User:Tohline/Appendix/Ramblings/Nonlinar Oscillation

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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. In support of this idea, I point to the discussion of "Turning-Points and the Onset of Instability" found in §6.8 of [ST83] — specifically, on p. 149 in the paragraph that follows eq. (6.8.11) — where we find the following statement: "… the eigenfunction at a critical point is simply the Lagrangian displacement <math>~\xi</math> that carries an equilibrium configuration on the low-density side of the critical point into an equilibrium configuration on the high-density side."

See related arguments made by:


Whitworth's (1981) Isothermal Free-Energy Surface
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Revised Attack

Equilibrium Structural Properties

As we have detailed in an accompanying chapter, some basic properties of pressure-truncated, <math>~n = 5</math> polytropic configurations are:

<math>~\theta</math>

<math>~=</math>

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

<math>~\frac{d\theta}{d\xi}</math>

<math>~=</math>

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

<math>~\ell^2 \equiv \frac{\xi^2}{3}</math>

    <math>~\Rightarrow</math>   

<math>~\xi^2 = 3\ell^2 \, ,</math>

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

<math>~=~</math>

<math> \biggl( \frac{n^3}{4\pi} \biggr)^{1/2} \theta^{(n-3)/2} (- \xi^2 \theta^') = \biggl( \frac{5^3}{4\pi} \biggr)^{1/2} \theta (- \xi^2 \theta^') </math>

 

<math>~=~</math>

<math> \biggl( \frac{3 \cdot 5^3}{4\pi} \biggr)^{1/2} \ell^3 (1+\ell^2)^{- 2} \, , </math>

<math> ~\frac{R_\mathrm{eq}}{R_\mathrm{SWS} } </math>

<math>~=~</math>

<math> \biggl( \frac{n}{4\pi} \biggr)^{1/2} \xi \theta^{(n-1)/2} = \biggl( \frac{5}{4\pi} \biggr)^{1/2} \xi \theta^{2} </math>

 

<math>~=~</math>

<math> \biggl( \frac{3 \cdot 5}{4\pi} \biggr)^{1/2} \ell (1+\ell^2)^{-1} \, , </math>

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

<math>~\equiv</math>

<math>~\biggl( \frac{n+1}{nG} \biggr)^{3/2} K^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]} = \biggl( \frac{6}{5G} \biggr)^{3/2} K^{5 / 3} P_\mathrm{e}^{- 1 / 6} \, ,</math>

<math>~R_\mathrm{SWS}</math>

<math>~\equiv</math>

<math>~ \biggl( \frac{n+1}{nG} \biggr)^{1/2} K^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]} = \biggl( \frac{6}{5G} \biggr)^{1/2} K^{5/6} P_\mathrm{e}^{- 1 / 3} \, ,</math>

<math>~0</math>

<math>~=~</math>

<math> \biggl( \frac{M}{M_\mathrm{SWS}} \biggr)^2 - 5 \biggl( \frac{M}{M_\mathrm{SWS}} \biggr)\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr) + \frac{2^2 \cdot 5 \pi}{3} \biggl( \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr)^4 \, . </math>

This last expression can either be viewed as a quadratic equation whose solution provides an analytic expression for m(r), or a quartic equation whose solution provides an analytic expression for r(m).

Roots of Quadratic Equation

Solving the quadratic equation, we find that,

<math>~\frac{M}{M_\mathrm{SWS}}</math>

<math>~=</math>

<math>~ \frac{5}{2}\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr) \biggl\{ 1 \pm \biggl[ 1 - \frac{2^4\pi}{3\cdot 5}\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr)^2 \biggr]^{1 / 2} \biggr\} \, . </math>

As has been summarized in Table 3 of an accompanying discussion, two extrema exist with the following coordinates:

  • Maximum radius:     <math>\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{SWS}}, \frac{M}{M_\mathrm{SWS}} \biggr)

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

  • Maximum mass:     <math>\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{SWS}}, \frac{M}{M_\mathrm{SWS}} \biggr)

= \biggl[ \biggl( \frac{3^2\cdot 5}{2^6 \pi} \biggr)^{1 / 2} , \biggl( \frac{3^4\cdot 5^3}{2^{10} \pi} \biggr)^{1 / 2} \biggr]</math>


Roots of Quartic Equation

Here we will adopt the shorthand notation,

<math>~\chi \equiv \frac{R_\mathrm{eq}}{R_\mathrm{SWS}}</math>      and      <math>~m \equiv \frac{M}{M_\mathrm{SWS}} \, .</math>

Drawing from the Wikipedia discussion of the quartic function, we will think in terms of the generic quartic form,

<math>~0</math>

<math>~=</math>

<math>~a\chi^4 + b\chi^3 + c\chi^2 + d\chi + e \,.</math>

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

<math>~a</math>

<math>~=</math>

<math>~\frac{2^2 \cdot 5 \pi}{3} </math>

<math>~b</math>

<math>~=</math>

<math>~0</math>

<math>~c</math>

<math>~=</math>

<math>~0</math>

<math>~d</math>

<math>~=</math>

<math>~-5m</math>

<math>~e</math>

<math>~=</math>

<math>~m^2</math>

We need to evaluate the following expressions:

