^{1} 3x \biggl[\frac{M_r(x)}{M_\mathrm{tot}} \biggr]_\mathrm{env}  \biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{env} dx
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 529: Line 1,116:
  <td align="left">
  <td align="left">
<math>
<math>
- \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1}  
- \chi^{-1} \biggl[ \frac{\nu}{q^3 \theta_i^3} \biggr]
\biggl[ \frac{\rho_e}{\rho_c} \biggr] \biggl\{ \frac{5}{2}
\int_{q}^{1} 3x \biggl\{ \frac{\nu}{C_1} \biggl[ \sin(B-b_\eta x) + xb_\eta \cos(B - b_\eta x) \biggr]\biggr\}   
\biggl[ 1  - \frac{\rho_e}{\rho_c}  \biggr] \biggl(\frac{1}{q^2}-1 \biggr) + \biggl[ \frac{\rho_e}{\rho_c}
\biggl\{ A \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5 \biggl[ \frac{\sin(b_\eta x - B)}{b_\eta x} \biggr] \biggr\}  dx
\biggr] \biggl( \frac{1}{q^5}-1\biggr) \biggr\}
</math>
</math>
   </td>
   </td>
Line 546: Line 1,132:
  <td align="left">
  <td align="left">
<math>
<math>
- \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1}  
- \chi^{-1} \biggl( \frac{\mu_e}{\mu_c} \biggr)  \biggl[ \frac{3\nu^2 A \theta_i^2}{b_\eta q^3 } \biggr] \frac{1}{C_1}
\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl\{ \frac{5}{2}\biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c}
\int_{q}^{1} [ xb_\eta \cos(b_\eta x-B) - \sin(b_\eta x- B)  ]\sin(b_\eta x - B)  dx
\biggr) \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr] \biggr\} \, .
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
<td align="left">
<math>
- \chi^{-1} \biggl( \frac{\mu_e}{\mu_c} \biggr) \biggl[ \frac{3\nu^2 A \theta_i^2}{b_\eta q^3 } \biggr]
\biggl[ - 3A\theta_i^2 \biggl( \frac{\mu_e}{\mu_c} \biggr) (b_\eta q)^{-3} \biggr]
\int_{q}^{1} [ xb_\eta \cos(b_\eta x-B) - \sin(b_\eta x- B)  ]\sin(b_\eta x - B)   dx
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
<td align="left">
<math>
- \chi^{-1} \biggl( \frac{\mu_e}{\mu_c} \biggr)^2  \biggl[ \frac{3^2 A^2 }{b_\eta^4 } \biggr] \biggl( \frac{\nu^2 \theta_i^4}{q^6} \biggr)
\int_{q}^{1} [\sin(b_\eta x- B) - xb_\eta \cos(b_\eta x-B) ]\sin(b_\eta x - B)  dx \, .
</math>
</math>
   </td>
   </td>
Line 555: Line 1,171:
</table>
</table>
</div>
</div>
So, when put together to obtain the total gravitational potential energy, we have,
But we also know, [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy5_1#The_Envelope|from above]], that,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 561: Line 1,177:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~W_\mathrm{grav} = W_\mathrm{grav}\biggr|_\mathrm{core} + W_\mathrm{grav}\biggr|_\mathrm{env}</math>
<math>~\chi_\mathrm{eq} = b_\eta^{-5} \biggl( \frac{3^4  \pi}{2^3} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^5 \frac{\nu^2 \theta_i^4}{q^6}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;&nbsp;&nbsp;&nbsp;<math>~~~~\Rightarrow ~~~~</math>&nbsp;&nbsp;&nbsp;&nbsp;
   </td>
   </td>
<td align="left">
  <td align="left">
<math>
<math>
- \frac{3}{5} E_\mathrm{norm} \cdot \chi^{-1} \biggl(\frac{\nu^2}{q} \biggr) f(\nu,q) \, ,
~\frac{\nu^2 \theta_i^4}{q^6} = \chi_\mathrm{eq} b_\eta^{5} \biggl( \frac{2^3}{3^4  \pi} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-5}  \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
where,
So we have,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 581: Line 1,196:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~f(\nu,q)</math>
<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
  <td align="left">
  <td align="left">
<math>
<math>
1+
- \frac{\chi_\mathrm{eq}}{\chi} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \biggl( \frac{2^3}{\pi} \biggr)^{1/2} A^2 b_\eta
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)  \biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c}
\int_{q}^{1} [\sin(b_\eta x- B) - xb_\eta \cos(b_\eta x-B) ]\sin(b_\eta x - B)  dx \, .
\biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr]  \, .
</math>
</math>
   </td>
   </td>
Line 597: Line 1,211:
</table>
</table>
</div>
</div>
(This result agrees with Tohline's earlier derivations in other sections of this H_Book, which may now be erased to avoid repetition.)


==Thermodynamic Energy Reservoir==


According to our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Step_4:__Throughout_the_core_.28.29|derivation of the properties of detailed force-balance <math>~(n_c, n_e) = (0, 0) </math> bipolytropes]], in this case the pressure throughout the core is defined by the dimensionless function,
The integral can be broken into two separate parts:
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~p_c(x)</math>
<math>~
\int_{q}^{1} \sin^2(b_\eta x - B) dx
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 612: Line 1,227:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{2\pi}{3} \biggr) \xi^2 \, ,</math>
<math>~
\frac{1}{4b_\eta} \biggl\{2 b_\eta x - \sin[2(b_\eta x-B)] \biggr\}_q^1 \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
and the pressure throughout the envelope is defined by the dimensionless function,
and,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~p_e(x)</math>
<math>~
- \int_{q}^{1} xb_\eta \cos(b_\eta x-B)\sin(b_\eta x - B) dx
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 628: Line 1,248:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>\frac{2\pi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} }
<math>~\frac{1}{8b_\eta} \biggl\{2b_\eta x \cos[2(b_\eta x - B)- \sin[2(b_\eta x-B)]  \biggr\}_q^1 \, .</math>
\biggl[ \frac{\rho_e}{\rho_0} (\xi^2 - \xi_i^2) -
2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \xi_i^3\biggl( \frac{1}{\xi} -
\frac{1}{\xi_i}\biggr)
\biggr] \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
 
(Note:  We have dropped integration constants that might result from carrying out an indefinite integral because such constants would disappear upon application of our specified limits of integration.)  When added together, they give,
where, for both functions,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 645: Line 1,259:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\xi</math>
<math>\int_q^1 \ldots ~dx</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \biggl( \frac{G\rho_0^2}{P_0} \biggr)^{1/2} R_\mathrm{edge} \biggr]_\mathrm{eq} x</math>
<math>
\frac{1}{8b_\eta} \biggl\{2b_\eta x \cos[2(b_\eta x - B)] - 3\sin[2(b_\eta x-B)]  + 4 b_\eta x \biggr\}_q^1
</math>
   </td>
   </td>
</tr>
</tr>
Line 663: Line 1,279:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[
<math>
\frac{G R_\mathrm{edge}^2}{P_0} \biggl( \frac{3 \nu M_\mathrm{tot}}{4\pi q^3 R_\mathrm{edge}^3} \biggr)^2
\frac{1}{8b_\eta} \biggl\{2b_\eta x \biggl[ 1 - 2\sin^2(b_\eta x - B) \biggr] - 3\sin[2(b_\eta x-B)] +  4 b_\eta x \biggr\}_q^1
\biggr]^{1/2}_\mathrm{eq} x</math>
</math>
   </td>
   </td>
</tr>
</tr>
Line 677: Line 1,293:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \biggl( \frac{3^2}{2^4 \pi^2} \biggr)
<math>
\frac{G M_\mathrm{tot}^2 }{P_0 R_\mathrm{edge}^4} \biggl( \frac{\nu}{q^3}\biggr)^2
\frac{1}{8b_\eta} \biggl[6 b_\eta x  - 3\sin[2(b_\eta x-B)] - 4b_\eta x \sin^2(b_\eta x - B) \biggr]_q^1 \, .
\biggr]^{1/2}_\mathrm{eq} x</math>
</math>
   </td>
   </td>
</tr>
</tr>
Line 685: Line 1,301:
</table>
</table>
</div>
</div>
So, defining the coefficient,
Hence,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~b_\xi</math>
<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
  <td align="left">
<td align="left">
<math>~\biggl( \frac{3}{2^3 \pi} \biggr)
<math>
\frac{G M_\mathrm{tot}^2 }{P_0 R_\mathrm{edge}^4} \biggl( \frac{\nu}{q^3}\biggr)^2\, ,</math>
- \frac{\chi_\mathrm{eq}}{\chi} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \biggl( \frac{1}{2^3\pi} \biggr)^{1/2} A^2
\biggl[6 b_\eta x  - 3\sin[2(b_\eta x-B)] - 4b_\eta x \sin^2(b_\eta x - B)  \biggr]_q^1 \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
such that,
 
This expression matches in detail the expression for the gravitational potential energy of the envelope derived in the context of our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|derivation of detailed force-balanced models of this bipolytrope]].
 
