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Radial Oscillations of a Zero-Zero Bipolytrope

Whitworth's (1981) Isothermal Free-Energy Surface
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Groundwork

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

Adiabatic Wave (or Radial Pulsation) Equation

LSU Key.png

<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>

 

 

<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>~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.

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>~=</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>~=</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>~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>

 

 

<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,

<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 + (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{\nu}{q^3}</math>

<math>~=</math>

<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>~0</math>

<math>~=</math>

<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. We will insist that the eigenfrequency of the envelope match the eigenfrequency of the core; and, following J. O. Murphy & R. Fiedler (1985b) (see the top paragraph of the right-hand column on p. 223 of their article), we seek solutions for which there is continuity in both the eigenfunction and its first derivative at the interface <math>~(\xi = 1)</math>.

Try Quadratic Core Eigenfunction

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>

Now, the eigenfunction for the envelope is,

<math>~x_\mathrm{env} = \xi^{c_0} \, ,</math>

where,

<math>~c_0</math>

<math>~=</math>

<math>~-1 \pm (1+\alpha)^{1 / 2} \, .</math>

The value of this function will match the value of its core counterpart at the interface <math>~(\xi=1)</math> if,

<math>~a\biggl[ 1-\frac{7}{5}\biggl( \frac{1}{g}\biggr)^2 \biggr]</math>

<math>~=</math>

<math>~1</math>

<math>~\Rightarrow ~~~ a</math>

<math>~=</math>

<math>~\biggl[ 1-\frac{7}{5}\biggl( \frac{1}{g}\biggr)^2 \biggr]^{-1} \, .</math>

Finally, the slope (first derivative) of the core eigenfunction will match the slope of the envelope eigenfunction at the interface if,

<math>~c_0 \xi^{c_0-1}\biggr|_\mathrm{\xi=1}</math>

<math>~=</math>

<math>~ -\frac{14a}{5g^2} \cdot \xi\biggr|_\mathrm{\xi=1} </math>

<math>~\Rightarrow ~~~ -\frac{14}{5c_0} </math>

<math>~=</math>

<math>~ \frac{g^2}{a} </math>

 

<math>~=</math>

<math>~ g^2-\frac{7}{5} </math>

 

<math>~=</math>

<math>~ -\frac{2}{5} + 2\biggl(\frac{\rho_e}{\rho_c}\biggr) - 3\biggl(\frac{\rho_e}{\rho_c}\biggr)^2 </math>

<math>~\Rightarrow ~~~ 3\biggl(\frac{\rho_e}{\rho_c}\biggr)^2 - 2\biggl(\frac{\rho_e}{\rho_c}\biggr) + \frac{2}{5}\biggl( 1-\frac{7}{c_0}\biggr) </math>

<math>~=</math>

<math>~ 0 \, .</math>

The solution to this quadratic equation gives,

<math>~ \biggl(\frac{\rho_e}{\rho_c}\biggr) </math>

<math>~=</math>

<math>~ \frac{1}{6} \biggl[ 2 \pm \sqrt{4-\frac{24}{5}\biggl( 1-\frac{7}{c_0}\biggr)} \biggr] </math>

 

<math>~=</math>

<math>~ \frac{1}{3} \biggl[ 1 \pm \sqrt{1-\frac{6}{5}\biggl( 1-\frac{7}{c_0}\biggr)} \biggr] </math>

In order for this condition to hold while also meeting the demands of the eigenfrequency, we need <math>~\alpha</math> to satisfy the relation,

<math>~\frac{2}{3}\cdot \frac{7 + \alpha}{ (\alpha-1) \pm \sqrt{\alpha+1} }</math>

<math>~=</math>

<math>~ \frac{1}{3} \biggl[ 1 \pm \sqrt{1-\frac{6}{5}\biggl( 1-\frac{7}{c_0}\biggr)} \biggr] </math>