<math>~p</math>

<math>~\equiv</math>

<math>~\frac{8ac-3b^2}{8a^2}</math>

 

<math>~=</math>

<math>~0</math>

<math>~q</math>

<math>~\equiv</math>

<math>~\frac{b^3 - 4abc + 8a^2d}{8a^3} </math>

 

<math>~=</math>

<math>~\frac{d}{a} = - \frac{3m}{2^2 \pi} </math>

<math>~\Delta_0</math>

<math>~\equiv</math>

<math>~c^2 - 3bd + 12ae</math>

 

<math>~=</math>

<math>~12ae = 2^4 \cdot 5~\pi m^2</math>

<math>~\Delta_1</math>

<math>~\equiv</math>

<math>~2c^3 - 9bcd + 27b^2e+27ad^2 - 72ace</math>

 

<math>~=</math>

<math>~27ad^2 = 3^3 \cdot 5^2 m^2 \cdot \frac{2^2 \cdot 5 \pi}{3} = 2^2 \cdot 3^2 \cdot 5^3 ~\pi m^2</math>

Note that the discriminant is,

<math>~\Delta</math>

<math>~=</math>

<math>~ 256a^3 e^3 - 192a^2bde^2 - 128a^2c^2e^2 + 144a^2 c d^2e - 27a^2d^4 </math>

 

 

<math>~ + 144 a b^2c e^2 - 6ab^2d^2e - 80 abc^2de + 18abcd^3 + 16ac^4e </math>

 

 

<math>~ - 4ac^3 d^2 - 27b^4e^2 + 18b^3cde - 4b^3d^3 - 4b^2c^3e + b^2c^2d^2 </math>

 

<math>~=</math>

<math>~ 256a^3 e^3 - 27a^2d^4 </math>

 

<math>~=</math>

<math>~ 2^8\biggl( \frac{2^2 \cdot 5 \pi}{3} \biggr)^3 m^6 - 3^3\biggl( \frac{2^2 \cdot 5 \pi}{3} \biggr)^2 5^4 m^4 </math>

 

<math>~=</math>

<math>~ \biggl( \frac{2^{14} \cdot 5^3 \pi^3}{3^3} \biggr) m^6 - \biggl( 2^4 \cdot 3 \cdot 5^6 \pi^2 \biggr) m^4 </math>

 

<math>~=</math>

<math>~ - \biggl( 2^4 \cdot 3 \cdot 5^6 \pi^2 \biggr) m^4 \biggl[1 - \biggl( \frac{1}{2^4 \cdot 3 \cdot 5^6 \pi^2 m^4} \biggr) \biggl( \frac{2^{14} \cdot 5^3 \pi^3}{3^3} \biggr) m^6 \biggr] </math>

 

<math>~=</math>

<math>~ - \biggl( 2^4 \cdot 3 \cdot 5^6 \pi^2 \biggr) m^4 \biggl[1 - \biggl( \frac{2^{10} \pi}{3^4\cdot 5^3} \biggr) m^2 \biggr] \, , </math>

and it will be negative (or, in the limit, zero) as long as <math>~m \le m_\mathrm{max}</math>, where <math>~m_\mathrm{max} \equiv [3^4\cdot 5^3/(2^{10}\pi)]^{1 / 2}</math> is the value that we already have associated, above, with the maximum allowed mass. Because the discriminant is always negative (or, at most, zero), then our quartic equation has two distinct real roots and two complex conjugate non-real roots.

Furthermore note that,

<math>~\Delta_1^2 - 4\Delta_0^3</math>

<math>~=</math>

<math>~ (2^2 \cdot 3^2 \cdot 5^3 ~\pi m^2)^2 - 2^2(2^4 \cdot 5~\pi m^2 )^3 = (2^4 \cdot 3^4 \cdot 5^6 ~\pi^2 m^4) \biggl[1 - \frac{(2^{14} \cdot 5^3~\pi^3 m^6 )}{2^4 \cdot 3^4 \cdot 5^6 ~\pi^2 m^4}\biggr] </math>

 

<math>~=</math>

<math>~ (2^4 \cdot 3^4 \cdot 5^6 ~\pi^2 m^4) \biggl[1 - \biggl(\frac{2^{10} ~\pi}{3^4 \cdot 5^3 }\biggr) m^2\biggr] \, , </math>

and it will never be negative, as long as <math>~m \le m_\mathrm{max}</math>.


For a given value of <math>~m</math>, then, the pair of real roots is:

<math>~\chi_\pm</math>

<math>~=</math>

<math>~ -\frac{b}{4a} + S \pm \frac{1}{2}\biggl[ -4S^2 - 2p - \frac{q}{S} \biggr]^{1/2} \, , </math>

where,

<math>~S</math>

<math>~\equiv</math>

<math>~ \frac{1}{2}\biggl[- \frac{2p}{3} + \frac{1}{3a}\biggl(Q + \frac{\Delta_0}{Q}\biggr) \biggr]^{1/2} \, , </math>

<math>~Q</math>

<math>~\equiv</math>

<math>~ \biggl[ \frac{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}}{2} \biggr]^{1/3} \, . </math>

Let's work through these expressions.

