SSC/FreeEnergy/Powerpoint

Supporting Derivations for Free-Energy PowerPoint Presentation

The derivations presented here are an extension of our accompanying free-energy synopsis. These additional details proved to be helpful while developing an overarching PowerPoint presentation.

General Free-Energy Expression

We're considering a free-energy function of the following form:

${\mathfrak{𝔊}}_{\mathrm{t}\mathrm{y}\mathrm{p}\mathrm{e}}^{*}$

$=$

$-a{x}^{-1}+b{x}^{-3/n}+c{x}^{-3/j}+{\mathfrak{𝔊}}_0,$

where,

$x\equiv \frac{R}{R_0}.$

As we have shown, setting,

$\frac{\partial {\mathfrak{𝔊}}_{\mathrm{t}\mathrm{y}\mathrm{p}\mathrm{e}}^{*}}{\partial x}$

$=$

$0,$

generates a mathematical statement of virial equilibrium, namely,

$\frac{b}{nc}\cdot x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{a}{3c}+\frac{1}{j}\cdot x_{\mathrm{e}\mathrm{q}}^{(j-3)/j}$

$=$

$0.$

And equilibrium configurations for which the second (as well as first) derivative of the free energy is zero are found at "critical" radii given by the expression,

$[x_{\mathrm{e}\mathrm{q}}^{(j-3)/j}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$

$=$

$\frac{a}{3^{2}c}\left[\frac{j^2(n-3)}{n-j}\right].$

Pressure-Truncated Polytropes

For pressure-truncated polytropes, set $j=-1$ and let $n$ be the chosen polytropic index. In this case, the statement of virial equilibrium is,

$\frac{b}{nc}\cdot x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{a}{3c}-x_{\mathrm{e}\mathrm{q}}^4$

$=$

$0;$

and the critical equilibrium configuration has,

$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$

$=$

$[\frac{a(n-3)}{3^{2}c(n+1)}{]}^{1/4}.$

Case M

Set $K$ and $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ constant and examine how the free-energy behaves as a function of the coordinates, $(R,P_e)$. In this case (see, for example, here),

$a$	$\equiv$	$\frac{3}{5}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2},$
$b$	$\equiv$	$n(\frac{4\pi }{3}{)}^{-1/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}},$
$c$	$\equiv$	$\frac{4\pi }{3}\left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right),$

where the structural form factors for pressure-truncated polytropes are precisely defined here. And (see, for example, here),

$R_0=R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$	$\equiv$	$\left[\right(\frac{G}{K}{)}^n{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{n-1}{]}^{1/(n-3)},$
$P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$	$\equiv$	$[\frac{K^{4n}}{G^{3(n+1)}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{2(n+1)}}{]}^{1/(n-3)}.$

If we set all three structural form-factors to unity, we have,

$\frac{a}{3c}$	$=$	$\frac{3}{2^2\cdot 5\pi }(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1},$
$\frac{b}{nc}$	$=$	$\left(\frac{3}{4\pi }{)}^{(n+1)/n}\right(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}.$

Virial Equilibrium

Figure 1

So the statement of virial equilibrium becomes,

$x_{\mathrm{e}\mathrm{q}}^4$	$=$	$\left[\right(\frac{3}{4\pi }{)}^{(n+1)/n}{x}_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{3}{2^2\cdot 5\pi }\left]\right(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}$
	$=$	$\frac{3}{2^2\cdot 5\pi }\left[5\right(\frac{3}{4\pi }{)}^{1/n}{x}_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-1\left]\right(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}$
$\Rightarrow \frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\frac{3}{2^2\cdot 5\pi x_{\mathrm{e}\mathrm{q}}^4}\left[5\right(\frac{3}{4\pi }{)}^{1/n}{x}_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-1].$

The light-blue dots in Figure 1 trace the equilibrium sequence that is defined by this virial equilibrium function in the case of $n=5$.

