SSC/Structure/BiPolytropes/Analytic51 - Revision history

Joel2: /* Step 8: Throughout the envelope (ηi ≤ η ≤ ηs) */

2026-03-25T23:36:12Z

Step 8: Throughout the envelope (ηi ≤ η ≤ ηs)

← Older revision		Revision as of 19:36, 25 March 2026
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	</table>		</table>

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

Joel2: /* Step 8: Throughout the envelope (ηi ≤ η ≤ ηs) */

2026-03-25T23:30:43Z

Step 8: Throughout the envelope (ηi ≤ η ≤ ηs)

← Older revision		Revision as of 19:30, 25 March 2026
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	----		----

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

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

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

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

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

Joel2: /* Step 8: Throughout the envelope (ηi ≤ η ≤ ηs) */

2026-03-25T22:56:13Z

Step 8: Throughout the envelope (ηi ≤ η ≤ ηs)

← Older revision		Revision as of 18:56, 25 March 2026
Line 684:		Line 684:
	<td align="right"><math>\phi</math></td>		<td align="right"><math>\phi</math></td>
	<td align="center"><math>=</math></td>		<td align="center"><math>=</math></td>
	<td align="left"><math>-\frac{A}{\eta} \cdot \sin(B_\mathrm{~~our~~}-\eta) \, ,</math></td>		<td align="left"><math>-\frac{A}{\eta} \cdot \sin(B_\mathrm{ours}-\eta) \, ,</math></td>
	</tr>		</tr>
	</table>		</table>
Line 736:		Line 736:

	<tr>		<tr>
	<td align="right"~~><math>\eta</math></td>~~		<td align="right" bgcolor="pink"><math>\eta = \beta r \, .</math></td>
	~~<td align~~="~~center~~"><math>=~~</math></td>~~
	~~<td align="left"><math>~~\beta r \, .</math></td>
	</tr>		</tr>
	</table>		</table>
Line 760:		Line 758:
	<table border="0" align="center" cellpadding="5">		<table border="0" align="center" cellpadding="5">
	<tr>		<tr>
	<td align="center"><math>\theta_\mathrm{efc} ~\rightarrow~ \phi \, ;</math></td>		<td align="center" bgcolor="pink"><math>\theta_\mathrm{efc} ~\rightarrow~ \phi \, ;</math></td>
	<td align="center">       </td>		<td align="center">       </td>
	<td align="center"><math>\~~beta~~ B_\mathrm{~~efc~~} ~~~\rightarrow~ -A~~ \, ;</math></td>		<td align="center" bgcolor="pink"><math>\eta_s ~\rightarrow~ B_\mathrm{ours} \, ;</math></td>
	<td align="center">       </td>		<td align="center">        and        </td>
	<td align="center"~~><math>\eta_s ~\rightarrow~ B_\mathrm{our} \, .</math></td>~~		<td align="center" bgcolor="pink"><math>B_\mathrm{efc} ~\rightarrow~ -\frac{A}{\beta } ~~\Rightarrow
	~~</tr>~~		~~ \frac{B_\mathrm{efc}}{r} = -\frac{A}{\eta }\, .</math></td>
	~~</table>~~
	~~where,~~

	~~We therefore conclude that <math>\beta r = \eta</math>, and <math>\beta r_s = B</math>. If, as we assume to be the case, <math>\theta_\mathrm{efc} = \phi</math>, it must also be the case that,~~

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

	~~<tr>~~
	~~<td align="right~~"><math>~~\frac{~~B_\mathrm{efc}~~}{r}</math></td>~~
	~~<td align="center"><math>=</math></td>~~
	~~<td align="left">~~
	~~<math>~~-\frac{A}{\~~eta~~ } ~~</math>~~
	~~</td>~~
	~~</tr>~~

