Difference between revisions of "User:Tohline/SSC/Stability/BiPolytrope0 0"
(→Attempt to Solve: First power-law solution described in detail) |
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</td> | </td> | ||
<td align="left"> | <td align="left"> | ||
<math>~\frac{3\omega^2 }{2\pi \gamma_\mathrm{g} G \ | <math>~\frac{3\omega^2 }{2\pi \gamma_\mathrm{g} G \rho_c} - 2 \alpha \, ,</math> | ||
</td> | </td> | ||
</tr> | </tr> | ||
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</table> | </table> | ||
</div> | </div> | ||
===A Specific Choice of the Density Ratio=== | |||
Now, let's focus on the ''specific'' model for which <math>~\rho_e/\rho_c = 1/2</math>. In this case, | Now, let's focus on the ''specific'' model for which <math>~\rho_e/\rho_c = 1/2</math>. In this case, | ||
<div align="center"> | <div align="center"> | ||
Line 1,126: | Line 1,128: | ||
2\xi \cdot \frac{dx}{d\xi} \biggr] | 2\xi \cdot \frac{dx}{d\xi} \biggr] | ||
- \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] x | - \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] x | ||
\biggr\} | |||
</math> | |||
</td> | |||
</tr> | |||
<tr> | |||
<td align="right"> | |||
| |||
</td> | |||
<td align="center"> | |||
<math>~=</math> | |||
</td> | |||
<td align="left"> | |||
<math>~\frac{x}{4\xi^3} \biggl\{ | |||
\biggl[ 2 + (2g^2 - 5)\xi - \xi^3 \biggr] \biggl[ | |||
\frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} \biggr] | |||
+\biggl[ 6 + 4(2g^2 - 5)\xi - 6\xi^3 \biggr] \biggl[ | |||
\frac{\xi}{x} \cdot \frac{dx}{d\xi} \biggr] | |||
- \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] | |||
\biggr\} \, . | \biggr\} \, . | ||
</math> | </math> | ||
</td> | |||
</tr> | |||
</table> | |||
</div> | |||
===Idea Involving Logarithmic Derivatives=== | |||
Notice that the term involving the first derivative of <math>~x</math> can be written as a logarithmic derivative; specifically, | |||
<div align="center"> | |||
<table border="0" cellpadding="5" align="center"> | |||
<tr> | |||
<td align="right"> | |||
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi} </math> | |||
</td> | |||
<td align="center"> | |||
<math>~=</math> | |||
</td> | |||
<td align="left"> | |||
<math>~\frac{d\ln x}{d\ln \xi} \, .</math> | |||
</td> | |||
</tr> | |||
</table> | |||
</div> | |||
Let's look at the second derivative of this quantity. | |||
<div align="center"> | |||
<table border="0" cellpadding="5" align="center"> | |||
<tr> | |||
<td align="right"> | |||
<math>~\frac{d}{d\xi} \biggl[ \frac{d\ln x}{d\ln \xi} \biggr]</math> | |||
</td> | |||
<td align="center"> | |||
<math>~=</math> | |||
</td> | |||
<td align="left"> | |||
<math>~ | |||
\frac{\xi}{x} \cdot \frac{d^2x}{d\xi^2} | |||
+ \frac{dx}{d\xi} \cdot \biggl[ \frac{1}{x} - \frac{\xi}{x^2} \cdot \frac{dx}{d\xi}\biggr] | |||
</math> | |||
</td> | |||
</tr> | |||
<tr> | |||
<td align="right"> | |||
| |||
</td> | |||
<td align="center"> | |||
<math>~=</math> | |||
</td> | |||
<td align="left"> | |||
<math>~ | |||
\frac{\xi}{x} \cdot \frac{d^2x}{d\xi^2} | |||
+ \frac{1}{x} \biggl[ 1 - \frac{d\ln x}{d\ln \xi} \biggr]\cdot \frac{dx}{d\xi} | |||
</math> | |||
</td> | |||
</tr> | |||
<tr> | |||
<td align="right"> | |||
<math>~\Rightarrow ~~~ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} | |||
</math> | |||
</td> | |||
<td align="center"> | |||
<math>~=</math> | |||
</td> | |||
<td align="left"> | |||
<math>~ | |||
\frac{d}{d\ln\xi} \biggl[ \frac{d\ln x}{d\ln \xi} \biggr] | |||