<math>~Q</math>

<math>~=</math>

<math>~(2 \cdot 3^2 \cdot 5^3 ~\pi m^2)^{1 / 3} \biggl\{ 1 + \biggl[1 - \biggl( \frac{m}{m_\mathrm{max}}\biggr)^2\biggr]^{1 / 2} \biggr\}^{1/3} \, , </math>

<math>~S^2</math>

<math>~=</math>

<math>~ \frac{1}{2^2 \cdot 3a}\biggl(Q + \frac{\Delta_0}{Q}\biggr) </math>

 

<math>~=</math>

<math>~ \frac{Q}{2^4 \cdot 5 \pi} + \frac{m^2}{Q} \, , </math>

<math>~\chi_\pm</math>

<math>~=</math>

<math>~

S \pm \frac{1}{2}\biggl[\frac{1}{S}\biggl( \frac{3m}{2^2\pi} \biggr) -4S^2   \biggr]^{1/2} 

</math>

 

<math>~=</math>

<math>~

S \pm \biggl( \frac{3m}{2^4\pi ~S} \biggr)^{1 / 2} \biggl[1 - \frac{2^4\pi ~S^3}{3m}   \biggr]^{1/2} 

</math>

 

<math>~=</math>

<math>~ S \biggl\{1 \pm \biggl[\frac{3m}{2^4\pi ~S^3} - 1 \biggr]^{1/2} \biggr\} \, . </math>

As a check, recognize that the two roots should be identical, and given by <math>~r_\mathrm{crit} = [3^2 \cdot 5/(2^6\pi)]^{1 / 2}</math>, when <math>~m = m_\mathrm{max}</math>. Let's see …

<math>~Q\biggr|_{m_\mathrm{max}}</math>

<math>~=</math>

<math>~\biggl[ 2 \cdot 3^2 \cdot 5^3 ~\pi \biggl( \frac{3^4\cdot 5^3}{2^{10}\pi}\biggr)\biggr]^{1 / 3} = \biggl[ \frac{3^6\cdot 5^6}{2^{9}} \biggr]^{1 / 3} = \frac{3^2\cdot 5^2}{2^{3}} </math>

<math>~\Rightarrow ~~~ S^2 \biggr|_{m_\mathrm{max}}</math>

<math>~=</math>

<math>~ \frac{1}{2^4 \cdot 5 \pi} \biggl(\frac{3^2\cdot 5^2}{2^{3}}\biggr) + \biggl( \frac{3^4\cdot 5^3}{2^{10}\pi}\biggr)\biggl(\frac{2^{3}}{3^2\cdot 5^2}\biggr) = \frac{3^2\cdot 5}{2^{6} \pi} = r_\mathrm{crit}^2 </math>

<math>~\Rightarrow~~~ \chi_\pm\biggr|_{m_\mathrm{max}}</math>

<math>~=</math>

<math>~ r_\mathrm{crit} \biggl\{1 \pm \biggl[\frac{3}{2^4\pi }\biggl( \frac{3^4\cdot 5^3}{2^{10}\pi}\biggr)^{1 / 2}\biggl( \frac{2^{6} \pi}{3^2\cdot 5} \biggr)^{3 / 2} - 1 \biggr]^{1/2} \biggr\} = r_\mathrm{crit} \, , </math>

Q.E.D.

Identifying Equal-Mass Pairs

Figure 1:   n = 5 Mass-Radius Sequence
n = 5 mass-radius equilibrium sequence
file = Dropbox/WorkFolder/Wiki edits/EmbeddedPolytropes/Workbook_n5.xlsx --- worksheet = Quartic

The mass-radius relationship for pressure-truncated, n = 5 polytropic configurations is displayed as a green solid curve, here on the right, in Figure 1. This sequence can be constructed either: (a) by choosing various values of the radius (between zero and the above-specified maximum radius) and, for each choice, determining the two corresponding values of the equilibrium mass from the pair of roots of the quadratic equation; or (b) by choosing various values of the mass (between zero and the above-specified maximum mass) and, for each choice, using the two real roots of the quartic equation to determine the two corresponding values of the equilibrium radius. The green curve shown in Figure 1 is identical to the orange-dashed curve, labeled n = 5, that is nested among six other polytropic equilibrium sequences in the righthand panel of Figure 3 in an accompanying discussion.

Here we are interested in comparing the relative distribution of mass inside various pairs of models that have identical total masses. We therefore will focus on method "b". Specifically, given any value of the mass, <math>~m < m_\mathrm{max}</math>, the roots of the quartic equation will give us the equilibrium radii of the two configurations that have the same, specified mass. From these two values of <math>~\chi</math>, we can, in turn, determine the corresponding pair of values of <math>~\ell_\pm</math> via the expression,

<math>~\chi_\pm</math>

<math>~=</math>

<math>~\biggl( \frac{3 \cdot 5}{4\pi} \biggr)^{1/2} \ell_\pm (1+\ell_\pm^2)^{-1} </math>

<math>~\Rightarrow ~~~0</math>

<math>~=</math>

<math>~\ell_\pm^2-\biggl( \frac{3 \cdot 5}{4\pi \chi^2_\pm} \biggr)^{1/2} \ell_\pm +1 \, . </math>

