Editing SSC/Structure/BiPolytropes/Analytic1.53/Pt2 (section)

==The Point-Source Model==

According to Chapter IX.3 (p. 332) of [<b>[[Appendix/References#C67|<font color="red">C67</font>]]</b>], in the so-called "point-source" model, <font color="darkgreen">"&hellip; it is assumed that the entire source of energy is liberated at the center of the star; analytically, the assumption is that <math>L_r =~\mathrm{constant}~= L</math>."</font>

===Handling Radiation Transport===

Here we begin with the [[PGE/FirstLawOfThermodynamics#Example_B|familiar expression for the radiation flux]],  
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\vec{F}_\mathrm{rad}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \frac{c}{3\rho\kappa_R} \nabla (a_\mathrm{rad}T^4) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>-\chi_\mathrm{rad} \nabla T \, ,</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#Shu92|<font color="red">Shu92</font>]]</b>], Vol. I, &sect;2, p. 17, Eq. (2.17)
  </td>
  <td align="left" colspan="2">and &nbsp; &nbsp;[<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>], &sect;3.4, p. 57, Eq. (67)
  </td>
</tr>
</table>
</div>
where [<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>] refers to 
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\chi_\mathrm{rad}</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\frac{4c a_\mathrm{rad} T^3}{3\kappa \rho} \, ,
</math>
  </td>
</tr>
</table>
[<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>], &sect;3.4, p. 57, Eq. (68)
</div>
as the coefficient of ''radiative'' conductivity.  When modeling spherically symmetric configurations, the radiation flux has only a radial component, that is, <math>\vec{F}_\mathrm{rad} = \hat{e}_r(F_r)</math>.  And, as pointed out in the context of Eq. (170) on p. 214 of [<b>[[Appendix/References#C67|<font color="red">C67</font>]]</b>] <font color="darkgreen">&hellip; the quantity <math>L_r \equiv 4\pi r^2 F_r</math>, which is the net amount of energy crossing a spherical surface of radius <math>r</math>, is generally introduced instead of <math>F_r</math>.</font>  We therefore have,
<div align="center">
<table border="0" cellpadding="5">

<tr>
  <td align="right"><math>\frac{L_r}{4\pi r^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- \frac{c}{3\rho\kappa_R} \frac{d}{dr} (a_\mathrm{rad}T^4)
  </math></td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~\frac{dT}{dr}
  </math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>- \frac{3}{4ca_\mathrm{rad}} \frac{\rho\kappa_R}{T^3} \frac{L_r}{4\pi r^2}</math></td>
</tr>
</table>
[<b>[[Appendix/References#C67|<font color="red">C67</font>]]</b>], Chapter V, Eq. (171)<br />
[<b>[[Appendix/References#Clayton68|<font color="red">Clayton68</font>]]</b>], &sect;6, Eq. (6-4a)<br />
[<b>[[Appendix/References#KW94|<font color="red">KW94</font>]]</b>], &sect;9.1, Eq. (9.6)<br />
[<b>[[Appendix/References#HK94|<font color="red">HK94</font>]]</b>], &sect;7.1, Eq. (7.8)<br />
[<b>[[Appendix/References#BLRY07|<font color="red">BLRY07</font>]]</b>], &sect;5.2, Eq. (5.15)
</div>

<table border="1" align="center" cellpadding="8" width="60%"><tr><td align="left">
Dimensional Analysis:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\frac{\kappa_R L}{ca_\mathrm{rad}}</math></td>
  <td align="center"><math>\sim</math></td>
  <td align="left"><math>
\frac{rT^4}{\rho} \sim m^{-1}\ell^4 ( {^\circ}K)^4 \, .
  </math></td>
</tr>
</table>
NOTE:  This is consistent with the opacity, <math>\kappa_R \sim (\ell^2 m^{-1})</math>.
</td></tr></table>

===Harrison's Approach===
Following {{ Harrison46full }}, we seek to solve this last expression in concert with solutions to a pair of additional key governing relations for spherically symmetric equilibrium configurations, namely,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="center">{{Math/EQ_SShydrostaticBalance01}}</td>
  <td align="center">; &nbsp; &nbsp; &nbsp;</td>
  <td align="center">{{Math/EQ_SSmassConservation01}}</td>
</tr>

<tr>
  <td align="center" colspan="3">
{{ Harrison46 }}, p. 196, Eq. (20)
  </td>
</tr>
</table>

while adopting (see [[SR/IdealGas#Consequential_Ideal_Gas_Relations|related discussion]])