==Thermodynamic Energy Reservoir==
 
===The Core===
 
From our [[User:Tohline/SSC/BipolytropeGeneralization_Version2#Separate_Thermodynamic_Energy_Reservoirs|introductory discussion of the free energy of bipolytropes]], the energy contained in the core's thermodynamic reservoir may be written as,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\xi </math>
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 713: Line 1,341:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{3}{2\pi} \cdot b_\xi \biggr)^{1/2} x \, ,</math>
<math>
\frac{2}{3({\gamma_c}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_c} \biggl[ \frac{2\pi P_{ic} \chi^3}{P_\mathrm{norm}} \biggr]_\mathrm{eq} 
\biggl[ q^3 s_\mathrm{core} \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="8" align="center">
 
<tr>
  <td align="right">
<math>~q^3 s_\mathrm{core} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\int_0^q  3\biggl[\frac{1 - p_c(x)}{1-p_c(q)} \biggr]  x^2 dx \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
and remembering that, at the interface, <math>~x \rightarrow x_i = q</math>, so <math>~\xi_i = (3b_\xi/2\pi)^{1/2} q</math>, the two dimensionless pressure functions become,
defines the relevant integral over the core's pressure distribution.  According to our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Profile|derivation of the properties of detailed force-balance <math>~(n_c, n_e) = (5, 1)</math> bipolytropes]] &#8212; see also the relevant derivations [[User:Tohline/SSC/BipolytropeGeneralization_Version2#Bipolytrope_Generalization|in our accompanying overview]] &#8212;  in this case the pressure throughout the core is defined by the dimensionless function,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~p_c(x)</math>
<math>~P^* \equiv \frac{P_\mathrm{core}(\xi)}{P_0} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 729: Line 1,380:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~b_\xi x^2 \, ,</math>
<math>~\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3} \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>\Rightarrow ~~~~ 1-p_c(x) = \frac{P_\mathrm{core}(x)}{P_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 1 + a_\xi x^2 \biggr)^{-3} \, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
and,
where, <math>~a_\xi</math> is defined above in connection with our [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy5_1#Mass_Profile|derivation of the mass profile]].
The desired integral over this pressure distribution therefore gives,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="8" align="center">
 
<tr>
  <td align="right">
<math>~q^3 s_\mathrm{core} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3\biggl( 1 + a_\xi q^2 \biggr)^{3} \int_0^q \frac{x^2 dx}{(1+a_\xi x^2)^3}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3\biggl( 1 + a_\xi q^2 \biggr)^{3} \biggl\{
\frac{\tan^{-1}[a_\xi^{1/2}q]}{2^3 a_\xi^{3/2}}
+ \frac{q}{2^3 a_\xi (a_\xi q^2 +1)}
- \frac{q}{2^2 a_\xi (a_\xi q^2 +1)^2}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3}{2^3 a_\xi^{3/2}} \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggl\{
\tan^{-1}[a_\xi^{1/2}q]
+ \frac{a_\xi^{1/2}q}{(a_\xi q^2 +1)}
- \frac{2a_\xi^{1/2}q}{(a_\xi q^2 +1)^2}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~p_e(x)</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 745: Line 1,460:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>b_\xi\biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} }  
<math>~
\biggl[ \frac{\rho_e}{\rho_0} (x^2 - q^2) -
\frac{3}{2^3 a_\xi^{3/2}} \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggl[
2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{1}{x} -  
\tan^{-1}[a_\xi^{1/2}q] - a_\xi^{1/2}q ~\frac{(1 - a_\xi q^2)}{(1 + a_\xi q^2)^2}
\frac{1}{q}\biggr)  
\biggr] \, .
\biggr] \, .
</math>
</math>
Line 755: Line 1,469:
</table>
</table>
</div>
</div>
The desired integrals over these pressure distributions therefore give,
 
Next, let's examine the factor in square brackets with an "eq" subscript.  From our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Profile|derivation of the properties of detailed force-balance <math>~(n_c, n_e) = (5, 1)</math> bipolytropes]], we know that,
<div align="center">
<math>P_{ic} = K_c \rho_0^{6/5} \biggl(1 + \frac{1}{3}\xi_i^2 \biggr)^{-3} \, ,</math>
</div>
and,
<div align="center">
<math>
\chi_\mathrm{eq} = \biggl(\frac{R_\mathrm{edge}}{R_\mathrm{norm}} \biggr)_\mathrm{eq}
= \frac{1}{q}\biggl(\frac{r_i}{R_\mathrm{norm}}\biggr)_\mathrm{eq}
= \frac{1}{q}\biggl[ \frac{K_c^{1/2} G^{-1/2} \rho_0^{-2/5}}{R_\mathrm{norm}}\biggr] \biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi_i \, .
</math>
</div>
 