But this is a quadratic equation, meaning that for <math>~\chi_+</math> there are two viable roots for <math>~\ell_+</math>, and for <math>~\chi_-</math> there are two viable roots for <math>~\ell_-</math>. We will deal with this by referring to the "plus" root as the "high" value, and by referring to the "minus" root as the "low" value. Specifically,

<math>~\xi_\pm\biggr|_\mathrm{high} = \sqrt{3} \ell_\pm\biggr|_\mathrm{high}</math>

<math>~=</math>

<math>~\sqrt{3} \biggl( \frac{3 \cdot 5}{2^4\pi \chi^2_\pm} \biggr)^{1/2} \biggl[ 1 + \biggl( 1 - \frac{2^4\pi \chi^2_\pm}{3 \cdot 5} \biggr)^{1 / 2} \biggr] \, , </math>

<math>~\xi_\pm\biggr|_\mathrm{low} = \sqrt{3} \ell_\pm\biggr|_\mathrm{low}</math>

<math>~=</math>

<math>~\sqrt{3} \biggl( \frac{3 \cdot 5}{2^4\pi \chi^2_\pm} \biggr)^{1/2} \biggl[ 1 - \biggl( 1 - \frac{2^4\pi \chi^2_\pm}{3 \cdot 5} \biggr)^{1 / 2} \biggr] \, . </math>

This seems to work because, if we plug in a single value for <math>~\chi_\pm</math> — for example, the degenerate case of <math>~\chi_\pm = r_\mathrm{crit}</math> — we get the pair of values of <math>~\xi</math> along the equilibrium sequence where the equilibrium radius has this selected value. Specifically, when <math>~\chi_\pm = r_\mathrm{crit}</math>, we find that, <math>~\xi_\mathrm{high} = 3</math> and <math>~\xi_\mathrm{low} = 1</math>.

In what follows, we will be especially interested in examining the region of parameter space in the vicinity of the marginally unstable case, as identified by the black-dashed rectangle drawn in Figure 1. Hence, we will only be interested the "high" roots.

Table 1:   Selected Pairings

Pairing (N)

<math>~\frac{M}{M_\mathrm{SWS}}</math>

<math>~\chi_+</math>

<math>~\chi_-</math>

"high" roots

<math>~{\tilde{C}}_+</math>

<math>~{\tilde{C}}_-</math>

<math>~\xi_+</math>

<math>~\xi_-</math>

A (11)

<math>~\biggl( \frac{3^4\cdot 5^3}{2^{10} \pi} \biggr)^{1 / 2}</math>

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

(degenerate)
<math>~3</math>

(degenerate)
<math>~4</math>

(degenerate)

B (12)

1.7696424

0.486212

0.458911

2.833124

3.180242

4.121273

3.889861

C (14)

1.7607720

0.495129

0.447886

2.718303

3.321996

4.217999

3.815538

D (16)

1.7519016

0.500918

0.439985

2.642460

3.425043

4.288919

3.767203

E (18)

1.7430312

0.505407

0.433378

2.582586

3.512395

4.349376

3.729518

F (20)

1.7341608

0.509128

0.427533

2.532015

3.590722

4.403816

3.698038

G (22)

1.7252904

0.512327

0.422206

2.487708

3.663068

4.454266

3.670737

H (24)

1.7164200

0.515138

0.417261

2.447976

3.731126

4.501855

3.646491

I (26)

1.7075496

0.517648

0.4126122

2.411770

3.795950

4.547286

3.624599

J (28)

1.6986793

0.519913

0.408203

2.378383

3.858252

4.591032

3.604591

NOTE:  The mass of a given configuration pair has been specified according to the expression,

<math>~\frac{M}{M_\mathrm{SWS}} = \biggl( \frac{3^4\cdot 5^3}{2^{10} \pi} \biggr)^{1 / 2}\biggl[1 - \frac{(N-11)}{400}\biggr] \, ,</math>

where, N is the integer that appears inside the parentheses in the first column of this table.

Inferred Displacement Function

Foundation

From our discussion, below, for any value of the truncation radius, <math>~\tilde\xi</math>, the fractional mass <math>~(0 \le m_\xi \le 1)</math> that lies interior to <math>~\xi</math> is given by the expression,

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

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

where,

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

      <math>~\Rightarrow</math>      

<math>~ {\tilde\xi}^2 = \frac{9}{\tilde{C} - 3} \, . </math>

And, when normalized to <math>~R_\mathrm{SWS}</math>, the corresponding radius is,

<math>~r_\mathrm{SWS}(\xi)</math>

<math>~=</math>

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

 

<math>~=</math>

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

 

<math>~=</math>

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

 

<math>~=</math>

<math>~\xi\biggl[ \frac{3^2 \cdot 5}{2^2 \pi} \biggr]^{1 / 2} \biggl[ \frac{\tilde{C} - 3}{3\tilde{C}} \biggr] \, .</math>

Now, this works fine in the sense that, for any choice of <math>~\tilde\xi</math>, and therefore <math>~\tilde{C}</math>, this pair of parametric relations can be used to generate a plot of <math>~r_\mathrm{SWS}</math> versus <math>~m_\xi</math> that correctly displays how the mass enclosed within a given radius varies with radial location throughout the spherical configuration. But, in order to compare one of these configurations to another, we really need to identify how this function varies across a Lagrangian mass grid that is the same for both configurations. The easiest way to accomplish this is to derive an expression for <math>~r_\mathrm{SWS}</math> that is directly a function of <math>~m_\xi</math>. Fortunately, this can be done analytically. First, we invert the mass expression to find,