<div align="center">
<span id="IdealGas:FormA"><font color="#770000">'''Form A'''</font></span><br />
of the Ideal Gas Equation of State,

{{ Template:Math/EQ_EOSideal0A }}
</div>
and adopting [https://en.wikipedia.org/wiki/Kramers'_opacity_law Kramers' opacity law], that is,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\kappa_R ~~ \rightarrow ~~ \kappa_\mathrm{Kramers}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\kappa_0 \rho T^{-7 / 2} \, .</math></td>
</tr>
</table>

<table border="1" align="center" cellpadding="8" width="60%"><tr><td align="left">
Dimensional Analysis:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\biggl[ \frac{\kappa_0 L}{ca_\mathrm{rad}} \biggr]</math></td>
  <td align="center"><math>\sim</math></td>
  <td align="left"><math>
\frac{rT^{15/2}}{\rho^2} \sim m^{-2}\ell^7 ( {^\circ}K)^{15/2}
  </math></td>
</tr>

<tr>
  <td align="right"><math>\frac{\mathfrak{R}}{\mu}</math></td>
  <td align="center"><math>\sim</math></td>
  <td align="left"><math>
t^{-2}\ell^2 ( {^\circ}K)^{-1}
  </math></td>
</tr>
</table>

----

We note as well that the leading coefficient in Kramers' opacity is,
<div align="center">
<table border="0" cellpadding="5">

<tr>
  <td align="right"><math>\kappa_0</math></td>
  <td align="center"><math>\approx</math></td>
  <td align="left"><math>
5 \times 10^{22} ~\mathrm{cm}^5~\mathrm{g}^{-2} ({^\circ}K)^{7 / 2} \, .
  </math></td>
</tr>
</table>
[<b>[[Appendix/References#Clayton68|<font color="red">Clayton68</font>]]</b>], &sect;3.3, Eq. (3-170)<br />
[<b>[[Appendix/References#KW94|<font color="red">KW94</font>]]</b>], &sect;17.2, Eq. (17.5)<br />
[<b>[[Appendix/References#HK94|<font color="red">HK94</font>]]</b>], &sect;4.4.2, Eq. (4.35)
</div>

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

===Does a Polytropic Relation Work?===

Let's examine whether a point-source model can be represented by a polytropic relation.

====Adopting Temperature (instead of enthalpy)====
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>K\rho^{1 + 1/n}  ~~~~\Rightarrow~~~~  \rho = \biggl(\frac{P}{K}\biggr)^{n/(n+1)} \, ;</math>
</tr>

<tr>
  <td align="right"><math>\biggl(\frac{\mathfrak{R}}{\mu}\biggr)\rho T</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>K\rho^{1 + 1/n} ~~~~\Rightarrow~~~~ T = \biggl(\frac{K}{\mathfrak{R}/\mu}\biggr)\rho^{1/n} \, ;</math>
</tr>

<tr>
  <td align="right"><math>T^{n}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl(\frac{K}{\mathfrak{R}/\mu}\biggr)^{n}\rho  
~~~~\Rightarrow~~~~  
T^n = \biggl(\frac{K}{\mathfrak{R}/\mu}\biggr)^{n}\biggl(\frac{P}{K}\biggr)^{n/(n+1)} 
~~~~\Rightarrow~~~~  
T^{(n+1)} = \biggl(\frac{K}{\mathfrak{R}/\mu}\biggr)^{(n+1)}\biggl(\frac{P}{K}\biggr) 
\, .</math>
</tr>
</table>
Hydrostatic balance is governed by the single 2<sup>nd</sup> order ODE,

<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\frac{1}{r^2} \frac{d}{dr} \biggl[ \frac{r^2}{\rho} \frac{dP}{dr} \biggr]</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>- 4\pi G\rho \, .</math>
</tr>
</table>
Normally in order to arrive at the Lane-Emden equation, <math>P</math> is converted to <math>\rho</math>; here, let's convert both <math>P</math> and <math>\rho</math> to <math>T</math>.  First, on the RHS we have,

<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\rho</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl(\frac{\mathfrak{R}/\mu}{K}\biggr)^n T^n \, ;</math>
</tr>
</table>
and second, on the LHS we have,