Dynamical Instability

Along the "Case M" equilibrium sequence, the transition from stable to unstable configurations occurs at,

$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$	$=$	$\left[\frac{(n-3)}{3(n+1)}\right]\frac{3}{2^2\cdot 5\pi }(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}$
$\Rightarrow \frac{2^2\cdot 5\pi }{3}\left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$	$=$	$\left[\frac{(n-3)}{3(n+1)}\right]$

which, in combination with the virial equilibrium condition gives,

$5(\frac{3}{4\pi }{)}^{1/n}[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n-3)/n}-1$	$=$	$\left[\frac{(n-3)}{3(n+1)}\right]$
$\Rightarrow [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$	$=$	$\left[\frac{4n}{3\cdot 5(n+1)}\right(\frac{4\pi }{3}{)}^{1/n}{]}^{n/(n-3)}.$

The location of this critical configuration along the $n=5$ equilibrium sequence is marked by the red circular dot in Figure 1.

Turning Point

Let's examine the curvature of the equilibrium sequence.

$\frac{d}{dx}\left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)$	$=$	$-\frac{3}{5\pi x^5}\left[5\right(\frac{3}{4\pi }{)}^{1/n}{x}^{(n-3)/n}-1]+\frac{3(n-3)}{2^{2}n\pi x^4}(\frac{3}{4\pi }{)}^{1/n}{x}^{-3/n}$
	$=$	$\frac{3}{5\pi x^5}+\frac{3}{4\pi }\left(\frac{3}{4\pi }{)}^{1/n}\right[\frac{(n-3)}{n}-4]\frac{x^{(n-3)/n}}{x^5}$
	$=$	$\frac{3}{5\pi x^5}-3\left(\frac{3}{4\pi }{)}^{(n+1)/n}\right[\frac{n+1}{n}]\frac{x^{(n-3)/n}}{x^5}.$

Setting this derivative to zero let's us identify the location of the turning point that identifies $P_{\mathrm{m}\mathrm{a}\mathrm{x}}.$

$[x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}{]}_{\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}}$	$=$	$\frac{1}{5\pi }\left[\frac{n}{n+1}\right](\frac{4\pi }{3}{)}^{(n+1)/n}$
$\Rightarrow [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}}$	$=$	$\left[\frac{4n}{15(n+1)}\right(\frac{4\pi }{3}{)}^{1/n}{]}^{n/(n-3)}.$

And, returning to the virial equilibrium expression, we find that, associated with this equilibrium radius,

$\frac{P_{\mathrm{m}\mathrm{a}\mathrm{x}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\frac{3}{2^2\cdot 5\pi x_{\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}}^4}\left[5\right(\frac{3}{4\pi }{)}^{1/n}{x}_{\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}}^{(n-3)/n}-1]$
$\Rightarrow 2^2\cdot 5\pi x_{\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}}^4\left(\frac{P_{\mathrm{m}\mathrm{a}\mathrm{x}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)$	$=$	$15\left(\frac{3}{4\pi }{)}^{1/n}\right[\frac{4n}{15(n+1)}\left(\frac{4\pi }{3}{)}^{1/n}\right]-3$
	$=$	$\left(\frac{n-3}{n+1}\right)$
$\Rightarrow \frac{P_{\mathrm{m}\mathrm{a}\mathrm{x}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\frac{1}{20\pi }\left(\frac{n-3}{n+1}\right)\left[\frac{15(n+1)}{4n}\right(\frac{3}{4\pi }{)}^{1/n}{]}^{4n/(n-3)}.$

Notice that, under the assumption that all three structural filling-factors are unity, $[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}}=[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$, that is, the location of the turning point coincides precisely with the point along the equilibrium sequence where the transition from stable to unstable equilibrium configurations occurs (marked by the red circular dot in Figure 1).