- \biggl[ 1 - \frac{d\ln x}{d\ln \xi} \biggr]\cdot \frac{d\ln x}{d\ln \xi} \, . | |||
</math> | |||
</td> | |||
</tr> | |||
</table> | |||
</div> | |||
Now, if we ''assume'' that the envelope's eigenfunction is a power-law of <math>~\xi</math>, that is, ''assume'' that, | |||
<div align="center"> | |||
<math>~x = a_0 \xi^{c_0} \, ,</math> | |||
</div> | |||
then the logarithmic derivative of <math>~x</math> is a constant, namely, | |||
<div align="center"> | |||
<math>~\frac{d\ln x}{d\ln\xi} = c_0 \, ,</math> | |||
</div> | |||
and the two key derivative terms will be, | |||
<div align="center"> | |||
<table border="0" cellpadding="5" align="center"> | |||
<tr> | |||
<td align="right"> | |||
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi} = c_0 \, ,</math> | |||
</td> | |||
<td align="center"> | |||
and | |||
</td> | |||
<td align="left"> | |||
<math>~\frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} = c_0(c_0-1) \, .</math> | |||
</td> | |||
</tr> | |||
</table> | |||
</div> | |||
Hence, in order for the wave equation for the envelope for the ''specific'' density ratio being considered here to be satisfied, we need, | |||
<div align="center"> | |||
<table border="0" cellpadding="5" align="center"> | |||
<tr> | |||
<td align="right"> | |||
<math>~0</math> | |||
</td> | |||
<td align="center"> | |||
<math>~=</math> | |||
</td> | |||
<td align="left"> | |||
<math>~ | |||
\biggl[ 2 + (2g^2 - 5)\xi - \xi^3 \biggr] \biggl[ | |||
c_0(c_0-1) \biggr] | |||
+\biggl[ 6 + 4(2g^2 - 5)\xi - 6\xi^3 \biggr] \biggl[ | |||
c_0 \biggr] | |||
- \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] | |||
</math> | |||
</td> | |||
</tr> | |||
<tr> | |||
<td align="right"> | |||
| |||
</td> | |||
<td align="center"> | |||
<math>~=</math> | |||
</td> | |||
<td align="left"> | |||
<math>~ | |||
\biggl[ 2 + (2g^2 - 5)\xi - \xi^3 \biggr] c_0(c_0-1) | |||
+c_0\biggl[ 6 + 4(2g^2 - 5)\xi - 6\xi^3 \biggr] | |||
- \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] | |||
</math> | |||
</td> | |||
</tr> | |||
<tr> | |||
<td align="right"> | |||
| |||
</td> | |||
<td align="center"> | |||
<math>~=</math> | |||
</td> | |||
<td align="left"> | |||
<math>~ \biggl[2c_0(c_0-1) + 6c_0 - 2\alpha \biggr] | |||
+ \biggl[(2g^2-5)(c_0^2 - c_0 + 4c_0 ) \biggr]\xi | |||
+ \biggl[ -c_0(c_0-1) -6c_0 + 2(\mathfrak{F}+\alpha) \biggr]\xi^3 | |||
</math> | |||
</td> | |||
</tr> | |||
<tr> | |||
<td align="right"> | |||
| |||
</td> | |||
<td align="center"> | |||
<math>~=</math> | |||
</td> | |||
<td align="left"> | |||
<math>~ 2\biggl[c_0^2 + 2c_0 - \alpha \biggr] | |||
+ \biggl[(2g^2-5)(c_0^2 + 3c_0 ) \biggr]\xi | |||
+ \biggl[ 2(\mathfrak{F}+\alpha) - c_0(c_0+5) \biggr]\xi^3 \, . | |||
</math> | |||
</td> | |||
</tr> | |||
</table> | |||
</div> | |||
This means that three algebraic relations must simultaneously be satisfied, namely: | |||
<div align="center"> | |||
<table border="0" cellpadding="5" align="center"> | |||
<tr> | |||
<td align="right"> | |||
<math>~\xi^{0}:</math> | |||
</td> | |||
<td align="left"> | |||
<math>~c_0^2 + 2c_0 - \alpha =0</math> | |||
</td> | |||
<td align="center"> | |||
<math>~\Rightarrow~</math> | |||
</td> | |||
<td align="left"> | |||
<math>~c_0 = -1 \pm (1+\alpha)^{1 / 2} \, ;</math> | |||
</td> | |||
</tr> | |||
<tr> | |||
<td align="right"> | |||
<math>~\xi^{1}:</math> | |||
</td> | |||
<td align="left"> | |||
<math>~g^2 = \frac{5}{2}</math> | |||
</td> | |||
<td align="center"> | |||
<math>~\Rightarrow~</math> | |||
</td> | |||
<td align="left"> | |||
