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Now, according to our [[accompanying discussion of the equilibrium mass and radius of a zero-zero polytrope|User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#MassRadius]], we know that,
Now, according to our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#MassRadius|accompanying discussion of the equilibrium mass and radius of a zero-zero polytrope]], we know that,
 
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Revision as of 21:21, 30 November 2016

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 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>

 

 

<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 = 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>

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