Case M Summary

Case M
Order-of-Magnitude Analysis: Assume ${\widetilde{\mathfrak{𝔣}}}_M={\widetilde{\mathfrak{𝔣}}}_W={\widetilde{\mathfrak{𝔣}}}_A=1$
Virial Equilibrium:	$\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$ $=$ $\frac{3}{2^2\cdot 5\pi }\left(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-4}\right[5\left(\frac{3}{4\pi }{)}^{1/n}\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-1]$
Dynamical Instability: $(n>3)$	$\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{|}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$ $=$ $\left[\frac{4n}{3\cdot 5(n+1)}\right(\frac{4\pi }{3}{)}^{1/n}{]}^{n/(n-3)}$
Turning Point $(P_{\mathrm{m}\mathrm{a}\mathrm{x}})$: $(n>3)$	$\frac{P_{\mathrm{m}\mathrm{a}\mathrm{x}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$ $=$ $\frac{1}{20\pi }\left(\frac{n-3}{n+1}\right)\left[\frac{15(n+1)}{4n}\right(\frac{3}{4\pi }{)}^{1/n}{]}^{4n/(n-3)}$ $\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{|}_{\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}}$ $=$ $\left[\frac{4n}{3\cdot 5(n+1)}\right(\frac{4\pi }{3}{)}^{1/n}{]}^{n/(n-3)}$

Case P

Set $K$ and $P_e$ constant and examine how the free-energy behaves as a function of the coordinates, $(R,M_{\mathrm{t}\mathrm{o}\mathrm{t}})$. In this case (see, for example, here),

$a$	$=$	$\frac{3}{5}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\left(\frac{n+1}{n}{)}^{n/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)/(n-3)},$
$b$	$=$	$n\left(\frac{4\pi }{3}{)}^{-1/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}}\right(\frac{n+1}{n}{)}^{3/(n-3)}(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{3(n-1)/[n(n-3)]},$
$c$	$=$	$\frac{4\pi }{3}\left(\frac{n+1}{n}{)}^{3/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(5-n)/(n-3)},$
$R_0=R_{\mathrm{S}\mathrm{W}\mathrm{S}}$	$\equiv$	$(\frac{n+1}{n}{)}^{1/2}{G}^{-1/2}{K}_n^{n/(n+1)}{P}_{\mathrm{e}}^{(1-n)/[2(n+1)]},$
$M_{\mathrm{S}\mathrm{W}\mathrm{S}}$	$\equiv$	$(\frac{n+1}{n}{)}^{3/2}{G}^{-3/2}{K}_n^{2n/(n+1)}{P}_{\mathrm{e}}^{(3-n)/[2(n+1)]}.$

where the structural form factors for pressure-truncated polytropes are precisely defined here. If we set all three structural form-factors to unity, we have,

$\frac{a}{3c}$	$=$	$\frac{3}{20\pi }\left(\frac{n+1}{n}\right)(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2$
$\frac{b}{nc}$	$=$	$\left[\right(\frac{3}{4\pi }\left)\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}){]}^{(n+1)/n}$

Virial Equilibrium

So, the statement of virial equilibrium becomes,

$\left[\right(\frac{3}{4\pi }\left)\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}){]}^{(n+1)/n}{x}_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{3}{20\pi }(\frac{n+1}{n}\left)\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2-x_{\mathrm{e}\mathrm{q}}^4$

$=$

$0.$

${\xi }_1$

Known Analytic Lane-Emden Functions
$n$	${\theta }_n(\xi )$	$-{\theta }_n^{\text{'}}$	${\xi }_1$
$0$	$1-\frac{{\xi }^2}{6}$	$\frac{\xi }{3}$	$\sqrt{6}$
$1$	$\frac{\sin \xi }{\xi }$	$\frac{\sin \xi }{{\xi }^2}-\frac{\cos \xi }{\xi }$	$\pi$
$5$	$[1+\frac{{\xi }^2}{3}{]}^{-1/2}$	$\frac{\xi }{3}[1+\frac{{\xi }^2}{3}{]}^{-3/2}$	$\mathrm{\infty }$