<math>~\Rightarrow ~~~ \frac{14 + 2\alpha}{\alpha + c_0 }</math>

<math>~=</math>

<math>~ 1 \pm \biggl[ \frac{5c_0}{5c_0}-\frac{6}{5}\biggl( \frac{c_0-7}{c_0}\biggr) \biggr]^{1 / 2} </math>

<math>~\Rightarrow ~~~ \biggl[ \frac{14 + \alpha -c_0}{\alpha + c_0 } \biggr]^2</math>

<math>~=</math>

<math>~ \frac{42-c_0}{5c_0} </math>

<math>~\Rightarrow ~~~ 5c_0 (14 + \alpha -c_0)^2</math>

<math>~=</math>

<math>~ (42-c_0)(\alpha + c_0)^2 \, , </math>

where, keep in mind,

<math>~c_0</math>

<math>~=</math>

<math>~-1 \pm (1+\alpha)^{1 / 2} \, .</math>

RESULT:  After examining a range of physically reasonable values of <math>~\alpha</math>, we do not find any values for which the left-hand-side of this condition matches the right-hand-side.


Try Quartic Core Eigenfunction

Let's begin with Sterne's quartic function and see if we can match it to the envelope's power-law eigenfunction. From Sterne's presentation, we have,

<math>~x_\mathrm{core}</math>

<math>~=</math>

<math>~a\biggl[ 1-\frac{18}{5}\biggl( \frac{\xi}{g}\biggr)^2 +\frac{99}{35} \biggl( \frac{\xi}{g}\biggr)^4 \biggr]</math>

and the associated eigenfrequency is obtained by setting,

<math>~\mathfrak{F} = \sigma^2 - 2\alpha = 36 \, .</math>

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

<math>~36 + 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{36 + 2\alpha}{ 3[(\alpha-1) \pm \sqrt{\alpha+1}] }</math>

The eigenfunction for the envelope is, as before. The value of this envelope function will match the value of its core counterpart at the interface <math>~(\xi=1)</math> if,

<math>~a\biggl[ 1-\frac{18}{5}\biggl( \frac{1}{g}\biggr)^2 +\frac{99}{35} \biggl( \frac{1}{g}\biggr)^4 \biggr]</math>

<math>~=</math>

<math>~1</math>

<math>~\Rightarrow ~~~ a</math>

<math>~=</math>

<math>~\biggl[ 1-\frac{18}{5}\biggl( \frac{1}{g}\biggr)^2 +\frac{99}{35} \biggl( \frac{1}{g}\biggr)^4 \biggr]^{-1} \, .</math>

Finally, the slope (first derivative) of the core eigenfunction will match the slope of the envelope eigenfunction at the interface if,

<math>~\biggl( \frac{c_0}{a}\biggr) \xi^{c_0-1}\biggr|_\mathrm{\xi=1}</math>

<math>~=</math>

<math>~\biggl[ -\frac{36}{5g^2} \cdot \xi + \frac{4\cdot 99}{35g^4} \cdot \xi^3 \biggr]_\mathrm{\xi=1} </math>

<math>~\Rightarrow ~~~c_0 \biggl[ 1-\frac{18}{5}\biggl( \frac{1}{g}\biggr)^2 +\frac{99}{35} \biggl( \frac{1}{g}\biggr)^4 \biggr]</math>

<math>~=</math>

<math>~ -\frac{36}{5g^2} + \frac{4\cdot 99}{35g^4} </math>

<math>~\Rightarrow ~~~c_0 \biggl[ 35g^4-7\cdot 18g^2 +99 \biggr]</math>

<math>~=</math>

<math>~ - 7\cdot 36 g^2 + 4\cdot 99 \, . </math>

Eureka Regarding Prasad's 1948 Paper

Envelope Solution Outline

Comment by J. E. Tohline on 5 December 2016: Yesterday, I stumbled on this key paper by Prasad (1948) while I was looking back through the published literature to catalog who has solved the polytropic wave equation numerically.