<math>~m_\xi^{2/3}</math>

<math>~=</math>

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

<math>~\Rightarrow ~~~ 3 + \xi^2</math>

<math>~=</math>

<math>~ m_\xi^{-2/3}\biggl(\frac{\tilde{C}}{3}\biggr)\xi^2 </math>

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

<math>~=</math>

<math>~ -3 </math>

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

<math>~=</math>

<math>~ 3^2 [ \tilde{C}~m_\xi^{-2/3} -3 ]^{-1} \, . </math>

Inserting this into the radial equation, then, gives,

<math>~r_\mathrm{SWS}(m_\xi)</math>

<math>~=</math>

<math>~\biggl[ \frac{3^2 \cdot 5}{2^2 \pi} \biggr]^{1 / 2} \biggl[ \frac{\tilde{C} - 3}{\tilde{C}} \biggr] \biggl[ \tilde{C}~m_\xi^{-2/3} -3 \biggr]^{-1 / 2} \, .</math>

Analytic, Marginally Unstable Eigenfunction

In terms of <math>~\xi</math>, we know that the eigenfunction of the marginally unstable model is,

<math>~x_P = \frac{\delta r}{r_0}</math>

<math>~=</math>

<math>~1 - \frac{\xi^2}{15} \, .</math>

We can now rewrite this eigenfunction in terms of the fractional mass, <math>~m_\xi</math>. Specifically, given that <math>~\tilde{C} = 4</math> in the marginally unstable configuration, we find that,

<math>~x_P </math>

<math>~=</math>

<math>~1 - \frac{3}{5} \biggl[ 4~m_\xi^{-2/3} -3 \biggr]^{-1} </math>

 

<math>~=</math>

<math>~\frac{2}{5} \biggl[ \frac{10~m_\xi^{-2/3} -9}{4~m_\xi^{-2/3} -3} \biggr] </math>

 

<math>~=</math>

<math>~\frac{2}{5} \biggl[ \frac{10 -9~m_\xi^{2/3}}{4 -3~m_\xi^{2/3}} \biggr] \, .</math>

Delta Profiles

Next, let's define a fractional difference in configuration profiles.

<math>~\mathfrak{x} </math>

<math>~\equiv</math>

<math>~ \frac{\Delta r_\mathrm{SWS}}{<r_\mathrm{SWS}>} = \frac{ r_2(m_\xi) - r_1(m_\xi) }{ r_2(m_\xi) + r_1(m_\xi)}</math>

 

<math>~=</math>

<math>~ \biggl\{\biggl[ \frac{\tilde{C}_2 - 3}{\tilde{C}_2} \biggr] \biggl[ \tilde{C}_2~m_\xi^{-2/3} -3 \biggr]^{-1 / 2} - \biggl[ \frac{\tilde{C}_1 - 3}{\tilde{C}_1} \biggr] \biggl[ \tilde{C}_1~m_\xi^{-2/3} -3 \biggr]^{-1 / 2} \biggr\} \biggl\{\biggl[ \frac{\tilde{C}_2 - 3}{\tilde{C}_2} \biggr] \biggl[ \tilde{C}_2~m_\xi^{-2/3} -3 \biggr]^{-1 / 2} + \biggl[ \frac{\tilde{C}_1 - 3}{\tilde{C}_1} \biggr] \biggl[ \tilde{C}_1~m_\xi^{-2/3} -3 \biggr]^{-1 / 2} \biggr\}^{-1} </math>

 

<math>~=</math>

<math>~ \biggl\{ \tilde{C}_1 ( \tilde{C}_2 - 3 ) \biggl[ \tilde{C}_1~m_\xi^{-2/3} -3 \biggr]^{1 / 2} - \tilde{C}_2 ( \tilde{C}_1 - 3 ) \biggl[ \tilde{C}_2~m_\xi^{-2/3} -3 \biggr]^{1 / 2} \biggr\} \biggl\{ \tilde{C}_1 ( \tilde{C}_2 - 3 ) \biggl[ \tilde{C}_1~m_\xi^{-2/3} -3 \biggr]^{1 / 2} + \tilde{C}_2 ( \tilde{C}_1 - 3 ) \biggl[ \tilde{C}_2~m_\xi^{-2/3} -3 \biggr]^{1 / 2} \biggr\}^{-1} </math>

 

<math>~=</math>

<math>~ \biggl\{ \tilde{C}_1 ( \tilde{C}_2 - 3 ) \biggl[ \tilde{C}_1 -3~m_\xi^{2/3} \biggr]^{1 / 2} - \tilde{C}_2 ( \tilde{C}_1 - 3 ) \biggl[ \tilde{C}_2 -3 ~m_\xi^{2/3}\biggr]^{1 / 2} \biggr\} \biggl\{ \tilde{C}_1 ( \tilde{C}_2 - 3 ) \biggl[ \tilde{C}_1 -3~m_\xi^{2/3} \biggr]^{1 / 2} + \tilde{C}_2 ( \tilde{C}_1 - 3 ) \biggl[ \tilde{C}_2 -3~m_\xi^{2/3} \biggr]^{1 / 2} \biggr\}^{-1} \, . </math>