<math>~2q^3 + 5q^2 - 1 = 0 \, ;</math> | |||
</td> | |||
</tr> | |||
<tr> | |||
<td align="right"> | |||
<math>~\xi^{3}:</math> | |||
</td> | |||
<td align="left"> | |||
<math>~2(\mathfrak{F}+\alpha) = c_0(c_0+5)</math> | |||
</td> | |||
<td align="center"> | |||
<math>~\Rightarrow~</math> | |||
</td> | |||
<td align="left"> | |||
<math>~\sigma^2 = \alpha + \frac{c_0}{2}\biggl( c_0+5 \biggr) \, .</math> | |||
</td> | </td> | ||
</tr> | </tr> |
Revision as of 03:14, 27 November 2016
Radial Oscillations of a Zero-Zero Bipolytrope
| Tiled Menu | Tables of Content | Banner Video | Tohline Home Page | |
Groundwork
In an accompanying discussion, we derived the so-called,
Adiabatic Wave (or Radial Pulsation) Equation
<math>~ \frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} + \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr] x = 0 </math> |
whose solution gives eigenfunctions that describe various radial modes of oscillation in spherically symmetric, self-gravitating fluid configurations. According to our accompanying derivation, if the initial, unperturbed equilibrium configuration is an <math>~(n_c, n_e) = (0,0)</math> bipolytrope, then we know that the relevant functional profiles are as follows for the core and envelope, separately. Note that, throughout, we will preferentially adopt as the dimensionless radial coordinate, the parameter,
<math>~\xi</math> |
<math>~\equiv</math> |
<math>~\frac{r}{r_i} \, ,</math> |
in which case,
<math>~\chi</math> |
<math>~=</math> |
<math>~ \chi_i \xi = q \biggl( \frac{G\rho_c^2 R^2}{P_c} \biggr)^{1 /2 }\xi \, .</math> |
The corresponding radial coordinate range is,
<math>~0 \le \xi \le 1 </math> for the core, and
<math>~1 \le \xi \le \frac{1}{q} </math> for the envelope.
Core
<math>~r_0</math> |
<math>~=</math> |
<math>~\biggl( \frac{P_c}{G\rho_c^2}\biggr)^{1 / 2} \chi = (qR) \xi \, ,</math> |
<math>~\rho_0</math> |
<math>~=</math> |
<math>~\rho_c \, ,</math> |
<math>~\frac{P_0}{P_c}</math> |
<math>~=</math> |
<math>~1 - \frac{2\pi}{3} \chi^2 = 1 - \frac{2\pi}{3} \biggl[ \frac{G\rho_c^2 R^2}{P_c} \biggr] q^2 \xi^2 = 1 - \frac{\xi^2}{g^2} \, ,</math> |
<math>~M_r</math> |
<math>~=</math> |
<math>~\frac{4\pi}{3} \biggl( \frac{P_c^3}{G^3 \rho_c^4} \biggr)^{1 / 2}\chi^3 = \frac{4\pi}{3} \biggl( \frac{P_c^3}{G^3 \rho_c^4} \biggr)^{1 / 2} \biggl( \frac{G\rho_c^2 R^2}{P_c} \biggr)^{3 /2 } (q\xi)^3 </math> |
|
<math>~=</math> |
<math>~ \frac{4\pi}{3} ( \rho_c R^3 ) (q\xi)^3 = \frac{4\pi}{3} (q\xi)^3 \rho_c \biggl[ \biggl( \frac{P_c}{G\rho_c^2} \biggr)^{1 / 2} \biggl( \frac{3}{2\pi} \biggr)^{1 / 2} \frac{1}{qg}\biggr]^3 </math> |
|
<math>~=</math> |
<math>~ \frac{4\pi}{3} (q\xi)^3 \biggl[ \biggl( \frac{P_c^3}{G^3\rho_c^4} \biggr)^{1 / 2} \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} \frac{1}{q^3g^3}\biggr] = \frac{4\pi}{3} \biggl[ \biggl(\frac{\pi}{6}\biggr)^{1 / 2} \nu g^3 M_\mathrm{tot} \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} \frac{1}{g^3}\biggr]\xi^3 </math> |
|
<math>~=</math> |
<math>~ M_\mathrm{tot} \nu \xi^3 \, , </math> |
where,
<math>~g^2(\nu,q)</math> |
<math>~\equiv</math> |
<math> \biggl\{ 1 + \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( 1-q \biggr) + \frac{\rho_e}{\rho_c} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] \biggr\} \, , </math> |
<math>~\frac{\rho_e}{\rho_c}</math> |
<math>~=</math> |
<math> \frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3}\biggr) \, . </math> |
Hence,
<math>~g_0</math> |
<math>~=</math> |