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

<tr>
  <td align="right"><math>\frac{1}{\rho} \frac{dP}{dr}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{K}{\mathfrak{R}/\mu}\biggr)^n T^{-n} \cdot \frac{d}{dr}\biggl[
K^{-n}(\mathfrak{R}/\mu)^{(n+1)} T^{(n+1)}
\biggr]
</math>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{\mathfrak{R}}{\mu}\biggr) T^{-n} \cdot \frac{d}{dr}\biggl[T^{(n+1)}
\biggr]
</math>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(n+1)\biggl(\frac{\mathfrak{R}}{\mu}\biggr) \frac{dT}{dr}
</math>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(n+1)\biggl(\frac{\mathfrak{R}}{\mu}\biggr) \biggl\{
- \frac{3}{4ca_\mathrm{rad}} \frac{\rho\kappa_R}{T^3} \frac{L}{4\pi r^2}
\biggr\}
</math>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (n+1)\biggl(\frac{\mathfrak{R}}{\mu}\biggr) \biggl[
\frac{3L}{16\pi ca_\mathrm{rad}} 
\biggr] \biggl[ \frac{\rho}{r^2T^3} \biggr]\kappa_0 \rho T^{-7 / 2} 
</math>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{r^2}{\rho} \frac{dP}{dr}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (n+1)\biggl(\frac{\mathfrak{R}}{\mu}\biggr) \biggl[
\frac{3\kappa_0 L}{16\pi ca_\mathrm{rad}} 
\biggr] \rho^2 T^{-13 / 2} 
</math>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (n+1)K^{-2n}\biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{2n+1} \biggl[
\frac{3\kappa_0 L}{16\pi ca_\mathrm{rad}} 
\biggr] T^{(4n-13 )/ 2} \, .
</math>
</tr>
</table>
So, the hydrostatic-balance condition becomes,

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

<tr>
  <td align="right">
<math>
- (n+1)K^{-2n}\biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{2n+1} \biggl[
\frac{3\kappa_0 L}{16\pi ca_\mathrm{rad}} 
\biggr] \cdot \frac{1}{r^2}\frac{dT^{(4n-13 )/ 2}}{dr} 
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>- 4\pi G\biggl(\frac{\mathfrak{R}/\mu}{K}\biggr)^n T^n </math></td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~
\frac{(n+1)}{4\pi G K^{2n}}\biggl(\frac{\mathfrak{R}/\mu}{K}\biggr)^{-n}\biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{2n+1} \biggl[
\frac{3\kappa_0 L}{16\pi ca_\mathrm{rad}} 
\biggr] \cdot \frac{1}{r^2}\frac{dT^{(4n-13 )/ 2}}{dr} 
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> T^n </math></td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~
\frac{\mathcal{A}}{r^2}\frac{dT^{(4n-13 )/ 2}}{dr} 
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> T^n </math></td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ {\mathcal{A}}^{-1}r^2 dr </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
T^{-n} \cdot dT^{(4n-13 )/ 2} \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\mathcal{A}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
\frac{(n+1)}{4\pi G K^n}\biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{n+1} \biggl[
\frac{3\kappa_0 L}{16\pi ca_\mathrm{rad}} 
\biggr] \, .  
</math>
  </td>
</tr>
</table>


<table border="1" align="center" cellpadding="8" width="60%"><tr><td align="left">
Dimensional Analysis:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\biggl[ \frac{\kappa_0 L}{ca_\mathrm{rad}} \biggr]</math></td>
  <td align="center"><math>\sim</math></td>
  <td align="left"><math>
m^{-2}\ell^7 ( {^\circ}K)^{15/2}
  </math></td>
</tr>

<tr>
  <td align="right"><math>\frac{\mathfrak{R}}{\mu}</math></td>
  <td align="center"><math>\sim</math></td>
  <td align="left"><math>
t^{-2}\ell^2 ( {^\circ}K)^{-1}
  </math></td>
</tr>

<tr>
  <td align="right">polytropic <math>K</math></td>
  <td align="center"><math>\sim</math></td>
  <td align="left"><math>
t^{-2} \ell^{2 + 3/n} m^{-1 / n}
  </math></td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\mathcal{A}</math></td>
  <td align="center"><math>\sim</math></td>
  <td align="left"><math>
G^{-1} K^{-n} \biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{n+1} \biggl[ m^{-2}\ell^7 ( {^\circ}K)^{15/2} \biggr]
  </math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>\sim</math></td>
  <td align="left"><math>
\biggl[\ell^{-3} m t^{2}\biggr] \biggl[ t^{-2} \ell^{2 + 3/n} m^{-1 / n} \biggr]^{-n} 
\biggl[ t^{-2}\ell^2 ( {^\circ}K)^{-1}\biggr]^{n+1} \biggl[ m^{-2}\ell^7 ( {^\circ}K)^{15/2} \biggr]
  </math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>\sim</math></td>
  <td align="left"><math>
\biggl[\ell^{-3} \biggr] \biggl[ \ell^{2 + 3/n}  \biggr]^{-n} 
\biggl[ \ell^2 \biggr]^{n+1} \biggl[ \ell^7\biggr] ( {^\circ}K)^{13/2-n} 
\sim
\ell^3( {^\circ}K)^{13/2-n} 
\, .
  </math></td>
</tr>
</table>