Next, let's define,

<math>~\Delta C_i \equiv C_i - 4 </math>      <math>~\Rightarrow</math>      <math>~C_i \equiv \Delta C_i + 4 \, ,</math>

in which case,

<math>~\mathfrak{x} </math>

<math>~=</math>

<math>~ \biggl\{ (\Delta C_1 + 4) ( \Delta C_2 + 1 ) \biggl[ \Delta C_1 + 4 -3~m_\xi^{2/3} \biggr]^{1 / 2} - ( \Delta C_2 + 4) ( \Delta C_1 + 1 ) \biggl[ \Delta C_2 + 4 -3 ~m_\xi^{2/3}\biggr]^{1 / 2} \biggr\} </math>

 

 

<math>~\times \biggl\{ (\Delta C_1 + 4) ( \Delta C_2 + 1 ) \biggl[ \Delta C_1 + 4 -3~m_\xi^{2/3} \biggr]^{1 / 2} + ( \Delta C_2 + 4 ) ( \Delta C_1 + 1 ) \biggl[ \Delta C_2 + 4 -3~m_\xi^{2/3} \biggr]^{1 / 2} \biggr\}^{-1} \, . </math>

Now define,

<math>~\beta \equiv (4 -3~m_\xi^{2/3})^{-1 / 2} \, ,</math>

in which case,

<math>~\mathfrak{x} </math>

<math>~=</math>

<math>~ \biggl\{ (1 + \Delta C_1/4 ) (1 + \Delta C_2 ) (1 + \beta^2 \Delta C_1 )^{1 / 2} - (1 + \Delta C_2/4 ) (1+ \Delta C_1 ) ( 1+\beta^2 \Delta C_2 )^{1 / 2} \biggr\} </math>

 

 

<math>~\times \biggl\{ (1 + \Delta C_1/4) (1 + \Delta C_2 ) ( 1 + \beta^2\Delta C_1 )^{1 / 2} + (1 + \Delta C_2/4 ) (1 + \Delta C_1) ( 1 + \beta^2\Delta C_2 )^{1 / 2} \biggr\}^{-1} </math>

 

<math>~\approx</math>

<math>~\frac{1}{2} \biggl[ (1 + \Delta C_1/4 + \Delta C_2 + \beta^2 \Delta C_1/2 ) - (1 + \Delta C_2/4 + \Delta C_1 +\beta^2 \Delta C_2/2 ) \biggr] </math>

 

<math>~=</math>

<math>~\frac{1}{2} \biggl[ \Delta C_1/4 + \Delta C_2 + \beta^2 \Delta C_1/2 - \Delta C_2/4 - \Delta C_1 -\beta^2 \Delta C_2/2 \biggr] \, . </math>

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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>~\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) \, . </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 \, . </math>

Finite Difference Representation of Radial Eigenfunction

Preamble

<math>~\tilde{C}</math>

<math>~\equiv</math>

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

<math>~\tilde{r}_\mathrm{edge}</math>

<math>~\equiv</math>

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

 

<math>~=</math>

<math>~ \biggl[ \frac{\pi}{2^3\cdot 3}\biggr]^{1/2} \biggl[ \frac{\tilde{C} }{ 3^2 }\biggr] ^3 </math>

 

<math>~=</math>

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

Note that when <math>~\tilde\xi = 3</math>,

<math>~\tilde{C}_3</math>

<math>~=</math>

<math>~ 4 \, , </math>

<math>~\tilde{r}_{e3}</math>

<math>~=</math>

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


Conjectures

A first-cut examination of the structure of the radial eigenfunction associated with the <math>~P_\mathrm{max}</math> turning point is given by simply subtracting one <math>~r_\xi(m_\xi)</math> profile from another at the same applied external pressure. (The specific choices of the two appropriate values of <math>~\tilde\xi</math> are discussed in the subsection titled, "Configuration Pairing", which follows.) The answer appears to be,

<math>~[ \Delta r_\xi ]_{21} = [r_\xi (m_\xi)]_2 - [r_\xi (m_\xi)]_1 </math>

<math>~=</math>

<math>~ [\tilde{r}_\mathrm{edge}]_2 \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C}_2 - 3 m_\xi^{2/3}}\biggr]^{1/2} - [\tilde{r}_\mathrm{edge}]_1 \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C}_1 - 3 m_\xi^{2/3}}\biggr]^{1/2} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{\pi}{2^3\cdot 3^{13}}\biggr]^{½} \biggl\{ {\tilde{C}_2}^3 \biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C}_2 - 3 m_\xi^{2/3}}\biggr]^{1/2} -

{\tilde{C}_1}^3

\biggl[\frac{3^2m_\xi^{2/3}}{\tilde{C}_1 - 3 m_\xi^{2/3}}\biggr]^{1 / 2} \biggr\} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{\pi}{2^3\cdot 3^{13}}\biggr]^{1 / 2} \biggl\{ {\tilde{C}_2}^3 \biggl[\frac{3^2m_\xi^{2/3}/\tilde{C}_1 }{\tilde{C}_2/\tilde{C}_1 - 3 m_\xi^{2/3}/\tilde{C}_1 }\biggr]^{1 / 2} -