Hence, the relevant factor may be rewritten as,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 761: Line 1,489:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\int_0^q  \biggl[\frac{1 - p_c(x)}{1-p_c(q)} \biggr]  x^2 dx</math>
<math>\frac{2\pi P_{ic} \chi_\mathrm{eq}^3}{P_\mathrm{norm}} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 767: Line 1,495:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{1}{1-b_\xi q^2} \biggr] \int_0^q (1-b_\xi x^2)x^2 dx</math>
<math>
2\pi \biggl[ \frac{K_c \rho_0^{6/5}}{P_\mathrm{norm}}\biggr] \biggl(1 + \frac{1}{3}\xi_i^2 \biggr)^{-3}
\biggl\{ \frac{1}{q}\biggl[ \frac{K_c^{1/2} G^{-1/2} \rho_0^{-2/5}}{R_\mathrm{norm}}\biggr] \biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi_i \biggr\}^3
</math>
   </td>
   </td>
</tr>
</tr>
Line 779: Line 1,510:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{1}{1-b_\xi q^2} \biggr] \biggl[ \frac{1}{3}\cdot q^3 - \biggl( \frac{b_\xi}{5} \biggr) q^5 \biggr] </math>
<math>
\biggl( \frac{3^3}{2\pi} \biggr)^{1/2} \biggl[ \frac{K_c^{5/2} G^{-3/2}}{P_\mathrm{norm} R_\mathrm{norm}^3 }\biggr] \biggl(1 + \frac{1}{3}\xi_i^2 \biggr)^{-3}
\biggl( \frac{\xi_i }{q}\biggr)^3
</math>
   </td>
   </td>
</tr>
</tr>
Line 791: Line 1,525:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{q^3}{3} \biggl[ \frac{1}{1-b_\xi q^2} \biggr] \biggl[ 1 - \biggl( \frac{3b_\xi}{5} \biggr) q^2 \biggr]
<math>
= \frac{q^3}{3} \biggl( \frac{P_0}{P_{ic}} \biggr) \biggl[ 1 - \biggl( \frac{3b_\xi}{5} \biggr) q^2 \biggr]
\biggl( \frac{3^6}{2\pi} \biggr)^{1/2} \biggl(1 + a_\xi q^2 \biggr)^{-3} a_\xi^{3/2} \, ,
\, ;</math>
</math>
  </td>
</tr>
</table>
</div>
where, the last expression has been obtained by employing the substitution, defined above, <math>~\xi_i = (3a_\xi)^{1/2}q</math>.  Finally, then, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{2}{3({\gamma_c}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_c}
\biggl\{\biggl( \frac{3^8}{2^7\pi} \biggr)^{1/2} \biggl[
\tan^{-1}[a_\xi^{1/2}q] - a_\xi^{1/2}q ~\frac{(1 - a_\xi q^2)}{(1 + a_\xi q^2)^2} \biggr]
\biggr\} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
As it should, the term inside the curly brackets precisely matches the analytic expression for the dimensionless thermal energy of the core, <math>~S^*_\mathrm{core}</math>, that has been derived elsewhere in conjunction with our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|discussion of the detailed force-balanced structure of this bipolytrope]].
===The Envelope===
Similarly, the energy contained in the envelope's thermodynamic reservoir may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\int_q^1  \biggl[1 - p_e(x) \biggr]  x^2 dx</math>
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 805: Line 1,571:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{3}(1-q^3) - b_\xi\biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \int_q^1 
<math>
\biggl[ \frac{\rho_e}{\rho_0} (x^2 - q^2) -
\frac{2}{3({\gamma_e}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_e} \biggl( \frac{P_{ie}}{P_{ic}} \biggr)
2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{1}{x} -  
\biggl[ \frac{2\pi P_{ic} \chi^3}{P_\mathrm{norm}} \biggr]_\mathrm{eq} 
\frac{1}{q}\biggr)
\biggl[ (1-q^3) s_\mathrm{env} \biggr] \, ,
\biggr]  x^2 dx</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="8" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~(1-q^3) s_\mathrm{env} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\int_q^1  3\biggl[ 1 - p_e(x) \biggr]  x^2 dx \, ,
</math>
  </td>
</tr>
</table>
</div>
defines the relevant integral over the envelope's pressure distribution.  According to our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Profile|derivation of the properties of detailed force-balance <math>~(n_c, n_e) = (5, 1)</math> bipolytropes]] &#8212; see also the relevant derivations [[User:Tohline/SSC/BipolytropeGeneralization_Version2#Bipolytrope_Generalization|in our accompanying overview]] &#8212;  the pressure throughout the envelope is defined by the dimensionless function,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~P^* \equiv \frac{P_\mathrm{env}(\eta)}{P_0} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 821: Line 1,611:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{3}(1-q^3) - b_\xi\biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} }
<math>~\theta_i^6 \phi^2(\eta) = \theta_i^6 \biggl( \frac{A}{\eta} \biggr)^2 \sin^2(\eta-B) \, ,</math>
\biggl[ \frac{\rho_e}{\rho_0} \biggl( \frac{x^5}{5} - \frac{q^2 x^3}{3} \biggr) -
2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{x^2}{2} -  
\frac{x^3}{3q}\biggr)  
\biggr]_q^1</math>
   </td>
   </td>
</tr>
</tr>
Line 831: Line 1,617:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>\Rightarrow ~~~~ 1-p_e(x) \equiv \frac{P_\mathrm{env}(x)}{P_{ie}}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 837: Line 1,624:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{3}(1-q^3) - \frac{b_\xi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \biggl\{
<math>~
\biggl[ \frac{\rho_e}{\rho_0} \biggl( \frac{3}{5} - q^2 \biggr) -
\biggl( \frac{P_{ic}}{P_{ie}} \biggr) \biggl( \frac{P_0}{P_{ic}}  \biggr) \frac{P_\mathrm{env}(x)}{P_0}
\biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^2\biggl( 3q - 2\biggr)  
\biggr]
</math>
</math>
   </td>
   </td>
Line 850: Line 1,635:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
- \biggl[ \frac{\rho_e}{\rho_0} \biggl( -\frac{2}{5} \biggr)q^5 -
\biggl( \frac{P_{ic}}{P_{ie}}  \biggr) \biggl( \theta_i^{-6} \biggr) \theta_i^6 \biggl( \frac{A}{b_\eta x} \biggr)^2 \sin^2(b_\eta x-B) \, ,
\biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^5
\biggr] \biggr\}
</math>
</math>
   </td>
   </td>
Line 869: Line 1,652:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{3}(1-q^3) - \frac{b_\xi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \biggl\{
<math>~
\biggl[ q^2(2-3q) +q^5\biggr]
\biggl( \frac{P_{ic}}{P_{ie}} \biggr) \biggl( \frac{A}{b_\eta} \biggr)^2 \frac{\sin^2(b_\eta x-B)}{x^2} \, ,
+ \frac{\rho_e}{\rho_0}\biggl[ \biggl( \frac{3}{5} - q^2  \biggr) + q^2\biggl( 3q - 2\biggr) +\frac{2q^5}{5} -q^5\biggr] \biggl\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where, <math>~b_\eta</math> has been defined above in connection with our [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy5_1#The_Envelope|derivation of the envelope's mass profile]].  The desired integral over this pressure distribution therefore gives,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~(1-q^3)s_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 884: Line 1,671:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{3}(1-q^3) - \frac{b_\xi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \biggl[
<math>~
(2q^2 - 3q^3 +q^5)
3 \biggl( \frac{P_{ic}}{P_{ie}} \biggr) \biggl( \frac{A}{b_\eta} \biggr)^2 \int_q^1 \sin^2(b_\eta x-B) dx
+ \frac{3}{5} \cdot \frac{\rho_e}{\rho_0} ( 1 - 5q^2 + 5q^3 - q^5 ) \biggr]
</math>
</math>
   </td>
   </td>
Line 899: Line 1,685:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{3}\biggl\{ (1-q^3) +  b_\xi \biggl(\frac{P_0}{P_{ie} } \biggr) \biggl[\frac{2}{5} q^5 \mathfrak{F}
<math>~
\biggr] \biggr\} \, ,
\frac{3}{4} \biggl( \frac{P_{ic}}{P_{ie}} \biggr) \biggl( \frac{A^2}{b_\eta^3} \biggr) \biggl[ 2b_\eta x -\sin[2(b_\eta x - B)] \biggr]_q^1 \, ,
</math>
</math>
   </td>
   </td>
Line 906: Line 1,692:
</table>
</table>
</div>
</div>
where,
where, as before, we have dropped the integration constant because it cancels upon insertion of the specified integration limits.  Therefore, we have,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 912: Line 1,698:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{F} </math>
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (-2q^2 + 3q^3 - q^5) +
\frac{2}{3({\gamma_e}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_e}
\frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr] \, .
\biggl[ \frac{2\pi P_{ic} \chi^3}{P_\mathrm{norm}} \biggr]_\mathrm{eq}  
\frac{3}{4} \biggl( \frac{A^2}{b_\eta^3} \biggr) \biggl[ 2b_\eta x -\sin[2(b_\eta x - B)] \biggr]_q^1  \, .
</math>
</math>
   </td>
   </td>
Line 926: Line 1,713:
</table>
</table>
</div>
</div>
Now, drawing from our above derivation steps and discussion, we know that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~b_\eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \frac{\theta_i^2 \xi_i}{q} \, ,</math>
  </td>
</tr>
</table>
</div>
and
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
  <td align="right">
<math>\frac{2\pi P_{ic} \chi_\mathrm{eq}^3}{P_\mathrm{norm}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{3^6}{2\pi} \biggr)^{1/2} \biggl(1 + a_\xi q^2 \biggr)^{-3} a_\xi^{3/2} = \biggl( \frac{3^3}{2\pi} \biggr)^{1/2} \biggl( \frac{\theta_i^2\xi_i }{q}\biggr)^3 \, .
</math>
  </td>
</tr>
</table>
</div>


Finally, then, we can write,


Finally, then, we have,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 935: Line 1,756:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 942: Line 1,763:
   <td align="left">
   <td align="left">
<math>
<math>
\frac{4\pi/3 }{({\gamma_c}-1)} \biggl[ \frac{P_{ic} \chi^{3\gamma_c}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3-3\gamma_c}  
\frac{2}{3({\gamma_e}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_e} \biggl\{
\biggl\{ \biggl( \frac{P_0}{P_{ic}} \biggr) \biggl[ q^3 - \biggl( \frac{3b_\xi}{5} \biggr) q^5 \biggr] \biggr\}  
\frac{3}{4} A^2\biggl[ \biggl( \frac{3^3}{2\pi} \biggr)^{1/2} \biggl( \frac{\theta_i^2\xi_i }{q}\biggr)^3 \biggr]
\biggl[ 3^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \frac{\theta_i^2 \xi_i}{q} \biggr]^{-3} \biggl[ 2b_\eta x -\sin[2(b_\eta x - B)] \biggr]_q^1  \biggr\}
</math>
</math>
   </td>
   </td>
Line 950: Line 1,772:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 957: Line 1,779:
   <td align="left">
   <td align="left">
<math>
<math>
\frac{4\pi/3 }{({\gamma_e}-1)} \biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3-3\gamma_e}  
\frac{2}{3({\gamma_e}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_e} \biggl\{
\biggl\{ (1-q^3) +  b_\xi \biggl(\frac{P_0}{P_{ie} } \biggr) \biggl[\frac{2}{5} q^5 \mathfrak{F}
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} A^2  \biggl( \frac{3^2}{2^5\pi} \biggr)^{1/2} \biggl[ 2b_\eta x -\sin[2(b_\eta x - B)] \biggr]_q^1 \biggr\} \ .
\biggr]  \biggr\} \, .
</math>
</math>
   </td>
   </td>
Line 965: Line 1,786:
</table>
</table>
</div>
</div>
 
The term inside the curly brackets precisely matches the analytic expression for the dimensionless thermal energy of the envelope, <math>~S^*_\mathrm{env}</math>, that has been derived elsewhere in conjunction with our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|discussion of the detailed force-balanced structure of this bipolytrope]].
 


