Revision as of 20:08, 1 December 2016

Radial Oscillations of a Zero-Zero Bipolytrope

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>~\equiv</math>

in which case,

<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>~ \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>
		<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>
		<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>~ \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.

Boundary Condition

In order to ensure finite pressure fluctuations at the surface of this bipolytropic configuration, we need the logarithmic derivative of <math>~x</math> to obey the following relation:

<math>~ \frac{d\ln x}{d\ln r_0}\biggr|_\mathrm{surface}</math>

<math>~\frac{1}{\gamma_g} \biggl( 4 - 3\gamma_g + \frac{\omega^2 R^3}{GM_\mathrm{tot}}\biggr) \, .</math>

Now, according to our accompanying discussion of the equilibrium mass and radius of a zero-zero polytrope, we know that,

<math>~\frac{R^3}{M_\mathrm{tot}}</math>	<math>~=</math>	<math>~ \biggl(\frac{P_c}{G\rho_c^2}\biggr)^{3 / 2} \biggl( \frac{3}{2\pi}\biggr)^{3 / 2} \frac{1}{(qg)^3} \biggl(\frac{G^3 \rho_c^4}{P_c^3}\bigg)^{1 / 2} \biggl(\frac{\pi}{2\cdot 3}\biggr)^{1 / 2} \nu g^3 </math>
	<math>~=</math>	<math>~ \biggl( \frac{3}{4\pi \rho_c} \biggr) \frac{\nu}{q^3} \, . </math>

Hence, a reasonable surface boundary condition is,

<math>~ \frac{d\ln x}{d\ln r_0}\biggr|_\mathrm{surface}</math>

<math>~\frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}} \biggl( \frac{\nu}{q^3}\biggr) - \biggl( 3 - \frac{4}{\gamma_\mathrm{g}}\biggr) \, .</math>

Attempt to Find Eigenfunction for the Envelope

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>~ \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>
		<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[ 4g^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[ 4g^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 + (4g^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 + (4g^2 - 5)\xi - \xi^3 \biggr] \biggl[ \xi^2 \cdot \frac{d^2x}{d\xi^2} \biggr] +\biggl[ 3 + (8g^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 + (4g^2 - 5)\xi - \xi^3 \biggr] \biggl[ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} \biggr] +\biggl[ 6 + 4(4g^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,

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,

then the logarithmic derivative of <math>~x</math> is a constant, namely,

and the two key derivative terms will be,

and

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 + (4g^2 - 5)\xi - \xi^3 \biggr] \biggl[ c_0(c_0-1) \biggr] +\biggl[ 6 + 4(4g^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 + (4g^2 - 5)\xi - \xi^3 \biggr] c_0(c_0-1) +c_0\biggl[ 6 + 4(4g^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[(4g^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[(4g^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}{4}</math>	<math>~\Rightarrow~</math>	<math>~q=\biggl(\frac{1}{2}\biggr)^{1 / 3} </math> and, hence, <math>~\nu = \frac{2}{3} \, ;</math>
<math>~\xi^{3}:</math>	<math>~2(\mathfrak{F}+\alpha) = c_0(c_0+5)</math>	<math>~\Rightarrow~</math>	<math>~\frac{2}{3}\cdot \sigma^2 = (\alpha-1) \pm \sqrt{\alpha+1} \, .</math>

More General Solution

Leaving the density ratio unspecified, let's try to write the wave equation for the envelope in the same form, and see if the logarithmic derivatives can be manipulated in a similar fashion.

<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>
		<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>
<math>~\Rightarrow ~~~\frac{\xi^3}{x} \cdot 0</math>	<math>~=</math>	<math>~ \frac{g^2\xi P_0}{P_c} \biggl[ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} \biggr] + \biggl\{ \frac{4g^2 \xi P_0}{P_c} - 2 \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[\biggl( 1-\frac{\rho_e}{\rho_c} \biggr) +\biggl(\frac{\rho_e}{\rho_c} \biggr) \xi^3 \biggr] \biggr\} \frac{\xi}{x} \cdot \frac{dx}{d\xi} </math>
		<math>~ +\xi^3 \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl\{ \biggl[ \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr] + 2\alpha\biggl( \frac{\rho_e}{\rho_c} -1\biggr) \cdot \frac{1}{\xi^3} \biggr\} </math>

