SSC/Stability/BiPolytropes/RedGiantToPN/Pt4 - Revision history

Joel2: /* Envelope Displacement Function */

2026-02-19T17:33:36Z

Envelope Displacement Function

@@ Line 2,122: / Line 2,122: @@
    </td>
    <td align="left">
-<math>\ln r_0^{3(\gamma_c/\gamma_e - 1) }
+<math>
 - \frac{2\gamma_c}{\gamma_e}
-\biggl[\frac{r_0\cdot dr_0}{45/(2\pi)-r_0^2} \biggr] </math>
+\int\frac{r_0\cdot dr_0}{45/(2\pi)-r_0^2}
    </td>
    <td align="right">&nbsp;</td>
@@ Line 2,137: / Line 2,139: @@
    </td>
    <td align="left">
-<math>\ln r_0^{3(\gamma_c/\gamma_e - 1) }
+<math>\ln \biggl[r_0\biggr]^{3(\gamma_c/\gamma_e - 1) }
-+ \frac{\gamma_c}{\gamma_e}\cdot \ln\biggl[45/(2\pi)-r_0^2 \biggr] </math>
++ \frac{\gamma_c}{\gamma_e}\cdot \ln\biggl[45/(2\pi)-r_0^2 \biggr]
    </td>
    <td align="right">&nbsp;</td>

2026-02-19T17:15:48Z

Envelope Displacement Function

@@ Line 2,076: / Line 2,076: @@
 <math>~3\biggl(\frac{\gamma_c}{\gamma_e}  -1\biggr) + \frac{\gamma_c}{\gamma_e}
 \biggl[\frac{2r_0^2}{r_0^2 - 45/(2\pi)} \biggr] </math>
    </td>
    <td align="right">&nbsp;</td>

2026-02-19T16:54:35Z

Envelope Displacement Function

@@ Line 2,059: / Line 2,059: @@
    <td align="left">
 <math>~3\biggl(\frac{\gamma_c}{\gamma_e}  -1\biggr) + \frac{\gamma_c}{\gamma_e}
-\biggl\{ \frac{d\ln x}{d\ln \xi}\biggr|_\mathrm{int} \biggr\}_\mathrm{core} \, .</math>
+\biggl\{ \frac{d\ln x}{d\ln \xi}\biggr|_\mathrm{int} \biggr\}_\mathrm{core}</math>
    </td>
    <td align="right">(-1.473523186)<sup>&dagger;</sup></td>
 </tr>
 </table>

2026-02-19T16:48:37Z

Guessing Game

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 ==Guessing Game==
 <ol type="A"><li>First Try
 <ol>
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 </ol>
 </li></ol>
 =Related Discussions=

2026-02-19T00:22:39Z

Guessing Game

← Older revision		Revision as of 20:22, 18 February 2026
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	<td align="left">		<td align="left">
	<math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\biggl[ 1+\frac{\xi^2}{3} \biggr] \cdot \frac{1}{3^{1 / 2}\xi}		<math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\biggl[ 1+\frac{\xi^2}{3} \biggr] \cdot \frac{1}{3^{1 / 2}\xi}
	+ 3^{1 / 2}\biggl[ 1+\frac{\xi^2}{3} \biggr]^{3/2} \biggl[- \frac{\xi^2}{3} \biggl(1 + \frac{\xi^2}{3} \biggr)^{-3/2}\biggr]</math>		+ 3^{1 / 2}\biggl[ 1+\frac{\xi^2}{3} \biggr]^{3/2} \biggl[- \frac{\xi}{3} \biggl(1 + \frac{\xi^2}{3} \biggr)^{-3/2}\biggr]</math>
	</td>		</td>
	</tr>		</tr>
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	<td align="center"><math>=</math></td>		<td align="center"><math>=</math></td>
	<td align="left">		<td align="left">
	<math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\biggl[ 1+\frac{\xi^2}{3} \biggr] \cdot \frac{1}{3^{1 / 2}\xi} - \frac{\xi^2}{3^{1 / 2}} \, .</math>		<math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\biggl[ 1+\frac{\xi^2}{3} \biggr] \cdot \frac{1}{3^{1 / 2}\xi} - \frac{\xi}{3^{1 / 2}} \, .</math>
	</td>		</td>
	</tr>		</tr>