Dynamical Instability

Along the "Case P" equilibrium sequence, the transition from stable to unstable configurations occurs at,

$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$	$=$	$\frac{(n-3)}{3(n+1)}\left[\frac{3}{20\pi }\right(\frac{n+1}{n}\left)\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2]$
	$=$	$\frac{(n-3)}{20\pi n}(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2=\frac{2^2\pi (n-3)}{3^2\cdot 5n}[\frac{3}{4\pi }\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}\right){]}^2,$

which, in combination with the "Case P" virial equilibrium expression gives,

$0$	$=$	$\left\{\right[\frac{3^2\cdot 5n}{2^2\pi (n-3)}{]}^{1/2}{x}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^2{\}}^{(n+1)/n}{x}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n-3)/n}-\frac{3}{20\pi }\left(\frac{n+1}{n}\right)\left\{\right[\frac{20\pi n}{(n-3)}\left]x_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4\right\}-x_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$
	$=$	$[\frac{3^2\cdot 5n}{2^2\pi (n-3)}{]}^{(n+1)/(2n)}{x}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(3n-1)/n}-x_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4\{\left[\frac{3(n+1)}{(n-3)}\right]+1\}$
	$=$	$[\frac{3^2\cdot 5n}{2^2\pi (n-3)}{]}^{(n+1)/(2n)}{x}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(3n-1)/n}-[\frac{4n}{(n-3)}]x_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$
$\Rightarrow x_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n+1)/n}$	$=$	$\left[\frac{(n-3)}{4n}\right][\frac{3^2\cdot 5n}{2^2\pi (n-3)}{]}^{(n+1)/(2n)}$

Turning Points

Let's simplify the notation, defining,

$m\equiv \frac{3}{4\pi }\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}\right).$

The statement of virial equilibrium becomes,

$m^{(n+1)/n}{x}^{(n-3)/n}-c_0{m}^2-x^4$

$=$

$0,$

where,

$c_0\equiv \left[\frac{4\pi (n+1)}{15n}\right].$

Differentiating gives,

$0$	$=$	$\left(\frac{n+1}{n}\right)m^{1/n}{x}^{(n-3)/n}dm+\left(\frac{n-3}{n}\right)m^{(n+1)/n}{x}^{-3/n}dx-2c_{0}mdm-4x^{3}dx$
	$=$	$[(n+1)m^{1/n}{x}^{(n-3)/n}-2c_{0}nm]dm+[(n-3)m^{(n+1)/n}{x}^{-3/n}-4n{x}^3]dx$
$\Rightarrow \frac{dm}{dx}$	$=$	$\frac{4n{x}^3-(n-3)m^{(n+1)/n}{x}^{-3/n}}{(n+1)m^{1/n}{x}^{(n-3)/n}-2c_{0}nm}.$

---

One turning point occurs where the numerator is zero, that is,

$4n{x}^3$	$=$	$(n-3)m^{(n+1)/n}{x}^{-3/n}$
$\Rightarrow 4n{x}^{3(n+1)/n}$	$=$	$(n-3)m^{(n+1)/n}$
$\Rightarrow \frac{m}{x^3}$	$=$	$[\frac{4n}{(n-3)}{]}^{n/(n+1)}.$

Plugging this into the virial equilibrium expression gives,

$0$	$=$	$\left\{\right[\frac{4n}{(n-3)}{]}^{n/(n+1)}{x}^3{\}}^{(n+1)/n}{x}^{(n-3)/n}-c_0\left\{\right[\frac{4n}{(n-3)}{]}^{n/(n+1)}{x}^3{\}}^2-x^4$
	$=$	$\left[\frac{4n}{(n-3)}\right]x^4-c_0[\frac{4n}{(n-3)}{]}^{2n/(n+1)}{x}^6-x^4$
$\Rightarrow c_0[\frac{4n}{(n-3)}{]}^{2n/(n+1)}{x}^2$	$=$	$\left[\frac{4n}{(n-3)}\right]-1$
$\Rightarrow x^2$	$=$	$\left[\frac{15n}{4\pi (n+1)}\right]\left[\frac{3(n+1)}{(n-3)}\right][\frac{(n-3)}{4n}{]}^{2n/(n+1)}$
	$=$	$\left[\frac{3^2\cdot 5n}{4\pi (n-3)}\right][\frac{(n-3)}{4n}{]}^{2n/(n+1)}.$

