SSC/Stability/Yabushita75: Difference between revisions

From jetwiki
Jump to navigation Jump to search
Line 309: Line 309:
   <td align="center"><math>=</math></td>
   <td align="center"><math>=</math></td>
   <td align="left">
   <td align="left">
<math>\biggl(\frac{2}{5}\biggr)\rho_0^{- 1 / 2}
<math>\biggl(\frac{2}{5}\biggr)^{3 / 2}\rho_0^{- 1 / 2}
\biggl[ \frac{c_s^2}{K_e} \biggr]^{3/2}  
\biggl[ \frac{c_s^2}{K_e} \biggr]^{3/2}  
\biggl[\rho_e\biggr]^{- 1 / 2}
\biggl[\rho_0 \biggl(\frac{\mu_e}{\mu_c}\biggr)e^{-\psi_i} \theta_i^{-3 / 2}\biggr]^{- 1 / 2}
\biggl( \xi^2 \frac{d\psi}{d\xi} \biggr)_i
\biggl( \xi^2 \frac{d\psi}{d\xi} \biggr)_i
</math>
</math>
   </td>
   </td>
   <td align="right" width="5%">&nbsp;</td>
   <td align="right" width="5%">&nbsp;</td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\biggl(\frac{2}{5}\biggr)^{3 / 2}\rho_0^{- 1 }
\biggl[ \biggl(\frac{\mu_e}{\mu_c}\biggr)^{5 / 2} \rho_0e^{-\psi_i} \biggr]
\biggl[\biggl(\frac{\mu_e}{\mu_c}\biggr)e^{-\psi_i} \theta_i^{-3 / 2}\biggr]^{- 1 / 2}
\biggl( \xi^2 \frac{d\psi}{d\xi} \biggr)_i
</math>
  </td>
  <td align="right" width="5%">&nbsp;</td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\biggl(\frac{2}{5}\biggr)^{3 / 2}
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{2} e^{-\psi_i / 2} \theta_i^{3 / 4}
\biggl( \xi^2 \frac{d\psi}{d\xi} \biggr)_i
</math>
  </td>
  <td align="right" width="5%">&nbsp;</td>
</tr>
<tr>
  <td align="right">
<math>\Rightarrow ~~~
\biggl(-\frac{d\theta}{d\eta} \biggr)_i
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\biggl(\frac{2}{5}\biggr)^{3 / 2}
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{2} e^{-\psi_i / 2} \theta_i^{3 / 4}
\biggl( \frac{d\psi}{d\xi} \biggr)_i
\biggl[ \biggl(\frac{2}{5}\biggr)^{1 / 2}
\biggl(\frac{\mu_e}{\mu_c}\biggr) e^{-\psi_i/ 2} 
\theta_i^{-1 / 4}
\biggr]^{-2}
</math>
  </td>
  <td align="right" width="5%">&nbsp;</td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\biggl(\frac{2}{5}\biggr)^{1 / 2}
e^{+ \psi_i / 2} \theta_i^{5 / 4}
\biggl( \frac{d\psi}{d\xi} \biggr)_i
\, .
</math>
  </td>
  <td align="right" width="5%">&nbsp;</td>
</tr>
</table>
<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="3">Summary Interface Relations</td>
</tr>
<tr>
  <td align="center" colspan="1">Our Derivations</td>
  <td align="center" colspan="1">After setting <math>(\mu_e/\mu_c) = 1</math> and <math>\theta_i = 1</math></td>
  <td align="center" colspan="1">Presented by {{ Yabushita75 }}</td>
</tr>
<tr>
  <td align="center" colspan="1">
<table border="0" align="center" cellpadding="8">
<tr>
  <td align="right"><math>\frac{\rho_e}{\rho_0}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{\mu_e}{\mu_c}\biggr)e^{-\psi_i} \theta_i^{-3 / 2}
\, ;
</math>
  </td>
  <td align="right" width="5%">(2.9)</td>
</tr>
<tr>
  <td align="right"><math>\frac{c_s^2}{K_e}
</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[
\biggl(\frac{\mu_e}{\mu_c}\biggr)^{5 / 2} \rho_0e^{-\psi_i}
\biggr]^{2 / 3}
\, ;
</math>
  </td>
  <td align="right" width="5%">(2.11)</td>
</tr>
<tr>
  <td align="right"><math>\frac{\eta_i}{\xi_i}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{2}{5}\biggr)^{1 / 2}
\biggl(\frac{\mu_e}{\mu_c}\biggr) e^{-\psi_i/ 2} 
\theta_i^{-1 / 4}
\, .
</math>
  </td>
  <td align="right" width="5%">(2.12)</td>
</tr>
<tr>
  <td align="right">
<math>
\biggl(-\frac{d\theta}{d\eta} \biggr)_i
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\biggl(\frac{2}{5}\biggr)^{1 / 2}
e^{+ \psi_i / 2} \theta_i^{5 / 4}
\biggl( \frac{d\psi}{d\xi} \biggr)_i
\, .
</math>
  </td>
  <td align="right" width="5%">&nbsp;</td>
</tr>
</table>
  </td>
  <td align="center" colspan="1">
Two
  </td>
  <td align="center" colspan="1">
Three
  </td>
</tr>
</tr>
</table>
</table>

Revision as of 18:25, 8 November 2023


Stability of a BiPolytrope with an Isothermal Core

This analysis pulls largely from 📚 S. Yabushita (1975, MNRAS, Vol. 172, pp. 441 - 453); the focus is on bipolytropes having (nc,ne)=(,32). In an accompanying discussion, we summarize the steps that Yabushita took — from 1968 and 1974, to 1975 — that led up to his discovery of an analytic description of the isothermal displacement function.

Equilibrium Structure

We will follow the accompanying formal recipe for building a bipolytropic model, using the step-by-step construction of Milne's (1930) configurations as a guide.

Step 1

The 📚 Yabushita (1975) bipolytrope has an isothermal core (nc=) and an ne=32 polytropic envelope.

Steps 2 & 3

Throughout the core, the properties of this bipolytrope can be expressed in terms of the Lane-Emden function, ψ(χ), which derives from a solution of the 2nd-order ODE,

Isothermal Lane-Emden Equation

1ξ2ddξ(ξ2dψdξ)=eψ

subject to the boundary conditions,

ψ=1       and       dψdξ=0       at       ξ=0.

The solution, ψ(ξ), extends to infinity, so the interface between the core and the envelope can be positioned anywhere within the range, 0<ξi<.

Step4: Throughout the core

Specify: cs2 and ρ0

r

=

[cs24πGρ0]1/2ξ

(2.2)

ρ

=

ρ0eψ

(2.2)

P

=

cs2ρ0eψ

(2.1)

Mr

=

[cs64πG3ρ0]1/2(ξ2dψdξ)

(2.3)
📚 Yabushita (1975), § 2, pp. 442-443

After adopting the substitute notation, cs2K1 and ρ0λc, it is clear that these last four parameter-profile expressions are identical to the ones that appear, respectively, as equations (2.2), (2.1) and (2.3) of 📚 Yabushita (1975).

Step 5: Interface Conditions

(ρ0μc)eψi

=

(ρeμe)θine

=

(ρeμe)θi3/2

cs2ρ0eψi

=

Keρe1+1/neθine+1

=

Keρe5/3θi5/2

[cs24πGρ0]1/2ξi

=

[(ne+1)Ke4πG]1/2ρe(1ne)/(2ne)ηi

=

[(5/2)Ke4πG]1/2ρe(1/6)ηi

[cs64πG3ρ0]1/2(ξ2dψdξ)i

=

4π[(ne+1)Ke4πG]3/2ρe(3ne)/(2ne)(η2dθdη)i

=

4π[(5/2)Ke4πG]3/2ρe1/2(η2dθdη)i

This means that,

ρeρ0 =
Comment by J. E. Tohline: There is a type-setting mistake in Yabushita75's Eq. (2.9); ρe appears as ρ0.
Comment by J. E. Tohline: There is a type-setting mistake in Yabushita75's Eq. (2.9); ρe appears as ρ0.

(μeμc)eψiθi3/2;

(2.9)
Ke[ρ0(μeμc)eψiθi3/2]5/3θi5/2 =

cs2ρ0eψi

 
Keρ02/3cs2 =

(μeμc)5/3e+2ψi/3

 
cs2Ke =

[(μeμc)5/2ρ0eψi]2/3;

(2.11)
ηiξi =

[cs24πGρ0]1/2[4πG(5/2)Ke]1/2[ρe]1/6

 
  =

(25)1/2[cs2Ke]1/2[ρ0(μeμc)eψiθi3/2]1/6ρ01/2

 
  =

(25)1/2[(μeμc)5/2ρ0eψi]1/3[(μeμc)1/6eψi/6θi1/4]ρ01/3

 
  =

(25)1/2(μeμc)eψi/2θi1/4.

(2.12)

And, finally,

(η2dθdη)i

=

14π[cs64πG3ρ0]1/2[4πG(5/2)Ke]3/2[ρe]1/2(ξ2dψdξ)i

 

 

=

(25)3/2ρ01/2[cs2Ke]3/2[ρ0(μeμc)eψiθi3/2]1/2(ξ2dψdξ)i

 

 

=

(25)3/2ρ01[(μeμc)5/2ρ0eψi][(μeμc)eψiθi3/2]1/2(ξ2dψdξ)i

 

 

=

(25)3/2(μeμc)2eψi/2θi3/4(ξ2dψdξ)i

 

(dθdη)i

=

(25)3/2(μeμc)2eψi/2θi3/4(dψdξ)i[(25)1/2(μeμc)eψi/2θi1/4]2

 

 

=

(25)1/2e+ψi/2θi5/4(dψdξ)i.

 
Summary Interface Relations
Our Derivations After setting (μe/μc)=1 and θi=1 Presented by 📚 Yabushita (1975)
ρeρ0 =

(μeμc)eψiθi3/2;

(2.9)
cs2Ke =

[(μeμc)5/2ρ0eψi]2/3;

(2.11)
ηiξi =

(25)1/2(μeμc)eψi/2θi1/4.

(2.12)

(dθdη)i

=

(25)1/2e+ψi/2θi5/4(dψdξ)i.

 

Two

Three

See Also


Tiled Menu

Appendices: | VisTrailsEquations | VisTrailsVariables | References | Ramblings | VisTrailsImages | myphys.lsu | ADS |