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Revision as of 03:56, 6 March 2017
Radial Oscillations of n = 5 Polytropic Spheres
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Background
General Form of the LAWE for Spherical Polytropes
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. Because this widely used form of the radial pulsation equation is not dimensionless but, rather, has units of inverse length-squared, we have found it useful to also recast it in the following dimensionless form:
<math>
\frac{d^2x}{d\chi_0^2} + \biggl[\frac{4}{\chi_0} - \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_0}{P_c}\biggr)^{-1} \biggl(\frac{g_0}{g_\mathrm{SSC}}\biggr) \biggr] \frac{dx}{d\chi_0} + \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{P_0}{P_c}\biggr)^{-1} \biggl(\frac{1}{\gamma_\mathrm{g}} \biggr)\biggl[\tau_\mathrm{SSC}^2 \omega^2 + (4 - 3\gamma_\mathrm{g})\biggl(\frac{g_0}{g_\mathrm{SSC}}\biggr) \frac{1}{\chi_0} \biggr] x = 0 ,
</math>
where,
<math>~g_\mathrm{SSC} \equiv \frac{P_c}{R\rho_c} \, ,</math> and <math>~\tau_\mathrm{SSC} \equiv \biggl[\frac{R^2 \rho_c}{P_c}\biggr]^{1/2} \, .</math>
In a separate discussion, we showed that specifically for isolated, polytropic configurations, this linear adiabatic wave equation (LAWE) can be rewritten as,
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<math>~0 </math> |
<math>~=</math> |
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4 - (n+1)V(\xi)}{\xi} \biggr] \frac{dx}{d\xi} + \biggl[\frac{\omega^2}{\gamma_g \theta} \biggl(\frac{n+1 }{4\pi G \rho_c} \biggr) - \biggl(3-\frac{4}{\gamma_g}\biggr) \cdot \frac{(n+1)V(x)}{\xi^2} \biggr] x </math> |
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|
<math>~=</math> |
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} + \frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma_g } - \frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x \, ,</math> |
where we have adopted the dimensionless frequency notation,
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<math>~\sigma_c^2</math> |
<math>~\equiv</math> |
<math>~\frac{3\omega^2}{2\pi G \rho_c} \, .</math> |
Specifically for n=5 Configurations
Here we focus on an analysis of the specific case of isolated, <math>~n=5</math> polytropic configurations, whose unperturbed equilibrium structure can be prescribed in terms of analytic functions. Our hope — as yet unfulfilled — is that we can discover an analytically prescribed eigenvector solution to the governing LAWE.
From our discussion of the equilibrium structure of isolated, <math>~n=5</math> polytropes, we know that,
|
<math>~\theta</math> |
<math>~=</math> |
<math>~\biggl( 1 + \frac{\xi^2}{3} \biggr)^{-1/2} =3^{1/2} ( 3 + \xi^2 )^{-1/2}\, .</math> |
Hence, we know as well that,
|
<math>~\frac{d\theta}{d\xi}</math> |
<math>~=</math> |
<math>~- \frac{\xi}{3}\biggl( 1 + \frac{\xi^2}{3} \biggr)^{-3/2} = - 3^{1/2}\xi ( 3 + \xi^2 )^{-3/2} \, .</math> |
The LAWE therefore becomes,
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<math>~0 </math> |
<math>~=</math> |
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} + \frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma_g } - \frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x </math> |
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|
<math>~=</math> |
<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{6}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} + \biggl[\frac{\sigma_c^2}{\gamma_g } \cdot \frac{1}{\theta} - \frac{6\alpha}{\xi} \cdot \frac{1}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x</math> |
|
|
<math>~=</math> |
<math>~\frac{d^2x}{d\xi^2} + \biggl[4 - \frac{6\xi^2}{(3+\xi^2)} \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + \biggl[\frac{\sigma_c^2}{3^{1/2} \gamma_g } \cdot (3+\xi^2)^{1/2} - \frac{6\alpha}{(3+\xi^2)}\biggr)\biggr] x \, .</math> |
Or,
| LAWE for <math>~n=5</math> Polytropes | |||
|---|---|---|---|
|
Numerical Integration of LAWE
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Search for Analytic Solutions to the LAWE
Setup Using Lagrangian Radial Coordinate
Individual Terms
From our accompanying discussion, we have, for pressure-truncated, <math>~n=5</math> polytropic spheres
|
<math> ~\frac{R_\mathrm{eq}}{R_\mathrm{norm}} </math> |
<math>~=~</math> |
<math>~\biggl[ \frac{4\pi}{(n+1)^n}\biggr]^{1/(n-3)} \tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{(1-n)/(n-3)} </math> |
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|
<math>~=~</math> |
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2} \tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{-2} \, , </math> |
which matches the expression derived in an ASIDE box found with our introduction of the Lane-Emden equation, and
|
<math> ~\frac{P_\mathrm{e}}{P_\mathrm{norm}} </math> |
<math>~=~</math> |
<math>~\biggl[ \frac{(n+1)^3}{4\pi}\biggr]^{(n+1)/(n-3)} \tilde\theta_n^{n+1}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)/(n-3)} </math> |
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<math>~=~</math> |
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3} \tilde\theta^{6}( -\tilde\xi^2 \tilde\theta' )^{6} \, , </math> |
where,
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<math>~R_\mathrm{norm}</math> |
<math>~\equiv</math> |
<math>~\biggl[ \biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{n-1} \biggr]^{1/(n-3)} = \biggl( \frac{G}{K} \biggr)^{5/2} M_\mathrm{tot}^{2} \, ,</math> |
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<math>~P_\mathrm{norm}</math> |
<math>~\equiv</math> |
<math>~\biggl[ \frac{K^{4n}}{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)}} \biggr]^{1/(n-3)} = \frac{K^{10}}{G^{9} M_\mathrm{tot}^{6} } \, ,</math> |
and, from our more detailed analysis,
|
<math> ~{\tilde\theta}_5 = 3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2} </math> |
and |
<math> ~\biggl(- {\tilde\xi}^2 {\tilde\theta}^'_5\biggr) = 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \, . </math> |
Hence,
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<math> ~\frac{R_\mathrm{eq}}{R_\mathrm{norm}} </math> |
<math>~=~</math> |
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2} \tilde\xi \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \biggr]^{-2} </math> |
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<math>~=~</math> |
<math>~\biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2} \tilde\xi \biggl[ 3^{-1} {\tilde\xi}^{-6} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \biggr] </math> |
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|
<math>~=~</math> |
<math>~ \biggl[ \frac{4\pi}{2^5\cdot 3^7}\biggr]^{1/2} {\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \, ,
</math> |
|
<math> ~\frac{P_\mathrm{e}}{P_\mathrm{norm}} </math> |
<math>~=~</math> |
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3} \biggl[ 3^{1 / 2} \biggl( 3 + {\tilde\xi}^2\biggr)^{-1/2} \biggr]^{6} \biggl[ 3^{1 / 2} {\tilde\xi}^3 \biggl( 3 + {\tilde\xi}^2\biggr)^{-3/2} \biggr]^{6} </math> |
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<math>~=~</math> |
<math>~\biggl[ \frac{2^3\cdot 3^3}{4\pi}\biggr]^{3} \biggl[ 3^{3} \biggl( 3 + {\tilde\xi}^2\biggr)^{-3} \biggr] \biggl[ 3^{3} {\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-9} \biggr] </math> |
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|
<math>~=~</math> |
<math>~\biggl[ \frac{2^3\cdot 3^5}{4\pi}\biggr]^{3} {\tilde\xi}^{18} \biggl( 3 + {\tilde\xi}^2\biggr)^{-12} \, . </math> |
Now, given that the structural form-factors for <math>~n=5</math> configurations are,
|
<math>~\mathfrak{f}_M</math> |
<math>~=</math> |
<math>~ ( 1 + \ell^2 )^{-3/2} = 3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2} </math> |
|
<math>~\mathfrak{f}_W</math> |
<math>~=</math> |
<math>~ \frac{5}{2^4} \cdot \ell^{-5} \biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr] </math> |
|
<math>~\mathfrak{f}_A</math> |
<math>~=</math> |
<math>~ \frac{3}{2^3} \ell^{-3} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ] \, , </math> |
we understand that the central density is,
|
<math>~\rho_c = \frac{\bar\rho}{ {\tilde\mathfrak{f}}_M }</math> |
<math>~=</math> |
<math>~ \biggl[3^{3 / 2} (3 + {\tilde\xi}^2)^{-3 / 2} \biggr]^{-1} \biggl[ \frac{3 M_\mathrm{tot}}{4 \pi R_\mathrm{eq}^3} \biggr] </math> |
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<math>~=</math> |
<math>~ \biggl( \frac{3}{4\pi}\biggr) \biggl[ \frac{2^5\cdot 3^6}{4\pi}\biggr]^{ 3 / 2} (3 + {\tilde\xi}^2)^{3 / 2} M_\mathrm{tot} \biggl[ R_\mathrm{norm} {\tilde\xi}^{-5} \biggl( 3 + {\tilde\xi}^2\biggr)^{3} \biggr]^{-3}
</math> |
|
|
<math>~=</math> |
<math>~ \biggl[ \frac{2^{5}\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} {\tilde\xi}^{15} (3 + {\tilde\xi}^2)^{-15 / 2} M_\mathrm{tot} R^{-3}_\mathrm{norm} </math> |
|
|
<math>~=</math> |
<math>~ \biggl[ \frac{2^{5}\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} M_\mathrm{tot}^{-5} \biggl( \frac{G}{K} \biggr)^{-15/2} </math> |
|
|
<math>~=</math> |
<math>~ \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} \, . </math> |
Now let's derive the prescription for the Lagrangian radial coordinate in the context of pressure-truncated, <math>~n=5</math> polytropes.