==Virial Theorem==
==Virial Theorem==
Line 1,236: Line 2,056:


=Related Discussions=
=Related Discussions=
* [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy0_0|Free-energy determination of equilibirum if BiPolytrope]] with <math>~n_c = 0</math> and <math>~n_e=0</math>.
* [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy0_0|Free-energy determination of equilibrium configurations for BiPolytropes]] with <math>~n_c = 0</math> and <math>~n_e=0</math>.
* [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy5_1|Free-energy determination of equilibirum if BiPolytrope]] with <math>~n_c = 5</math> and <math>~n_e=1</math>.
* [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy5_1#Free_Energy_of_BiPolytrope_with|Free-energy determination of equilibrium configurations for BiPolytropes]] with <math>~n_c = 5</math> and <math>~n_e=1</math>.
* [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#BiPolytrope_with_nc_.3D_0_and_ne_.3D_0|Analytic solution of Detailed-Force-Balance BiPolytrope]] with <math>~n_c = 0</math> and <math>~n_e=0</math>.
* [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#BiPolytrope_with_nc_.3D_0_and_ne_.3D_0|Analytic solution of Detailed-Force-Balance BiPolytrope]] with <math>~n_c = 0</math> and <math>~n_e=0</math>.
* [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#BiPolytrope_with_nc_.3D_5_and_ne_.3D_1|Analytic solution of Detailed-Force-Balance BiPolytrope]] with <math>~n_c = 5</math> and <math>~n_e=1</math>.
* [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#BiPolytrope_with_nc_.3D_5_and_ne_.3D_1|Analytic solution of Detailed-Force-Balance BiPolytrope]] with <math>~n_c = 5</math> and <math>~n_e=1</math>.
* [[User:Tohline/SSC/BipolytropeGeneralization|Old ''Bipolytrope Generalization'' derivations]].
* [[User:Tohline/SSC/BipolytropeGeneralization|Old ''Bipolytrope Generalization'' derivations]].
=See Also=
<ul>
<li>[[User:Tohline/SphericallySymmetricConfigurations/IndexFreeEnergy#Index_to_Free-Energy_Analyses|Index to a Variety of Free-Energy and/or Virial Analyses]]</li>
</ul>


{{LSU_HBook_footer}}
{{LSU_HBook_footer}}

Latest revision as of 23:20, 10 July 2016

Free Energy of BiPolytrope with <math>~(n_c, n_e) = (5, 1)</math>

Whitworth's (1981) Isothermal Free-Energy Surface
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Here we present a specific example of the equilibrium structure of a bipolytrope as determined from a free-energy analysis. The example is a bipolytrope whose core has a polytropic index, <math>~n_c = 5</math>, and whose envelope has a polytropic index, <math>~n_e = 1</math>. The details presented here build upon an overview of the free energy of bipolytropes that has been presented elsewhere.

Preliminaries

Mass Profile

The Core

The core has <math>~n_c = 5 \Rightarrow \gamma_c = 1+1/n_c = 6/5</math>. Referring to the general relation as established in our accompanying overview, and using <math>~\rho_0</math> to represent the central density, we can write,

<math>(\mathrm{For}~0 \leq x \leq q)</math>       <math>~M_r </math>

<math>~=</math>

<math> M_\mathrm{tot} \biggl( \frac{\nu}{q^3} \biggr) \biggl( \frac{\rho_0} {{\bar\rho}_\mathrm{core}}\biggr)_\mathrm{eq} \int_0^{x} 3 \biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{core} x^2 dx \, . </math>

Drawing on the derivation of detailed force-balance models of <math>~(n_c, n_e) = (5, 1)</math> bipolytropes, the density profile throughout the core is,

<math>~\biggl[ \frac{\rho(\xi)}{\rho_0} \biggr]_\mathrm{core}</math>

<math>~=</math>

<math>~\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} \, ,</math>

where the dimensionless radial coordinate is,

<math>~\xi</math>

<math>~=</math>

<math>~\biggl[ \frac{G \rho_0^{4/5}}{K_c} \biggr]^{1/2} \biggl( \frac{2\pi}{3} \biggr)^{1/2} r \, .</math>

Switching to the normalizations that have been adopted in the broad context of our discussion of configurations in virial equilibrium and inserting the adiabatic index of the core <math>~(\gamma_c = 6/5)</math> into all normalization parameters, we have,

<math>~R_\mathrm{norm} = \biggl[ \biggl(\frac{G}{K_c} \biggr) M_\mathrm{tot}^{2-\gamma} \biggr]^{1/(4-3\gamma)}</math>

<math>~\Rightarrow</math>

<math>~R_\mathrm{norm} = \biggl( \frac{G^5 M_\mathrm{tot}^4}{K_c^5} \biggr)^{1/2} \, ,</math>

<math>~\rho_\mathrm{norm} = \frac{3}{4\pi} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{1/(4-3\gamma)}</math>

<math>~\Rightarrow</math>

<math>~\rho_\mathrm{norm} = \frac{3}{4\pi} \biggl( \frac{K_c^{3}}{G^3 M_\mathrm{tot}^2} \biggr)^{5/2} \, .</math>

Hence, we can rewrite,

<math>~\xi</math>

<math>~=</math>

<math>~\biggl( \frac{r}{R_\mathrm{norm}} \biggr) \biggl( \frac{\rho_0}{\rho_\mathrm{norm}} \biggr)^{2/5} \biggl[ \frac{G }{K_c} \biggr]^{1/2} \biggl( \frac{2\pi}{3} \biggr)^{1/2} R_\mathrm{norm} \rho_\mathrm{norm}^{2/5}</math>

 

<math>~=</math>

<math>~r^* (\rho_0^*)^{2/5} \biggl[ \frac{G }{K_c} \biggr]^{1/2} \biggl( \frac{2\pi}{3} \biggr)^{1/2} \biggl( \frac{G^5 M_\mathrm{tot}^4}{K_c^5} \biggr)^{1/2} \biggl( \frac{3}{4\pi} \biggr)^{2/5} \biggl( \frac{K_c^{3}}{G^3 M_\mathrm{tot}^2} \biggr)</math>

 

<math>~=</math>

<math> ~r^* (\rho_0^*)^{2/5} \biggl[ \biggl( \frac{2\pi}{3} \biggr)^{5} \biggl( \frac{3}{4\pi} \biggr)^{4} \biggr]^{1/10} = r^* (\rho_0^*)^{2/5} \biggl[ \frac{\pi}{2^3 \cdot 3}\biggr]^{1/10} \, . </math>

Now, following the same approach as was used in our introductory discussion and appreciating that our aim here is to redefine the coordinate, <math>~\xi</math>, in terms of normalized parameters evaluated in the equilibrium configuration, we will set,

<math>~r^*</math>

<math>~\rightarrow~</math>

<math> ~ x \chi_\mathrm{eq} \, ; </math>

<math>~\rho_0^*</math>

<math>~\rightarrow~</math>

<math> \biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{core} \biggl( \frac{{\bar\rho}_\mathrm{core}}{\rho_\mathrm{norm}} \biggr) = \biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{core} \frac{\nu M_\mathrm{tot}/(q^3 R_\mathrm{edge}^3)_\mathrm{eq}}{M_\mathrm{tot}/R_\mathrm{norm}^3} = \frac{\nu}{q^3} \biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{core} \chi_\mathrm{eq}^{-3} \, . </math>

Then we can set,

<math>~\xi</math>

<math>~=</math>

<math>~(3a_\xi)^{1/2} x \, ,</math>

in which case,

<math>~\biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{core}</math>

<math>~=</math>

<math>~\biggl( 1 + a_\xi x^2 \biggr)^{-5/2} \, ,</math>

where the coefficient,

<math>~(3a_\xi)^{1/2}</math>

<math>~\equiv</math>

<math>~ \chi_\mathrm{eq} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \chi_\mathrm{eq}^{-3} \biggr]^{2/5} \biggl( \frac{\pi}{2^3 \cdot 3}\biggr)^{1/10} =\chi_\mathrm{eq}^{-1/5} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq}^{2/5} \biggl( \frac{\pi}{2^3 \cdot 3}\biggr)^{1/10} </math>

<math>\Rightarrow~~~~a_\xi</math>

<math>~\equiv</math>

<math>~ \frac{1}{3} \biggl\{ \chi_\mathrm{eq}^{-1/5} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq}^{2/5} \biggl( \frac{\pi}{2^3 \cdot 3}\biggr)^{1/10} \biggr\}^2 = \chi_\mathrm{eq}^{-2/5} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq}^{4/5} \biggl( \frac{\pi}{2^3 \cdot 3^6}\biggr)^{1/5} \, . </math>

We therefore have,

<math>~M_r\biggr|_\mathrm{core} </math>

<math>~=</math>

<math> M_\mathrm{tot} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0} Template:\bar\rho\biggr)_\mathrm{core} \biggr]_\mathrm{eq} \int_0^{x} 3 \biggl( 1 + a_\xi x^2 \biggr)^{-5/2} x^2 dx </math>

 

<math>~=</math>

<math> M_\mathrm{tot} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0} Template:\bar\rho\biggr)_\mathrm{core} \biggr]_\mathrm{eq} \biggl[ x^3\biggl( 1 + a_\xi x^2 \biggr)^{-3/2} \biggr] \, . </math>

Note that, when <math>~x \rightarrow q</math>, <math>~M_r|_\mathrm{core} \rightarrow M_\mathrm{core} = \nu M_\mathrm{tot}</math>. Hence, this last expression gives,

<math>~\nu M_\mathrm{tot}</math>

<math>~=</math>

<math> M_\mathrm{tot} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0} Template:\bar\rho\biggr)_\mathrm{core} \biggr]_\mathrm{eq} \biggl[ q^3\biggl( 1 + a_\xi q^2 \biggr)^{-3/2} \biggr] </math>

<math>\Rightarrow~~~~\biggl[\biggl( \frac{\rho_0} Template:\bar\rho\biggr)_\mathrm{core} \biggr]_\mathrm{eq}</math>

<math>~=</math>

<math> \biggl( 1 + a_\xi q^2 \biggr)^{3/2} \, . </math>

Hence, finally,

<math>~M_r\biggr|_\mathrm{core} </math>

<math>~=</math>

<math> \nu M_\mathrm{tot} \biggl( \frac{x^3}{q^3} \biggr) \biggl[ \frac{ 1 + a_\xi x^2 }{ 1 + a_\xi q^2 } \biggr]^{-3/2} \, ; </math>

and, after the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, has been determined from the free-energy analysis, the coefficient, <math>~a_\xi</math>, can be determined via the relation,