Now, the [[SSC/Structure/Polytropes#Lane-Emden_Equation|characteristic length scale for polytropic configurations]] is given by the expression,

<div align="center">
<math>~
a_\mathrm{n}^2 \equiv \biggl[\frac{(n+1)K}{4\pi G} \cdot \rho_c^{(1-n)/n} \biggr] \, .
</math>
</div>
If we divide <math>\mathcal{A}</math> by <math>a_n^3</math>, the resulting expression should give us the characteristic temperature of the envelope.  Specifically, we find that,

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

<tr>
  <td align="right"><math>\frac{\mathcal{A}}{a_n^3}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{(n+1)}{4\pi G K^n}\biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{n+1} \biggl[
\frac{3\kappa_0 L}{16\pi ca_\mathrm{rad}} 
\biggr] 
\times
\biggl[\frac{(n+1)K_n}{4\pi G} \cdot \rho_c^{(1-n)/n} \biggr]^{-3 / 2}   
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(n+1)^{- 1 / 2}
\rho_c^{3(n-1)/2n}   
\biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{n+1} \biggl[
\frac{3\kappa_0 L}{16\pi ca_\mathrm{rad}} 
\biggr] (4\pi G)^{ 1 / 2} K^{-n - 3 / 2}</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>\sim</math></td>
  <td align="left">
<math>
\biggl[m \ell^{-3}\biggr]^{3(n-1)/2n}   
\biggl[ t^{-2}\ell^2 ( {^\circ}K)^{-1}\biggr]^{n+1} \biggl[
m^{-2}\ell^7 ( {^\circ}K)^{15/2} 
\biggr] \biggl[ \ell^3 m^{-1} t^{-2} \biggr]^{ 1 / 2} 
\biggl[ t^{-2} \ell^{2 + 3/n} m^{-1 / n} \biggr]^{-n - 3 / 2}</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>\sim</math></td>
  <td align="left">
<math>
t^{-2n - 2 - 1 + 2n + 3}
m^{ 3(n-1)/2n - 2 - 1 / 2 + 1 + 3/2n}
\ell^{ -9(n-1)/2n + 2n + 2 + 7 + 3 / 2}
\biggl[ \ell^{(2n + 3)/n} \biggr]^{-(2n + 3) / 2}
( {^\circ}K)^{13/2 - n }  
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>\sim</math></td>
  <td align="left">
<math>
\ell^{ (9 + 4n^2 + 12n)/2n }
\ell^{-(2n + 3)^2 / 2n}
( {^\circ}K)^{13/2 - n }  
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>\sim</math></td>
  <td align="left">
<math>
( {^\circ}K)^{(13-2n)/2 }  \, .
</math>
  </td>
</tr>
</table>


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

Quite generally we can write,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\frac{1}{\alpha}~\frac{dT^\alpha}{dr}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
T^{\alpha - 1} \frac{dT}{dr} \, .
</math>
  </td>
</tr>
</table>
Rewriting the hydrostatic-balance condition, we find that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>{\mathcal{A}}^{-1}r^2 </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
T^{-n} \cdot \frac{d}{dr}\biggl[T^{(4n-13 )/ 2}\biggr]
=
T^{-n} \biggl(\frac{4n-13}{2}  \biggr) T^{(4n-15 )/ 2} \cdot \frac{dT}{dr}
=
\biggl(\frac{4n-13}{2}  \biggr) T^{(2n-15 )/ 2} \cdot \frac{dT}{dr} \, .
</math>
  </td>
</tr>
</table>
Associating the exponents,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\alpha - 1</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{2n-15}{2} ~~~\Rightarrow ~~~ \alpha = \frac{2n-13}{2} \, , 
</math>
  </td>
</tr>
</table>
we can write,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>{\mathcal{A}}^{-1}r^2 </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{4n-13}{2}  \biggr) \biggl(\frac{2}{2n-13}  \biggr)\frac{d}{dr} \biggl[ T^{(2n-13)/2 } \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ 3d(r^3)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\mathcal{A} \biggl(\frac{4n-13}{2n-13}  \biggr) d\biggl[ T^{(2n-13)/2 } \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ 0</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
d\biggl[ \mathcal{A} \biggl(\frac{13-4n}{13 - 2n}  \biggr) T^{(2n-13)/2 } - 3r^3\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~</math> constant</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\mathcal{A}}{a_n^3} \biggl(\frac{13-4n}{13 - 2n}  \biggr) T^{(2n-13)/2 } - 3\biggl( \frac{r}{a_n}\biggr) ^3  \, .
</math>
  </td>
</tr>
</table>