{\tilde{C}_1}^3

\biggl[\frac{3^2m_\xi^{2/3}/\tilde{C}_1 }{1 - 3 m_\xi^{2/3}/\tilde{C}_1 }\biggr]^{1 / 2} \biggr\} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{\pi}{2^3\cdot 3^{13}}\biggr]^{1 / 2} \biggl\{ {\tilde{C}_2}^3 \biggl[\frac{3/u }{\tilde{C}_2/\tilde{C}_1 - 1/u }\biggr]^{1 / 2} -

{\tilde{C}_1}^3

\biggl[\frac{3/u }{1 - 1/u }\biggr]^{1 / 2} \biggr\} </math>

 

<math>~=</math>

<math>~

{\tilde{C}_1}^3 \biggl[ \frac{\pi}{2^3\cdot 3^{12}}\biggr]^{1 / 2} \biggl\{ k_{21}^3

\biggl[\frac{1 }{k_{21} u - 1 }\biggr]^{1 / 2} - \biggl[\frac{1 }{u - 1 }\biggr]^{1 / 2} \biggr\} \, , </math>

where,

<math>~u \equiv \tilde{C}_1/(3 m_\xi^{2/3}) \, ,</math>

<math>~k_{21} \equiv \frac{ \tilde{C}_2 }{ \tilde{C}_1 } \, .</math>

After examining the form of this last expression, it is clear that we can also write,

<math>~[r_\xi (m_\xi)]_1</math>

<math>~=</math>

<math>~

{\tilde{C}_1}^3 \biggl[ \frac{\pi}{2^3\cdot 3^{12}}\biggr]^{1 / 2} 

\biggl[\frac{1 }{u - 1 }\biggr]^{1 / 2} \, , </math>

in which case, the lopsided fractional eigenfunction takes the form,

<math>~\frac{[ \Delta r_\xi ]_{21} }{[r_\xi (m_\xi)]_1} </math>

<math>~=</math>

<math>~ k_{21}^3 \biggl[\frac{u - 1 }{k_{21} u - 1 }\biggr]^{1 / 2} - 1 \, . </math>

And the centered fractional eigenfunction is,

<math>~\frac{[ \Delta r_\xi ]_{32} }{[r_\xi (m_\xi)]_1} </math>

<math>~=</math>

<math>~ \biggl\{ k_{31}^3 \biggl[\frac{u - 1 }{k_{31} u - 1 }\biggr]^{1 / 2} - 1 \biggr\} - \biggl\{ k_{21}^3 \biggl[\frac{u - 1 }{k_{21} u - 1 }\biggr]^{1 / 2} - 1 \biggr\} </math>

 

<math>~=</math>

<math>~ k_{31}^3 \biggl[\frac{u - 1 }{k_{31} u - 1 }\biggr]^{1 / 2} - k_{21}^3 \biggl[\frac{u - 1 }{k_{21} u - 1 }\biggr]^{1 / 2} \, . </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{2\cdot 3^4}{\pi}\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>~\ell</math>

      <math>~\rightarrow</math>      

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

and look for pairs of values, <math>~\epsilon_\pm</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] +(16- 3^{3/2}p_0)\, . </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] -\delta </math>

 

<math>~=~</math>

<math>~ 9\epsilon^4 +\epsilon^3 (20-\delta ) +\epsilon^2 ( 12 - 3\delta ) -\epsilon (3\delta ) -\delta \, . </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,

<math>~0</math>

<math>~=</math>

<math>~ax^4 + bx^3 + cx^2 + dx + e \,.</math>

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>~-\delta</math>

We need to evaluate the following expressions:

<math>~p</math>

<math>~\equiv</math>

<math>~\frac{8ac-3b^2}{8a^2}</math>

 

<math>~=</math>

<math>~\frac{2^3 \cdot 3^3(4-\delta )-3(20-\delta )^2}{2^3\cdot 3^4}</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\delta</math>

 

<math>~=</math>

<math>~144</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 \delta+3^7\delta^2 + 2^3\cdot 3^5(4-\delta)\delta</math>

 

<math>~=</math>

<math>~3^3(128 + 32\delta + \delta^2) \, .</math>

Note:   <math>~\Delta_1^2 - 4\Delta_0^3</math>

<math>~=</math>

<math>~3^6(2^{13}\delta + 2^8\cdot 5 \delta^2 + 2^6\delta^3 + \delta^4) \, .</math>

For a given value of <math>~\delta</math>, then, the pair of real roots is:

<math>~\epsilon_\pm</math>

<math>~=</math>

<math>~ -\frac{b}{4a} + S \pm \frac{1}{2}\biggl[ -4S^2 - 2p - \frac{q}{S} \biggr]^{1/2} \, , </math>

where,

<math>~S</math>

<math>~\equiv</math>

<math>~ \frac{1}{2}\biggl[- \frac{2p}{3} + \frac{1}{3a}\biggl(Q + \frac{\Delta_0}{Q}\biggr) \biggr]^{1/2} \, , </math>

<math>~Q</math>

<math>~\equiv</math>

<math>~ \biggl[ \frac{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}}{2} \biggr]^{1/3} \, . </math>

We have used an Excel spreadsheet to evaluate these expressions. The following table identifies <math>~\epsilon_\pm</math> pairs (the middle two columns of numbers) for twenty different values of the external pressure; more specifically, for twenty values of <math>~0 \le \delta \le 0.69938</math>, equally spaced between the two limits. The corresponding pairs of <math>~\tilde\xi_\pm</math> are also listed (rightmost pair of columns).