	~~<tr>~~
	~~<td align="right"><math>~~\Rightarrow ~~~ B_\mathrm{efc}~~</math></td>~~
	~~<td align="center"><math>=</math></td>~~
	~~<td align~~=~~"left">~~
	~~<math>~~-\frac{A}{\~~beta~~ } \, .</math>
	</td>
	</tr>		</tr>
	</table>		</table>

Joel2: /* Step 8: Throughout the envelope (ηi ≤ η ≤ ηs) */

2026-03-25T18:20:23Z

Step 8: Throughout the envelope (ηi ≤ η ≤ ηs)

@@ Line 684: / Line 684: @@
    <td align="right"><math>\phi</math></td>
    <td align="center"><math>=</math></td>
-   <td align="left"><math>-\frac{A}{\eta} \cdot \sin(B-\eta) \, ,</math></td>
+   <td align="left"><math>-\frac{A}{\eta} \cdot \sin(B_\mathrm{our}-\eta) \, ,</math></td>
 </tr>
 </table>
@@ Line 741: / Line 741: @@
 </tr>
 </table>
+<tr>
 where,

Joel2: /* Step 8: Throughout the envelope (ηi ≤ η ≤ ηs) */

2026-03-25T17:54:07Z

Step 8: Throughout the envelope (ηi ≤ η ≤ ηs)

← Older revision		Revision as of 13:54, 25 March 2026
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	</table>		</table>

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

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

	<tr>		<tr>
	<td align="right"><math>~~\theta_\mathrm{efc}</math></td>~~		<td align="right"><math>\beta </math></td>
	~~<td align="center"><math>=</math></td>~~
	~~<td align="left"><math>\frac{B_\mathrm{efc}}{r} \cdot \sin[~~\beta~~(r_s - r)] \, ,</math></td>~~
	~~</tr>~~
	~~</table>~~
	~~where,~~

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

	~~<tr>~~
	~~<td align="right"><math>\beta ~r~~</math></td>
	<td align="center"><math>=</math></td>		<td align="center"><math>=</math></td>
	<td align="left">		<td align="left">
	<math>\frac{3\theta_c^2}{\alpha } \biggl[a_\mathrm{efc}\biggr]^{-1} r</math>		<math>\frac{3\theta_c^2}{\alpha } \biggl[a_\mathrm{efc}\biggr]^{-1} </math>
	</td>		</td>
	</tr>		</tr>
Line 727:		Line 717:
	<td align="left">		<td align="left">
	<math>3 \theta_c^2 \biggl(\frac{\mu_e}{\mu_c}\biggr)		<math>3 \theta_c^2 \biggl(\frac{\mu_e}{\mu_c}\biggr)
	\biggl\{\sqrt{3} \biggl[ \frac{K_c }{G \rho_0^{4/5}} \biggr]^{1 / 2} \biggl( \frac{3}{2\pi}\biggr)^{1 / 2}\biggr\}^{-1} r		\biggl\{\sqrt{3} \biggl[ \frac{K_c }{G \rho_0^{4/5}} \biggr]^{1 / 2} \biggl( \frac{3}{2\pi}\biggr)^{1 / 2}\biggr\}^{-1}
	</math>		</math>
	</td>		</td>
Line 736:		Line 726:
	<td align="center"><math>=</math></td>		<td align="center"><math>=</math></td>
	<td align="left">		<td align="left">
	<math>\biggl[ \frac{G \rho_0^{4/5}}{K_c} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{2}_i (2\pi)^{1 / 2}~r \, .</math>		<math>\biggl[ \frac{G \rho_0^{4/5}}{K_c} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{2}_i (2\pi)^{1 / 2} \, .</math>
	</td>		</td>
	</tr>		</tr>
	</table>		</table>

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

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

			<tr>
			<td align="right"><math>\eta</math></td>
			<td align="center"><math>=</math></td>
			<td align="left"><math>\beta r \, .</math></td>
			</tr>
			</table>



			where,

	We therefore conclude that <math>\beta r = \eta</math>, and <math>\beta r_s = B</math>. If, as we assume to be the case, <math>\theta_\mathrm{efc} = \phi</math>, it must also be the case that,		We therefore conclude that <math>\beta r = \eta</math>, and <math>\beta r_s = B</math>. If, as we assume to be the case, <math>\theta_\mathrm{efc} = \phi</math>, it must also be the case that,

