User:Tohline/SSC/Structure/BiPolytropes/Analytic0 0
BiPolytrope with <math>n_c = 0</math> and <math>n_e=0</math>
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Here we construct a bipolytrope in which both the core and the envelope have uniform densities, that is, the structure of both the core and the envelope will be modeled using an <math>n = 0</math> polytropic index. It should be possible for the entire structure to be described by closed-form, analytic expressions. Generally, we will follow the general solution steps for constructing a bipolytrope that we have outlined elsewhere. [On 1 February 2014, J. E. Tohline wrote: This particular system became of interest to me during discussions with Kundan Kadam about the relative stability of bipolytropes.]
Step 4: Throughout the core (<math>0 \le \chi \le \chi_i</math>)
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Specify: <math>~P_0</math> and <math>\rho_0 ~\Rightarrow</math> |
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<math>~\rho</math> |
<math>~=</math> |
<math>~\rho_0</math> |
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<math>~P</math> |
<math>~=</math> |
<math>P_0 - \frac{2}{3} \pi G \rho_0^2 r^2</math> |
<math>~=</math> |
<math>P_0 \biggl( 1 - \frac{2\pi}{3}\chi^2 \biggr)</math> |
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<math>~r</math> |
<math>~=</math> |
<math>\biggl[ \frac{P_0}{G \rho_0^2} \biggr]^{1/2} \chi</math> |
<math>~=</math> |
<math>\biggl[ \frac{P_0}{G \rho_0^2} \biggr]^{1/2} \chi</math> |
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<math>~M_r</math> |
<math>~=</math> |
<math>\frac{4\pi}{3} \rho_0 r^3</math> |
<math>~=</math> |
<math>\frac{4\pi}{3} \rho_0 \biggl[ \frac{P_0}{G \rho_0^2} \biggr]^{3/2} \chi^3 = \frac{4\pi}{3} \biggl[ \frac{P_0^3}{G^3 \rho_0^4} \biggr]^{1/2} \chi^3</math> |
Step 5: Interface Conditions
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Specify: <math>~\chi_i</math> and <math>~\rho_e/\rho_0</math>, and demand … |
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<math>~P_{ei}</math> |
<math>~=</math> |
<math>~P_{ci}</math> |
<math>~=</math> |
<math>P_0 \biggl( 1 - \frac{2\pi}{3}\chi_i^2 \biggr)</math> |
Step 6: Envelope Solution (<math>~\chi > \chi_i</math>)
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<math>~\rho</math> |
<math>~=</math> |
<math>~\rho_e</math> |
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<math>~P</math> |
<math>~=</math> |
<math>P_{ei} + \biggl(\frac{2}{3} \pi G \rho_e\biggr) \biggl[ 2(\rho_0 - \rho_e) r_i^3\biggl( \frac{1}{r} - \frac{1}{r_i}\biggr) - \rho_e(r^2 - r_i^2) \biggr]</math> |
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<math>~=</math> |
<math>P_{ei} + \frac{2\pi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) P_0 \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \chi_i^3\biggl( \frac{1}{\chi} - \frac{1}{\chi_i}\biggr) - \frac{\rho_e}{\rho_0} (\chi^2 - \chi_i^2) \biggr]</math> |
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<math>~\frac{P}{P_0}</math> |
<math>~=</math> |
<math>1 - \frac{2\pi}{3}\chi_i^2 + \frac{2\pi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \chi_i^2 \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( \frac{1}{\xi} - 1\biggr) - \frac{\rho_e}{\rho_0} (\xi^2 - 1) \biggr]</math> |
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<math>~M_r</math> |
<math>~=</math> |
<math>\frac{4\pi}{3} \biggl[ \rho_0 r_i^3 + \rho_e(r^3 - r_i^3) \biggr]</math> |
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<math>~=</math> |
<math>\frac{4\pi}{3} \biggl[ \frac{P_0^3}{G^3 \rho_0^4} \biggr]^{1/2} \biggl[\chi_i^3 +\frac{\rho_e}{\rho_0} \biggl( \chi^3 - \chi_i^3 \biggr) \biggr]</math> |
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<math>~=</math> |
<math>\frac{4\pi}{3} \biggl[ \frac{P_0^3}{G^3 \rho_0^4} \biggr]^{1/2} \chi_i^3\biggl[1 +\frac{\rho_e}{\rho_0} \biggl( \xi^3 - 1\biggr) \biggr]</math> |
Step 7: Surface Boundary Condition
At the surface (that is, at <math>r = R</math> and <math>M_r = M_\mathrm{tot}</math>), <math>P/P_0 = 0</math> and <math>\xi = \xi_s = R/r_i = 1/q</math>. Also, we can write,
<math> \chi_i = q\biggl[ \frac{G\rho_0^2 R^2}{P_0} \biggr]^{1/2} \, ; </math>
and, from earlier derivations,
<math> R^3 = \frac{3M_\mathrm{tot}}{4\pi \bar\rho} = \frac{3M_\mathrm{tot}}{4\pi \rho_0} \biggl( \frac{\nu}{q^3} \biggr) \, ; </math>
<math> \frac{\rho_e}{\rho_0} = \frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3}\biggr) \, . </math>
Therefore, setting the pressure to zero at the surface means,
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<math>~\frac{3}{2\pi}\chi_i^{-2}</math> |
<math>~=</math> |
<math>1 - \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( q - 1\biggr) - \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr]</math> |
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<math>\Rightarrow ~~~~~\biggl( \frac{3}{2\pi} \biggr) q^{-2} \biggl[ \frac{P_0}{G\rho_0^2 R^2} \biggr]</math> |
<math>~=</math> |
<math>1 + \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1- q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] </math> |
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<math>\Rightarrow ~~~~~\biggl( \frac{3}{2\pi} \biggr) \biggl( \frac{4\pi}{3} \biggr)^{2/3} \nu^{-2/3} \biggl[ \frac{P_0^3}{G^3\rho_0^4 M_\mathrm{tot}^2} \biggr]^{1/3}</math> |
<math>~=</math> |
<math>1 + \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1- q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] </math> |
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<math>\Rightarrow ~~~~~\biggl\{ \biggl( \frac{6}{\pi} \biggr) \frac{1}{\nu^{2} } \biggl[ \frac{P_0^3}{G^3\rho_0^4 M_\mathrm{tot}^2} \biggr] \biggr\}^{1/3}</math> |
<math>~=</math> |
<math>1 + \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1- q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] </math> |
It therefore seems prudent to define a function,
<math> g(\nu,q) \equiv \biggl\{ 1 + \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] \biggr\}^{1/2}\, , </math>
in which case the expressions for the equilibrium radius and equilibrium total mass are, respectively,
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<math> \biggl[ \frac{G\rho_0^2}{P_0} \biggr]^{1/2} R</math> |
<math>~=</math> |
<math>\biggl( \frac{3}{2\pi} \biggr)^{1/2} \frac{1}{q g} \, ;</math> |
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<math> \biggl[ \frac{G^3\rho_0^4}{P_0^3} \biggr]^{1/2} M_\mathrm{tot}</math> |
<math>~=</math> |
<math>\biggl( \frac{6}{\pi} \biggr)^{1/2} \frac{1}{\nu g^3} \, .</math> |
Related Discussions
- Analytic solution with <math>n_c = 5</math> and <math>n_e=1</math>.
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© 2014 - 2021 by Joel E. Tohline |