The associated mass is,

$\left(\frac{M_{\mathrm{m}\mathrm{a}\mathrm{x}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}\right)=\left(\frac{4\pi }{3}\right)m$	$=$	$\frac{4\pi }{3}\left[\frac{n-3}{4n}{]}^{-n/(n+1)}\right[\frac{3^2\cdot 5n}{4\pi (n-3)}{]}^{3/2}[\frac{(n-3)}{4n}{]}^{3n/(n+1)}$
	$=$	$\frac{4\pi }{3}\left[\frac{n-3}{4n}{]}^{2n/(n+1)}\right[\frac{3^2\cdot 5n}{4\pi (n-3)}{]}^{3/2}.$

Notice that, for $n=3$,

$(\frac{M_{\mathrm{m}\mathrm{a}\mathrm{x}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}_{n=3}$

$=$

$[\frac{3^4\cdot 5^3}{2^8\pi }{]}^{1/2}.$

---

Another turning point occurs where the denominator is zero, that is,

$(n+1)m^{1/n}{x}^{(n-3)/n}$	$=$	$2c_{0}nm$
$\Rightarrow (n+1)x^{(n-3)/n}$	$=$	$2c_{0}n{m}^{(n-1)/n}$
$\Rightarrow \frac{x^{n-3}}{m^{n-1}}$	$=$	$[\frac{2c_{0}n}{(n+1)}{]}^n$
	$=$	$[\frac{8\pi }{15}{]}^n$
$\Rightarrow m$	$=$	$\left[x^{(n-3)/(n-1)}\right(\frac{15}{8\pi }{)}^{n/(n-1)}].$

Plugging this into the virial equilibrium expression gives,

$x^4$	$=$	$\left[x^{(n-3)/(n-1)}\right(\frac{15}{8\pi }{)}^{n/(n-1)}{]}^{(n+1)/n}{x}^{(n-3)/n}-c_0\left[x^{(n-3)/(n-1)}\right(\frac{15}{8\pi }{)}^{n/(n-1)}{]}^2$
	$=$	$(\frac{15}{8\pi }{)}^{(n+1)/(n-1)}{x}^{(n-3)(n+1)/[n(n-1)]}\cdot x^{(n-3)/n}-c_0(\frac{15}{8\pi }{)}^{2n/(n-1)}{x}^{2(n-3)/(n-1)}$
	$=$	$(\frac{15}{8\pi }{)}^{(n+1)/(n-1)}{x}^{2(n-3)/(n-1)}-[\frac{(n+1)}{2n}\left]\right(\frac{15}{8\pi }{)}^{(n+1)/(n-1)}{x}^{2(n-3)/(n-1)}$
	$=$	$\left(\frac{15}{8\pi }{)}^{(n+1)/(n-1)}{x}^{2(n-3)/(n-1)}\right[1-\frac{(n+1)}{2n}]$
$\Rightarrow x^{2(n+1)/(n-1)}$	$=$	$\frac{(n-1)}{2n}(\frac{15}{8\pi }{)}^{(n+1)/(n-1)}$
$\Rightarrow \frac{R_{\mathrm{m}\mathrm{a}\mathrm{x}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}$	$=$	$\left[\frac{(n-1)}{2n}{]}^{(n-1)/[2(n+1)]}\right(\frac{15}{8\pi }{)}^{1/2}.$

