Editing Appendix/Ramblings/OriginOfPlanetaryNebulae (section)

===Zero-Age Main Sequence (ZAMS) Configuration===
As a check, we note that in the initial (ZAMS) equilibrium configuration <math>(\xi_i=0)</math> there is no "core" &#8212; that is, <math>M_\mathrm{core} = 0</math> &#8212; so its structure should match that of an [[SSC/Structure/Polytropes#n_=_1_Polytrope|isolated <math>n = 1</math> polytrope]].  Specifically, given that, <math>M_\odot = 1.99\times 10^{33}~\mathrm{g}</math>, <math>R_\odot = 6.96\times 10^{10}~\mathrm{cm}</math>, and <math>G = 6.674\times 10^{-8}~\mathrm{cm^3~g^{-1}~s^{-2}}</math> &hellip; 
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right"><math>M_\mathrm{tot}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{4}{\pi} \rho_c R^3</math></td>
<td align="center">&nbsp; &nbsp; &nbsp; <math>\Rightarrow</math> &nbsp; &nbsp; &nbsp; </td>
  <td align="right"><math>\rho_c\biggr|_\mathrm{initial}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{\pi}{4} \cdot \frac{M_\odot}{R_\odot^3} = 4.64~\mathrm{g~cm^{-3}} \, .</math></td>
</tr>
</table>
Also,
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right"><math>\frac{\rho_c}{\bar\rho}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{\pi^2}{3}</math></td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right"><math>P_c\biggr|_\mathrm{initial}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{\pi G}{8} \biggl( \frac{M_\odot^2}{R_\odot^4} \biggr) = 4.42 \times 10^{15}~\mathrm{g~cm^{-1}~s^{-2}} \, .</math></td>
</tr>
</table>

<table border="1" align="center" cellpadding="5" width="80%"><tr><td align="left">
The sun's luminosity is, <math>L_\odot = 4\pi R_\odot^2 \sigma_\mathrm{SB} T^4 = 3.846\times 10^{33}~\mathrm{ergs~s^{-1}}</math>, where, in terms of {{ Math/C_RadiationConstant }} and {{ Math/C_SpeedOfLight }}, the Stefan-Boltzmann constant is,
<div align="center">
<math>\sigma_\mathrm{SB} = \frac{a_\mathrm{rad} c}{4} = 5.671\times 10^{-5} ~\mathrm{erg~cm^{-2}~s^{-1}~K^{-4}}\, .</math>
</div>
Hence, in the context of our ZAMS toy model, the configuration's surface temperature is,

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

<tr>
  <td align="right"><math>T\biggr|_\mathrm{initial}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\biggl[ \frac{L_\odot}{4\pi R_\odot^2 \sigma_\mathrm{SB} } \biggr]^{1 / 4} = 5777~\mathrm{K} \, .</math>
  </td>
</tr>
</table>

----

In the context of our examination of a <math>1 M_\odot</math> evolutionary track published by {{ Iben67full }} &#8212; [[#IbenFigure|see below]] &#8212; we can write quite generally that,

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

<tr>
  <td align="right"><math>L</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>4\pi R^2 \sigma_\mathrm{SB}T_e^4</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \biggl(\frac{R}{R_\odot}\biggr)^2 </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{L}{L_\odot}\biggl( \frac{T_e}{5777 K}\biggr)^{-4}</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \log\biggl(\frac{R}{R_\odot}\biggr) </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\tfrac{1}{2}\log\biggl(\frac{L}{L_\odot} \biggr)
- 2 \log\biggl( \frac{T_e}{5777 K}\biggr)</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\tfrac{1}{2}\log\biggl(\frac{L}{L_\odot} \biggr)
- 2 \log T_e + 7.523 \, .</math>
  </td>
</tr>
</table>

----

Note as well that if an evolving star radiates at an average luminosity, <math>L_\mathrm{avg}/L_\odot</math>, over a specified time interval in units of <math>10^9~\mathrm{yrs}</math>, <math>(\Delta t)_9</math>, during the specified interval of time it will have burned through an amount of mass, <math>\Delta M/M_\odot</math>, given by the expression,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>0.007(\Delta M)c^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>(\Delta t)L_\mathrm{avg}</math></td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{(\Delta M)}{M_\odot}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(\Delta t)_9 \cdot \frac{L_\mathrm{avg}}{L_\odot} \biggl[
\frac{L_\odot}{0.007 M_\odot c^2} 
\cdot 3.156\times 10^{16}~\mathrm{s}
\biggr]
=
0.01 \biggl[ (\Delta t)_9 \cdot \frac{L_\mathrm{avg}}{L_\odot}  \biggr] \, ,
</math>
  </td>
</tr>
</table>
where we have assumed that the efficiency of burning hydrogen to helium is 0.7%.
</td></tr></table>

