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.

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>~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>~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>~\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>~\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>~\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

Setup

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> — where <math>~p_0 = (2^8/3^3)^{1/2}</math>.

Because we are hunting for equilibrium configurations near <math>~P_\mathrm{max}</math>, it makes sense to make the variable substitution,

<math>~\rightarrow</math>

<math>~\sqrt{3}(1+\epsilon) \, ,</math>

and look for pairs of values of <math>~|\epsilon| \ll 1</math> (both real, but one positive and the other negative).

<math>~0</math>	<math>~=~</math>	<math>~3^2(1+\epsilon)^4 - 3^{3/2}p_0(1+\epsilon)^3 + 6(1+\epsilon)^2 + 1</math>
	<math>~=~</math>	<math>~ 3^2\biggl[1 + 4\epsilon + 6\epsilon^2 + 4\epsilon^3 + \epsilon^4\biggr] - 3^{3/2}p_0\biggl[ 1+ 3\epsilon + 3\epsilon^2 + \epsilon^3 )\biggr] + 6\biggl[1 + 2\epsilon + \epsilon^2\biggr] + 1 </math>
	<math>~=~</math>	<math>~ \epsilon^4 \biggl[ 9 \biggr] +\epsilon^3 \biggl[ 36 - 3^{3/2}p_0 \biggr] +\epsilon^2 \biggl[ 54 - 3^{5/2}p_0 + 6 \biggr] +\epsilon \biggl[36 - 3^{5/2}p_0 + 12 \biggr] +1 \, . </math>

And, because we will only be examining values of the external pressure that are less than <math>~P_\mathrm{max}</math>, and we know that at the point of maximum pressure, <math>~3^{3/2}p_0 = 16</math>, it makes sense to make the substitution,

<math>~3^{3/2}p_0 ~~~\rightarrow ~~~ (16+\delta) \, .</math>

Hence, for a fixed choice of <math>~\delta </math> (reasonably small, and positive), we seek two real roots (one positive and the other negative) of the quartic relation,

<math>~0</math>	<math>~=~</math>	<math>~ 9\epsilon^4 +\epsilon^3 \biggl[ 36 - (16+\delta ) \biggr] +\epsilon^2 \biggl[ 60 - 3(16+\delta ) \biggr] +\epsilon \biggl[48 - 3(16+\delta ) \biggr] +1 </math>
	<math>~=~</math>	<math>~ 9\epsilon^4 +\epsilon^3 (20-\delta ) +\epsilon^2 ( 12 - 3\delta ) -\epsilon (3\delta ) +1 \, . </math>

What are the reasonable limits on <math>~\delta</math>? Well, first note that,

<math>~p_0</math>	<math>~=</math>	<math>~\frac{(1+\ell^2)^2}{\ell^3}</math>
<math>~\Rightarrow ~~~ 16+\delta </math>	<math>~=</math>	<math>~3^{3/2}\biggl[\frac{(1+\ell^2)^2}{\ell^3}\biggr]</math>
<math>~\Rightarrow ~~~ \delta </math>	<math>~=</math>	<math>~3^{3/2}\biggl[\frac{(1+\ell^2)^2}{\ell^3}\biggr] - 2^4 \, .</math>

Now, according to our accompanying discussion, the relevant limits on <math>~\ell</math> are <math>~\sqrt{3}</math> (set by the maximum pressure turning point) and 2.223175 (set by the transition to dynamical instability). The corresponding values of <math>~\delta </math> are: 0 (by design) and 0.69938.

Quartic Solution

Here, we will draw from the Wikipedia discussion of the quartic function. The generic form is,

Relating this to our specific quartic function, we should ultimately make the following assignments:

<math>~a</math>	<math>~=</math>	<math>~9</math>
<math>~b</math>	<math>~=</math>	<math>~20 - \delta</math>
<math>~c</math>	<math>~=</math>	<math>~12 - 3\delta = 3(4-\delta )</math>
<math>~d</math>	<math>~=</math>	<math>~-3 \delta</math>
<math>~e</math>	<math>~=</math>	<math>~1</math>

We need to evaluate the following expressions:

<math>~p</math>	<math>~\equiv</math>	<math>~\frac{8ac-3b^2}{3a^2}</math>
	<math>~=</math>	<math>~\frac{2^3 \cdot 3^3(4-\delta )-3(20-\delta )^2}{3^5}</math>
<math>~q</math>	<math>~\equiv</math>	<math>~\frac{b^3 - 4abc + 8a^2d}{8a^3}</math>
	<math>~=</math>	<math>~\frac{(20 - \delta )^3 - 2^2\cdot 3^3(4-\delta ) (20 - \delta ) - 2^3\cdot 3^5\delta }{2^3 \cdot 3^6}</math>
<math>~\Delta_0</math>	<math>~\equiv</math>	<math>~c^2 - 3bd + 12ae</math>
	<math>~=</math>	<math>~3^2(4-\delta )^2 + 3^2\delta (20 - \delta ) + 2^2\cdot 3^3</math>
<math>~\Delta_1</math>	<math>~\equiv</math>	<math>~2c^3 - 9bcd + 27b^2e+27ad^2 - 72ace</math>
	<math>~=</math>	<math>~2\cdot 3^3(4-\delta)^3 + 3^4(20-\delta)(4-\delta)\delta + 3^3(20-\delta)^2+3^7\delta^2 - 2^3\cdot 3^5(4-\delta)</math>

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