And the associated mass is,

$\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{|}_{\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}}$	$=$	$\frac{4\pi }{3}\left(\frac{15}{8\pi }{)}^{n/(n-1)}\right\{\left[\frac{(n-1)}{2n}{]}^{(n-1)/[2(n+1)]}\right(\frac{15}{8\pi }{)}^{1/2}{\}}^{(n-3)/(n-1)}$
	$=$	$\frac{4\pi }{3}\left(\frac{15}{8\pi }{)}^{3/2}\right[\frac{(n-1)}{2n}{]}^{(n-3)/[2(n+1)]}.$

Case P Summary

Case P
Order-of-Magnitude Analysis: Assume ${\widetilde{\mathfrak{𝔣}}}_M={\widetilde{\mathfrak{𝔣}}}_W={\widetilde{\mathfrak{𝔣}}}_A=1$
Virial Equilibrium:	$\left[\right(\frac{3}{4\pi }\left)\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}\left){]}^{(n+1)/n}\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-3)/n}-\frac{3}{20\pi }\left(\frac{n+1}{n}\right)(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2-(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4$ $=$ $0$
Dynamical Instability: $(n>3)$	$\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{|}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$ $=$ $\left[\frac{(n-3)}{4n}{]}^{n/(n+1)}\right[\frac{3^2\cdot 5n}{2^2\pi (n-3)}{]}^{1/2}$
Turning Point $(M_{\mathrm{m}\mathrm{a}\mathrm{x}})$: $(n>3)$	$\frac{M_{\mathrm{m}\mathrm{a}\mathrm{x}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}$ $=$ $\frac{4\pi }{3}\left[\frac{n-3}{4n}{]}^{2n/(n+1)}\right[\frac{3^2\cdot 5n}{4\pi (n-3)}{]}^{3/2}$ $\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{|}_{\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}}$ $=$ $\left[\frac{(n-3)}{4n}{]}^{n/(n+1)}\right[\frac{3^2\cdot 5n}{2^2\pi (n-3)}{]}^{1/2}$
Turning Point $(R_{\mathrm{m}\mathrm{a}\mathrm{x}})$: $(n>1)$	$\frac{R_{\mathrm{m}\mathrm{a}\mathrm{x}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}$ $=$ $\left[\frac{(n-1)}{2n}{]}^{(n-1)/[2(n+1)]}\right(\frac{15}{8\pi }{)}^{1/2}$ $\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{|}_{\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}}$ $=$ $\frac{4\pi }{3}\left(\frac{15}{8\pi }{)}^{3/2}\right[\frac{(n-1)}{2n}{]}^{(n-3)/[2(n+1)]}$

Figure 2
Stahler (1983) Title Page Stahler (1983) Title Page	Case P equilibrium sequences with key configurations highlighted
Stahler (1983) Figure 17 (edited) Stahler (1983) Figure 17 (edited)

The right-hand panel of Figure 2 presents substantial segments of Case P virial equilibrium sequences for a range of polytropic indexes (n = 1, 2, 2.8, 3, 3.5, 4, 5). For each sequence, the location of the $R_{\mathrm{m}\mathrm{a}\mathrm{x}}$ and $M_{\mathrm{m}\mathrm{a}\mathrm{x}}$ turning points — if they exist — are denoted by a yellow or red circular dot, respectively. The point along each $(n\ge 3)$ sequence at which the transition from dynamically stable to dynamically unstable structures occurs coincides with the location of $M_{\mathrm{m}\mathrm{a}\mathrm{x}}$ (i.e., with the red circular dot).

For display purposes, all normalized masses $(M_{\mathrm{t}\mathrm{o}\mathrm{t}}/M_{\mathrm{S}\mathrm{W}\mathrm{S}})$ have been further normalized to the maximum mass on the n = 3 sequence.