|
<math>~r_0 \equiv a_5 \xi</math> |
<math>~=</math> |
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \rho_c^{-2/5} \xi</math> |
|
|
<math>~=</math> |
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \xi \biggl\{ \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} \biggr\}^{-2/5} </math> |
|
|
<math>~=</math> |
<math>~\biggl[\frac{3K}{2\pi G} \biggr]^{1 / 2} \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr] \biggl( \frac{G^3M_\mathrm{tot}^2}{K^3} \biggr) \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi </math> |
|
|
<math>~=</math> |
<math>~ R_\mathrm{norm} \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi </math> |
Also,
|
<math>~m_0 \equiv M(r_0)</math> |
<math>~=</math> |
<math>~\biggl[ 4\pi a_n^3 \rho_c \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr] \, ,</math> |
|
|
<math>~=</math> |
<math>~2^2\pi \biggl\{ R_\mathrm{norm} \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2} \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3 \biggl\{ \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} \biggr\} \biggl\{ 3^{1 / 2} \xi^3 \biggl( 3 + \xi^2\biggr)^{-3/2} \biggr\} </math> |
|
|
<math>~=</math> |
<math>~ 3^{1 / 2} \biggl[ 2^4 \pi^2\biggr]^{1 / 2} \biggl[ \frac{\pi^3}{2^9\cdot 3^{21}}\biggr]^{1 / 2} \biggl[ \frac{2^5\cdot 3^{20}}{\pi^5}\biggr]^{ 1 / 2} \biggl\{ \tilde\xi^{-6} (3 + {\tilde\xi}^2)^{3} \biggr\}^3 \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2} R_\mathrm{norm}^3 \biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} </math> |
|
|
<math>~=</math> |
<math>~ \biggl\{ \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} M_\mathrm{tot} \biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} \, . </math> |
Hence,
|
<math>~g_0 = \frac{Gm_0}{r_0^2}</math> |
<math>~=</math> |
<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2} \biggl\{ \tilde\xi^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr\} \biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} \biggl\{ \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \biggr\}^{-2} </math> |
|
|
<math>~=</math> |
<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^2}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr] \biggl[ \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9} \xi ( 3 + \xi^2 )^{-3/2} \, ; </math> |
|
<math>~\frac{g_0 }{r_0} </math> |
<math>~=</math> |
<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr] \biggl\{ \tilde\xi^{9} (3 + {\tilde\xi}^2)^{-9 / 2} \biggr\} \biggl\{ \biggl[ \frac{\pi}{2^3\cdot 3^{7}}\biggr]^{1 / 2} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \biggr\}^{-1} \xi ( 3 + \xi^2 )^{-3/2} </math> |
|
|
<math>~=</math> |
<math>~ \frac{GM_\mathrm{tot}}{R_\mathrm{norm}^3}\biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2} \biggl[ \tilde\xi (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} ( 3 + \xi^2 )^{-3/2} \, ; </math> |
|
<math>~\frac{\rho_0}{P_0} = \frac{\rho_0}{K\rho_0^{1+1/n}} </math> |
<math>~=</math> |
<math>~ \biggl[K^5 \rho_c \theta^5 \biggr]^{-1/5} </math> |
|
|
<math>~=</math> |
<math>~ \theta^{-1} \biggl\{ K^5 \biggl[ \frac{2\cdot 3^{4}}{\pi}\biggr]^{ 5 / 2} \biggl[{\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{15} \biggl( \frac{K^3}{G^3M_\mathrm{tot}^2} \biggr)^{5/2}\biggr\}^{-1/5} </math> |
|
|
<math>~=</math> |
<math>~ \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2} \biggl\{ \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]^{1 / 2} \cancelto{\mathrm{mistake}}{\biggl[{\tilde\xi}^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr]^{-3} } \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggr\} </math> |
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|
<math>~=</math> |
<math>~ \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1 / 2} \biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \cancelto{\mathrm{mistake}}{\biggl[ {\tilde\xi} (3 + {\tilde\xi}^2)^{-1 / 2} \biggr]^{9} } ( 3 + \xi^2 )^{1 / 2} \, ; </math> |
|
|
<math>~=</math> |
<math>~ \biggl[ 3^{-1} ( 3 + \xi^2 ) \biggr]^{1/2} \biggl\{ \biggl[ \frac{\pi}{2\cdot 3^{4}}\biggr]^{1 / 2} \biggl[{\tilde\xi}^{-3} (3 + {\tilde\xi}^2)^{3 / 2} \biggr] \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggr\} </math> |
|
|
<math>~=</math> |
<math>~ \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2} ( 3 + \xi^2 )^{1/2} </math> |
|
<math>~\frac{g_0\rho_0}{P_0} </math> |
<math>~=</math> |
<math>~ \biggl( \frac{G^3M_\mathrm{tot}^2}{K^5} \biggr)^{1/2}\biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{3 / 2} ( 3 + \xi^2 )^{1/2} </math> |
|
|
|
<math>~ \times ~ \biggl( \frac{G^2M_\mathrm{tot}^2}{R_\mathrm{norm}^4} \biggr)^{1 / 2}\biggl[ \frac{2^6\cdot 3^{14}}{\pi^2}\biggr]^{1 / 2} \biggl[ \frac{(3 + {\tilde\xi}^2)}{{\tilde\xi}^2 } \biggr]^{-9 / 2} \xi ( 3 + \xi^2 )^{-3/2} </math> |
|
|
<math>~=</math> |
<math>~ \biggl( \frac{G^5 M_\mathrm{tot}^4}{K^5} \biggr)^{1 / 2} R_\mathrm{norm}^{-2} \biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1} </math> |
|
|
<math>~=</math> |
<math>~ \biggl( \frac{K^5}{G^5 M_\mathrm{tot}^4} \biggr)^{1 / 2} \biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1} \, . </math> |
The Wave Equation
Starting from our Key Adiabatic Wave Equation
The adiabatic wave equation therefore becomes,
|
<math>~0</math> |
<math>~=</math> |
<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 </math> |
|
|
<math>~=</math> |
<math>~ \frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl\{ \biggl[ \frac{2^7\cdot 3^{7}}{\pi} \biggr]^{1 / 2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \frac{1}{\xi} - \biggl[ \frac{{\tilde\xi}^2 }{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{2^5\cdot 3^{9}}{\pi}\biggr]^{1 / 2} \xi ( 3 + \xi^2 )^{-1} \biggr\} \frac{dx}{dr_0} </math> |
|
|
|
<math>~ + \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2} \biggl[ \frac{\pi}{2\cdot 3^{5}}\biggr]^{1 / 2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-3/2} ( 3 + \xi^2 )^{1 / 2} \biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr] + \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{3/2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{15/2} ( 3 + \xi^2 )^{-3/2} \biggr\} x </math> |
|
|
<math>~=</math> |
<math>~ \frac{d^2x}{dr_0^2} + \frac{1}{R_\mathrm{norm}} \biggl[ \frac{2^3\cdot 3^{7}}{\pi} \biggr]^{1 / 2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{4}{\xi} - \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{dr_0} </math> |
|
|
|
<math>~ + \frac{6(4 - 3\gamma_\mathrm{g})}{\gamma_g R_\mathrm{norm}^2} ( 3 + \xi^2 )^{1 / 2} \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr] \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6} \biggl\{ \frac{R_\mathrm{norm}^3}{GM_\mathrm{tot}} \biggl[\frac{\omega^2}{(4 - 3\gamma_\mathrm{g})} \biggr] \biggl[ \frac{2^3\cdot 3^{7}}{\pi}\biggr]^{-3/2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2} + ( 3 + \xi^2 )^{-3/2} \biggr\} x </math> |
|
|
<math>~=</math> |
<math>~ \frac{d^2x}{dr_0^2} + \frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{3} \biggl[ \frac{4}{\xi} - \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{dr_0} + \frac{6}{\gamma_g R_*^2} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6} \biggl\{ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}} \biggl[ \frac{{\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{-15/2}( 3 + \xi^2 )^{1 / 2} + \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) } \biggr\} x </math> |
where,
<math>R_* \equiv R_\mathrm{norm} \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1/2} \, .</math>
Recognizing that,
|
<math>~r_0</math> |
<math>~=</math> |
<math>~ R_* \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi \, , </math> |
we can write,
|
<math>~0</math> |
<math>~=</math> |
<math>~\frac{1}{R_*^2} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)} \biggr]^{6} \biggl\{ \frac{d^2x}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{d\xi} + \frac{6}{\gamma_g } \biggl[\sigma^2 ( 3 + \xi^2 )^{1 / 2} + \frac{(4 - 3\gamma_\mathrm{g})}{( 3 + \xi^2 ) } \biggr] x \biggr\} \, , </math> |
where,
|
<math>~\sigma^2</math> |
<math>~\equiv</math> |
<math>~ \frac{\omega^2 R_*^3}{GM_\mathrm{tot}} \biggl( \frac{3 + {\tilde\xi}^2}{{\tilde\xi}^2} \biggr)^{15/2} = \frac{\sigma_c^2}{2\cdot 3^{3/2}}\, .</math> |
Finally, if — because we are specifically considering the case of <math>~n=5</math> — we set <math>~\gamma_\mathrm{g} = 1 + 1/n = 6/5</math>, we have,
|
<math>~0</math> |
<math>~=</math> |
<math>~ \frac{d^2x}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 \xi}{ ( 3 + \xi^2 )} \biggr] \frac{dx}{d\xi} + \biggl[\frac{5\sigma_c^2}{2\cdot 3^{3/2}} ( 3 + \xi^2 )^{1 / 2} + \frac{2}{( 3 + \xi^2 ) }\biggr] x </math> |
|
|
<math>~=</math> |
<math>~\frac{1}{( 3 + \xi^2 ) } \biggl\{ ( 3 + \xi^2 )\frac{d^2x}{d\xi^2} + \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi} + \biggl[\frac{5\sigma_c^2}{2\cdot 3^{3/2}} ( 3 + \xi^2 )^{3 / 2} + 2 \biggr] x \biggr\} \, , </math> |
which matches exactly the form of the LAWE derived above, if in that expression, <math>~\gamma_g</math> is also forced to align with our specification of the polytropic index, that is, if <math>~\gamma_g = (n+1)/n = 6/5</math> and, in turn, <math>~\alpha = (3-4/\gamma) = -1/3</math>.