C. Prasad (1948, MNRAS, 108, 414-416) has examined a closely related problem and, as it turns out, the mathematical approach that he used to solve that problem analytically is gratifyingly useful to me here. If, as above, we restrict our investigation to configurations for which,

<math>~g^2 = \mathcal{B} \, ,</math>

and if we multiply through by <math>~x/(\mathcal{A}\xi^2)</math>, our governing wave equation becomes,

<math>~0</math>

<math>~=</math>

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

 

<math>~=</math>

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

where,

<math>\mathcal{D} \equiv \frac{1}{\mathcal{A}} \biggl(\frac{\rho_e}{\rho_c}\biggr)^2 = \biggl(\frac{\rho_e}{\rho_c}\biggr)^2\biggl[2\biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl(1 - \frac{\rho_e}{\rho_c} \biggr)\biggr]^{-1} = \biggl[2 \biggl(\frac{\rho_c}{\rho_e}-1 \biggr)\biggr]^{-1} \, .</math>

This wave equation is very similar to equation (2) of Prasad (1948). If, following Prasad's guidance, we then assume a series solution of the form,

<math>~x</math>

<math>~=</math>

<math>~ \xi^{c_0} \sum_0^\infty a_k \xi^k \, , </math>

the indicial equation gives,

<math>~c_0 = -1 \pm \sqrt{1+\alpha} \, .</math>

This is precisely the value of the exponent, <math>~c_0</math>, that we derived — in a more stumbling fashion — above and, as is shown by the following framed image, it is identical to the exponent derived by Prasad (1948).

Equation and accompanying text extracted† from C. Prasad (1948)

"Radial Oscillations of a Particular Stellar Model"

Monthly Notices of the Royal Astronomical Society, vol. 108, pp. 414-416 © Royal Astronomical Society

Prasad (1948)
†Displayed here exactly as presented in the original publication.

Using equation (7) from Prasad (1948) as a guide, we hypothesize that the eigenfrequency of the jth mode in the envelope is given by the relation,

<math>~ \sigma_j^2 \biggr|_\mathrm{env} </math>

<math>~=</math>

<math>~(c_0 + 3j)(c_0 + 3j+5) + 2\alpha \, ,</math>

where,

<math>~ \sigma_j^2\biggr|_\mathrm{env} \equiv \biggl(\mathfrak{F} + 2\alpha\biggr)\frac{\rho_c}{\rho_e} = \frac{3\omega^2}{2\pi \gamma_\mathrm{g} G\rho_e} \, . </math>

And guided by equation (6) from Prasad (1948), we hypothesize that successive coefficients in the (truncated) series that defines the radial structure of each mode is governed by the recurrence relation,

<math>~\frac{1}{\mathcal{D}}\cdot \frac{a_{k+3}}{a_k}</math>

<math>~=</math>

<math>~ \frac{(c_0+k)(c_0+k+5) - (\sigma_j^2 - 2\alpha)}{(c_0 + k + 3)(c_0+k+5) - \alpha} \, . </math>

Example Envelope Eigenvectors

Mode j = 0

Here we assume that the series defining the eigenfunction has only one term. This should match our earlier restricted solution. Specifically,

<math>~x_{j=0}</math>

<math>~=</math>

<math>~a_0 \xi^{c_0}</math>

<math>~\Rightarrow ~~~\frac{1}{\xi} \cdot \frac{d x_{j=0}}{d\xi}</math>

<math>~=</math>

<math>~a_0 c_0 \xi^{c_0-2} \, ;</math>

<math>~\Rightarrow ~~~\frac{d^2 x_{j=0}}{d\xi^2}</math>

<math>~=</math>

<math>~a_0 c_0(c_0-1) \xi^{c_0-2} \, .</math>

In this case, the wave equation becomes,

<math>~0</math>

<math>~=</math>

<math>~a_0 \xi^{c_0-2}\biggl\{ \biggl[1 - \mathcal{D} \xi^3\biggr] c_0(c_0-1) + \biggl[ 3 - 6\mathcal{D} \xi^3 \biggr] c_0 + \biggl[ \mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr) \biggl( \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 -\alpha \biggr] \biggr\} \, . </math>