<math>~\frac{G(M_\mathrm{tot} \nu \xi^3)}{(qR\xi)^2} = \biggl( \frac{GM_\mathrm{tot} }{R^2 } \biggr) \frac{\nu \xi}{q^2} </math> |
|
<math>~=</math> |
<math>~ G \biggl[\biggl( \frac{P_c^3}{G^3\rho_c^4} \biggr)^{1 / 2} \biggl(\frac{6}{\pi}\biggr)^{1 / 2} \frac{1}{\nu g^3} \biggr] \biggl[\biggl(\frac{G\rho_c^2}{P_c} \biggr)^{ 1 / 2} \biggl(\frac{2\pi}{3} \biggr)^{1 / 2} qg \biggr]^2 \frac{\nu \xi}{q^2} </math> |
|
<math>~=</math> |
<math>~ (P_c G)^{1 / 2} \biggl(\frac{2^3\pi}{3} \biggr)^{1 / 2} \frac{\xi}{g} </math> |
<math>~\frac{\rho_0}{P_0}</math> |
<math>~=</math> |
<math>~ \frac{\rho_c}{P_c} \biggl[ 1 - \frac{\xi^2}{g^2} \biggr]^{-1} = \frac{\rho_c}{P_c} \biggl( \frac{g^2}{g^2 - \xi^2} \biggr) \, ;</math> |
and the wave equation for the core becomes,
<math>~0</math> |
<math>~=</math> |
<math>~ \frac{1}{(qR)^2} \cdot \frac{d^2x}{d\xi^2} + \biggl[\frac{4qR}{r_0} - \biggl(\frac{qR g_0 \rho_0}{P_0}\biggr) \biggr] \frac{1}{(qR)^2} \cdot \frac{dx}{d\xi} + \biggl(\frac{\rho_0}{P_0} \biggr)\biggl[ \frac{\omega^2}{\gamma_\mathrm{g} } + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g} } \biggr)\frac{g_0}{r_0} \biggr] x </math> |
|
<math>~=</math> |
<math>~ \frac{1}{(qR)^2} \biggl\{ \frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - q\biggl(\frac{P_c}{G\rho_c^2} \biggr)^{1 / 2}\biggl(\frac{3}{2\pi}\biggr)^{1 / 2} \frac{1}{qg} (P_c G)^{1 / 2} \biggl(\frac{2^3\pi}{3} \biggr)^{1 / 2} \frac{\xi}{g} \frac{\rho_c}{P_c} \biggl( \frac{g^2}{g^2 - \xi^2} \biggr) \biggr] \frac{dx}{d\xi} \biggr\} </math> |
|
|
<math>~ + \frac{\rho_c}{P_c} \biggl( \frac{g^2}{g^2 - \xi^2} \biggr) \biggl[ \frac{\omega^2}{\gamma_\mathrm{g} } + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g} } \biggr)(P_c G)^{1 / 2} \biggl(\frac{2^3\pi}{3} \biggr)^{1 / 2} \frac{\xi}{g} \cdot \frac{1}{qR\xi}\biggr] x </math> |
|
<math>~=</math> |
<math>~ \frac{1}{(qR)^2} \biggl\{ \frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \biggl( \frac{2\xi}{g^2 - \xi^2} \biggr) \biggr] \frac{dx}{d\xi} \biggr\} </math> |
|
|
<math>~ + \frac{\rho_c}{P_c} \biggl( \frac{g^2}{g^2 - \xi^2} \biggr) \biggl[ \frac{\omega^2}{\gamma_\mathrm{g} } + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g} } \biggr)(P_c G)^{1 / 2} \biggl(\frac{2^3\pi}{3} \biggr)^{1 / 2} \frac{1}{qg} \biggl(\frac{G\rho_c^2}{P_c} \biggr)^{1 / 2} \biggl( \frac{2\pi}{3} \biggr)^{1 / 2} qg \biggr] x </math> |
|
<math>~=</math> |
<math>~ \frac{1}{(qR)^2(g^2 - \xi^2)} \biggl\{ (g^2 - \xi^2)\frac{d^2x}{d\xi^2} + ( 4g^2 - 6\xi^2 ) \frac{1}{\xi} \cdot \frac{dx}{d\xi} + \frac{q^2 g^2 R^2 \rho_c}{P_c} \biggl[ \frac{\omega^2}{\gamma_\mathrm{g} } + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g} } \biggr) \frac{4\pi G\rho_c}{3} \biggr] x \biggr\} </math> |
|
<math>~=</math> |
<math>~ \frac{1}{(qR)^2(g^2 - \xi^2)} \biggl\{ (g^2 - \xi^2)\frac{d^2x}{d\xi^2} + ( 4g^2 - 6\xi^2 ) \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 2\biggl[ \frac{3\omega^2}{\gamma_\mathrm{g}4\pi G\rho_c} + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g} } \biggr) \biggr] x \biggr\} \, . </math> |
Envelope
<math>~r_0</math> |
<math>~=</math> |
<math>~ (qR) \xi \, ,</math> |
<math>~\rho_0</math> |
<math>~=</math> |
<math>~\rho_e \, ,</math> |
<math>~\frac{P_0}{P_c}</math> |
<math>~=</math> |
<math> 1 - \frac{2\pi}{3}\chi_i^2 + \frac{2\pi}{3} \biggl(\frac{\rho_e}{\rho_c}\biggr) \chi_i^2 \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{1}{\xi} - 1\biggr) - \frac{\rho_e}{\rho_c} (\xi^2 - 1) \biggr] </math> |
|
<math>~=</math> |