Starting from the HRW66 Radial Pulsation Equation
More directly, if we begin with the HRW66 radial pulsation equation that is already tuned to polytropic configurations, the wave equation appropriate to <math>~n=5</math> polytropes is,
|
<math>~0</math> |
<math>~=</math> |
<math>~ \frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 (-\theta^'_5)}{\theta_5} \biggr]\frac{d X}{d\xi} + \frac{5(-\theta_5^') }{6\theta_5 \xi} \bigg[ \frac{\xi (s^')^2}{\theta^'_5} + \frac{12}{5} \biggr] X </math> |
|
|
<math>~=</math> |
<math>~ \frac{d^2 X}{d\xi^2} + \biggl[ \frac{4}{\xi} - \frac{6 \xi}{(3 + \xi^2)} \biggr]\frac{d X}{d\xi} + \frac{1}{(3 + \xi^2)} \bigg[ -\frac{5(s^')^2(3 + \xi^2)^{3 / 2}}{2 \cdot 3^{3 / 2}} + 2 \biggr] X </math> |
|
|
<math>~=</math> |
<math>~\frac{1}{(3+\xi^2)} \biggl\{ (3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi} + \bigg[ -\frac{5(s^')^2}{2 \cdot 3^{3 / 2}} \cdot (3 + \xi^2)^{3 / 2} + 2 \biggr] X \biggr\} \, , </math> |
which is identical to the brute-force derivation just presented, allowing for the mapping,
<math>\sigma^2 ~~ \Leftrightarrow ~~ -\frac{(s^')^2}{2 \cdot 3^{3 / 2}} \, .</math>
Finally, remembering that the HRW66 dimensionless frequency definition is,
|
<math>~(s^')^2</math> |
<math>~=</math> |
<math>~-\omega^2 \biggl[\frac{n+1 }{4\pi G \rho_c} \biggr] \, ,</math> |
we recognize that, specifically for the case of <math>~n=5</math>, we can make the substitution, <math>~(s^')^2 \rightarrow -\sigma_c^2</math>, in which case the LAWE becomes,
|
<math>~0</math> |
<math>~=</math> |
<math>~\frac{1}{(3+\xi^2)} \biggl\{ (3+\xi^2)\frac{d^2 X}{d\xi^2} + \biggl[ \frac{2(6-\xi^2)}{\xi}\biggr]\frac{d X}{d\xi} + \bigg[ \frac{5\sigma_c^2}{2 \cdot 3^{3 / 2}} \cdot (3 + \xi^2)^{3 / 2} + 2 \biggr] X \biggr\} \, , </math> |
which matches exactly the form of the LAWE derived above, if in that expression, <math>~\gamma_g</math> is also forced to align with our specification of the polytropic index, that is, if <math>~\gamma_g = (n+1)/n = 6/5</math> and, in turn, <math>~\alpha = (3-4/\gamma) = -1/3</math>.
New Independent Variable
Guided by our conjecture regarding the proper shape of the radial eigenfunction, let's switch the dependent variable to,
|
<math>~u \equiv 1 + \frac{3}{\xi^2}</math> |
<math>~\Rightarrow</math> |
<math>~3 + \xi^2 = \frac{3u}{(u-1)} \, ,</math> |
and |
<math>~\xi = 3^{1 / 2} (u-1)^{-1 / 2} \, .</math> |
This implies that,
|
<math>~\frac{d}{d\xi}</math> |
<math>~~~\rightarrow ~~~</math> |
<math>~-\frac{2}{\sqrt{3}}(u-1)^{3 / 2} \frac{d}{du} \, ,</math> |
and,
|
<math>~\frac{d^2}{d\xi^2}</math> |
<math>~~~\rightarrow ~~~</math> |
<math>~\frac{4}{3}(u-1)^3 \frac{d^2}{du^2} + 2(u-1)^{2} \frac{d}{du} \, .</math> |
Hence, the governing wave equation becomes,
|
<math>~0</math> |
<math>~=</math> |
<math>~( 3 + \xi^2 )\frac{d^2x}{d\xi^2} + \biggl[ \frac{2(6 - \xi^2) }{ \xi} \biggr] \frac{dx}{d\xi} + \biggl[5\sigma^2 ( 3 + \xi^2 )^{3 / 2} + 2 \biggr] x </math> |
|
|
<math>~=</math> |
<math>~\frac{3u}{(u-1)} \biggl[\frac{4}{3}(u-1)^3 \frac{d^2x}{du^2} + 2(u-1)^{2} \frac{dx}{du}\biggr] + 4(2u-3)(u-1)\frac{dx}{du} + \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2} + 2 \biggr\} x </math> |
|
|
<math>~=</math> |
<math>~4u(u-1)^2 \frac{d^2x}{du^2} + (14u-12)(u-1)\frac{dx}{du} + \biggl\{ 5\sigma^2 \biggl[ \frac{3u}{(u-1)} \biggr]^{3 / 2} + 2 \biggr\} x \, . </math> |
Material that appears after this point in our presentation is under development and therefore
may contain incorrect mathematical equations and/or physical misinterpretations.
| Go Home |
If we assume that <math>~\sigma^2 = 0</math>, then the governing relation is,
|
<math>~0</math> |
<math>~=</math> |
<math>~4u(u-1)^2 \frac{d^2x}{du^2} + (14u-12)(u-1)\frac{dx}{du} + 2 x \, . </math> |
Now, again, guided by our conjecture, let's guess an eigenfunction of the form:
First Guess (n5)
|
<math>~x</math> |
<math>~=</math> |
<math>~ A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2} \, , </math> |
in which case,
|
<math>~\frac{dx}{du}</math> |
<math>~=</math> |
<math>~ \frac{A^3}{2} \biggl[ (u - 1)^{-1 / 2} (A u - 1 )^{-1 / 2} - A(u - 1)^{1 / 2} (A u - 1 )^{-3 / 2} \biggr] </math> |
|
|
<math>~=</math> |
<math>~ \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \, ; </math> |
|
<math>~\frac{d^2x}{du^2}</math> |
<math>~=</math> |
<math>~ \biggl[ \frac{A^3(A-1)}{2} \biggr] \biggl\{ -\frac{1}{2}(u-1)^{-3 / 2} (Au-1)^{-3 / 2} -\frac{3A}{2} (u-1)^{-1 / 2} (Au-1)^{-5 / 2} \biggr\} </math> |
|
|
<math>~=</math> |
<math>~ -\frac{1}{2} \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[ (Au-1) +3A (u-1)\biggr] </math> |
|
|
<math>~=</math> |
<math>~ \biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] \, . </math> |
So the governing relation becomes:
|
<math>~0</math> |
<math>~=</math> |
<math>~4u(u-1)^2 \biggl\{ \biggl[ \frac{A^3(A-1)}{4} \biggr] (u-1)^{-3 / 2} (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] \biggr\} </math> |
|
|
|
<math>~ + (14u-12)(u-1) \biggl\{ \biggl[ \frac{A^3(A-1)}{2} \biggr] (u-1)^{-1 / 2} (Au-1)^{-3 / 2} \biggr\} + 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2} </math> |
|
|
<math>~=</math> |
<math>~u(u-1)^{1 / 2} A^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] </math> |
|
|
|
<math>~ + (7u-6)(u-1)^{1 / 2} A^3(A-1) (Au-1)^{-3 / 2} + 2 A^3 (u - 1)^{1 / 2} (A u - 1 )^{-1 / 2} </math> |
|
|
<math>~=</math> |
<math>~(u-1)^{1 / 2} \biggl\{ uA^3(A-1) (Au-1)^{-5 / 2}\biggl[(3A+1) - 4Au \biggr] + (7u-6) A^3(A-1) (Au-1)^{-3 / 2} + 2 A^3 (A u - 1 )^{-1 / 2} \biggr\} </math> |
|
|
<math>~=</math> |
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u(A-1) \biggl[(3A+1) - 4Au \biggr] + (7u-6) (A-1) (Au-1) + 2 (A u - 1 )^{2} \biggr\} </math> |
|
|
<math>~=</math> |
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ - 4u^2 A(A-1) + u(A-1) (3A+1) + (7u-6) [A(A-1)u +1 - A] + 2 (A^2u^2 - 2Au +1) \biggr\} </math> |
|
|
<math>~=</math> |
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ u^2 \biggl[ - 4A(A-1) +7A(A-1) +2A^2 \biggr] + u\biggl[ (A-1) (3A+1) - 7(A-1) -6A(A-1) - 4A \biggr] + 2(3A-2) \biggr\} </math> |