<math>~\chi_\mathrm{eq}^{2} </math>

<math>~=</math>

<math>~\biggl( \frac{\pi}{2^3 \cdot 3^6}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{4} \biggl( 1 + a_\xi q^2 \biggr)^{6} a_\xi^{-5} \, . </math>

MORE USEFUL:

Letting, <math>\ell \equiv \xi/\sqrt{3}</math>,

<math>~a_\xi</math>

<math>~=</math>

<math>~\biggl( \frac{\ell_i}{q} \biggr)^2 \, ,</math>

<math>~\tilde\mathfrak{f}_\mathrm{Mcore}</math>

<math>~=</math>

<math>~\biggl( \frac{\bar\rho}{\rho_c} \biggr)_\mathrm{core} = (1 + \ell_i^2)^{-3/2} \, . </math>

The Envelope

The envelope has <math>~n_e = 1 \Rightarrow \gamma_e = 1+1/n_e = 2</math>. Again, referring to the general relation as established in our accompanying overview, and continuing to use <math>~\rho_0</math> to represent the central density, we can write,

<math>(\mathrm{For}~q \leq x \leq 1)</math>       <math>~M_r </math>

<math>~=</math>

<math> M_\mathrm{tot} \biggl\{\nu + \biggl( \frac{1-\nu}{1-q^3} \biggr) \int_{x_i}^{x} 3 \biggl[ \frac{\rho(x)}{\bar\rho} \biggr]_\mathrm{env} x^2 dx \biggr\} \, . </math>

Drawing on the derivation of detailed force-balance models of <math>~(n_c, n_e) = (5, 1)</math> bipolytropes, the density profile throughout the envelope is,

<math>~\biggl[ \frac{\rho(\eta)}{\rho_0} \biggr]_\mathrm{env}</math>

<math>~=</math>

<math>~A \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5 \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr] \, ,</math>

where definitions of the constants <math>~A</math> and <math>~B</math> are given in an accompanying table of parameter values, and the dimensionless radial coordinate is,

<math>~\eta</math>

<math>~=</math>

<math>~\biggl[ \frac{G \rho_0^{4/5}}{K_c} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2} r \, .</math>

Using the same radial and mass-density normalizations as defined, above, for the core, we can write,

<math>~\eta</math>

<math>~=</math>

<math>~r^* (\rho_0^*)^{2/5}\biggl[ \frac{G}{K_c} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2} R_\mathrm{norm} \rho_\mathrm{norm}^{2/5}</math>

 

<math>~=</math>

<math> ~r^* (\rho_0^*)^{2/5} \biggl( \frac{3}{4\pi} \biggr)^{2/5} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2} \, . </math>

Next, we set,

<math>~r^*</math>

<math>~\rightarrow~</math>

<math> ~ x \chi_\mathrm{eq} \, ; </math>

<math>~\rho_0^*</math>

<math>~\rightarrow~</math>

<math> \biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{env} \biggl( \frac{{\bar\rho}_\mathrm{env}}{\rho_\mathrm{norm}} \biggr) = \biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{env} \frac{(1-\nu) M_\mathrm{tot}/[(1-q^3) R_\mathrm{edge}^3]_\mathrm{eq}}{M_\mathrm{tot}/R_\mathrm{norm}^3} = \frac{1-\nu}{1-q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{env} \chi_\mathrm{eq}^{-3} \, . </math>

Hence, we can write,

<math>~\eta = b_\eta x \, ,</math>

where,

<math>~b_\eta</math>

<math>~\equiv</math>

<math> ~\chi_\mathrm{eq}^{-1/5} \biggl[ \frac{1-\nu}{1-q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{env} \biggr]^{2/5} \biggl( \frac{3}{4\pi} \biggr)^{2/5} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2} \, . </math>

In which case,

<math>~\biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{env}</math>

<math>~=</math>

<math>~A \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5 \biggl[ \frac{\sin(b_\eta x - B)}{b_\eta x} \biggr] \, ,</math>

so,

<math>~M_r \biggr|_\mathrm{env}</math>

<math>~=</math>

<math> M_\mathrm{tot} \biggl\{\nu + \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq} \int_{q}^{x} 3 A \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5 \biggl[ \frac{\sin(b_\eta x - B)}{b_\eta x} \biggr] x^2 dx \biggr\} </math>

 

<math>~=</math>

<math>\nu M_\mathrm{tot} + M_\mathrm{tot} \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq} \biggl( \frac{\mu_e}{\mu_c} \biggr) \frac{3A\theta_i^5 }{b_\eta} \int_{q}^{x} \sin(b_\eta x - B) x dx </math>

 

<math>~=</math>

<math>\nu M_\mathrm{tot} + M_\mathrm{tot} \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq} \biggl( \frac{\mu_e}{\mu_c} \biggr) \frac{3A\theta_i^5 }{b_\eta} \biggl[ - \frac{\sin(B-b_\eta x) + b_\eta x \cos(B - b_\eta x)}{b_\eta^2} \biggr]_q^x </math>

 

<math>~=</math>

<math>\nu M_\mathrm{tot} + M_\mathrm{tot} \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq} \biggl( \frac{\mu_e}{\mu_c} \biggr) \frac{3A\theta_i^5 }{b_\eta^3} \biggl[ C_1-\sin(B-b_\eta x) -xb_\eta \cos(B - b_\eta x) \biggr] \, , </math>

where, <math>~C_1</math> is a constant obtained by evaluating the integral at the interface <math>~(x = x_i = q)</math>, specifically,

<math> C_1 \equiv \sin(B-b_\eta q)+ b_\eta q \cos(B - b_\eta q) \, . </math>

Now, this expression can be significantly simplified by drawing on earlier results of this section as well as on attributes of the corresponding detailed force-balanced model. First, independent of the specific density profiles that define the structure of a bipolytrope, the ratio of the mean densities of the two structural regions is,

<math>~\frac{\bar\rho_e}{\bar\rho_c}</math>

<math>~=</math>

<math>~\frac{q^3(1-\nu)}{\nu(1-q^3)} \, .</math>

Hence the bracketed pre-factor of the second term of the expression for <math>~M_r|_\mathrm{env}</math> may be rewritten as,

<math>~\biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq} </math>

<math>~=</math>

<math>~\biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{core} \biggr]_\mathrm{eq} \, .</math>

But, from the above derivation of the mass profile in the core, we know that,

<math>\biggl[\biggl( \frac{\rho_0} Template:\bar\rho\biggr)_\mathrm{core} \biggr]_\mathrm{eq}</math>

<math>~=</math>

<math> \biggl( 1 + a_\xi q^2 \biggr)^{3/2} = \biggl( 1 + \frac{1}{3} \xi_i^2 \biggr)^{3/2} = \theta_i^{-3} \, , </math>

where the final step comes from knowledge of the expression for <math>~\theta_i</math> drawn from the detailed force-balanced model (see, for example, the associated Parameter Values table). Hence, we can write,

<math>~M_r \biggr|_\mathrm{env}</math>

<math>~=</math>

<math>\nu M_\mathrm{tot} + M_\mathrm{tot} \biggl( \frac{\mu_e}{\mu_c} \biggr) \frac{3\nu A\theta_i^2 }{(b_\eta q)^3} \biggl[ C_1-\sin(B-b_\eta x) -xb_\eta \cos(B - b_\eta x) \biggr] \, , </math>

and note that the expression for the coefficient, <math>~b_\eta</math>, becomes simpler as well, specifically,

<math>~b_\eta</math>

<math>~=</math>

<math> ~\chi_\mathrm{eq}^{-1/5} \biggl( \frac{3}{4\pi} \biggr)^{2/5} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2} \biggl( \frac{\nu}{q^3 \theta_i^3} \biggr)^{2/5} </math>

<math>~\Rightarrow ~~~~\chi_\mathrm{eq} </math>

<math>~=</math>

<math> ~b_\eta^{-5} \biggl( \frac{3^4 \pi}{2^3} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^5 \frac{\nu^2 \theta_i^4}{q^6} \, . </math>

Next — and, again, drawing from knowledge of the internal structure of the detailed force-balanced model, in particular, realizing that,

<math>b_\eta q = \eta_i = 3^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 \xi_i \, ,</math>

— note that the constant, <math>~C_1</math>, can be rewritten as,

<math>~C_1</math>

<math>~=</math>

<math>~b_\eta q \cos(b_\eta q - B) - \sin(b_\eta q- B) </math>

 

<math>~=</math>

<math>~\eta_i \cos(\eta_i - B) - \sin(\eta_i - B) </math>

 

<math>~=</math>

<math>~\frac{\eta_i^2}{A} \biggl( \frac{d\phi}{d\eta} \biggr)_i = - \frac{1}{A} \biggl( \frac{\mu_e}{\mu_c} \biggr)^2 3^{1/2} \theta_i^4 \xi_i^3</math>

 