The coefficients of the <math>~\xi^0</math> terms will sum to zero if the above-defined indicial exponent condition is satisfied; that is, by setting,

<math>~c_0</math>

<math>~=</math>

<math>~-1 \pm (1+\alpha)^{1 / 2} \, .</math>

In order for the coefficients of the <math>~\xi^3</math> terms to sum to zero, we need,

<math>~\biggl(\frac{\rho_c}{\rho_e}\biggr) \biggl( \mathfrak{F} + 2\alpha \biggr) - 2\alpha</math>

<math>~=</math>

<math>~c_0(c_0-1) +6c_0 </math>

<math>~\Rightarrow ~~~ \sigma^2_{j=0}\biggr|_\mathrm{env}</math>

<math>~=</math>

<math>~c_0(c_0+5) + 2\alpha \, .</math>

Mode j = 1

Here we assume that the series defining the eigenfunction has two terms: <math>~k=0</math> and <math>~k=3</math>. Specifically,

<math>~x_{j=1}</math>

<math>~=</math>

<math>~a_0 \xi^{c_0} + a_3 \xi^{c_0+3}</math>

<math>~\Rightarrow ~~~\frac{1}{\xi}\cdot \frac{d x_{j=1}}{d\xi}</math>

<math>~=</math>

<math>~a_0 c_0 \xi^{c_0-2} + a_3 (c_0+3) \xi^{c_0+1} \, ;</math>

<math>~\Rightarrow ~~~\frac{d^2 x_{j=1}}{d\xi^2}</math>

<math>~=</math>

<math>~a_0 c_0(c_0-1) \xi^{c_0-2} + a_3 (c_0+3)(c_0+2) \xi^{c_0+1} \, .</math>

In this case, after factoring out <math>~a_0\xi^{c_0-2}</math>, the wave equation becomes,

<math>~0</math>

<math>~=</math>

<math>~ \biggl[1 - \mathcal{D} \xi^3\biggr] \biggl[ c_0(c_0-1) + \frac{a_3}{a_0} (c_0+3)(c_0+2) \xi^{3} \biggr] + \biggl[ 3 - 6\mathcal{D} \xi^3 \biggr] \biggl[c_0 + \frac{a_3}{a_0} (c_0+3) \xi^{3}\biggr] </math>

 

 

<math>~ + \biggl[ \mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr) \biggl( \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 -\alpha \biggr] \biggl[1 + \frac{a_3}{a_0} \xi^{3} \biggr] \, . </math>

Again, the coefficients of the <math>~\xi^0</math> terms will sum to zero if the above-defined indicial exponent condition is satisfied; that is, by setting,

<math>~c_0</math>

<math>~=</math>

<math>~-1 \pm (1+\alpha)^{1 / 2} \, .</math>

In order for the coefficients of the <math>~\xi^3</math> terms to sum to zero, we need,

<math>~0</math>

<math>~=</math>

<math>~ -\mathcal{D} c_0(c_0-1)+ \frac{a_3}{a_0} (c_0+3)(c_0+2) -6\mathcal{D}c_0 + \frac{3a_3}{a_0} (c_0+3) + \mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr) \biggl( \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr)-\frac{\alpha a_3}{a_0} </math>

<math>~\Rightarrow~~~ \frac{a_3}{a_0} \biggl[ (c_0+3)(c_0+2) + 3(c_0+3)-\alpha\biggr] </math>