|
|
<math>~=</math> |
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr] + u\biggl[ 3A^2-2A-1-7A+7 -6A^2+6A -4A \biggr] + 2(3A-2) \biggr\} \, . </math> |
|
|
<math>~=</math> |
<math>~A^3(u-1)^{1 / 2} (Au-1)^{-5 / 2} \biggl\{ Au^2 \biggl[ 5A-3 \biggr] + u\biggl[ -3A^2 -7A +6\biggr] + 2(3A-2) \biggr\} \, . </math> |
Second Guess (n5)
|
<math>~x</math> |
<math>~=</math> |
<math>~ (u - 1)^{b / 2} (A u - 1 )^{-a / 2} \, , </math> |
in which case,
|
<math>~\frac{dx}{du}</math> |
<math>~=</math> |
<math>~ \frac{b}{2}(u-1)^{b/2-1} (A u - 1 )^{-a / 2} - \frac{aA}{2}(u - 1)^{b / 2} (A u - 1 )^{-a / 2-1} </math> |
|
|
<math>~=</math> |
<math>~x \biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2} (A u - 1 )^{-1} \biggr] </math> |
|
<math>~\Rightarrow ~~~ \frac{(u-1)}{x} \frac{dx}{du}</math> |
<math>~=</math> |
<math>~ (A u - 1 )^{-1} \biggl[ \frac{b}{2} (A u - 1 ) - \frac{aA}{2} (u-1) \biggr] </math> |
|
|
<math>~=</math> |
<math>~\frac{1 }{2(A u - 1 )} \biggl[ b (A u - 1 ) - aA (u-1) \biggr] </math> |
|
|
<math>~=</math> |
<math>~ \frac{1 }{2(A u - 1 )} \biggl[ (aA - b) + A(b - a)u \biggr] \, ; </math> |
and,
|
<math>~\frac{d^2x}{du^2}</math> |
<math>~=</math> |
<math>~ \biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2} (A u - 1 )^{-1} \biggr]\frac{dx}{du} + x \frac{d}{du}\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2} (A u - 1 )^{-1} \biggr] </math> |
|
|
<math>~=</math> |
<math>~ x\biggl[ \frac{b}{2}(u-1)^{-1} - \frac{aA}{2} (A u - 1 )^{-1} \biggr]^2 + x \biggl[ -\frac{b}{2}(u-1)^{-2} + \frac{aA^2}{2} (A u - 1 )^{-2} \biggr] </math> |
|
|
<math>~=</math> |
<math>~ \frac{x}{4(u-1)^2 (Au-1)^2} \biggl\{ \biggl[ b(Au-1) - aA (u - 1 ) \biggr]^2 + \biggl[ 2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr] \biggr\} </math> |
|
<math>~\Rightarrow ~~~ \frac{(1-u)^2}{x}\frac{d^2x}{du^2}</math> |
<math>~=</math> |
<math>~ \frac{1}{4 (Au-1)^2} \biggl\{ \biggl[ b(Au-1) - aA (u - 1 ) \biggr]^2 + \biggl[ 2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr] \biggr\} </math> |
Hence, the governing wave equation becomes,
|
<math>~0</math> |
<math>~=</math> |
<math>~2u \biggl\{ \frac{(u-1)^2}{x} \frac{d^2x}{du^2} \biggr\} + (7u-6)\biggl\{ \frac{(u-1)}{x} \frac{dx}{du} \biggl\} + 1 </math> |
|
|
<math>~=</math> |
<math>~ \frac{2u}{4 (Au-1)^2} \biggl\{ \biggl[ (aA - b) + A(b - a)u \biggr]^2 + \biggl[ 2aA^2 (u-1)^{2} -2b (A u - 1 )^{2} \biggr] \biggr\} </math> |
|
|
|
<math>~ + \frac{(7u-6) }{2(A u - 1 )} \biggl[ (aA - b) + A(b - a)u \biggr] + 1 </math> |
|
|
<math>~=</math> |
<math>~ \frac{1}{4 (Au-1)^2} \biggl\{ 2u\biggl[ (aA - b)^2 + 2(aA - b)A(b - a)u + A^2(b - a)^2u^2 \biggr] + 2u\biggl[ 2aA^2 (u^2 - 2u + 1) -2b (A^2 u^2 - 2Au + 1 ) \biggr] </math> |
|
|
|
<math>~ + 2(A u - 1 )(7u-6) \biggl[ (aA - b) + A(b - a)u \biggr] + 4 (Au-1)^2 \biggr\} </math> |
|
|
<math>~=</math> |
<math>~ \frac{1}{4 (Au-1)^2} \biggl\{ 2u\biggl[ (aA - b)^2 + 2(aA - b)A(b - a)u + A^2(b - a)^2u^2 \biggr] + 2u\biggl[ 2A^2(a-b)u^2 + 4A(b - aA) u + 2(aA^2 -b) \biggr] </math> |
|
|
|
<math>~ + 2\biggl[7Au^2 - (6A+7)u +6 \biggr]\biggl[ (aA - b) + A(b - a)u \biggr] + (4A^2u^2-8Au + 4) \biggr\} </math> |
If <math>~b=a</math>,
|
<math>~0</math> |
<math>~=</math> |
<math>~ 2u\biggl[ (aA - b)^2 \biggr] + 2u\biggl[ 4A(b - aA) u + 2(aA^2 -b) \biggr] </math> |
|
|
|
<math>~ + 2\biggl[7Au^2 - (6A+7)u +6 \biggr]\biggl[ (aA - b) \biggr] + (4A^2u^2-8Au + 4) </math> |
|
|
<math>~=</math> |
<math>~ 2a^2u (A - 1)^2 + 2au [ 4A(1 - A) u + 2(A^2 -1) ] </math> |
|
|
|
<math>~ + 2a(A - 1) \biggl[7Au^2 - (6A+7)u +6 \biggr] + (4A^2u^2-8Au + 4) </math> |
|
|
<math>~=</math> |
<math>~ 2Au^2 [4a (1 - A) + 7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 + 2a(A^2 -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1] </math> |
This should then match the "first guess" algebraic condition if we set <math>~a=1</math>. Let's see.
|
<math>~0</math> |
<math>~=</math> |
<math>~ 2Au^2 [4 (1 - A) + 7(A - 1) + 2A] + 2u [ (A - 1)^2 + 2(A^2 -1) - (A - 1) (6A+7) - 4A] + 4[ 3(A-1) + 1] </math> |
|
|
<math>~=</math> |
<math>~ 2Au^2 [4 - 4A + 7A - 7 + 2A] + 2u [ (A^2 - 2A + 1) + 2A^2 -2 + (1-A ) (6A+7) -4A] + 4[ 3A-2] </math> |
|
|
<math>~=</math> |
<math>~ 2Au^2 [5A - 3] + 2u [ - 3A^2 - 7A + 6 ] + 4[ 3A-2] \, . </math> |
And we see that this expression does match the one derived earlier.
Going back a bit, before setting <math>~a=1</math>, we have the expression:
|
<math>~0</math> |
<math>~=</math> |
<math>~ 2Au^2 [4a (1 - A) + 7a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 + 2a(A^2 -1) - a(A - 1) (6A+7) - 4A] + 4[ 3a(A-1) + 1] </math> |
|
|
<math>~=</math> |
<math>~ 2Au^2 [ 3aA -3a + 2A] + 2u [ a^2 (A - 1)^2 + 2a(A^2 -1) - a(6A^2+A-7) - 4A] + 4[ 3a(A-1) + 1] </math> |
|
|
<math>~=</math> |
<math>~ 2Au^2 [ 3a(A - 1) + 2A] + 2u [ a^2 (A - 1)^2 + a( -4A^2-A+5) - 4A] + 4[ 3a(A-1) + 1] \, . </math> |
Now, in order for all three expressions inside the square-bracket pairs to be zero, we need, first,
|
<math>~3a(A - 1) + 2A</math> |
<math>~=</math> |
<math>~0</math> |
|
<math>~\Rightarrow ~~~ a</math> |
<math>~=</math> |
<math>~\frac{2A}{3(1-A)} \, ;</math> |
and, third, by simple visual comparison with the first expression,
|
<math>~3a(A-1) + 1</math> |
<math>~=</math> |
<math>~3a(A-1) + 2A</math> |
|
<math>~\Rightarrow A</math> |
<math>~=</math> |
<math>~\frac{1}{2} </math> |
|
<math>~\Rightarrow ~~~ a</math> |
<math>~=</math> |
<math>~\frac{2}{3} \, ;</math> |
which forces the second expression to the value,
|
<math>~a^2 (A - 1)^2 + a( -4A^2-A+5) - 4A</math> |
<math>~=</math> |
<math>~\biggl(\frac{2}{3}\biggr)^2 \biggl(-\frac{1}{2} \biggr)^2 + \frac{2}{3}\biggl[ -1-\frac{1}{2} +5 \biggr] - 2</math> |
|
|
<math>~=</math> |
<math>~\frac{1}{9} + \frac{7}{3} - 2</math> |
|
|
<math>~=</math> |
<math>~\frac{4}{9} \, ,</math> |
which is not zero. Hence our pair of unknown parameters — <math>~a </math> and <math>~A</math> — do not simultaneously satisfy all three conditions. (Not really a surprise.)