<math>~=</math>

<math> ~- \frac{1}{A} \biggl( \frac{\mu_e}{\mu_c} \biggr)^2 3^{1/2} \theta_i^4 \biggl[ 3^{-1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i^{-2} b_\eta q\biggr]^3 </math>

 

<math>~=</math>

<math> ~- \frac{1}{3A\theta_i^2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} (b_\eta q)^3 \, , </math>

which means,

<math>~M_r \biggr|_\mathrm{env}</math>

<math>~=</math>

<math>\nu M_\mathrm{tot} - \nu M_\mathrm{tot} \biggl\{ 1 - \frac{1}{C_1} \biggl[ \sin(B-b_\eta x) + xb_\eta \cos(B - b_\eta x) \biggr] \biggr\} </math>

 

<math>~=</math>

<math> \frac{\nu M_\mathrm{tot}}{C_1} \biggl[ \sin(B-b_\eta x) + xb_\eta \cos(B - b_\eta x) \biggr] \, . </math>

MORE USEFUL:

Letting, <math>\ell \equiv \xi/\sqrt{3}</math>,

<math>~b_\eta = \eta_s</math>

       and       

<math>~b_\eta q = \eta_i = 3\biggl( \frac{\mu_e}{\mu_c} \biggr) \ell_i (1 + \ell_i^2)^{-1} \, ,</math>

<math>~\tilde\mathfrak{f}_\mathrm{Menv} \equiv \biggl( \frac{\bar\rho}{\rho_c}\biggr)_\mathrm{env}</math>

<math>~=</math>

<math>~ \frac{q^3(1-\nu)}{\nu(1-q^3)} \cdot \tilde\mathfrak{f}_\mathrm{Mcore} </math>

Gravitational Potential Energy

The Core

Borrowing from our derivation, above, of the mass distribution in this type of bipolytrope, the expression for the gravitational potential energy in the core that has been outlined in our accompanying overview may be written as,

<math>~W_\mathrm{grav}\biggr|_\mathrm{core}</math>

<math>~=</math>

<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl[ \frac{\nu}{q^3} \biggl(\frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq} \int_0^{q} 3x \biggl[\frac{M_r(x)}{M_\mathrm{tot}} \biggr]_\mathrm{core} \biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{core} dx </math>

 

<math>~=</math>

<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl[ \frac{\nu}{q^3} \biggl( 1 + a_\xi q^2 \biggr)^{3/2} \biggr]_\mathrm{eq} \int_0^{q} 3x \biggl\{ \nu \biggl( \frac{x^3}{q^3} \biggr) \biggl[ \frac{ 1 + a_\xi x^2 }{ 1 + a_\xi q^2 } \biggr]^{-3/2} \biggr\} \biggl( 1 + a_\xi x^2 \biggr)^{-5/2} dx </math>

 

<math>~=</math>

<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl[ 3\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggr]_\mathrm{eq} \int_0^{q} x^4 \biggl( 1 + a_\xi x^2 \biggr)^{-4} dx </math>

 

<math>~=</math>

<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl[ 3\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggr]_\mathrm{eq} \biggl\{ \frac{a_\xi^{1/2} q(3a_\xi^2 q^4 - 8a_\xi q^2 - 3) + 3(a_\xi q^2 +1)^3 \tan^{-1}(a_\xi^{1/2} q)}{48 a_\xi^{5/2}(a_\xi q^2 + 1)^3} \biggr\} </math>

 

<math>~=</math>

<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl[\biggl(\frac{3}{2^4}\biggr) a_\xi^{-5/2}\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggr]_\mathrm{eq} \biggl[ a_\xi^{1/2} q(a_\xi^2 q^4 - \frac{8}{3}a_\xi q^2 - 1) (a_\xi q^2 +1)^{-3} + \tan^{-1}(a_\xi^{1/2} q) \biggr] \, . </math>

MORE USEFUL:

<math>~- \chi \biggl[ \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr]_\mathrm{core}</math>

<math>~=</math>

<math> \biggl[\biggl(\frac{3}{2^4}\biggr) \biggl( \frac{q}{\ell_i}\biggr)^{5}\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + \ell_i^2 \biggr)^{3} \biggr]_\mathrm{eq} \biggl[ \ell_i (\ell_i^4 - \frac{8}{3}\ell_i^2 - 1) (\ell_i^2 +1)^{-3} + \tan^{-1}\ell_i \biggr] </math>

 

<math>~=</math>

<math>\frac{3}{5} \biggl[\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + \ell_i^2 \biggr)^{3} \biggr]_\mathrm{eq} \biggl(\frac{5}{2^4}\biggr) \biggl( \frac{q}{\ell_i}\biggr)^{5} \biggl[ \ell_i (\ell_i^4 - \frac{8}{3}\ell_i^2 - 1) (\ell_i^2 +1)^{-3} + \tan^{-1}\ell_i \biggr] \, . </math>

But, also from our above discussion of the mass profile, we can write,

<math>~a_\xi^{-5/2} \biggl( \frac{\nu}{q^3} \biggr)^2 (1 + a_\xi q^2)^3</math>

<math>~=</math>

<math>~\chi_\mathrm{eq} \biggl( \frac{2^3 \cdot 3^6}{\pi} \biggr)^{1/2} \, .</math>

Hence,

<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{core} </math>

<math>~=</math>

<math> - \frac{\chi_\mathrm{eq}}{\chi} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ a_\xi^{1/2} q(a_\xi^2 q^4 - \frac{8}{3}a_\xi q^2 - 1) (a_\xi q^2 +1)^{-3} + \tan^{-1}(a_\xi^{1/2} q) \biggr] \, . </math>

After making the substitution, <math>~(a_\xi^{1/2} q) \rightarrow x_i</math>, this expression agrees with a result for the dimensionless energy, <math>~W^*_\mathrm{core}</math>, derived by Tohline in the context of detailed force-balanced bipolytropes.

The Envelope

Again, borrowing from our derivation, above, of the mass distribution in this type of bipolytrope, the expression for the gravitational potential energy in the envelope that has been outlined in our accompanying overview may be written as,

<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>

<math>~=</math>

<math> - \chi^{-1} \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq} \int_{q}^{1} 3x \biggl[\frac{M_r(x)}{M_\mathrm{tot}} \biggr]_\mathrm{env} \biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{env} dx </math>

 

<math>~=</math>

<math> - \chi^{-1} \biggl[ \frac{\nu}{q^3 \theta_i^3} \biggr] \int_{q}^{1} 3x \biggl\{ \frac{\nu}{C_1} \biggl[ \sin(B-b_\eta x) + xb_\eta \cos(B - b_\eta x) \biggr]\biggr\} \biggl\{ A \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5 \biggl[ \frac{\sin(b_\eta x - B)}{b_\eta x} \biggr] \biggr\} dx </math>

 

<math>~=</math>

<math> - \chi^{-1} \biggl( \frac{\mu_e}{\mu_c} \biggr) \biggl[ \frac{3\nu^2 A \theta_i^2}{b_\eta q^3 } \biggr] \frac{1}{C_1} \int_{q}^{1} [ xb_\eta \cos(b_\eta x-B) - \sin(b_\eta x- B) ]\sin(b_\eta x - B) dx </math>

 

<math>~=</math>

<math> - \chi^{-1} \biggl( \frac{\mu_e}{\mu_c} \biggr) \biggl[ \frac{3\nu^2 A \theta_i^2}{b_\eta q^3 } \biggr] \biggl[ - 3A\theta_i^2 \biggl( \frac{\mu_e}{\mu_c} \biggr) (b_\eta q)^{-3} \biggr] \int_{q}^{1} [ xb_\eta \cos(b_\eta x-B) - \sin(b_\eta x- B) ]\sin(b_\eta x - B) dx </math>

 

<math>~=</math>

<math> - \chi^{-1} \biggl( \frac{\mu_e}{\mu_c} \biggr)^2 \biggl[ \frac{3^2 A^2 }{b_\eta^4 } \biggr] \biggl( \frac{\nu^2 \theta_i^4}{q^6} \biggr) \int_{q}^{1} [\sin(b_\eta x- B) - xb_\eta \cos(b_\eta x-B) ]\sin(b_\eta x - B) dx \, . </math>

But we also know, from above, that,

<math>~\chi_\mathrm{eq} = b_\eta^{-5} \biggl( \frac{3^4 \pi}{2^3} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^5 \frac{\nu^2 \theta_i^4}{q^6}</math>

    <math>~~~~\Rightarrow ~~~~</math>    

<math> ~\frac{\nu^2 \theta_i^4}{q^6} = \chi_\mathrm{eq} b_\eta^{5} \biggl( \frac{2^3}{3^4 \pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-5} \, . </math>

So we have,

<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>

<math>~=</math>

<math> - \frac{\chi_\mathrm{eq}}{\chi} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \biggl( \frac{2^3}{\pi} \biggr)^{1/2} A^2 b_\eta \int_{q}^{1} [\sin(b_\eta x- B) - xb_\eta \cos(b_\eta x-B) ]\sin(b_\eta x - B) dx \, . </math>