Joel2: /* Step 8: Throughout the envelope (ηi ≤ η ≤ ηs) */

2026-03-17T20:34:21Z

Step 8: Throughout the envelope (ηi ≤ η ≤ ηs)

← Older revision		Revision as of 16:34, 17 March 2026
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	<math>		<math>
	\biggl(\frac{2}{\pi}\biggr)^{1 / 2}\biggl\{ \biggl( \frac{K_c}{G } \biggr)^{3/2}\rho_0^{-1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggr\}\theta^{-1}_i A		\biggl(\frac{2}{\pi}\biggr)^{1 / 2}\biggl\{ \biggl( \frac{K_c}{G } \biggr)^{3/2}\rho_0^{-1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggr\}\theta^{-1}_i A
	\biggl[\sin(\eta - B) - (\eta )\cos(\eta - B) \biggr]_{\eta_i}^{\eta}		\biggl[\sin(\eta - B) - \eta \cos(\eta - B) \biggr]_{\eta_i}^{\eta}
	</math>		</math>
	</td>		</td>

Joel2: /* Step 6: Envelope Solution */

2026-03-17T20:31:10Z

Step 6: Envelope Solution

← Older revision		Revision as of 16:31, 17 March 2026
Line 346:		Line 346:

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

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

Joel2: /* Step 6: Envelope Solution */

2026-03-17T20:25:32Z

Step 6: Envelope Solution

← Older revision		Revision as of 16:25, 17 March 2026
Line 345:		Line 345:


			<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left" bgcolor="lightgreen">
			For purposes of determining the envelope mass, below, note that,

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

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

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

Line 363:		Line 377:

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

	From Step 5, above, we know the value of the function, <math>\phi</math> and its first derivative at the interface; specifically,		From Step 5, above, we know the value of the function, <math>\phi</math> and its first derivative at the interface; specifically,

Joel2: /* Step 8: Throughout the envelope (ηi ≤ η ≤ ηs) */

2026-03-17T18:48:18Z

Step 8: Throughout the envelope (ηi ≤ η ≤ ηs)

@@ Line 600: / Line 600: @@
 \biggl(\frac{2}{\pi}\biggr)^{1 / 2}\biggl\{ \biggl( \frac{K_c}{G } \biggr)^{3/2}\rho_0^{-1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggr\}\theta^{-1}_i A
 \int_{(\upsilon_i + B)}^{(\upsilon+B)} (\upsilon + B)\sin(\upsilon)~ d\upsilon
 </math>
    </td>

Joel2: /* Step 8: Throughout the envelope (ηi ≤ η ≤ ηs) */

2026-03-17T17:42:38Z

Step 8: Throughout the envelope (ηi ≤ η ≤ ηs)

← Older revision		Revision as of 13:42, 17 March 2026
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	<math>		<math>
	\biggl(\frac{2}{\pi}\biggr)^{1 / 2}\biggl\{ \biggl( \frac{K_c}{G } \biggr)^{3/2}\rho_0^{-1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggr\}\theta^{-1}_i A		\biggl(\frac{2}{\pi}\biggr)^{1 / 2}\biggl\{ \biggl( \frac{K_c}{G } \biggr)^{3/2}\rho_0^{-1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggr\}\theta^{-1}_i A
	\int_{(\upsilon_i + B)}^{(\upsilon+B)} (\upsilon + B)~~\biggl[~~ \sin(\~~eta - B~~) ~~\biggr]~~		\int_{(\upsilon_i + B)}^{(\upsilon+B)} (\upsilon + B)\sin(\upsilon)~ d\upsilon
	~ d\~~eta~~
	</math>		</math>
	</td>		</td>