Setup Using Lagrangian Mass Coordinate
Alternative Terms
Let's change the independent coordinate from <math>~r_0</math> to <math>~m_0</math>. In particular, the derivative operation will change as follows:
|
<math>~\frac{d}{dr_0}</math> |
<math>~~\rightarrow~~</math> |
<math>~\biggl( \frac{dm_0}{dr_0} \biggr)\frac{d}{dm_0} = \biggl( \frac{dm_0}{d\xi} \cdot \frac{d\xi}{dr_0} \biggr)\frac{d}{dm_0} \, ,</math> |
so what is the expression for the leading coefficient? From above, we have,
|
<math>~r_0</math> |
<math>~=</math> |
<math>~ R_* \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi </math> |
|
<math>~\Rightarrow ~~~ \xi</math> |
<math>~=</math> |
<math>~ \frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3} r_0 \, . </math> |
Also, from above, we know that,
|
<math>~m_0</math> |
<math>~=</math> |
<math>~ M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} \biggl\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \biggr\} </math> |
|
<math>~\Rightarrow ~~~ \frac{dm_0}{d\xi}</math> |
<math>~=</math> |
<math>~ M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} \biggl\{ 3\xi^2 ( 3 + \xi^2 )^{-3/2} - 3 \xi^4 ( 3 + \xi^2 )^{-5/2}\biggr\} </math> |
|
|
<math>~=</math> |
<math>~ M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3\xi^2 (3 + \xi^2)^{-5/2} \biggl\{ ( 3 + \xi^2 ) - \xi^2 \biggr\} </math> |
|
|
<math>~=</math> |
<math>~ M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2} </math> |
|
<math>~\Rightarrow ~~~ \frac{dm_0}{dr_0}</math> |
<math>~=</math> |
<math>~ M_\mathrm{tot} \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2} \frac{1}{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3} </math> |
|
|
<math>~=</math> |
<math>~ \frac{M_\mathrm{tot} }{R_*} \biggl[ \frac{ {\tilde\xi}^2}{(3 + {\tilde\xi}^2)}\biggr]^{3 / 2} 3^2\xi^2 (3 + \xi^2)^{-5/2} \, . </math> |
To simplify expressions, let's borrow from an accompanying derivation and define,
<math>\tilde{C} \equiv \frac{3^2}{{\tilde\xi}^2} \biggl( 1 + \frac{ {\tilde\xi}^2}{3} \biggr) = 3 \biggl[ \frac{( 3 + {\tilde\xi}^2 )}{ {\tilde\xi}^2} \biggr] \, .</math>
Then we have,
|
<math>~\frac{m_0}{M_\mathrm{tot}}</math> |
<math>~=</math> |
<math>~ \biggl[ \frac{\tilde{C}}{ 3}\biggr]^{3 / 2} \biggl[ \frac{\xi^2}{ ( 3 + \xi^2 )} \biggr]^{3/2} </math> |
|
<math>~\Rightarrow ~~~\biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3}</math> |
<math>~=</math> |
<math>~ \frac{\xi^2}{ ( 3 + \xi^2 )} </math> |
|
<math>~\Rightarrow ~~~( 3 + \xi^2 )\biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3}</math> |
<math>~=</math> |
<math>~ \xi^2 </math> |
|
<math>~\Rightarrow ~~~3 m_*</math> |
<math>~=</math> |
<math>~ \xi^2 (1-m_*) </math> |
|
<math>~\Rightarrow ~~~\xi^2 </math> |
<math>~=</math> |
<math>~ \frac{3m_*}{(1-m_*)} \, , </math> |
where,
<math>~m_* \equiv \biggl[ \frac{ 3}{\tilde{C}}\biggr] \biggl[\frac{m_0}{M_\mathrm{tot}}\biggr]^{2 / 3} \, .</math>
In summary:
|
<math>~ \frac{\xi^2}{ ( 3 + \xi^2 )} = m_* \, ; </math> |
while, |
<math>~ \frac{ {\tilde\xi}^2}{ ( 3 + {\tilde\xi}^2 )} = \frac{3}{\tilde{C}} \, ; </math> |
|
<math>~r_0</math> |
<math>~=</math> |
<math>~ R_* \biggl[ \frac{(3 + {\tilde\xi}^2)}{ {\tilde\xi}^2}\biggr]^{3} \xi = R_* \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggr[ \frac{3m_*}{ (1-m_*) }\biggr]^{1 / 2} \, ; </math> |
|
<math>~\frac{g_0\rho_0}{P_0} </math> |
<math>~=</math> |
<math>~ \frac{6}{R_*} \biggl[ \frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2) }\biggr]^{9} \frac{\xi}{ ( 3 + \xi^2 )} = \frac{6}{R_*} \biggl[ \frac{ 3 }{ \tilde{C} }\biggr]^{9} \frac{m_*}{ \xi } = \frac{6}{R_*} \biggl[ \frac{ 3 }{ \tilde{C} }\biggr]^{9} m_* \biggl[ \frac{(1-m_*)}{3m_*} \biggr]^{1 / 2} \, ; </math> |
|
<math>~\frac{g_0 }{r_0} </math> |
<math>~=</math> |
<math>~ \frac{GM_\mathrm{tot}}{R_*^3} \biggl[ \frac{ {\tilde\xi}^2 }{ (3 + {\tilde\xi}^2)}\biggr]^{15/2} \frac{1}{\xi^3} \biggl[ \frac{ \xi^2 }{ ( 3 + \xi^2 ) }\biggr]^{3/2} = \frac{GM_\mathrm{tot}}{R_*^3} \biggl[ \frac{3 }{ \tilde{C} }\biggr]^{15/2} (1-m_*)^{3 / 2} \, ; </math> |
|
<math>~\frac{\rho_0}{\gamma_g P_0} </math> |
<math>~=</math> |
<math>~ \frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \, . </math> |
So, the wave equation may be written as,
|
<math>~0</math> |
<math>~=</math> |
<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 </math> |
|
|
<math>~=</math> |
<math>~ \frac{d^2x}{dr_0^2} + \biggl\{ \frac{4}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2} - \frac{6}{R_*} \biggl[ \frac{ 3 }{ \tilde{C} }\biggr]^{9} m_* \biggl[ \frac{(1-m_*)}{3m_*} \biggr]^{1 / 2} \biggr\} \frac{dx}{dr_0} </math> |
|
|
|
<math>~ + \frac{6R_* }{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \biggl\{ \omega^2 + (4 - 3\gamma_\mathrm{g})\frac{GM_\mathrm{tot}}{R_*^3} \biggl[ \frac{3 }{ \tilde{C} }\biggr]^{15/2} (1-m_*)^{3 / 2} \biggr\} x </math> |
|
|
<math>~=</math> |
<math>~ \frac{d^2x}{dr_0^2} + \frac{1}{R_*} \biggl( \frac{ 3}{ \tilde{C} }\biggr)^{3} \biggl\{ 4 - 6\biggl[ \frac{ 3 }{ \tilde{C} }\biggr]^{6} m_* \biggr\} \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}\frac{dx}{dr_0} </math> |
|
|
|
<math>~ + \frac{6(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot \frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \biggl\{\sigma^2 + (1-m_*)^{3 / 2} \biggr\} x </math> |
|
|
<math>~=</math> |
<math>~\frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl\{ R_*^2 \biggl( \frac{ \tilde{C} }{3 }\biggr)^{3} \frac{d^2x}{dr_0^2} + R_* \biggl[ 4 - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] \biggr[ \frac{ (1-m_*) }{3m_*}\biggr]^{1 / 2}\frac{dx}{dr_0} </math> |
|
|
|
<math>~ + \frac{6(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x \biggr\} </math> |
|
|
<math>~=</math> |
<math>~\frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2} R_*^2 \biggl( \frac{ \tilde{C} }{3 }\biggr)^{3} \frac{d^2x}{dr_0^2} + R_* \biggl[ 4 - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*) \frac{dx}{dr_0} </math> |
|
|
|
<math>~ + \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2} \biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x \biggr\} \, , </math> |
where,
<math>~\sigma^2 \equiv (4 - 3\gamma_\mathrm{g})^{-1} \frac{R_*^3}{GM_\mathrm{tot}} \biggl[ \frac{ \tilde{C} }{3 } \biggr]^{15/2} \omega^2 \, .</math>
Now, let's look at the differential operators, after defining.