====Adopting Enthalpy (instead of Temperature)====

For [[SR#Barotropic_Structure|polytropic configurations]] the enthalpy, <math>H</math>, can easily be adopted in place of temperature via the relation,
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>\biggl(\frac{\mathfrak{R}}{\mu}\biggr) T</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{H}{(n+1)} \, .</math></td>
</tr>
</table>
Hence,

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

<tr>
  <td align="right"><math>P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>K\rho^{1 + 1/n}\, ;</math></td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>H</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>K\rho^{1/n}\, ;</math></td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp;</td>
  <td align="right"><math>H^{n+1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>K^n P\, .</math></td>
</table>
And the radiation-transport equation can be rewritten in the form,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\frac{L_r}{4\pi r^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- \frac{ca_\mathrm{rad}}{3\rho\kappa_R (n+1)^4}  
\biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{-4} 
\frac{dH^4}{dr}
</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- \frac{ca_\mathrm{rad}}{3\kappa_0 (n+1)^4}  
\biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{-4} \biggl[\rho^{-2} T^{7 / 2}\biggr]
\frac{dH^4}{dr}
</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- \frac{ca_\mathrm{rad}}{3\kappa_0 (n+1)^4}  
\biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{-4} 
\biggl\{\biggl(\frac{K}{H}\biggr)^{2n}  \biggl[ \biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{-1} \frac{H}{(n+1)}\biggr]^{7 / 2}\biggr\}
4 H^3 \frac{dH}{dr}
</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- \frac{4ca_\mathrm{rad} K^{2n}}{3\kappa_0 (n+1)^{15 / 2}}  
\biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{-15 / 2} 
H^{(13 - 4n) / 2}
\frac{dH}{dr}
</math></td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ r^2 \frac{dH}{dr} </math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- \biggl[ \frac{3\kappa_0 L (n+1)^{15 / 2}}{16\pi ca_\mathrm{rad} K^{2n}}  
\biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{15 / 2}  \biggr]
H^{(4n - 13 ) / 2} \ .
</math></td>
</tr>
</table>

In terms of the enthalpy, the hydrostatic-balance expression becomes,

<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>- 4\pi G\biggl[\frac{H^n}{K^n}\biggr]</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{1}{r^2} \frac{d}{dr} \biggl\{r^2\biggl[\frac{K^n}{H^n}\biggr] 
\frac{d}{dr} \biggl[K^{-n} H^{n+1}  \biggr]
\biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ - 4\pi GH^n</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{K^n}{r^2} \frac{d}{dr} \biggl\{
\biggl[\frac{r^2}{H^n}\biggr] 
\frac{d}{dr} \biggl[H^{n+1}  \biggr]
\biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{(n+1)K^n}{r^2} \frac{d}{dr} \biggl\{
r^2 \frac{dH}{dr} 
\biggr\} \, .</math>
  </td>
</tr>
</table>
Combining these two equations gives,

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

<tr>
  <td align="right"><math>r^2 H^n</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\mathcal{B} \cdot \frac{d}{dr} \biggl\{ H^{(4n - 13 ) / 2} \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{(4n - 13)\mathcal{B}}{2} \cdot H^{(4n - 15 ) / 2} \frac{dH}{dr} 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \biggl[\frac{2}{(4n - 13)\mathcal{B}}\biggr] r^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
H^{(2n - 15 ) / 2} \frac{dH}{dr} 
</math>
  </td>
</tr>
</table>
where,