<math>~=</math>

<math>~ \mathcal{D}\biggl[ c_0(c_0-1) +6c_0 -\biggl(\frac{\rho_c}{\rho_e}\biggr) \biggl( \mathfrak{F} + 2\alpha \biggr) + 2\alpha \biggr] </math>

<math>~\Rightarrow~~~ \frac{1}{\mathcal{D}} \cdot \frac{a_3}{a_0} \biggl[ (c_0+3)(c_0+5) -\alpha\biggr] </math>

<math>~=</math>

<math>~ \biggl[ c_0(c_0+5) -\biggl(\frac{\rho_c}{\rho_e}\biggr) \biggl( \mathfrak{F} + 2\alpha \biggr) + 2\alpha \biggr] </math>

<math>~\Rightarrow~~~ \frac{1}{\mathcal{D}} \cdot \frac{a_3}{a_0} </math>

<math>~=</math>

<math>~\frac{ c_0(c_0+5) - (\sigma^2_{j=1} - 2\alpha) }{ (c_0+3)(c_0+5) -\alpha} \, . </math>

In addition, we must also examine what condition is required for the <math>~\xi^6</math> terms to sum to zero. We have,

<math>~0</math>

<math>~=</math>

<math>~\mathcal{D}\frac{a_3}{a_0}\biggl[ - (c_0+3)(c_0+2) -6(c_0+3) + \biggl(\frac{\rho_c}{\rho_e}\biggr) \biggl( \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr) \biggr] </math>

<math>~\Rightarrow ~~~ \sigma^2_{j=1}\biggr|_\mathrm{env}</math>

<math>~=</math>

<math>~ (c_0+3)(c_0+8) +2\alpha \, . </math>


Mode j = 2

Here we assume that the series defining the eigenfunction has three terms: <math>~k=0</math>, <math>~k=3</math>, and <math>~k=6</math>. Specifically,

<math>~x_{j=2}</math>

<math>~=</math>

<math>~a_0 \xi^{c_0} + a_3 \xi^{c_0+3} + a_6 \xi^{c_0+6}</math>

<math>~\Rightarrow ~~~\frac{1}{\xi}\cdot \frac{d x_{j=2}}{d\xi}</math>

<math>~=</math>

<math>~\frac{1}{\xi} \biggl[ a_0 c_0 \xi^{c_0-1} + a_3 (c_0+3) \xi^{c_0+2} + a_6(c_0+6)\xi^{c_0+5} \biggr] \, ;</math>

<math>~\Rightarrow ~~~\frac{d^2 x_{j=2}}{d\xi^2}</math>

<math>~=</math>

<math>~a_0 c_0(c_0-1) \xi^{c_0-2} + a_3 (c_0+3)(c_0+2) \xi^{c_0+1} + a_6(c_0+6)(c_0+5) \xi^{c_0+4} \, .</math>

In this case, after factoring out <math>~a_0\xi^{c_0-2}</math>, the wave equation becomes,

<math>~0</math>

<math>~=</math>

<math>~ \biggl[1 - \mathcal{D} \xi^3\biggr] \biggl[ c_0(c_0-1) + \frac{a_3}{a_0} (c_0+3)(c_0+2) \xi^{3} + \frac{a_6}{a_0}(c_0+6)(c_0+5) \xi^{6} \biggr] </math>

 

 

<math>~ + \biggl[ 3 - 6\mathcal{D} \xi^3 \biggr] \biggl[ c_0 + \frac{a_3}{a_0} (c_0+3) \xi^{3} + \frac{a_6}{a_0}(c_0+6)\xi^{6} \biggr] + \biggl[ \mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr) \biggl( \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr)\xi^3 -\alpha \biggr] \biggl[1 + \frac{a_3}{a_0} \xi^{3} + \frac{a_6}{a_0} \xi^{6} \biggr] \, . </math>

Again, the coefficients of the <math>~\xi^0</math> terms will sum to zero if,

<math>~c_0</math>

<math>~=</math>

<math>~-1 \pm (1+\alpha)^{1 / 2} \, .</math>

In order for the coefficients of the <math>~\xi^3</math> terms to sum to zero, we need,