<math>~c_0 \equiv 3^{1 / 2} R_* \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} ~~~~\Rightarrow ~~~~R_* = c_0 3^{-1 / 2} \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} \, .</math>
We find,
|
<math>~dr_0</math> |
<math>~=</math> |
<math>~ c_0 ~d[ m_*^{1 / 2} (1-m_*)^{-1 / 2} ] </math> |
|
|
<math>~=</math> |
<math>~ c_0 ~\biggl[\frac{1}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-1 / 2} + \frac{1}{2} ~m_*^{1 / 2} (1 - m_*)^{-3 / 2} \biggr] dm_* </math> |
|
|
<math>~=</math> |
<math>~ \frac{c_0}{2} ~m_*^{-1 / 2}( 1 - m_*)^{-3 / 2}~ dm_* </math> |
|
<math>~\frac{d}{dr_0}</math> |
<math>~=</math> |
<math>~ \frac{2}{c_0} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*} </math> |
|
<math>~\Rightarrow ~~~ R_*\frac{dx}{dr_0}</math> |
<math>~=</math> |
<math>~ \frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \, . </math> |
Also,
|
<math>~\frac{d^2}{dr_0^2}</math> |
<math>~=</math> |
<math>~ \biggl( \frac{2}{c_0} \biggr)^{2}~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*} \biggl[ m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{d}{dm_*} \biggr] </math> |
|
|
<math>~=</math> |
<math>~ \biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2} +\biggl( \frac{2}{c_0} \biggr)^{2}~m_*^{1 / 2}( 1 - m_*)^{3 / 2} \biggl[ \frac{1}{2} m_*^{-1 / 2}( 1 - m_*)^{3 / 2} - \frac{3}{2} m_*^{1 / 2}( 1 - m_*)^{1 / 2}~ \biggr] ~ \frac{d}{dm_*} </math> |
|
|
<math>~=</math> |
<math>~ \biggl( \frac{2}{c_0} \biggr)^{2}~m_* ( 1 - m_*)^{3 }~ \frac{d^2}{dm_*^2} +\frac{1}{2} \biggl( \frac{2}{c_0} \biggr)^{2}~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{d}{dm_*} </math> |
|
<math>~\Rightarrow ~~~ R_*^2 \biggl( \frac{ \tilde{C} }{3 }\biggr)^{3} \frac{d^2x}{dr_0^2}</math> |
<math>~=</math> |
<math>~ \biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr] \biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr] </math> |
So, the wave equation becomes,
|
<math>~0</math> |
<math>~=</math> |
<math>~\frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{3} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2} \biggl[\frac{2^2}{3} \biggl(\frac{ \tilde{C} }{3} \biggr)^{-3} \biggr] \biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr] </math> |
|
|
|
<math>~ + \biggl[ 4 - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}}\biggl( \frac{ \tilde{C} }{ 3}\biggr)^{-3} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \biggr] + \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \cdot m_*^{1 / 2} \biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x \biggr\} </math> |
|
|
<math>~=</math> |
<math>~\frac{1 }{R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl[ \frac{1}{3m_*(1-m_*)}\biggr]^{1 / 2} \biggl\{ [ 3m_*(1-m_*) ]^{1 / 2} \biggl[\frac{2^2}{3} \biggr] \biggl[ ~m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} +\frac{1}{2} ~ ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} \biggr] </math> |
|
|
|
<math>~ + \biggl[ 4 - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*) \biggl[ \frac{2}{3^{1 / 2}} ~m_*^{1 / 2}( 1 - m_*)^{3 / 2}~ \frac{dx}{dm_*} \biggr] + \frac{18(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} m_*^{1 / 2} \biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x \biggr\} </math> |
|
|
<math>~=</math> |
<math>~\frac{2 }{3R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl\{ 2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} + ( 1 - m_*)^{2} ( 1 - 4m_*) \frac{dx}{dm_*} </math> |
|
|
|
<math>~ + \biggl[ 4 - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*} + \frac{9(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}\biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x \biggr\} </math> |
|
|
<math>~=</math> |
<math>~\frac{2 }{3R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl\{ 2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} </math> |
|
|
|
<math>~ + \biggl[ 5 - 4m_* - 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*)^2 \frac{dx}{dm_*} + \frac{9(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2}\biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x \biggr\} </math> |
|
|
<math>~=</math> |
<math>~\frac{2 }{3R_*^2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl\{ 2m_* ( 1 - m_*)^{3 }~ \frac{d^2x}{dm_*^2} + (5 - \mathcal{A} m_*) (1-m_*)^2 \frac{dx}{dm_*} + \mathcal{B} \biggl[ \frac{\sigma^2}{(1-m_*)^{1 / 2}} + (1-m_*) \biggr] x \biggr\} \, , </math> |
where,
|
<math>~\mathcal{A}</math> |
<math>~\equiv</math> |
<math>~4 + 6\biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} \, ,</math> |
|
<math>~\mathcal{B}</math> |
<math>~\equiv</math> |
<math>~\frac{3^{5/2}(4 - 3\gamma_\mathrm{g}) }{\gamma_g } \biggl( \frac{ \tilde{C} }{ 3}\biggr)^{3} \, .</math> |
Try Again
This time, let's adopt the notation used in a related chapter in our Ramblings appendix. Specifically, the parametric relationship between <math>~m_\xi</math> and <math>~r_\xi</math> in pressure-truncated, <math>~n=5</math> polytropes is,
|
<math>~m_\xi \equiv \frac{m_0}{ M_\mathrm{tot} } = \frac{M_r(\xi)}{M_\mathrm{tot}}</math> |
<math>~=</math> |
<math>~ \biggl(\frac{\xi}{\tilde\xi}\biggr)^3 \biggl(3 + \xi^2 \biggr)^{-3/2} \biggl(3 + {\tilde\xi}^2 \biggr)^{3/2} </math> |
|
|
<math>~=</math> |
<math>~ \biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3 / 2}\biggl[ \frac{( 3+\xi^2)}{ {\xi}^2} \biggr]^{- 3 / 2} \, ,</math> |
|
<math>~r_\xi \equiv \frac{r_0}{R_\mathrm{norm}} = \biggl(\frac{\xi}{\tilde\xi} \biggr) \frac{R_\mathrm{eq}}{R_\mathrm{norm}}</math> |
<math>~=</math> |
<math>~\xi \biggl\{ \biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \tilde\xi^{-6} \biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{3}\biggr\} </math> |
|
|
<math>~=</math> |
<math>~ \biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1/2} \biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr]^{3} \xi \, . </math> |
And we are in the fortunate situation of being able to eliminate <math>~\xi</math> to obtain the direct relation,
|
<math>~ r_\xi (m_\xi) </math> |
<math>~=</math> |
<math>~\tilde{r}_\mathrm{edge} \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C} - 3 m_\xi^{2/3}}\biggr]^{1/2} \, , </math> |
where,
|
<math>~\tilde{C}</math> |
<math>~\equiv</math> |
<math>~ \frac{3^2}{\tilde\xi^2}\biggl( 1 + \frac{\tilde\xi^2}{3} \biggr) = 3 \biggl[ \frac{( 3+\tilde\xi^2)}{ {\tilde\xi}^2} \biggr] \, , </math> |
|
<math>~\tilde{r}_\mathrm{edge}</math> |
<math>~\equiv</math> |
<math>~\biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} {\tilde\xi}^{-6} \biggl(1+\frac{\tilde\xi^2}{3}\biggr)^3 = \biggl[ \frac{\pi}{2^3\cdot 3^7}\biggr]^{1 / 2} \biggl[ \frac{\tilde{C}}{ 3} \biggr]^{3} \, . </math> |
If we furthermore define,
<math>m_* \equiv \frac{3}{\tilde{C}} \cdot m_\xi^{2 / 3} \, ,</math>
then,
|
<math>~ r_\xi (m_*) </math> |
<math>~=</math> |
<math>~ 3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{1/2} \, . </math> |
Hence,
|
<math>~ \frac{dr_0}{R_\mathrm{norm}} = dr_\xi </math> |
<math>~=</math> |
<math>~3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl\{ \frac{1}{2} (1-m_*)^{- 1 / 2} m_*^{-1 / 2} + \frac{1}{2}m_*^{1 / 2}(1-m_*)^{-3 / 2} \biggr\} dm_* </math> |
|
|
<math>~=</math> |
<math>~ \biggl( \frac{3^{1 / 2}}{2} \biggr) \tilde{r}_\mathrm{edge} m_*^{-1 / 2} (1-m_*)^{-3 / 2} dm_* </math> |
|
<math>\Rightarrow ~~~ R_\mathrm{norm} \cdot \frac{d}{dr_0} </math> |
<math>~=</math> |
<math>~ \frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*} \, . </math> |
We therefore also have,
|
<math>~ R^2_\mathrm{norm} \cdot \frac{d^2}{dr_0^2} </math> |
<math>~=</math> |