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

<tr>
  <td align="right"><math>\mathcal{B}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math> 
\biggl[ \frac{3\kappa_0 L (n+1)^{17 / 2}}{64\pi^2 G ca_\mathrm{rad} K^{n}}  
\biggl(\frac{\mathfrak{R}}{\mu}\biggr)^{15 / 2} \biggr] \, .
</math>
  </td>
</tr>
</table>
As above, quite generally we can write,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\frac{1}{\alpha}~\frac{dH^\alpha}{dr}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
H^{\alpha - 1} \frac{dH}{dr} \, .
</math>
  </td>
</tr>
</table>
So, associating the exponents, we appreciate that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\alpha - 1</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{2n-15}{2} ~~~\Rightarrow ~~~ \alpha = \frac{2n-13}{2} \, . 
</math>
  </td>
</tr>
</table>
Hence, we have,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\biggl[\frac{2}{(4n - 13)\mathcal{B}}\biggr] r^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{2}{(2n - 13)}\frac{d}{dr}\biggl[ H^{(2n - 13 ) / 2}  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ r^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{(4n - 13)\mathcal{B}}{(2n - 13)}\frac{d}{dr}\biggl[ H^{(2n - 13 ) / 2}  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ d\biggl[\frac{r^3}{a_n^3}\biggr]</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{3(4n - 13)\mathcal{B}}{(2n - 13)a_n^3}\cdot d\biggl[ H^{(2n - 13 ) / 2}  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ d\biggl[\xi^3\biggr]</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\biggl[\frac{3(4n - 13)}{(2n - 13)}\biggr] \biggl[\frac{\mathcal{B}}{a_n^3} \cdot H_\mathrm{norm}^{(2n-13)/2}\biggr]
d\biggl[ \frac{H}{H_\mathrm{norm}}  \biggr]^{(2n - 13 ) / 2} \, ;
</math>
  </td>
</tr>
</table>
so, if we adopt the definition,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>H_\mathrm{norm}</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
\biggl[\frac{(2n - 13)}{3(4n - 13)} \cdot \frac{a_n^3}{\mathcal{B}} \biggr]^{2/(2n-13)} \, ,
</math>
  </td>
</tr>
</table>
the relation becomes,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>d\biggl[\xi^3\biggr]</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
d\biggl[ \frac{H}{H_\mathrm{norm}}  \biggr]^{(2n - 13 ) / 2} \, .
</math>
  </td>
</tr>
</table>

===Power-Law Density Distribution===

In an [[SSC/Structure/PowerLawDensity#Power-Law_Density_Distributions|accompanying discussion]], we have demonstrated that power-law density distributions can provide analytic solutions of the Lane-Emden equation, although the associated boundary conditions do not naturally conform to the boundary conditions that are suitable to astrophysical configurations.  We have just shown that the point-source envelope configuration appears to admit a power-law temperature (alternatively, enthalpy) solution. Via the polytropic relation, <math>H = K\rho^{1 / n}</math>, we can convert to the density-radius relation,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>d\biggl[\xi^3\biggr]</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\biggl[ \frac{K\rho_c^{1/n}}{H_\mathrm{norm}}  \biggr]^{(2n - 13 ) / 2} d\biggl[ \frac{\rho}{\rho_c} \biggr]^{(2n - 13 ) / 2n}
</math>
  </td>
</tr>
</table>
which, upon integration gives,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">constant</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\biggl[ \frac{\rho}{\rho_c} \biggr]^{(2n - 13 ) / 2n} - \xi^3 \, ,
</math>
  </td>
</tr>
</table>
if we adopt the definition,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\rho_c</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left"><math>
\biggl(\frac{H_\mathrm{norm}}{K}\biggr)^n \, .
  </math>
  </td>
</tr>
</table>

Setting the integration constant to zero, our result gives,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>\frac{\rho}{\rho_c}</math></td>
  <td align="center"><math>\propto</math></td>
  <td align="left"><math>
\xi^{6n/(2n-13)} \, .
</math>
  </td>
</tr>
</table>
In astrophysically relevant configurations, the exponent on <math>\xi</math> must be negative, which means that we are confined to models for which <math>n < \tfrac{7}{2}</math>.

Now, from our [[SSC/Structure/PowerLawDensity#Derivation|associated discussion of power-law density distributions]] in polytropes, we discovered that hydrostatic balance can be established at all radial positions within a spherically symmetric configuration for power-law density distributions of the form,
<div align="center">
<math>
\frac{\rho}{\rho_c} \propto \xi^{- 2n/(n-1)}
</math>
</div>
This matches our just-derived point-source model if,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>6n/(2n-13)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>- 2n/(n-1)</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ 6(n-1)</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>2(13 - 2n)</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ n</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{16}{5} \, ,</math>
  </td>
</tr>
</table>
which ''is'' less than <math>\tfrac{7}{2}</math>, so it is an astrophysically viable result.