The integral can be broken into two separate parts:

<math>~ \int_{q}^{1} \sin^2(b_\eta x - B) dx </math>

<math>~=</math>

<math>~ \frac{1}{4b_\eta} \biggl\{2 b_\eta x - \sin[2(b_\eta x-B)] \biggr\}_q^1 \, , </math>

and,

<math>~ - \int_{q}^{1} xb_\eta \cos(b_\eta x-B)\sin(b_\eta x - B) dx </math>

<math>~=</math>

<math>~\frac{1}{8b_\eta} \biggl\{2b_\eta x \cos[2(b_\eta x - B)] - \sin[2(b_\eta x-B)] \biggr\}_q^1 \, .</math>

(Note: We have dropped integration constants that might result from carrying out an indefinite integral because such constants would disappear upon application of our specified limits of integration.) When added together, they give,

<math>\int_q^1 \ldots ~dx</math>

<math>~=</math>

<math> \frac{1}{8b_\eta} \biggl\{2b_\eta x \cos[2(b_\eta x - B)] - 3\sin[2(b_\eta x-B)] + 4 b_\eta x \biggr\}_q^1 </math>

 

<math>~=</math>

<math> \frac{1}{8b_\eta} \biggl\{2b_\eta x \biggl[ 1 - 2\sin^2(b_\eta x - B) \biggr] - 3\sin[2(b_\eta x-B)] + 4 b_\eta x \biggr\}_q^1 </math>

 

<math>~=</math>

<math> \frac{1}{8b_\eta} \biggl[6 b_\eta x - 3\sin[2(b_\eta x-B)] - 4b_\eta x \sin^2(b_\eta x - B) \biggr]_q^1 \, . </math>

Hence,

<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>

<math>~=</math>

<math> - \frac{\chi_\mathrm{eq}}{\chi} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \biggl( \frac{1}{2^3\pi} \biggr)^{1/2} A^2 \biggl[6 b_\eta x - 3\sin[2(b_\eta x-B)] - 4b_\eta x \sin^2(b_\eta x - B) \biggr]_q^1 \, . </math>

This expression matches in detail the expression for the gravitational potential energy of the envelope derived in the context of our derivation of detailed force-balanced models of this bipolytrope.

Thermodynamic Energy Reservoir

The Core

From our introductory discussion of the free energy of bipolytropes, the energy contained in the core's thermodynamic reservoir may be written as,

<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>

<math>~=</math>

<math> \frac{2}{3({\gamma_c}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_c} \biggl[ \frac{2\pi P_{ic} \chi^3}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \biggl[ q^3 s_\mathrm{core} \biggr] \, , </math>

where,

<math>~q^3 s_\mathrm{core} </math>

<math>~\equiv</math>

<math>~ \int_0^q 3\biggl[\frac{1 - p_c(x)}{1-p_c(q)} \biggr] x^2 dx \, , </math>

defines the relevant integral over the core's pressure distribution. According to our derivation of the properties of detailed force-balance <math>~(n_c, n_e) = (5, 1)</math> bipolytropes — see also the relevant derivations in our accompanying overview — in this case the pressure throughout the core is defined by the dimensionless function,

<math>~P^* \equiv \frac{P_\mathrm{core}(\xi)}{P_0} </math>

<math>~=</math>

<math>~\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3} \, ,</math>

<math>\Rightarrow ~~~~ 1-p_c(x) = \frac{P_\mathrm{core}(x)}{P_0} </math>

<math>~=</math>

<math>~\biggl( 1 + a_\xi x^2 \biggr)^{-3} \, ,</math>

where, <math>~a_\xi</math> is defined above in connection with our derivation of the mass profile. The desired integral over this pressure distribution therefore gives,

<math>~q^3 s_\mathrm{core} </math>

<math>~=</math>

<math>~ 3\biggl( 1 + a_\xi q^2 \biggr)^{3} \int_0^q \frac{x^2 dx}{(1+a_\xi x^2)^3} </math>

 

<math>~=</math>

<math>~ 3\biggl( 1 + a_\xi q^2 \biggr)^{3} \biggl\{ \frac{\tan^{-1}[a_\xi^{1/2}q]}{2^3 a_\xi^{3/2}} + \frac{q}{2^3 a_\xi (a_\xi q^2 +1)} - \frac{q}{2^2 a_\xi (a_\xi q^2 +1)^2} \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{3}{2^3 a_\xi^{3/2}} \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggl\{ \tan^{-1}[a_\xi^{1/2}q] + \frac{a_\xi^{1/2}q}{(a_\xi q^2 +1)} - \frac{2a_\xi^{1/2}q}{(a_\xi q^2 +1)^2} \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{3}{2^3 a_\xi^{3/2}} \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggl[ \tan^{-1}[a_\xi^{1/2}q] - a_\xi^{1/2}q ~\frac{(1 - a_\xi q^2)}{(1 + a_\xi q^2)^2} \biggr] \, . </math>

Next, let's examine the factor in square brackets with an "eq" subscript. From our derivation of the properties of detailed force-balance <math>~(n_c, n_e) = (5, 1)</math> bipolytropes, we know that,

<math>P_{ic} = K_c \rho_0^{6/5} \biggl(1 + \frac{1}{3}\xi_i^2 \biggr)^{-3} \, ,</math>

and,

<math> \chi_\mathrm{eq} = \biggl(\frac{R_\mathrm{edge}}{R_\mathrm{norm}} \biggr)_\mathrm{eq} = \frac{1}{q}\biggl(\frac{r_i}{R_\mathrm{norm}}\biggr)_\mathrm{eq} = \frac{1}{q}\biggl[ \frac{K_c^{1/2} G^{-1/2} \rho_0^{-2/5}}{R_\mathrm{norm}}\biggr] \biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi_i \, . </math>

Hence, the relevant factor may be rewritten as,

<math>\frac{2\pi P_{ic} \chi_\mathrm{eq}^3}{P_\mathrm{norm}} </math>

<math>~=</math>

<math> 2\pi \biggl[ \frac{K_c \rho_0^{6/5}}{P_\mathrm{norm}}\biggr] \biggl(1 + \frac{1}{3}\xi_i^2 \biggr)^{-3} \biggl\{ \frac{1}{q}\biggl[ \frac{K_c^{1/2} G^{-1/2} \rho_0^{-2/5}}{R_\mathrm{norm}}\biggr] \biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi_i \biggr\}^3 </math>

 

<math>~=</math>

<math> \biggl( \frac{3^3}{2\pi} \biggr)^{1/2} \biggl[ \frac{K_c^{5/2} G^{-3/2}}{P_\mathrm{norm} R_\mathrm{norm}^3 }\biggr] \biggl(1 + \frac{1}{3}\xi_i^2 \biggr)^{-3} \biggl( \frac{\xi_i }{q}\biggr)^3 </math>

 

<math>~=</math>

<math> \biggl( \frac{3^6}{2\pi} \biggr)^{1/2} \biggl(1 + a_\xi q^2 \biggr)^{-3} a_\xi^{3/2} \, , </math>

where, the last expression has been obtained by employing the substitution, defined above, <math>~\xi_i = (3a_\xi)^{1/2}q</math>. Finally, then, we have,

<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>

<math>~=</math>

<math> \frac{2}{3({\gamma_c}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_c} \biggl\{\biggl( \frac{3^8}{2^7\pi} \biggr)^{1/2} \biggl[ \tan^{-1}[a_\xi^{1/2}q] - a_\xi^{1/2}q ~\frac{(1 - a_\xi q^2)}{(1 + a_\xi q^2)^2} \biggr] \biggr\} \, . </math>

As it should, the term inside the curly brackets precisely matches the analytic expression for the dimensionless thermal energy of the core, <math>~S^*_\mathrm{core}</math>, that has been derived elsewhere in conjunction with our discussion of the detailed force-balanced structure of this bipolytrope.