where,

<math>~\frac{g^2\xi P_0}{P_c}</math>	<math>~=</math>	<math> \xi \biggl\{ 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] \biggr\} </math>
	<math>~=</math>	<math> \xi \biggl\{ \biggl[ 2\biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggr] \frac{1}{\xi} +\biggl[ g^2 - 1 - 2\biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) + \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \biggr] - \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^2 \biggr\} </math>
	<math>~=</math>	<math> \biggl[ 2\biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggr] +\biggl[ g^2 - 1 - 2\biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) + \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \biggr]\xi - \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3 </math>
	<math>~=</math>	<math> \biggl[ 2\biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggr] +\biggl[ g^2 - 1 - 2\biggl(\frac{\rho_e}{\rho_c}\biggr) + 3\biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \biggr]\xi - \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3 \, . </math>

Hence, the wave equation becomes,

<math>~0</math>	<math>~=</math>	<math>~ \biggl[ \mathcal{A} + (g^2-\mathcal{B}) \xi - \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3\biggr] \biggl[ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} \biggr] + \biggl\{ 4\biggl[ \mathcal{A} + (g^2-\mathcal{B}) \xi - \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3\biggr] - \mathcal{A} - 2\biggl(\frac{\rho_e}{\rho_c} \biggr)^2 \xi^3 \biggr\} \frac{\xi}{x} \cdot \frac{dx}{d\xi} </math>
		<math>~ + \biggl[ \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl( \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 + 2\alpha\biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl( \frac{\rho_e}{\rho_c} -1\biggr) \biggr] </math>
	<math>~=</math>	<math>~ \biggl[ \mathcal{A} + (g^2-\mathcal{B}) \xi - \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3\biggr] \biggl[ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} \biggr] + \biggl\{ 3\mathcal{A} + 4(g^2-\mathcal{B}) \xi - 6\biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3 \biggr\} \frac{\xi}{x} \cdot \frac{dx}{d\xi} </math>
		<math>~ + \biggl[ \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl( \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 -\alpha \mathcal{A} \biggr] \, , </math>

where,

<math>~\mathcal{A}</math>	<math>~\equiv</math>	<math>~2\biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \, ; </math>
<math>~\mathcal{B}</math>	<math>~\equiv</math>	<math>~1 + 2\biggl(\frac{\rho_e}{\rho_c}\biggr) - 3\biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \, . </math>

As before, if we assume a power-law solution, the wave equation for the envelope becomes,

<math>~0</math>	<math>~=</math>	<math>~ \biggl[ \mathcal{A} + (g^2-\mathcal{B}) \xi - \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3\biggr] \biggl[ c_0(c_0-1) \biggr] + \biggl\{ 3\mathcal{A} + 4(g^2-\mathcal{B}) \xi - 6\biggl(\frac{\rho_e}{\rho_c}\biggr)^2 \xi^3 \biggr\} c_0 </math>
		<math>~ + \biggl[ \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl( \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 -\alpha \mathcal{A} \biggr] </math>
	<math>~=</math>	<math>~\xi^0 \biggl[ \mathcal{A}c_0(c_0-1) + 3\mathcal{A}c_0 -\alpha\mathcal{A} \biggr] + \xi^1 \biggl[ (g^2-\mathcal{B})c_0(c_0-1) +4(g^2-\mathcal{B})c_0 \biggr] + \xi^3 \biggl[\biggl(\frac{\rho_e}{\rho_c}\biggr)^2c_0(1-c_0) - 6\biggl(\frac{\rho_e}{\rho_c}\biggr)^2c_0 +\biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl( \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr) \biggr] </math>
	<math>~=</math>	<math>~\xi^0 \biggl[ c_0(c_0-1) + 3c_0 -\alpha \biggr]\mathcal{A} + \xi^1 \biggl[ (g^2-\mathcal{B})(c_0^2+3c_0)\biggr] + \xi^3 \biggl[( \mathfrak{F} + 2\alpha ) - \biggl(\frac{\rho_e}{\rho_c}\biggr)(5c_0 +c_0^2 + 2\alpha) \biggr]\biggl(\frac{\rho_e}{\rho_c}\biggr) \, . </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^{3}:</math>	<math>~(\mathfrak{F}+2\alpha) = \biggl(\frac{\rho_e}{\rho_c}\biggr)(5c_0 +c_0^2 + 2\alpha)</math>	<math>~\Rightarrow~</math>	<math>~\sigma^2 \equiv \frac{3\omega^2}{2\pi G\rho_c \gamma_\mathrm{g}}= 3\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ (\alpha-1) \pm \sqrt{\alpha+1} \biggr] \, ;</math>
<math>~\xi^{1}:</math>	<math>~g^2 = 1 + 2\biggl(\frac{\rho_e}{\rho_c}\biggr) - 3\biggl(\frac{\rho_e}{\rho_c}\biggr)^2</math>	<math>~\Rightarrow~</math>	<math>~q^3 = \frac{(\rho_e/\rho_c)}{2[1-(\rho_e/\rho_c) ]} </math>
		and, hence,	<math>~\nu = \frac{1}{3[1-(\rho_e/\rho_c) ]} \, .</math>