2026-02-19T00:07:10Z

Guessing Game

← Older revision		Revision as of 20:07, 18 February 2026
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	<td align="center"><math>=</math></td>		<td align="center"><math>=</math></td>
	<td align="left">		<td align="left">
	<math>\frac{1}{\~~eta~~} + 3^{1 / 2}\theta^{-3} \biggl(\frac{d\theta}{d\xi}\biggr)</math>		<math>\biggl[3^{-1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\theta^{-2} \biggr]\cdot \frac{1}{\xi}
			+ 3^{1 / 2}\theta^{-3} \biggl(\frac{d\theta}{d\xi}\biggr)</math>
	</td>		</td>
	</tr>		</tr>
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	<td align="center"><math>=</math></td>		<td align="center"><math>=</math></td>
	<td align="left">		<td align="left">
	<math>\frac{1}{\~~eta~~} + 3^{1 / 2}\biggl[ 1+\frac{\xi^2}{3} \biggr]^{3/2} \biggl[- \frac{\xi^2}{3} \biggl(1 + \frac{\xi^2}{3} \biggr)^{-3/2}\biggr]</math>		<math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\biggl[ 1+\frac{\xi^2}{3} \biggr] \cdot \frac{1}{3^{1 / 2}\xi}
			+ 3^{1 / 2}\biggl[ 1+\frac{\xi^2}{3} \biggr]^{3/2} \biggl[- \frac{\xi^2}{3} \biggl(1 + \frac{\xi^2}{3} \biggr)^{-3/2}\biggr]</math>
	</td>		</td>
	</tr>		</tr>
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	<td align="center"><math>=</math></td>		<td align="center"><math>=</math></td>
	<td align="left">		<td align="left">
	<math>\frac{1}{\~~eta~~} - \frac{\xi^2}{3^{1 / 2}} </math>		<math>\biggl(\frac{\mu_e}{\mu_c}\biggr)^{-1}\biggl[ 1+\frac{\xi^2}{3} \biggr] \cdot \frac{1}{3^{1 / 2}\xi} - \frac{\xi^2}{3^{1 / 2}} \, .</math>
	</td>		</td>
	</tr>		</tr>

2026-02-18T23:56:34Z

Guessing Game

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 <table border="1" width="80%" cellpadding="8"><tr><td align="left">
 &hellip; note, in particular, that
 </td></tr></table>
 </div>

2026-02-18T23:36:12Z

Guessing Game

← Older revision		Revision as of 19:36, 18 February 2026
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	<math>r_0 = \biggl( \frac{3}{2\pi}\biggr)^{1 / 2} \xi = \biggl[(2\pi)^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)\theta^2 \biggr]^{-1} \eta \, ;</math>		<math>r_0 = \biggl( \frac{3}{2\pi}\biggr)^{1 / 2} \xi = \biggl[(2\pi)^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)\theta^2 \biggr]^{-1} \eta \, ;</math>
	</div>		</div>
	and from [[SSC/Stability/BiPolytropes/RedGiantToPN/Pt4#Steps\|an accompanying series of analytic expressions]], we also deduce that,		and from [[SSC/Stability/BiPolytropes/RedGiantToPN/Pt4#Steps\|an accompanying series of analytic expressions]]
			<div align="center">
			<table border="1" width="80%" cellpadding="8"><tr><td align="left">
			… note, in particular, that
			</td></tr></table>
			</div>
			we also deduce that,
	</div>		</div>
	<div align="center">		<div align="center">

2026-02-18T23:23:13Z

Guessing Game

← Older revision		Revision as of 19:23, 18 February 2026
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	<div align="center">		<div align="center">
	<math>r_0 = \biggl( \frac{3}{2\pi}\biggr)^{1 / 2} \xi = \biggl[(2\pi)^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)\theta^2 \biggr]^{-1} \eta \, ;</math>		<math>r_0 = \biggl( \frac{3}{2\pi}\biggr)^{1 / 2} \xi = \biggl[(2\pi)^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)\theta^2 \biggr]^{-1} \eta \, ;</math>
	</div~~><br /~~>		</div>
	and from ~~the series of~~ [[SSC/Stability/BiPolytropes/RedGiantToPN/Pt4#Steps\|analytic expressions ~~shown here~~]], we also deduce that,		and from [[SSC/Stability/BiPolytropes/RedGiantToPN/Pt4#Steps\|an accompanying series of analytic expressions]], we also deduce that,
			</div>
			<div align="center">
			<math>r_\mathrm{surf} = \biggl[(2\pi)^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)\theta^2 \biggr]^{-1}
			\biggl[ \eta + \frac{\pi}{2} + \tan^{-1}(\Lambda)\biggr] </math>
			<math>\Rightarrow</math>
			<math>\frac{r_0}{r_\mathrm{surf}} = \eta \cdot \biggl[ \eta + \frac{\pi}{2} + \tan^{-1}(\Lambda)\biggr]^{-1} \, . </math>
	</div>		</div>

2026-02-18T23:08:34Z

Guessing Game

← Older revision		Revision as of 19:08, 18 February 2026
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	<li>Pick a value of <math>(\mu_e/\mu_c)</math>, and a value of the interface location, <math>\xi</math>; the corresponding value of <math>\eta</math> is,		<li>Pick a value of <math>(\mu_e/\mu_c)</math>, and a value of the interface location, <math>\xi</math>; the corresponding value of <math>\eta</math> is,
	<div align="center">		<div align="center">
	<math>\eta = \biggl[3^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)\theta^2 \biggr]\xi</math>		<math>\eta = \biggl[3^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)\theta^2 \biggr]\xi \, .</math>
	</div>		</div>

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	<li>We can immediately deduce that,		<li>We can immediately deduce that,
	<div align="center">		<div align="center">
	<math>r_0 = \biggl( \frac{3}{2\pi}\biggr)^{1 / 2} \xi</math>		<math>r_0 = \biggl( \frac{3}{2\pi}\biggr)^{1 / 2} \xi = \biggl[(2\pi)^{1 / 2}\biggl(\frac{\mu_e}{\mu_c}\biggr)\theta^2 \biggr]^{-1} \eta \, ;</math>
			</div><br />
			and from the series of [[SSC/Stability/BiPolytropes/RedGiantToPN/Pt4#Steps\|analytic expressions shown here]], we also deduce that,
	</div>		</div>