<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2^2}{3} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*}\biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d}{dm_*}\biggr] </math> |
|
|
<math>~=</math> |
<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2^2}{3} \biggr) m_*^{1 / 2} (1-m_*)^{3 / 2} \biggl\{ \biggl[ m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{d^2}{dm_*^2}\biggr] + \biggl[ \frac{1}{2} m_*^{-1 / 2} (1-m_*)^{3 / 2} + \frac{3}{2}m_*^{1 / 2} (1-m_*)^{1 / 2}\biggr] \frac{d}{dm_*} \biggr\} </math> |
|
|
<math>~=</math> |
<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ \biggl[ 2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2}\biggr] + \biggl[ (1-m_*)^{3 } + 3m_* (1-m_*)^{2}\biggr] \frac{d}{dm_*} \biggr\} </math> |
|
|
<math>~=</math> |
<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 2m_* (1-m_*)^{3} \frac{d^2}{dm_*^2} + (1-m_*)^{2} ( 1 + 2m_* ) \frac{d}{dm_*} \biggr\} \, . </math> |
So the wave equation may be written,
|
<math>~0</math> |
<math>~=</math> |
<math>~ R_\mathrm{norm}^2 \cdot \frac{d^2x}{dr_0^2} + \biggl[\frac{4R_\mathrm{norm}}{r_0} - \biggl(\frac{g_0 \rho_0 R_\mathrm{norm}}{P_0}\biggr) \biggr] R_\mathrm{norm} \cdot \frac{dx}{dr_0} + \biggl(\frac{\rho_0 R_\mathrm{norm}}{\gamma_\mathrm{g} P_0} \biggr)\biggl[R_\mathrm{norm} \omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0 R_\mathrm{norm}}{r_0} \biggr] x </math> |
|
|
<math>~=</math> |
<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2} + (1-m_*)^{2} ( 1 + 2m_* ) \frac{dx}{dm_*} \biggr\} </math> |
|
|
|
<math>~ +\frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr) \biggl\{ \frac{4}{r_\xi} - \biggl[\frac{6R_\mathrm{norm}}{R_*} \biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{9} m_* \biggl[ \frac{(1-m_*)}{3m_*} \biggr]^{1 / 2} \biggr] \biggr\} m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*} </math> |
|
|
|
<math>~ + \frac{6R_* R_\mathrm{norm}}{\gamma_g GM_\mathrm{tot} }\biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \biggl\{ R_\mathrm{norm} \omega^2 + (4 - 3\gamma_\mathrm{g}) \frac{GM_\mathrm{tot} R_\mathrm{norm}}{R_*^3} \biggl[ \frac{3 }{ \tilde{C} }\biggr]^{15/2} (1-m_*)^{3 / 2} \biggr\} x \, . </math> |
Keeping in mind that,
<math>~\frac{R_*}{R_\mathrm{norm}} = \biggl[ \frac{\pi}{2^3 \cdot 3^7} \biggr]^{1 / 2} = {\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \, ,</math>
we therefore have,
|
<math>~0</math> |
<math>~=</math> |
<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2} + (1-m_*)^{2} ( 1 + 2m_* ) \frac{dx}{dm_*} \biggr\} </math> |
|
|
|
<math>~ +\frac{1}{ \tilde{r}_\mathrm{edge}} \biggl( \frac{2}{3^{1 / 2}} \biggr) \biggl\{ 4 \biggl[3^{1 / 2} \tilde{r}_\mathrm{edge} \biggl[\frac{m_*}{1-m_*}\biggr]^{1/2} \biggr]^{-1} - 6 \biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{9} \biggl[{\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \biggr]^{-1} m_* \biggl[ \frac{(1-m_*)}{3m_*} \biggr]^{1 / 2} \biggr\} m_*^{1 / 2} (1-m_*)^{3 / 2} \frac{dx}{dm_*} </math> |
|
|
|
<math>~ + 6 \biggl( \frac{ 3}{ \tilde{C} } \biggr)^{9 / 2} \biggl[{\tilde{r}}_\mathrm{edge} \biggl( \frac{3}{\tilde{C}} \biggr)^3 \biggr]^{-2} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \biggl\{ \biggl[ \frac{R_*^3}{\gamma_g GM_\mathrm{tot} } \biggr] \omega^2 + \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g} \biggl[ \frac{3 }{ \tilde{C} }\biggr]^{15/2} (1-m_*)^{3 / 2} \biggr\} x </math> |
|
|
<math>~=</math> |
<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2} + (1-m_*)^{2} ( 1 + 2m_* ) \frac{dx}{dm_*} \biggr\} </math> |
|
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<math>~ +\frac{1}{ \tilde{r}_\mathrm{edge}^2} \biggl( \frac{2^3}{3} \biggr) \biggl[ 1 - \frac{3}{2} \biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* \biggr] (1-m_*)^{2} \frac{dx}{dm_*} + \frac{6}{ {\tilde{r}}_\mathrm{edge}^2 } \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \biggl[ \frac{3}{(1-m_*)}\biggr]^{1 / 2} \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g} \biggl[ \sigma^2 + (1-m_*)^{3 / 2} \biggr] x </math> |
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<math>~=</math> |
<math>~ \frac{1}{ {\tilde{r}}^2_\mathrm{edge}} \biggl( \frac{2}{3} \biggr) \biggl\{ 2m_* (1-m_*)^{3} \frac{d^2x}{dm_*^2} + \biggl[ 5 - 6 \biggl( \frac{ 3 }{ \tilde{C} }\biggr)^{6} m_* + 2m_* \biggr] (1-m_*)^{2} \frac{dx}{dm_*} + 3^{5 / 2} \biggl( \frac{3 }{ \tilde{C} }\biggr)^{6} \frac{(4 - 3\gamma_\mathrm{g})}{\gamma_g} \biggl[ \frac{\sigma^2 }{(1-m_*)^{1 / 2}} + (1-m_*) \biggr] x \biggr\} \, , </math> |
where, as before,
<math>\sigma^2 \equiv \biggl( \frac{ \tilde{C} }{3 } \biggr)^{15/2} \biggl[ \frac{R_*^3}{(4 - 3\gamma_g) GM_\mathrm{tot} } \biggr] \omega^2 \, .</math>
Take Another Approach Using Logarithmic Derivatives
Change Independent Variable
Returning to the LAWE for n = 3 polytropes, as given, above, and repeated here,
| LAWE for <math>~n=5</math> Polytropes | |||
|---|---|---|---|
|
let's make the substitution,
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<math>~u \equiv (3 + \xi^2)^{1/2}</math> |
<math>~\Rightarrow</math> |
<math>~\xi^2 = u^2-3 \, .</math> |
We must therefore also make the operator substitution,
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<math>~\frac{d}{d\xi}</math> |
<math>~=</math> |
<math>~\frac{du}{d\xi} \cdot \frac{d}{du}</math> |
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|
<math>~=</math> |
<math>~\biggl[ \xi (3+\xi^2)^{-1/2} \biggr] \frac{d}{du} = \biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2} \frac{d}{du}</math> |
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<math>~\Rightarrow~~~ \frac{1}{\xi} \cdot \frac{dx}{d\xi}</math> |
<math>~=</math> |
<math>~\frac{1}{u}\cdot \frac{dx}{du} \, ;</math> |
and,
|
<math>~\frac{d^2}{d\xi^2}</math> |
<math>~=</math> |
<math>~\biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2} \frac{d}{du} \biggl\{ \biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2} \frac{d}{du} \biggr\}</math> |
|
|
<math>~=</math> |
<math>~\biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2} \biggl\{ \frac{3}{u^3} \biggl[ 1 - \frac{3}{u^2} \biggr]^{-1/2} \frac{d}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr]^{1/2} \frac{d^2}{du^2}\biggr\}</math> |
|
|
<math>~=</math> |
<math>~ \frac{3}{u^3} \frac{d}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr] \frac{d^2}{du^2}</math> |
|
<math>~\Rightarrow ~~~ \frac{d^2x}{d\xi^2}</math> |
<math>~=</math> |
<math>~ \frac{3}{u^3} \frac{dx}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr] \frac{d^2x}{du^2} \, .</math> |
The rewritten LAWE is therefore,
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<math>~0</math> |
<math>~=</math> |
<math>~u^2 \biggl\{ \frac{3}{u^3} \frac{dx}{du} + \biggl[ 1 - \frac{3}{u^2} \biggr] \frac{d^2x}{du^2} \biggr\} + 2\biggl[9 - u^2 \biggr] \frac{1}{u} \cdot \frac{dx}{du} + \biggl[\Omega^2 u^3 - 6\alpha \biggr] x </math> |
|
|
<math>~=</math> |
<math>~(u^2-3) \frac{d^2x}{du^2} + (21 - 2u^2 ) \frac{1}{u} \cdot \frac{dx}{du} + (\Omega^2 u^3 - 6\alpha ) x \, ,</math> |
where we have adopted the shorthand notation,
<math>~\Omega^2 \equiv \frac{\sigma_c^2}{3^{1/2} \gamma_g } \, .</math>
Look at Logarithmic Derivative
Multiplying through by <math>~(u^2/x)</math> gives,
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<math>~0</math> |
<math>~=</math> |
<math>~(u^2-3) \frac{u^2}{x} \cdot \frac{d^2x}{du^2} + (21 - 2u^2 ) \frac{d\ln x}{d\ln u} + (\Omega^2 u^5 - 6\alpha u^2 ) \, .</math> |
Now, in the context of a separate derivation, we showed that, quite generally we can make the substitution,
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<math>~\frac{u^2}{x} \cdot \frac{d^2x}{du^2} </math> |