Detailed Force-Balance Models

Structural Form Factors

The following table of structural form factors has been drawn from here,

Structural Form Factors for Isolated Polytropes

Structural Form Factors for Pressure-Truncated Polytropes

${\mathfrak{𝔣}}_M$ $=$ $[-\frac{3{\theta }^{\text{'}}}{\xi }{]}_{{\xi }_1}$ ${\mathfrak{𝔣}}_W$ $=$ $\frac{3^2\cdot 5}{5-n}[\frac{{\theta }^{\text{'}}}{\xi }{]}_{{\xi }_1}^2$ ${\mathfrak{𝔣}}_A$ $=$ $\frac{3(n+1)}{(5-n)}[{\theta }^{\text{'}}{]}_{{\xi }_1}^2$

${\widetilde{\mathfrak{𝔣}}}_M$ $=$ $(-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }})$ ${\widetilde{\mathfrak{𝔣}}}_W$ $=$ $\frac{3\cdot 5}{(5-n){\tilde{\xi }}^2}[{\tilde{\theta }}^{n+1}+3({\tilde{\theta }}^{\text{'}}{)}^2-{\widetilde{\mathfrak{𝔣}}}_M\tilde{\theta }]$ ${\widetilde{\mathfrak{𝔣}}}_A$ $=$ $\frac{1}{(5-n)}\{6{\tilde{\theta }}^{n+1}+(n+1)[3({\tilde{\theta }}^{\text{'}}{)}^2-{\widetilde{\mathfrak{𝔣}}}_M\tilde{\theta }]\}$

and here,

Structural Form Factors for Pressure-Truncated n = 5 Polytropes

${\widetilde{\mathfrak{𝔣}}}_M$ $=$ $(1+{\ell }^2{)}^{-3/2}$ ${\widetilde{\mathfrak{𝔣}}}_W$ $=$ $\frac{5}{2^4}\cdot {\ell }^{-5}\left[\ell \right({\ell }^4-\frac{8}{3}{\ell }^2-1)(1+{\ell }^2{)}^{-3}+{\tan }^{-1}(\ell )]$ ${\widetilde{\mathfrak{𝔣}}}_A$ $=$ $\frac{3}{2^3}{\ell }^{-3}[{\tan }^{-1}(\ell )+\ell ({\ell }^2-1)(1+{\ell }^2{)}^{-2}]$ where,       $\ell \equiv \frac{\tilde{\xi }}{\sqrt{3}}$

Case M Equilibrium Conditions

Employing the renormalization factors,

$\frac{R_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$[\frac{4\pi }{(n+1{)}^n}{]}^{1/(n-3)},$
$\frac{P_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$[\frac{(n+1{)}^3}{4\pi }{]}^{(n+1)/(n-3)},$

we find from detailed force-balance analyses that the equilibrium radius and corresponding external pressure for "Case M" configurations are,

$\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}\cdot \frac{R_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$

$=$

$[\frac{4\pi }{(n+1{)}^n}{]}^{1/(n-3)}\tilde{\xi }(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{(1-n)/(n-3)},$

which matches the expression derived in an ASIDE box found with our introduction of the Lane-Emden equation, and

$\frac{P_{\mathrm{e}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\frac{P_{\mathrm{e}}}{P_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}\cdot \frac{P_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$

$=$

$[\frac{(n+1{)}^3}{4\pi }{]}^{(n+1)/(n-3)}{\tilde{\theta }}_n^{n+1}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{2(n+1)/(n-3)}.$

There are two turning points: One associated with a maximum in $P_e$ and one associated with a maximum in $R_{\mathrm{e}\mathrm{q}}$. According to Kimura's discussion, the first of these occurs in the configuration for which,

$\frac{{\tilde{\theta }}^{n+1}}{({\tilde{\theta }}^{\text{'}}{)}^2}$

$=$

$\frac{(n-3)}{2}$

…

For n = 5, this occurs when $\tilde{\xi }=3$

This point along the equilibrium sequence is identified by the dark green circular dot in Figure 3, below. The second occurs in the configuration for which,

$\frac{\tilde{\xi }{\tilde{\theta }}^n}{(-{\tilde{\theta }}^{\text{'}})}$

$=$

$\frac{(n-3)}{(n-1)}$

…

For n = 5, this occurs when $\tilde{\xi }=\sqrt{15}\approx 3.87298$

This point along the equilibrium sequence is identified by the yellow circular dot in Figure 3, below. In addition, we have identified the point of dynamical instability.