<math> 1 - \frac{1}{g^2}\biggl\{ 1 - \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{1}{\xi} - 1\biggr) - \frac{\rho_e}{\rho_c} (\xi^2 - 1) \biggr] \biggr\} </math> |
<math>~\Rightarrow ~~~ \frac{g^2 P_0}{P_c}</math> |
<math>~=</math> |
<math> g^2 - 1 + \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{1}{\xi} - 1\biggr) - \frac{\rho_e}{\rho_c} (\xi^2 - 1) \biggr] \, , </math> |
<math>~M_r</math> |
<math>~=</math> |
<math>\frac{4\pi}{3} \biggl[ \frac{P_c^3}{G^3 \rho_c^4} \biggr]^{1/2} \chi_i^3\biggl[1 +\frac{\rho_e}{\rho_c} \biggl( \xi^3 - 1\biggr) \biggr]</math> |
|
<math>~=</math> |
<math>M_\mathrm{tot} \frac{4\pi}{3} \biggl[\biggl( \frac{\pi}{6}\biggr)^{1 / 2}\nu g^3 \biggr] \biggl[ \biggr(\frac{3}{2\pi}\biggr)\frac{1}{g^2} \biggr]^{3 /2} \biggl[1 +\frac{\rho_e}{\rho_c} \biggl( \xi^3 - 1\biggr) \biggr] </math> |
|
<math>~=</math> |
<math> \nu M_\mathrm{tot} \biggl[1 +\frac{\rho_e}{\rho_c} \biggl( \xi^3 - 1\biggr) \biggr] \, . </math> |
Hence,
<math>~g_0</math> |
<math>~=</math> |
<math>~ \frac{G M_\mathrm{tot}\nu }{ R^2 q^2\xi^2} \biggl[1 +\frac{\rho_e}{\rho_c} \biggl( \xi^3 - 1\biggr) \biggr] \, , </math> |
and, after multiplying through by <math>~(q^2 R^2 g^2P_0/P_c)</math>, the wave equation for the envelope becomes,
<math>~0</math> |
<math>~=</math> |
<math>~\frac{q^2 g^2 R^2 P_0}{P_c} \biggl\{ \frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} \biggr\} + \frac{q^2 g^2 R^2 \rho_0}{P_c} \biggl[ \frac{\omega^2 }{\gamma_\mathrm{g}} + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr)\frac{g_0}{r_0} \biggr] x </math> |
|
<math>~=</math> |
<math>~\frac{g^2 P_0}{P_c} \biggl\{ \frac{d^2x}{d\xi^2} + \biggl[4 - \biggl(\frac{qRg_0 \rho_e}{P_0}\biggr) \xi\biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} \biggr\} + \frac{q^2 g^2 R^2 \rho_e}{P_c} \biggl[ \frac{\omega^2 }{\gamma_\mathrm{g}} + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr)\frac{g_0}{r_0} \biggr] x </math> |
|
<math>~=</math> |
<math>~ \frac{g^2 P_0}{P_c} \biggl[ \frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr] - \biggl(\frac{qg^2Rg_0 \rho_e}{P_c}\biggr) \frac{dx}{d\xi} </math> |
|
|
<math>~ + 2\biggl(\frac{\rho_e}{\rho_c}\biggr) \frac{3}{4\pi G \rho_c} \biggl\{ \frac{\omega^2 }{\gamma_\mathrm{g}} + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr)\biggl(\frac{4\pi G \rho_c}{3}\biggr) \biggl[ \frac{1}{\xi^3} + \frac{\rho_e}{\rho_c}\biggl(1-\frac{1}{\xi^3}\biggr) \biggr] \biggr\} x </math> |
|
<math>~=</math> |
<math>~ \frac{g^2 P_0}{P_c} \biggl[ \frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr] - 2 \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[1 +\frac{\rho_e}{\rho_c} \biggl( \xi^3 - 1\biggr) \biggr] \frac{1}{\xi^2} \cdot \frac{dx}{d\xi} </math> |
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|
<math>~ + 2\biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl\{ \frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}} + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr) \biggl[ \frac{1}{\xi^3} + \frac{\rho_e}{\rho_c}\biggl(1-\frac{1}{\xi^3}\biggr) \biggr] \biggr\} x </math> |
Check1
If <math>~\rho_e/\rho_c = 1</math>, this envelope wave equation should match seamlessly into the core wave equation. Let's see if it does. First,
<math>~g^2(\nu,q)|_{\rho_e=\rho_c}</math> |
<math>~=</math> |
<math> 1 + \biggl(\frac{1}{q^2} - 1\biggr) =\frac{1}{q^2} \, , </math> |
<math>~\frac{g^2 P_0}{P_c} \biggr|_{\rho_e = \rho_c}</math> |
<math>~=</math> |
<math>~ g^2 - \xi^2 = \frac{1}{q^2} - \xi^2 \, . </math> |
Hence, for the envelope,