<math>~=</math> |
<math>~ \frac{d}{d\ln u} \biggl[ \frac{d\ln x}{d\ln u} \biggr] + \biggl[ \frac{d\ln x}{d\ln u}-1 \biggr]\cdot \frac{d\ln x}{d\ln u} \, . </math> |
Hence, if we assume that the displacement function can be expressed as a power-law in <math>~u</math>, such that,
<math>\frac{d\ln x}{d\ln u} = c_0 \, ,</math>
then the LAWE for <math>~n=5</math> polytropes simplifies as follows,
|
<math>~0</math> |
<math>~=</math> |
<math>~(u^2-3) c_0(c_0-1) + (21 - 2u^2 ) c_0 + (\Omega^2 u^5 - 6\alpha u^2 ) \, .</math> |
This polynomial equation will be satisfied if, simultaneously, we set:
- <math>\Omega^2 = 0 \, ;</math>
- <math>c_0^2 -3c_0 -6\alpha = 0 </math> <math>~\Rightarrow</math> <math>c_0 = \frac{3}{2}\biggl[1 \pm \biggl(1+\frac{8\alpha}{3} \biggr)^{1/2} \biggl]\, ;</math>
- <math>~\alpha = 20/3 \, .</math>
This gives us some hope that a more general solution of the following form will work:
|
<math>~x</math> |
<math>~=</math> |
<math>~u^{c_0} \biggl[ a + bu + cu^2 + du^3 + \cdots\biggr] \, .</math> |
This means that, for example,
|
<math>~\frac{dx}{du}</math> |
<math>~=</math> |
<math>~ c_0 u^{c_0-1} \biggl[ a + bu + cu^2 + du^3 \biggr] + u^{c_0} \biggl[ b + 2cu + 3du^2 \biggr] </math> |
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<math>~\Rightarrow ~~~\frac{d\ln x}{d\ln u}</math> |
<math>~=</math> |
<math>~ \frac{c_0(a + bu + cu^2 + du^3) + bu + 2cu^2 + 3du^3}{a + bu + cu^2 + du^3} </math> |
and,
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<math>~\frac{d^2x}{du^2}</math> |
<math>~=</math> |
<math>~ c_0(c_0-1) u^{c_0-2} \biggl[ a + bu + cu^2 + du^3 \biggr] + 2c_0 u^{c_0-1} \biggl[ b + 2cu + 3du^2 \biggr] + u^{c_0} \biggl[ 2c + 6du \biggr] </math> |
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<math>~\Rightarrow~~~ \frac{u^2}{x} \cdot \frac{d^2x}{du^2}</math> |
<math>~=</math> |
<math>~ \frac{c_0(c_0-1) ( a + bu + cu^2 + du^3 ) + 2c_0 ( bu + 2cu^2 + 3du^3 ) + ( 2cu^2 + 6du^3 ) }{ a + bu + cu^2 + du^3} </math> |
So the LAWE becomes,
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<math>~- (\Omega^2 u^5 - 6\alpha u^2 ) (a + bu + cu^2 + du^3)</math> |
<math>~=</math> |
<math>~(u^2-3) [c_0(c_0-1) ( a + bu + cu^2 + du^3 ) + 2c_0 ( bu + 2cu^2 + 3du^3 ) + ( 2cu^2 + 6du^3 )] + (21 - 2u^2 ) [c_0(a + bu + cu^2 + du^3) + bu + 2cu^2 + 3du^3] \,. </math> |
This is cute, but I don't see any way that this approach will provide an avenue to cancel the <math>~\Omega^2 u^5</math> term.
Yet Another Guess
Let's try,
|
<math>~x</math> |
<math>~=</math> |
<math>~e^{a + b\ln\xi + c(\ln\xi)^2} \, ,</math> |
and examine the specific case of <math>~\sigma_c^2 = 0</math>, and, <math>~\gamma = (n+1)/n = 6/5 ~~\Rightarrow~~ \alpha = (3-20/6) = -1/3</math>. Under these conditions, the LAWE for <math>~n=5</math> polytropes becomes,
|
<math>~0</math> |
<math>~=</math> |
<math>~(3+\xi^2) \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi} + 2x </math> |
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|
<math>~=</math> |
<math>~(3+\xi^2) \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2} + \biggl[12 - 2\xi^2 \biggr] \frac{\xi}{x} \cdot \frac{dx}{d\xi} + 2\xi^2 \, . </math> |
And the derivatives give,
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<math>~\frac{dx}{d\xi}</math> |
<math>~=</math> |
<math>~x \frac{d}{d\xi}\biggl[ a + b\ln\xi + c(\ln\xi)^2 \biggr]</math> |
|
|
<math>~=</math> |
<math>~x \biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr]</math> |
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<math>~\Rightarrow ~~~ \frac{\xi}{x} \cdot \frac{dx}{d\xi}</math> |
<math>~=</math> |
<math>~b + 2c\ln\xi \, ;</math> |
and,
|
<math>~\frac{d^2x}{d\xi^2}</math> |
<math>~=</math> |
<math>~ \biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr] \frac{dx}{d\xi} + x \frac{d}{d\xi}\biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr]</math> |
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<math>~\Rightarrow ~~~ \frac{\xi^2}{x} \cdot \frac{d^2x}{d\xi^2}</math> |
<math>~=</math> |
<math>~ \xi \biggl[ \frac{b}{\xi}+ \frac{2c\ln\xi}{\xi} \biggr] \frac{\xi}{x} \cdot \frac{dx}{d\xi} + \xi^2 \frac{d}{d\xi}\biggl[ \frac{b+ 2c\ln\xi}{\xi} \biggr]</math> |
|
|
<math>~=</math> |
<math>~ \biggl[ b + 2c\ln\xi \biggr]^2 + \xi \frac{d}{d\xi}\biggl[ b+ 2c\ln\xi \biggr] + (b+ 2c\ln\xi) \xi^2 \biggl[ - \frac{1}{\xi^2} \biggr] </math> |
|
|
<math>~=</math> |
<math>~ (b + 2c\ln\xi )^2 + 2c- (b+ 2c\ln\xi) </math> |
|
|
<math>~=</math> |
<math>~[ b^2 + 2c - b] + [4bc - 2c] \ln\xi+4c^2 (\ln\xi)^2 \, . </math> |
Hence the "fundamental mode" LAWE becomes,
|
<math>~0</math> |
<math>~=</math> |
<math>~(3+\xi^2) \biggl[ ( b^2 + 2c - b ) + (4bc - 2c) \ln\xi+4c^2 (\ln\xi)^2 \biggr] + (12 - 2\xi^2 ) \biggl[ b + 2c\ln\xi \biggr] \, . + 2\xi^2 </math> |
Now, this expression cannot be satisfied for arbitrary <math>~\xi</math>. But, here we seek a solution only at the surface for the specific model, <math>~\xi = 3</math>. Plugging this value into the expression gives,
|
<math>~0</math> |
<math>~=</math> |
<math>~12 \biggl[ ( b^2 + 2c - b ) + (4bc - 2c) \ln 3+4c^2 (\ln 3)^2 \biggr] + (12 - 18 ) \biggl[ b + 2c\ln 3 \biggr] + 18 </math> |
|
|
<math>~=</math> |
<math>~2 \biggl[ ( b^2 + 2c - b ) + (4bc - 2c) \ln 3+4c^2 (\ln 3)^2 \biggr] -\biggl[ b + 2c\ln 3 \biggr] + 3 \, . </math> |
It appears as though one perfectly satisfactory solution is, <math>~c = 0</math>, in which case, we need,
|
<math>~0</math> |
<math>~=</math> |
<math>~2 b^2 - 3b + 3 </math> |
|
<math>~\Rightarrow~~~b</math> |
<math>~=</math> |
<math>~ \frac{3}{4}\biggl[1 \pm \sqrt{1-\frac{8}{3} } \biggr] \, . </math> |
Thus, <math>~b</math> is an complex number.
Related Discussions
- Radial Oscillations of Uniform-density sphere
- Radial Oscillations of Isolated Polytropes
- Setup
- n = 1: Attempt at Formulating an Analytic Solution
- n = 3: Numerical Solution to compare with M. Schwarzschild (1941)
- n = 5: Attempt at Formulating an Analytic Solution
- In an accompanying Chapter within our "Ramblings" Appendix, we have played with the adiabatic wave equation for polytropes, examining its form when the primary perturbation variable is an enthalpy-like quantity, rather than the radial displacement of a spherical mass shell. This was done in an effort to mimic the approach that has been taken in studies of the stability of Papaloizou-Pringle tori.
- <math>~n=3</math> … M. Schwarzschild (1941, ApJ, 94, 245), Overtone Pulsations of the Standard Model: This work is referenced in §38.3 of [KW94]. It contains an analysis of the radial modes of oscillation of <math>~n=3</math> polytropes, assuming various values of the adiabatic exponent.
- <math>~n=2</math> … C. Prasad & H. S. Gurm (1961, MNRAS, 122, 409), Radial Pulsations of the Polytrope, n = 2
- <math>~n=\tfrac{3}{2}</math> … D. Lucas (1953, Bul. Soc. Roy. Sci. Liege, 25, 585) … Citation obtained from the Prasad & Gurm (1961) article.
- <math>~n=1</math> … L. D. Chatterji (1951, Proc. Nat. Inst. Sci. [India], 17, 467) … Citation obtained from the Prasad & Gurm (1961) article.
- Composite Polytropes … M. Singh (1968, MNRAS, 140, 235-240), Effect of Central Condensation on the Pulsation Characteristics
- Summary of Known Analytic Solutions … R. Stothers (1981, MNRAS, 197, 351-361), Analytic Solutions of the Radial Pulsation Equation for Rotating and Magnetic Star Models
- Interesting Composite! … C. Prasad (1948, MNRAS, 108, 414-416), Radial Oscillations of a Particular Stellar Model
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© 2014 - 2021 by Joel E. Tohline |