Most notably, the dimensionless radius that appears in the classic Lane-Emden equation is given by the expression,

<div align="center">
<math>
\xi \equiv \frac{r}{a_\mathrm{n = 1}} ,
</math>
</div>
where,
<div align="center">
<math>
a_\mathrm{n=1} \equiv \biggl[\frac{(n+1)K_n}{4\pi G} \cdot \rho_c^{(1-n)/n} \biggr]^{1/2} 
=
\biggl[\frac{K_\mathrm{env}}{2\pi G} \biggr]^{1/2}
\, .
</math>
</div>
And, given that the surface (<math>r = R_\odot</math>) of the isolated <math>n=1</math> polytrope occurs at <math>\xi = \xi_1 = \pi</math>, we find that,
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>a_\mathrm{n=1}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{R_\odot}{\pi}</math></td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ K_\mathrm{env}\biggr|_\mathrm{initial}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\biggl( \frac{2}{\pi} \biggr) G R_\odot^2 = 2.06 \times 10^{14}~\mathrm{cm^5~g^{-1}~s^{-2}} \, .</math></td>
</tr>
</table>
Double-check:
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>P_c\biggr|_\mathrm{initial}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
K_\mathrm{env}\biggr|_\mathrm{initial}\biggl[ \rho_c\biggr|_\mathrm{initial}\biggr]^2
=
4.42\times 10^{15} ~\mathrm{g~cm^{-1}~s^{-2}} \, .
</math>
  </td>
</tr>
</table>
Yes!


Now, the instant the "core" shows up, we demand that its central density be a factor of <math>(\mu_e/\mu_c)^{-1}</math> larger than the density at the inner edge of the envelope; at the same time, we demand that the pressures be the same.  Hence the relevant polytropic equation of state for the core must be,
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
K_\mathrm{core}\biggl[ \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1} \rho \biggr]^{6 / 5} \, .
</math>
  </td>
</tr>
</table>
In our toy model, this means that,
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>K_\mathrm{core}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
P \biggl[ \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1} \rho \biggr]^{-6 / 5} 
=
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{6 / 5} 4.42\times 10^{15} ~\mathrm{g~cm^{-1}~s^{-2}} 
\biggl[ 4.64~\mathrm{g~cm^{-3}} \biggr]^{-6 / 5}
=
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{6 / 5} 7.008\times 10^{14} \mathrm{g^{-1 / 5}~cm^{13 / 5}~s^{-2}}
\, .
</math>
  </td>
</tr>
</table>
Hence, the ratio,
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>\frac{K_\mathrm{env}}{K_\mathrm{core}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2.06 \times 10^{14}~\mathrm{cm^5~g^{-1}~s^{-2}}
\biggl\{\biggl(\frac{\mu_e}{\mu_c}\biggr)^{6 / 5} 7.008\times 10^{14} \mathrm{g^{-1 / 5}~cm^{13 / 5}~s^{-2}}\biggr\}^{-1}
=
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-6 / 5}0.293 ~\mathrm{g^{-4 / 5}~cm^{12 / 5}}
\, ,
</math>
  </td>
</tr>
</table>
which may be rewritten as,
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>\frac{K_\mathrm{env}}{K_\mathrm{core}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{4/ 5}0.293 \biggl[\mathrm{g~cm^{-3}} \biggr]^{- 4 / 5}\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
=\biggl[ \underbrace{ \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1} \frac{\pi}{4}\cdot \frac{M_\odot}{R_\odot^3} }_\mathrm{central~ density~of~n = 5 ~ core} \biggr]^{- 4 / 5}
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2} \, .
</math>
  </td>
</tr>
</table>

This matches the [[SSC/Structure/BiPolytropes/Analytic51#Step_5:_Interface_Conditions|interface condition expression]] that relates <math>K_\mathrm{core}</math> to <math>K_\mathrm{env}</math>,
<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>\frac{K_e}{K_c}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\rho_c^{-4 / 5} \biggl(\frac{\mu_e}{\mu_c}\biggr)^{-2} \theta_i^{-4} \, ,
</math>
  </td>
</tr>
</table>
where we appreciate that, in the initial bipolytropic configuration (i.e., when <math>\xi_i = 0</math>), <math>\theta_i = 1</math>.