<math>~0</math> |
<math>~=</math> |
<math>~ \frac{g^2 P_0}{P_c} \biggl[ \frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr] - 2 \biggl[1 + \biggl( \xi^3 - 1\biggr) \biggr] \frac{1}{\xi^2} \cdot \frac{dx}{d\xi} </math> |
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<math>~ + 2 \biggl\{ \frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}} + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr) \biggl[ \frac{1}{\xi^3} + \biggl(1-\frac{1}{\xi^3}\biggr) \biggr] \biggr\} x </math> |
|
<math>~=</math> |
<math>~ \biggl( \frac{1}{q^2} - \xi^2 \biggr) \biggl[ \frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr] - 2\xi \cdot \frac{dx}{d\xi} + 2 \biggl\{ \frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}} + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr) \biggr\} x </math> |
|
<math>~=</math> |
<math>~ \biggl( \frac{1}{q^2} - \xi^2 \biggr) \frac{d^2x}{d\xi^2} + \biggl\{ 4\biggl( \frac{1}{q^2} - \xi^2 \biggr) - 2\xi^2 \biggr\} \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 2 \biggl[ \frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}} + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr) \biggr] x </math> |
|
<math>~=</math> |
<math>~ \biggl( \frac{1}{q^2} - \xi^2 \biggr) \frac{d^2x}{d\xi^2} + \biggl( \frac{4}{q^2} - 6\xi^2 \biggr) \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 2 \biggl[ \frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}} + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g}} \biggr) \biggr] x \, . </math> |
Whereas, for the core,
<math>~0</math> |
<math>~=</math> |
<math>~ \biggl(\frac{1}{q^2} - \xi^2 \biggr)\frac{d^2x}{d\xi^2} + \biggl( \frac{4}{q^2} - 6\xi^2 \biggr) \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 2\biggl[ \frac{3\omega^2}{\gamma_\mathrm{g}4\pi G\rho_c} + \biggl( \frac{4 - 3\gamma_\mathrm{g}}{\gamma_\mathrm{g} } \biggr) \biggr] x \, , </math> |
which matches exactly.
Attempt to Solve
Adopting some of the notation used by T. E. Sterne (1937) and enunciated in our accompanying discussion of the uniform-density sphere, we'll define,
<math>~\alpha</math> |
<math>~\equiv</math> |
<math>~3 - 4/\gamma_\mathrm{g} \, ,</math> |
<math>~\mathfrak{F}</math> |
<math>~\equiv</math> |
<math>~\frac{3\omega^2 }{2\pi \gamma_\mathrm{g} G \rho_c} - 2 \alpha \, ,</math> |
in which case the wave equation for the core becomes,
<math>~0</math> |
<math>~=</math> |
<math>~ \frac{1}{(qR)^2(g^2 - \xi^2)} \biggl\{ (g^2 - \xi^2)\frac{d^2x}{d\xi^2} + ( 4g^2 - 6\xi^2 ) \frac{1}{\xi} \cdot \frac{dx}{d\xi} + \mathfrak{F} x \biggr\} \, , </math> |
and the wave equation for the envelope becomes,
<math>~0</math> |
<math>~=</math> |
<math>~ \frac{g^2 P_0}{P_c} \biggl[ \frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr] - 2 \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[1 +\frac{\rho_e}{\rho_c} \biggl( \xi^3 - 1\biggr) \biggr] \frac{1}{\xi^2} \cdot \frac{dx}{d\xi} </math> |
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<math>~ + \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl\{ \mathfrak{F} + 2\alpha \biggl[1 - \frac{1}{\xi^3} - \frac{\rho_e}{\rho_c}\biggl(1-\frac{1}{\xi^3}\biggr) \biggr] \biggr\} x \, . </math> |
A Specific Choice of the Density Ratio
Now, let's focus on the specific model for which <math>~\rho_e/\rho_c = 1/2</math>. In this case,
<math>~g^2(\nu,q) \biggr|_{\rho_e/\rho_c=1/2}</math> |
<math>~=</math> |
<math> 1 + \frac{1}{2} \biggl[ 1-q + \frac{1}{2} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] </math> |
|