The Envelope

Similarly, the energy contained in the envelope's thermodynamic reservoir may be written as,

<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>

<math>~=</math>

<math> \frac{2}{3({\gamma_e}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_e} \biggl( \frac{P_{ie}}{P_{ic}} \biggr) \biggl[ \frac{2\pi P_{ic} \chi^3}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \biggl[ (1-q^3) s_\mathrm{env} \biggr] \, , </math>

where,

<math>~(1-q^3) s_\mathrm{env} </math>

<math>~\equiv</math>

<math>~ \int_q^1 3\biggl[ 1 - p_e(x) \biggr] x^2 dx \, , </math>

defines the relevant integral over the envelope's pressure distribution. According to our derivation of the properties of detailed force-balance <math>~(n_c, n_e) = (5, 1)</math> bipolytropes — see also the relevant derivations in our accompanying overview — the pressure throughout the envelope is defined by the dimensionless function,

<math>~P^* \equiv \frac{P_\mathrm{env}(\eta)}{P_0} </math>

<math>~=</math>

<math>~\theta_i^6 \phi^2(\eta) = \theta_i^6 \biggl( \frac{A}{\eta} \biggr)^2 \sin^2(\eta-B) \, ,</math>

<math>\Rightarrow ~~~~ 1-p_e(x) \equiv \frac{P_\mathrm{env}(x)}{P_{ie}} </math>

<math>~=</math>

<math>~ \biggl( \frac{P_{ic}}{P_{ie}} \biggr) \biggl( \frac{P_0}{P_{ic}} \biggr) \frac{P_\mathrm{env}(x)}{P_0} </math>

 

<math>~=</math>

<math>~ \biggl( \frac{P_{ic}}{P_{ie}} \biggr) \biggl( \theta_i^{-6} \biggr) \theta_i^6 \biggl( \frac{A}{b_\eta x} \biggr)^2 \sin^2(b_\eta x-B) \, , </math>

 

<math>~=</math>

<math>~ \biggl( \frac{P_{ic}}{P_{ie}} \biggr) \biggl( \frac{A}{b_\eta} \biggr)^2 \frac{\sin^2(b_\eta x-B)}{x^2} \, , </math>

where, <math>~b_\eta</math> has been defined above in connection with our derivation of the envelope's mass profile. The desired integral over this pressure distribution therefore gives,

<math>~(1-q^3)s_\mathrm{env}</math>

<math>~=</math>

<math>~ 3 \biggl( \frac{P_{ic}}{P_{ie}} \biggr) \biggl( \frac{A}{b_\eta} \biggr)^2 \int_q^1 \sin^2(b_\eta x-B) dx </math>

 

<math>~=</math>

<math>~ \frac{3}{4} \biggl( \frac{P_{ic}}{P_{ie}} \biggr) \biggl( \frac{A^2}{b_\eta^3} \biggr) \biggl[ 2b_\eta x -\sin[2(b_\eta x - B)] \biggr]_q^1 \, , </math>

where, as before, we have dropped the integration constant because it cancels upon insertion of the specified integration limits. Therefore, we have,

<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>

<math>~=</math>

<math> \frac{2}{3({\gamma_e}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_e} \biggl[ \frac{2\pi P_{ic} \chi^3}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \frac{3}{4} \biggl( \frac{A^2}{b_\eta^3} \biggr) \biggl[ 2b_\eta x -\sin[2(b_\eta x - B)] \biggr]_q^1 \, . </math>

Now, drawing from our above derivation steps and discussion, we know that,

<math>~b_\eta</math>

<math>~=</math>

<math>~3^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \frac{\theta_i^2 \xi_i}{q} \, ,</math>

and

<math>\frac{2\pi P_{ic} \chi_\mathrm{eq}^3}{P_\mathrm{norm}} </math>

<math>~=</math>

<math> \biggl( \frac{3^6}{2\pi} \biggr)^{1/2} \biggl(1 + a_\xi q^2 \biggr)^{-3} a_\xi^{3/2} = \biggl( \frac{3^3}{2\pi} \biggr)^{1/2} \biggl( \frac{\theta_i^2\xi_i }{q}\biggr)^3 \, . </math>

Finally, then, we can write,

<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>

<math>~=</math>

<math> \frac{2}{3({\gamma_e}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_e} \biggl\{ \frac{3}{4} A^2\biggl[ \biggl( \frac{3^3}{2\pi} \biggr)^{1/2} \biggl( \frac{\theta_i^2\xi_i }{q}\biggr)^3 \biggr] \biggl[ 3^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \frac{\theta_i^2 \xi_i}{q} \biggr]^{-3} \biggl[ 2b_\eta x -\sin[2(b_\eta x - B)] \biggr]_q^1 \biggr\} </math>

 

<math>~=</math>

<math> \frac{2}{3({\gamma_e}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_e} \biggl\{ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} A^2 \biggl( \frac{3^2}{2^5\pi} \biggr)^{1/2} \biggl[ 2b_\eta x -\sin[2(b_\eta x - B)] \biggr]_q^1 \biggr\} \ . </math>

The term inside the curly brackets precisely matches the analytic expression for the dimensionless thermal energy of the envelope, <math>~S^*_\mathrm{env}</math>, that has been derived elsewhere in conjunction with our discussion of the detailed force-balanced structure of this bipolytrope.

Virial Theorem

As has been shown in our accompanying overview, the condition for equilibrium based on a free-energy analysis — that is, the virial theorem — is,

<math>~\mathcal{A}</math>

<math>~=</math>

<math>~\mathcal{B}_\mathrm{core} \chi_\mathrm{eq}^{4-3\gamma_c} + \mathcal{B}_\mathrm{env} \chi_\mathrm{eq}^{4-3\gamma_e} </math>

 

<math>~=</math>

<math>~\frac{4\pi}{3} \biggl[ \frac{P_i R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} [ q^3 s_\mathrm{core} + (1-q^3) s_\mathrm{env} ] \, . </math>

For <math>~(n_c, n_e) = (0, 0) </math> bipolytropes, the relevant coefficient functions are,

<math>~\mathcal{A}</math>

<math>~=</math>

<math>~\frac{1}{5} \biggl(\frac{\nu^2}{q}\biggr) f \, ,</math>

<math>~q^3 s_\mathrm{core}</math>

<math>~=</math>

<math>~ q^3 \biggl(\frac{P_0}{P_{ic}} \biggr) \biggl[ 1 - \frac{3}{5}q^2 b_\xi\biggr] \, , </math>

<math>~(1-q^3) s_\mathrm{env}</math>

<math>~=</math>

<math>~ (1-q^3) + \biggl(\frac{P_0}{P_{ie}} \biggr) \frac{2}{5} q^5 \mathfrak{F} b_\xi \, , </math>

where,

<math>~f</math>

<math>~\equiv</math>

<math> 1+ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr] \, , </math>

<math>~\mathfrak{F} </math>

<math>~\equiv</math>

<math>~ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (-2q^2 + 3q^3 - q^5) + \frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr] \, , </math>

<math>~\frac{P_{ic}}{P_0}</math>

<math>~=</math>

<math>~1- p_c(q) = 1 - b_\xi q^2 \, ,</math>

<math>~b_\xi</math>

<math>~\equiv</math>

<math>~\biggl( \frac{3}{2^3 \pi} \biggr) \frac{G M_\mathrm{tot}^2 }{P_0 R_\mathrm{edge}^4} \biggl( \frac{\nu}{q^3}\biggr)^2\, .</math>

Plugging these expressions into the equilibrium condition shown above, and setting the interface pressures equal to one another, gives,

<math>~\frac{1}{5} \biggl(\frac{\nu^2}{q}\biggr) f</math>

<math>~=</math>

<math>~\frac{4\pi}{3} \biggl[ \frac{P_i R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} \biggl\{ q^3 \biggl(\frac{P_0}{P_{i}} \biggr) \biggl[ 1 - \frac{3}{5}q^2 b_\xi\biggr] + (1-q^3) + \biggl(\frac{P_0}{P_{i}} \biggr) \frac{2}{5} q^5 \mathfrak{F} b_\xi \biggr\} </math>

 

<math>~=</math>

<math>~\frac{4\pi}{3} \biggl[ \frac{P_0 R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} \biggl\{ q^3 \biggl[ 1 - \frac{3}{5}q^2 b_\xi\biggr] + (1-q^3)( 1- b_\xi q^2) + \frac{2}{5} q^5 \mathfrak{F} b_\xi \biggr\} </math>

 

<math>~=</math>

<math>~\frac{4\pi}{3} \biggl[ \frac{P_0 R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} \biggl\{ 1 - b_\xi \biggl[ \frac{3}{5}q^5 + q^2(1-q^3) - \frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\} </math>

 

<math>~=</math>

<math>~\frac{4\pi}{3} \biggl[ \frac{P_0 R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} \biggl[ \frac{1}{b_\xi} - q^2 + \frac{2}{5} q^5( 1+\mathfrak{F} ) \biggr] b_\xi </math>

 

<math>~=</math>

<math>~\frac{1}{2} \biggl[ \frac{1}{b_\xi} - q^2 + \frac{2}{5} q^5( 1+\mathfrak{F} ) \biggr] \biggl( \frac{\nu}{q^3}\biggr)^2 </math>

<math>\Rightarrow~~~~\frac{1}{b_\xi}</math>

<math>~=</math>

<math>~ \frac{2}{5}q^5 f + \biggl[q^2 - \frac{2}{5} q^5( 1+\mathfrak{F} ) \biggr] </math>

<math>\Rightarrow~~~~\biggl( \frac{2^3 \pi}{3} \biggr) \frac{P_0 R_\mathrm{edge}^4}{G M_\mathrm{tot}^2 } \biggl( \frac{q^3}{\nu}\biggr)^2</math>

<math>~=</math>

<math>~ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} ) </math>

<math>\Rightarrow ~~~~ \frac{P_0 R_\mathrm{edge}^4}{G M_\mathrm{tot}^2 } </math>

<math>~=</math>

<math>~ \biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2 \biggl\{ q^2 + \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ 2q^2(1-q) + \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-3q^2 + 2q^3) \biggr] \biggr\} \, .</math>

This exactly matches the equilibrium relation that was derived from our detailed force-balance analysis of <math>~(n_c, n_e) = (0, 0)</math> bipolytropes.

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Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation