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===Configuration Pairing===
===Configuration Pairing===


Now, let's identify two equilibrium states that sit very near the <math>~P_\mathrm{max}</math> turning point on the two separate branches of the equilibrium sequence and that have identical external pressures.
Now, let's identify two <math>~n=5</math> equilibrium states that sit very near the <math>~P_\mathrm{max}</math> turning point on the two separate branches of the equilibrium sequence and that have identical external pressures. We know [[User:Tohline/SSC/FreeEnergy/PowerPoint#Case_M_Equilibrium_Conditions|from separate discussions]] that, in both cases,


<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>
~\frac{P_\mathrm{e}}{P_\mathrm{norm}}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2\cdot 3^3}{\pi}\biggr]^{3}
\tilde\xi^{12} \tilde\theta_n^{6}( - \tilde\theta' )^{6}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~ \biggl[ \frac{\pi}{2\cdot 3^3}\biggr]^{1/2}\biggl[\frac{P_\mathrm{e}}{P_\mathrm{norm}}\biggr]^{1/6}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\tilde\xi^{2} \tilde\theta_n( - \tilde\theta' )
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
3\ell^2 (1+\ell^2)^{-1/2} \frac{\ell}{\sqrt{3}} (1+\ell^2)^{-3/2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\sqrt{3}\ell^3  (1+\ell^2)^{-2}
</math>
  </td>
</tr>
</table>
</div>
We can therefore write,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~(1+\ell^2)^{2}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
p_0\ell^3 
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~\ell^4 - p_0\ell^3 + 2\ell^2 + 1</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
0 \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>p_0 \equiv \biggl[ \frac{\pi}{2\cdot 3^2}\biggr]^{-1/2}\biggl[\frac{P_\mathrm{e}}{P_\mathrm{norm}}\biggr]^{-1/6}</math>
</div>
So, in essence, we seek two real roots of this quartic equation that are near <math>~P_\mathrm{max}</math>, that is, that are near <math>~\ell = \sqrt{3}</math>.




{{LSU_HBook_footer}}
{{LSU_HBook_footer}}

Revision as of 17:15, 25 August 2016


Radial Oscillations in Pressure-Truncated n = 5 Polytropes

[Comment by Joel Tohline on 24 August 2016]  Over the past few weeks, I have been putting together a powerpoint presentation that summarizes what I've learned, especially over the last several years, about turning points — and their relative positioning with respect to points of dynamical instability — along equilibrium sequences. One key finding, which is illustrated in Figure 3 of that discussion, is that the transition from stable to unstable systems along the n = 5 sequence occurs after, rather than at, the pressure maximum of the sequence. This means that, in the immediate vicinity of the pressure maximum, two stable equilibrium configurations exist with the same <math>~(K, M_\mathrm{tot}, P_e) </math> but different radii. Perhaps this means that, in the absence of dissipation, and without the need for a driving mechanism, a permanent oscillation between these two states can be activated.

Upon further thought, it occurred to me that a careful examination of the internal structure of both models — especially relative to one another — might reveal what the eigenvector of that (nonlinear) oscillation might be.


Whitworth's (1981) Isothermal Free-Energy Surface
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Review of Internal Structure

Run of Mass

According to Chandrasekhar (Chapter IV, equation 67, p.97), the mass interior to <math>~\xi</math> is,

<math>~M(\xi)</math>

<math>~=</math>

<math>~4\pi a_n^3 \rho_c (-\xi^2 \theta^') \, .</math>

For a pressure-truncated polytrope, the total mass is,

<math>~M_\mathrm{tot}</math>

<math>~=</math>

<math>~4\pi a_n^3 \rho_c (-\tilde\xi^2 \tilde\theta^') \, ,</math>

which means that, as a function of <math>~\xi</math> in a pressure-truncated polytrope, the relative mass is,

<math>~m_\xi \equiv \frac{M(\xi)}{M_\mathrm{tot}}</math>

<math>~=</math>

<math>~\biggl[4\pi a_n^3 \rho_c (-\xi^2 \theta^') \biggr] \biggl[ 4\pi a_n^3 \rho_c (-\tilde\xi^2 \tilde\theta^') \biggr]^{-1}</math>

 

<math>~=</math>

<math>~\frac{(-\xi^2 \theta^')}{(-\tilde\xi^2 \tilde\theta^')} \, .</math>

Thus, for an <math>~n = 5</math> system we have,

<math>~m_\xi</math>

<math>~=</math>

<math>~ \biggl(\frac{\xi}{\tilde\xi}\biggr)^2 \biggl[ \frac{\xi }{ 3}\biggl(1 + \frac{\xi^2}{3}\biggr)^{-3/2} \biggr] \biggl[ \frac{\tilde\xi }{ 3}\biggl(1 + \frac{\tilde\xi^2}{3}\biggr)^{-3/2} \biggr]^{-1}</math>

 

<math>~=</math>

<math>~ \biggl(\frac{\xi}{\tilde\xi}\biggr)^3 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3/2} \biggl(1 + \frac{\tilde\xi^2}{3}\biggr)^{3/2} \, ;</math>

and, for the configuration at the pressure maximum <math>~(\tilde\xi = 3)</math>, in particular, we have,

<math>~m_0</math>

<math>~=</math>

<math>~\biggl(\frac{2\xi}{3}\biggr)^3 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3/2} \, .</math>

Corresponding Lagrangian Radial Coordinate

For any pressure-truncated polytrope, the fractional radial-coordinate running through the equilibrium configuration is,

<math>~\frac{r(\xi)}{R_\mathrm{eq}}</math>

<math>~=</math>

<math>~\frac{\xi}{\tilde\xi}</math>

<math>~\Rightarrow~~~ r(\xi)</math>

<math>~=</math>

<math>~\biggl(\frac{\xi}{\tilde\xi}\biggr) R_\mathrm{eq}</math>

<math>~\Rightarrow~~~r_\xi \equiv \frac{r(\xi)}{R_\mathrm{norm}}</math>

<math>~=</math>

<math>~\biggl(\frac{\xi}{\tilde\xi}\biggr) \biggl[ \frac{4\pi}{(n+1)^n}\biggr]^{1/(n-3)} \tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{(1-n)/(n-3)} \, . </math>

For <math>~n=5</math> configurations, this means,

<math>~r_\xi</math>

<math>~=</math>

<math>~\biggl(\frac{\xi}{\tilde\xi}\biggr) \biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2} \tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{-2} </math>

 

<math>~=</math>

<math>~\xi \biggl\{ \biggl[ \frac{4\pi}{2^5\cdot 3^5}\biggr]^{1/2} \tilde\xi^{-4} \biggl[ \frac{\tilde\xi}{3} \biggl( 1+\frac{\tilde\xi^2}{3} \biggr)^{-3/2}\biggr]^{-2} \biggr\} </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>

and, for the configuration at the pressure maximum <math>~(\tilde\xi = 3)</math>, in particular, this gives,

<math>~r_0</math>

<math>~=</math>

<math>~\xi \biggl\{ \biggl[ \frac{4\pi}{2^5\cdot 3}\biggr]^{1/2} \biggl(\frac{2}{3}\biggr)^6\biggr\} </math>

 

<math>~=</math>

<math>~\xi \biggl[ \frac{2^9 \pi}{3^{13}}\biggr]^{1/2} </math>

Exploration

n = 5 Mass-Radius Relation

So, for any <math>~\tilde\xi</math> configuration, 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</math>

<math>~=</math>

<math>~ \biggl(\frac{\xi}{\tilde\xi}\biggr)^3 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3/2} \biggl(1 + \frac{\tilde\xi^2}{3}\biggr)^{3/2} \, ,</math>

<math>~r_\xi</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>

And this can be inverted analytically in the case of <math>~\tilde\xi = 3</math>. Specifically,

<math>~m_0</math>

<math>~=</math>

<math>~\biggl(\frac{2\xi}{3}\biggr)^3 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3/2} </math>

<math>~\Rightarrow~~~ m_0^{2/3}</math>

<math>~=</math>

<math>~\biggl(\frac{2^2\xi^2}{3^2}\biggr) \biggl(1 + \frac{\xi^2}{3}\biggr)^{-1} </math>

<math>~\Rightarrow~~~ 2^2\xi^2 </math>

<math>~=</math>

<math>~ 3^2\biggl(1 + \frac{\xi^2}{3}\biggr) m_0^{2/3} </math>

<math>~\Rightarrow~~~ \xi^2 (2^2 - 3 m_0^{2/3}) </math>

<math>~=</math>

<math>~ 3^2m_0^{2/3} </math>

<math>~\Rightarrow~~~ \xi^2 </math>

<math>~=</math>

<math>~ \frac{3^2m_0^{2/3}}{(2^2 - 3 m_0^{2/3})} \, . </math>

Hence, the radius-mass relationship in the configuration at the <math>~P_\mathrm{max}</math> turning point is,

<math>~ r_0 (m_0) </math>

<math>~=</math>

<math>~\biggl[ \frac{2^9 \pi}{3^{13}}\biggr]^{1/2} \biggl[\frac{3^2m_0^{2/3}}{2^2 - 3 m_0^{2/3}}\biggr]^{1/2} \, . </math>

Actually, the inversion can be performed analytically for any choice of <math>~\tilde\xi</math> to obtain,

<math>~ r_\xi (m_\xi) </math>

<math>~=</math>

<math>~\biggl[ \frac{2^9 \pi}{3^{13}}\biggr]^{1/2} \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) \, . </math>

Configuration Pairing

Now, let's identify two <math>~n=5</math> equilibrium states that sit very near the <math>~P_\mathrm{max}</math> turning point on the two separate branches of the equilibrium sequence and that have identical external pressures. We know from separate discussions that, in both cases,

<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>

 

<math>~=~</math>

<math>~\biggl[ \frac{2\cdot 3^3}{\pi}\biggr]^{3} \tilde\xi^{12} \tilde\theta_n^{6}( - \tilde\theta' )^{6} </math>

<math>~\Rightarrow~~~ \biggl[ \frac{\pi}{2\cdot 3^3}\biggr]^{1/2}\biggl[\frac{P_\mathrm{e}}{P_\mathrm{norm}}\biggr]^{1/6}</math>

<math>~=~</math>

<math>~ \tilde\xi^{2} \tilde\theta_n( - \tilde\theta' ) </math>

 

<math>~=~</math>

<math>~ 3\ell^2 (1+\ell^2)^{-1/2} \frac{\ell}{\sqrt{3}} (1+\ell^2)^{-3/2} </math>

 

<math>~=~</math>

<math>~ \sqrt{3}\ell^3 (1+\ell^2)^{-2} </math>

We can therefore write,

<math>~(1+\ell^2)^{2}</math>

<math>~=~</math>

<math>~ p_0\ell^3 </math>

<math>~\Rightarrow~~~\ell^4 - p_0\ell^3 + 2\ell^2 + 1</math>

<math>~=~</math>

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

where,

<math>p_0 \equiv \biggl[ \frac{\pi}{2\cdot 3^2}\biggr]^{-1/2}\biggl[\frac{P_\mathrm{e}}{P_\mathrm{norm}}\biggr]^{-1/6}</math>

So, in essence, we seek two real roots of this quartic equation that are near <math>~P_\mathrm{max}</math>, that is, that are near <math>~\ell = \sqrt{3}</math>.


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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