<math>~=</math> |
<math>\frac{1}{4q^2}\biggl\{ 4q^2 + \biggl[ 2q^2 - 2q^3 + 1-q^2 \biggr] \biggr\} </math> |
|
<math>~=</math> |
<math>~ \biggl[ \frac{1+5q^2 - 2q^3 }{4q^2} \biggr] \, ; </math> |
<math>~\frac{g^2 P_0}{P_c}\biggr|_{\rho_e/\rho_c=1/2}</math> |
<math>~=</math> |
<math> g^2 - 1 + \frac{1}{2} \biggl[ \biggl( \frac{1}{\xi} - 1\biggr) - \frac{1}{2} \biggl(\xi^2 - 1 \biggr) \biggr] </math> |
|
<math>~=</math> |
<math> g^2 - 1 - \frac{1}{4} \biggl[ \xi^2 + 1 - \frac{2}{\xi} \biggr] </math> |
|
<math>~=</math> |
<math> g^2 - \frac{\xi^2}{4} \biggl[ 1 + \frac{5}{\xi^2} - \frac{2}{\xi^3} \biggr] \, . </math> |
Note that this last expression goes to zero at the surface of the bipolytrope, that is, at <math>~\xi = 1/q</math>. For this specific case, the wave equation for the envelope becomes,
<math>~0</math> |
<math>~=</math> |
<math>~ \frac{g^2 P_0}{P_c} \biggl[ \frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr] - \biggl[1 +\frac{1}{2} \biggl( \xi^3 - 1\biggr) \biggr] \frac{1}{\xi^2} \cdot \frac{dx}{d\xi} + \frac{1}{2} \biggl\{ \mathfrak{F} + 2\alpha \biggl[1 - \frac{1}{\xi^3} + \frac{1}{2}\biggl(-1 + \frac{1}{\xi^3}\biggr) \biggr] \biggr\} x </math> |
|
<math>~=</math> |
<math>~ \biggl\{ g^2 - \frac{\xi^2}{4} \biggl[ 1 + \frac{5}{\xi^2} - \frac{2}{\xi^3} \biggr] \biggr\} \biggl[ \frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr] - \frac{1}{2}\biggl[1 + \xi^3 \biggr] \frac{1}{\xi^2} \cdot \frac{dx}{d\xi} + \frac{1}{2} \biggl\{ \mathfrak{F} + \alpha \biggl[1 - \frac{1}{\xi^3} \biggr] \biggr\} x </math> |
|
<math>~=</math> |
<math>~\frac{1}{4\xi^3} \biggl\{ \biggl[ 2g^2\xi^3 - \xi^5 \biggl( 1 + \frac{5}{\xi^2} - \frac{2}{\xi^3} \biggr) \biggr] \biggl[ \frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr] - 2\xi (1 + \xi^3 ) \frac{dx}{d\xi} + 2 \xi^3 \biggl[ \mathfrak{F} + \alpha \biggl(1 - \frac{1}{\xi^3} \biggr) \biggr] x \biggr\} </math> |
|
<math>~=</math> |
<math>~\frac{1}{4\xi^3} \biggl\{ \biggl[ 2g^2\xi^3 - \xi^5 - 5\xi^3 + 2\xi^2 \biggr] \biggl[ \frac{d^2x}{d\xi^2} + \frac{4}{\xi} \cdot \frac{dx}{d\xi} \biggr] - 2\xi (1 + \xi^3 ) \frac{dx}{d\xi} + \biggl[ 2 \xi^3 (\mathfrak{F} + \alpha) - 2\alpha \biggr] x \biggr\} </math> |
|
<math>~=</math> |
<math>~\frac{1}{4\xi^3} \biggl\{ \biggl[ 2 + (2g^2 - 5)\xi - \xi^3 \biggr] \biggl[ \xi^2 \cdot \frac{d^2x}{d\xi^2} + 4\xi \cdot \frac{dx}{d\xi} \biggr] - 2(1 + \xi^3 ) \biggl[ \xi \cdot \frac{dx}{d\xi} \biggr] - \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] x \biggr\} </math> |
|
<math>~=</math> |
<math>~\frac{1}{4\xi^3} \biggl\{ \biggl[ 2 + (2g^2 - 5)\xi - \xi^3 \biggr] \biggl[ \xi^2 \cdot \frac{d^2x}{d\xi^2} \biggr] +\biggl[ 3 + (4g^2 - 10)\xi - 3\xi^3 \biggr] \biggl[ 2\xi \cdot \frac{dx}{d\xi} \biggr] - \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] x \biggr\} </math> |
|
<math>~=</math> |
<math>~\frac{x}{4\xi^3} \biggl\{ \biggl[ 2 + (2g^2 - 5)\xi - \xi^3 \biggr] \biggl[ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} \biggr] +\biggl[ 6 + 4(2g^2 - 5)\xi - 6\xi^3 \biggr] \biggl[ \frac{\xi}{x} \cdot \frac{dx}{d\xi} \biggr] - \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] \biggr\} \, . </math> |
Idea Involving Logarithmic Derivatives
Notice that the term involving the first derivative of <math>~x</math> can be written as a logarithmic derivative; specifically,
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi} </math> |
<math>~=</math> |
<math>~\frac{d\ln x}{d\ln \xi} \, .</math> |
Let's look at the second derivative of this quantity.