Surface Boundary Condition

Given that, with this solution, the ratio,

<math>~\frac{1}{3[1-(\rho_e/\rho_c) ]} \biggl\{ \frac{2[1-(\rho_e/\rho_c) ]}{(\rho_e/\rho_c)} \biggr\} = \frac{2}{3(\rho_e/\rho_c) } \, ,</math>

we see that the desired surface boundary condition is,

<math>~ \frac{d\ln x}{d\ln r_0}\biggr\|_\mathrm{surface}</math>	<math>~=</math>	<math>~\frac{3\omega^2 }{4\pi G\rho_c \gamma_\mathrm{g}} \cdot \frac{2}{3(\rho_e/\rho_c) } - \biggl( 3 - \frac{4}{\gamma_\mathrm{g}}\biggr) </math>
	<math>~=</math>	<math>~\frac{\sigma^2}{3(\rho_e/\rho_c) } - \alpha </math>
	<math>~=</math>	<math>~c_0 \, . </math>

But, for our identified solution, this is the logarithmic derivative of <math>~x</math> throughout the envelope as well as at the surface. So the boundary condition is automatically satisfied.

Match to a Core Eigenfunction

If we define,

<math>~\eta \equiv \frac{\xi}{g} \, ,</math>

the above wave equation for the core becomes,

<math>~ (1 - \eta^2)\frac{d^2x}{d\eta^2} + ( 4 - 6\eta^2 ) \frac{1}{\eta} \cdot \frac{dx}{d\eta} + \mathfrak{F} x \, . </math>

Not surprisingly, this is identical in form to the eigenvalue problem first presented by Sterne (1937) in connection with an examination of radial oscillations in uniform-density spheres. For the core of our zero-zero bipolytrope, we can therefore adopt any one of the polynomial eigenfunctions and corresponding eigenfrequencies derived by Sterne. Let's begin with Sterne's quadratic function and see if we can match it to the envelope's power-law eigenfunction. Keeping in mind that the overall normalization is arbitrary, from Sterne's presentation, we have,

<math>~x_\mathrm{core}</math>	<math>~=</math>	<math>~a\biggl[ 1-\frac{7}{5}\eta^2 \biggr]</math>
	<math>~=</math>	<math>~a\biggl[ 1-\frac{7}{5}\biggl( \frac{\xi}{g}\biggr)^2 \biggr] \, ,</math>

and the associated eigenfrequency is obtained by setting,

<math>~\mathfrak{F} = \sigma^2 - 2\alpha = 14 \, .</math>

In this case, then, the eigenfrequency for the envelope will match the eigenfrequency of the core if,

<math>~14 + 2\alpha</math>	<math>~=</math>	<math>~ 3\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ (\alpha-1) \pm \sqrt{\alpha+1} \biggr] </math>
<math>~\Rightarrow ~~~ \biggl( \frac{\rho_e}{\rho_c} \biggr) </math>	<math>~=</math>	<math>~\frac{2}{3}\cdot \frac{7 + \alpha}{ (\alpha-1) \pm \sqrt{\alpha+1} }</math>