<math>~0</math>

<math>~=</math>

<math>~ -\mathcal{D}c_0(c_0-1) + \frac{a_3}{a_0} (c_0+3)(c_0+2) - 6\mathcal{D}c_0 + 3\cdot \frac{a_3}{a_0} (c_0+3) + \mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr) \biggl( \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr) - \alpha\cdot \frac{a_3}{a_0} </math>

<math>~\Rightarrow~~~ \frac{1}{\mathcal{D}} \cdot \frac{a_3}{a_0} \biggl[ (c_0+3)(c_0+2) + 3 (c_0+3) - \alpha\biggr] </math>

<math>~=</math>

<math>~ 

c_0(c_0-1)  + 6c_0 +2\alpha

- \biggl(\frac{\rho_c}{\rho_e}\biggr) \biggl( \mathfrak{F} + 2\alpha \biggr) </math>

<math>~\Rightarrow~~~ \frac{1}{\mathcal{D}} \cdot \frac{a_3}{a_0} </math>

<math>~=</math>

<math>~ \frac{ c_0(c_0+5) +2\alpha - \sigma^2_{j=2} }{ (c_0+3)(c_0+5) - \alpha } </math>

In order for the coefficients of the <math>~\xi^6</math> terms to sum to zero, we need,

<math>~0</math>

<math>~=</math>

<math>~ \frac{a_6}{a_0}(c_0+6)(c_0+5) - \mathcal{D} \cdot \frac{a_3}{a_0} (c_0+3)(c_0+2) +3 \cdot \frac{a_6}{a_0}(c_0+6) - 6\mathcal{D}\cdot \frac{a_3}{a_0} (c_0+3) -\alpha \cdot \frac{a_6}{a_0} + \mathcal{D}\biggl(\frac{\rho_c}{\rho_e}\biggr) \biggl( \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr) \cdot \frac{a_3}{a_0} </math>

<math>~\Rightarrow ~~~ \frac{a_6}{a_0} \biggl[ (c_0+6)(c_0+5) +3 (c_0+6) -\alpha \biggr] </math>

<math>~=</math>

<math>~ \mathcal{D} \cdot \frac{a_3}{a_0} \biggl[ (c_0+3)(c_0+2) +6 (c_0+3) +2\alpha - \sigma^2_{j=2} \biggr] </math>

<math>~\Rightarrow ~~~\frac{1}{\mathcal{D}} \cdot \frac{a_6}{a_3} </math>

<math>~=</math>

<math>~ \frac{ (c_0+3)(c_0+8) - (\sigma^2_{j=2}-2\alpha) }{(c_0+6)(c_0+8) -\alpha } \, . </math>

And the frequency determined from setting to zero the sum of coefficients of the <math>~\xi^9</math> terms is,

<math>~0</math>

<math>~=</math>

<math>~ -\mathcal{D} \cdot \frac{a_6}{a_0}(c_0+6)(c_0+5) - 6\mathcal{D} \cdot \frac{a_6}{a_0}(c_0+6) + \mathcal{D} \biggl(\frac{\rho_c}{\rho_e}\biggr) \biggl( \mathfrak{F} + 2\alpha -2\alpha\frac{\rho_e}{\rho_c} \biggr)\cdot \frac{a_6}{a_0} </math>

 

<math>~=</math>

<math>~ -\mathcal{D} \cdot \frac{a_6}{a_0} \biggl[ (c_0+6)(c_0+5) + 6(c_0+6) - (\sigma^2_{j=2} - 2\alpha) \biggr] </math>

<math>~\Rightarrow ~~~ (\sigma^2_{j=2} - 2\alpha) </math>

<math>~=</math>

<math>~ 

(c_0+6)(c_0+11) \, .

</math>

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