Continue Search for Marginally Unstable (5,1) Bipolytropes

This Ramblings Appendix chapter — see also, various trials — provides some detailed trial derivations in support of the accompanying, thorough discussion of this topic.

Key Differential Equation

In an accompanying discussion, we derived the so-called,

Linear 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. After adopting an appropriate set of variable normalizations — as detailed here — this becomes,

<math>~ \frac{d^2x}{dr*^2} + \biggl\{ 4 -\biggl(\frac{\rho^*}{P^*}\biggr)\frac{ M_r^*}{(r^*)}\biggr\}\frac{1}{r^*} \frac{dx}{dr*} + \biggl(\frac{\rho^*}{ P^* } \biggr)\biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}} ~-~\frac{\alpha_\mathrm{g} M_r^*}{(r^*)^3}\biggr\} x \, , </math>

where, <math>~\alpha_g \equiv (3 - 4/\gamma_g)</math>.

Applied to the Core

As we have already summarized in an accompanying discussion, throughout the core we have,

<math>~\biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi \, ;</math>

<math>~\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2} \, ;</math>

<math>~ 2 \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \, . </math>

So the relevant core LAWE becomes,

<math>~0</math>	<math>~=</math>	<math>~ \biggl( \frac{2\pi}{3} \biggr) \frac{d^2x}{d\xi^2} + \biggl( \frac{2\pi}{3} \biggr) \biggl\{ 4 - \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2} \biggl[ 2 \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3 / 2} \biggr]\biggr\}\frac{1}{\xi} \frac{dx}{d\xi} + \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2}\biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}} ~-~\biggl( \frac{2\pi}{3} \biggr)\frac{\alpha_\mathrm{g} }{\xi^2} \biggl[ 2 \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3 / 2} \biggr] \biggr\} x </math>
<math>~\Rightarrow ~~~ 0</math>	<math>~=</math>	<math>~ \frac{1}{2}\cdot \frac{d^2x}{d\xi^2} + \biggl[ 2 - \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \biggr] \frac{1}{\xi} \frac{dx}{d\xi} + \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2}\biggl[ \frac{\sigma_c^2}{2\gamma_\mathrm{g}} ~-~\alpha_\mathrm{g} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3 / 2} \biggr] x \, . </math>

Now, following our separate discussion of an analytic solution to this LAWE, we try,

<math>~x_P\biggr\|_\mathrm{core}</math>	<math>~\equiv</math>	<math>~1 - \frac{\xi^2}{15}</math>
<math>~\Rightarrow~~~\frac{dx_P}{d\xi}\biggr\|_\mathrm{core}</math>	<math>~\equiv</math>	<math>~- \frac{2\xi}{15} </math>
<math>~\Rightarrow~~~\frac{d\ln x_P}{d\ln \xi}\biggr\|_\mathrm{core}</math>	<math>~\equiv</math>	<math>~- \frac{2\xi^2}{15} \biggl[ \frac{(15 - \xi^2)}{15} \biggr]^{-1} = - \frac{2\xi^2}{(15 - \xi^2)} \, .</math>

Plugging this trial function into the relevant LAWE gives,

LAWE	<math>~=</math>	<math>~ \frac{1}{2} \biggl( -\frac{2}{3\cdot 5}\biggr)^2 + \biggl( -\frac{2}{3\cdot 5}\biggr)\biggl[ 2 - \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \biggr] + \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2}\biggl[ \frac{\sigma_c^2}{2\gamma_\mathrm{g}} ~-~\alpha_\mathrm{g} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3 / 2} \biggr] \biggl[1 - \frac{\xi^2}{15}\biggr] </math>
	<math>~=</math>	<math>~ \biggl( \frac{2}{3^2 \cdot 5^2}\biggr) - \biggl( \frac{2^2}{3\cdot 5}\biggr) + \biggl( \frac{2}{3\cdot 5}\biggr)\biggl[ \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \biggr] + \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2}\biggl[ \frac{\sigma_c^2}{2\gamma_\mathrm{g}} ~-~\alpha_\mathrm{g} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3 / 2} \biggr] \biggl[1 - \frac{\xi^2}{15}\biggr] </math>

Now, if we set <math>~\sigma_c^2 = 0</math> and <math>~\gamma_g = \gamma_c = \tfrac{6}{5} ~~\Rightarrow ~~ \alpha_g = -1/3</math>, we have,

LAWE	<math>~=</math>	<math>~ \biggl( \frac{2}{3^2 \cdot 5^2}\biggr) - \biggl( \frac{2^2}{3\cdot 5}\biggr) + \biggl( \frac{2}{3\cdot 5}\biggr)\biggl[ \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \biggr] + \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2}\biggl[ \cancelto{0}{\frac{\sigma_c^2}{2\gamma_\mathrm{g}}} ~+~\frac{1}{3} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3 / 2} \biggr] \biggl[1 - \frac{\xi^2}{15}\biggr] </math>
	<math>~=</math>	<math>~\frac{2}{3^2\cdot 5^2}\biggl(1 - 30 \biggr) \biggl( \frac{2}{3^2 \cdot 5^2}\biggr) + \biggl( \frac{2}{3\cdot 5}\biggr)\biggl[ \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \biggr] + \frac{1}{3} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} - \frac{\xi^2}{3^2\cdot 5} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} </math>

Applied to the Envelope

And as we have also summarized in the same accompanying discussion, throughout the envelope we have,

<math>~r^*</math>	<math>~=</math>	<math>~\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta \, ;</math>
<math>~\frac{\rho^}{P^}</math>	<math>~=</math>	<math>~ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi(\eta)^{-1} \, ; </math>
<math>~\frac{M_r^}{r^}</math>	<math>~=</math>	<math>~ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-1}_i \biggl( \frac{2}{\pi} \biggr)^{1/2} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr) \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta \biggr]^{-1} = 2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i \eta \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr) \, . </math>

So the relevant envelope LAWE becomes,

<math>~ \frac{d^2x}{dr*^2} + \biggl\{ 4 -\biggl(\frac{\rho^*}{P^*}\biggr)\frac{ M_r^*}{(r^*)}\biggr\}\frac{1}{r^*} \frac{dx}{dr*} + \biggl(\frac{\rho^*}{ P^* } \biggr)\biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}} ~-~\frac{\alpha_\mathrm{g} M_r^*}{(r^*)^3}\biggr\} x \, , </math>

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Continue Search for Marginally Unstable (5,1) Bipolytropes

Key Differential Equation

Applied to the Core

Applied to the Envelope

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