Related Discussions

@@ Line 271: / Line 271: @@
 </table>
 </div>
-and the wave equation for the core becomes,
+<span id="CoreWaveEq">and the wave equation for the core becomes,</span>
 <!--
@@ Line 969: / Line 969: @@
 -->
-==Attempt to Solve==
+==Attempt to Find Eigenfunction for the Envelope==
 Adopting some of the notation used by [http://adsabs.harvard.edu/abs/1937MNRAS..97..582S T. E. Sterne (1937)] and enunciated in our [[User:Tohline/SSC/UniformDensity#Setup_as_Presented_by_Sterne_.281937.29|accompanying discussion of the uniform-density sphere]], we'll define,
@@ Line 1,900: / Line 1,900: @@
 </table>
 </div>
+===Surface Boundary Condition===
 Given that, with this solution, the ratio,
@@ Line 1,961: / Line 1,963: @@
 </div>
 But, for our identified solution, this is the logarithmic derivative of <math>~x</math> throughout the envelope as well as at the surface.  So the boundary condition is automatically satisfied.
+==Match to a Core Eigenfunction==
+If we define,
+<div align="center">
+<math>~\eta \equiv \frac{\xi}{g} \, ,</math>
+</div>
+the [[#CoreWaveEq|above wave equation for the core]] becomes,
+<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>~
+(1 - \eta^2)\frac{d^2x}{d\eta^2} +
+( 4 - 6\eta^2 )  \frac{1}{\eta} \cdot \frac{dx}{d\eta}
++ \mathfrak{F} x \, .
+</math>
+  </td>
+</tr>
+</table>
+</div>
+Not surprisingly, this is identical in form to the eigenvalue problem first presented by [[User:Tohline/SSC/UniformDensity#Setup_as_Presented_by_Sterne_.281937.29|Sterne (1937)]] in connection with an examination of radial oscillations in uniform-density spheres.  For the core of our zero-zero bipolytrope, we can therefore adopt any one of the [[User:Tohline/SSC/UniformDensity#Sterne.27s_General_Solution|polynomial eigenfunctions and corresponding eigenfrequencies]] derived by Sterne.  Let's begin with Sterne's quadratic function and see if we can match it to the envelope's power-law eigenfunction. Keeping in mind that the overall normalization is arbitrary, from Sterne's presentation, we have,
+<div align="center">
+<table border="0" cellpadding="5" align="center">
+<tr>
+  <td align="right">
+<math>~x_\mathrm{core}</math>
+  </td>
+  <td align="center">
+<math>~=</math>
+  </td>
+  <td align="left">
+<math>~a\biggl[ 1-\frac{7}{5}\eta^2 \biggr]</math>
+  </td>
+</tr>
+<tr>
+  <td align="right">
+&nbsp;
+  </td>
+  <td align="center">
+<math>~=</math>
+  </td>
+  <td align="left">
+<math>~a\biggl[ 1-\frac{7}{5}\biggl( \frac{\xi}{g}\biggr)^2 \biggr] \, ,</math>
+  </td>
+</tr>
+</table>
+</div>
+and the associated eigenfrequency is obtained by setting,
+<div align="center">
+<math>~\mathfrak{F} = \sigma^2 - 2\alpha =  14 \, .</math>
+</div>
+In this case, then, the eigenfrequency for the envelope will match the eigenfrequency of the core if,
+<div align="center">
+<table border="0" cellpadding="5" align="center">
+<tr>
+  <td align="right">
+<math>~14 + 2\alpha</math>
+  </td>
+  <td align="center">
+<math>~=</math>
+  </td>
+  <td align="left">
+<math>~ 3\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ (\alpha-1) \pm \sqrt{\alpha+1} \biggr] </math>
+  </td>
+</tr>
+<tr>
+  <td align="right">
+<math>~\Rightarrow ~~~ \biggl( \frac{\rho_e}{\rho_c} \biggr) </math>
+  </td>
+  <td align="center">
+<math>~=</math>
+  </td>
+  <td align="left">
+<math>~\frac{2}{3}\cdot  \frac{7 + \alpha}{ (\alpha-1) \pm \sqrt{\alpha+1} }</math>
+  </td>
+</tr>
+</table>
+</div>
 =Related Discussions=

Difference between revisions of "User:Tohline/SSC/Stability/BiPolytrope0 0"

Revision as of 20:08, 1 December 2016

Contents

Radial Oscillations of a Zero-Zero Bipolytrope

Groundwork

Core

Envelope

Check1

Boundary Condition

Attempt to Find Eigenfunction for the Envelope

A Specific Choice of the Density Ratio

Idea Involving Logarithmic Derivatives

More General Solution

Surface Boundary Condition

Match to a Core Eigenfunction

Related Discussions

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