Table 1

Sets of Paired Models from Quartic Solution

P_e/P_norm   delta	  eps_+   eps_-	         xi_+ 	 xi_-
160.867	    0.00000	0.00000  0.00000	3.00000	3.00000
158.664	    0.03681	0.05747	-0.05338	3.17241	2.83986
156.497	    0.07362	0.08253	-0.07435	3.24759	2.77695
154.363	    0.11043	0.10227	-0.09000	3.30681	2.72999
152.264	    0.14724	0.11927	-0.10291	3.35781	2.69127
150.198	    0.18405	0.13452	-0.11407	3.40355	2.65780
148.164	    0.22086	0.14852	-0.12398	3.44556	2.62805
146.163	    0.25767	0.16159	-0.13296	3.48476	2.60111
144.193	    0.29448	0.17391	-0.14120	3.52174	2.57640
142.254	    0.33129	0.18563	-0.14883	3.55690	2.55351
140.345	    0.36809	0.19685	-0.15596	3.59055	2.53212
138.466	    0.40490	0.20764	-0.16266	3.62291	2.51203
136.617	    0.44171	0.21805	-0.16898	3.65415	2.49305
134.796	    0.47852	0.22814	-0.17498	3.68442	2.47505
133.003	    0.51533	0.23794	-0.18070	3.71382	2.45791
131.239	    0.55214	0.24748	-0.18615	3.74245	2.44154
129.501	    0.58895	0.25680	-0.19138	3.77039	2.42586
127.790	    0.62576	0.26590	-0.19640	3.79770	2.41081
126.106	    0.66257	0.27481	-0.20122	3.82444	2.39634
124.447	    0.69938	0.28355	-0.20587	3.85065	2.38238

Two Example Eigenfunctions

The following figure is fundamentally a reproduction of Figure 3 from an accompanying discussion. It presents the "Case M" equilibrium sequence from both an order-of-magnitude analysis (marked by light-blue squares) and a detailed force-balance analysis (light-green triangles). The dark green circular dot identifies the configuration at the pressure maximum of the sequence — <math>~P_\mathrm{max}/P_\mathrm{norm} = 160.867</math> — and the red circular dot identifies the location along the sequence where the transition from stable to dynamically unstable configurations occurs — <math>~P_e/P_\mathrm{norm} = 124.447</math>. (All pressures have been normalized to <math>~P_\mathrm{max}</math> in the figure.)

Figure 1
Case M equilibrium sequences


At any <math>~P_e</math> between these two limiting values, a pair of stable equilibrium configurations exist; approximately twenty example pairings are listed in Table 1. The horizontal, black dashed line in the figure has been drawn at <math>~P_e/P_\mathrm{norm} = 158.664</math>. The pair of equilibrium configurations associated with this pressure is identified graphically by the two points at which this dashed line intersects the detailed force-balance equilibrium sequence; as is detailed in the second row of Table 1, the configurations correspond to models having <math>~\tilde\xi = 2.83986</math> (right intersection) and <math>~\tilde\xi = 3.17241</math> (left intersection). The left-hand panel of Figure 2 shows how the Lagrangian radial coordinate varies with mass, <math>~r_\xi(m_\xi)</math>, throughout the interior of these two equilibrium configurations (the locus of green and orange dots, respectively); for reference, the profile of the configuration at <math>~P_\mathrm{max}</math> is presented as well (locus of black dots). The right-hand panel of Figure 2 shows the same paired configuration profiles, but relative to the profile of the configuration at <math>~P_\mathrm{max}</math>.

Figure 2
Case M eigenfunction#1


The horizontal, black dot-dash line in Figure 1 has been drawn at <math>~P_e/P_\mathrm{norm} = 124.447</math>. The pair of equilibrium configurations associated with this pressure is identified graphically by the two points at which this dot-dash line intersects the detailed force-balance equilibrium sequence; as is detailed in the last row of Table 1, the configurations correspond to models having <math>~\tilde\xi = 2.83986</math> (right intersection) and <math>~\tilde\xi = 3.17241</math> (left intersection), which is the configuration that marks the onset of a dynamical instability. The left-hand panel of Figure 3 shows how the Lagrangian radial coordinate varies with mass, <math>~r_\xi(m_\xi)</math>, throughout the interior of these two paired equilibrium configurations (the locus of green and orange dots, respectively); again, for reference, the profile of the configuration at <math>~P_\mathrm{max}</math> is presented as well (locus of black dots). The right-hand panel of Figure 3 shows the same paired configuration profiles, but relative to the profile of the configuration at <math>~P_\mathrm{max}</math>.


Figure 3
Case M eigenfunction#2



Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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