<math>~\frac{d}{d\xi} \biggl[ \frac{d\ln x}{d\ln \xi} \biggr]</math> |
<math>~=</math> |
<math>~ \frac{\xi}{x} \cdot \frac{d^2x}{d\xi^2} + \frac{dx}{d\xi} \cdot \biggl[ \frac{1}{x} - \frac{\xi}{x^2} \cdot \frac{dx}{d\xi}\biggr] </math> |
|
<math>~=</math> |
<math>~ \frac{\xi}{x} \cdot \frac{d^2x}{d\xi^2} + \frac{1}{x} \biggl[ 1 - \frac{d\ln x}{d\ln \xi} \biggr]\cdot \frac{dx}{d\xi} </math> |
<math>~\Rightarrow ~~~ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} </math> |
<math>~=</math> |
<math>~ \frac{d}{d\ln\xi} \biggl[ \frac{d\ln x}{d\ln \xi} \biggr] - \biggl[ 1 - \frac{d\ln x}{d\ln \xi} \biggr]\cdot \frac{d\ln x}{d\ln \xi} \, . </math> |
Now, if we assume that the envelope's eigenfunction is a power-law of <math>~\xi</math>, that is, assume that,
<math>~x = a_0 \xi^{c_0} \, ,</math>
then the logarithmic derivative of <math>~x</math> is a constant, namely,
<math>~\frac{d\ln x}{d\ln\xi} = c_0 \, ,</math>
and the two key derivative terms will be,
<math>~\frac{\xi}{x} \cdot \frac{dx}{d\xi} = c_0 \, ,</math> |
and |
<math>~\frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} = c_0(c_0-1) \, .</math> |
Hence, in order for the wave equation for the envelope for the specific density ratio being considered here to be satisfied, we need,
<math>~0</math> |
<math>~=</math> |
<math>~ \biggl[ 2 + (2g^2 - 5)\xi - \xi^3 \biggr] \biggl[ c_0(c_0-1) \biggr] +\biggl[ 6 + 4(2g^2 - 5)\xi - 6\xi^3 \biggr] \biggl[ c_0 \biggr] - \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] </math> |
|
<math>~=</math> |
<math>~ \biggl[ 2 + (2g^2 - 5)\xi - \xi^3 \biggr] c_0(c_0-1) +c_0\biggl[ 6 + 4(2g^2 - 5)\xi - 6\xi^3 \biggr] - \biggl[ 2\alpha - 2 \xi^3 (\mathfrak{F} + \alpha) \biggr] </math> |
|
<math>~=</math> |
<math>~ \biggl[2c_0(c_0-1) + 6c_0 - 2\alpha \biggr] + \biggl[(2g^2-5)(c_0^2 - c_0 + 4c_0 ) \biggr]\xi + \biggl[ -c_0(c_0-1) -6c_0 + 2(\mathfrak{F}+\alpha) \biggr]\xi^3 </math> |
|
<math>~=</math> |
<math>~ 2\biggl[c_0^2 + 2c_0 - \alpha \biggr] + \biggl[(2g^2-5)(c_0^2 + 3c_0 ) \biggr]\xi + \biggl[ 2(\mathfrak{F}+\alpha) - c_0(c_0+5) \biggr]\xi^3 \, . </math> |
This means that three algebraic relations must simultaneously be satisfied, namely:
<math>~\xi^{0}:</math> |
<math>~c_0^2 + 2c_0 - \alpha =0</math> |
<math>~\Rightarrow~</math> |
<math>~c_0 = -1 \pm (1+\alpha)^{1 / 2} \, ;</math> |
<math>~\xi^{1}:</math> |
<math>~g^2 = \frac{5}{2}</math> |
<math>~\Rightarrow~</math> |
<math>~2q^3 + 5q^2 - 1 = 0 \, ;</math> |
<math>~\xi^{3}:</math> |
<math>~2(\mathfrak{F}+\alpha) = c_0(c_0+5)</math> |
<math>~\Rightarrow~</math> |
<math>~\sigma^2 = \alpha + \frac{c_0}{2}\biggl( c_0+5 \biggr) \, .</math> |
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