$(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^4$	$=$	$\frac{(n-3)}{3^2(n+1)}\cdot \frac{a}{c}=\frac{1}{3^3}\cdot \frac{3}{2^2\pi }(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\frac{3}{5}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}$
	$=$	$\frac{1}{2^2\cdot 3\cdot 5\pi }(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\frac{5}{2^4}(1+{\ell }^2{)}^3\cdot {\ell }^{-5}\left[\ell \right({\ell }^4-\frac{8}{3}{\ell }^2-1)(1+{\ell }^2{)}^{-3}+{\tan }^{-1}\ell ]$
$\Rightarrow \left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^4$	$=$	$\frac{1}{2^6\cdot 3\cdot \pi {\ell }^5}\left[\ell \right({\ell }^4-\frac{8}{3}{\ell }^2-1)+(1+{\ell }^2{)}^3{\tan }^{-1}\ell ]$

But the equilibrium condition for n = 5 configurations is,

$\left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^4$	$=$	$[\frac{(n+1{)}^3}{4\pi }{]}^{(n+1)/(n-3)}{\tilde{\theta }}_n^{n+1}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{2(n+1)/(n-3)}\left\{\right[\frac{4\pi }{(n+1{)}^n}{]}^{1/(n-3)}\tilde{\xi }(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{(1-n)/(n-3)}{\}}^4$
	$=$	$[\frac{2\cdot 3^3}{\pi }{]}^3{\tilde{\theta }}_n^6{\tilde{\xi }}^{12}(-{\tilde{\theta }}^{\prime }{)}^6\left\{\frac{1}{6}\right[\frac{\pi }{2\cdot 3^3}{]}^{1/2}{\tilde{\xi }}^{-3}(-{\tilde{\theta }}^{\prime }{)}^{-2}{\}}^4$
	$=$	$\frac{1}{2^4\cdot 3^4}\left[\frac{2\cdot 3^3}{\pi }\right]{\tilde{\theta }}_n^6(-{\tilde{\theta }}^{\prime }{)}^{-2}$
	$=$	$\frac{1}{2^3\cdot 3\pi }(1+{\ell }^2{)}^{-3}[\frac{\ell }{3^{1/2}}(1+{\ell }^2{)}^{-3/2}{]}^{-2}$
	$=$	$\frac{1}{2^3\pi }\cdot {\ell }^{-2}$

Putting the two expressions together gives,

$\frac{1}{2^3\pi }\cdot {\ell }^{-2}$	$=$	$\frac{1}{2^6\cdot 3\cdot \pi {\ell }^5}\left[\ell \right({\ell }^4-\frac{8}{3}{\ell }^2-1)+(1+{\ell }^2{)}^3{\tan }^{-1}\ell ]$
$\Rightarrow 2^3\cdot 3\cdot {\ell }^3$	$=$	${\ell }^5-\frac{8}{3}{\ell }^3-\ell +(1+{\ell }^2{)}^3{\tan }^{-1}\ell$
$\Rightarrow \left[\frac{(1+{\ell }^2{)}^3}{\ell }\right]{\tan }^{-1}\ell$	$=$	$1+\left[\frac{2^4\cdot 5}{3}\right]{\ell }^2-{\ell }^4$

This agrees with the expression derived in a separate ASIDE; as was pointed out in that context, the root of this equation is: $\ell \approx 2.223175$, that is, $\tilde{\xi }\approx 3.85065$. This point along the equilibrium sequence is identified by the red circular dot in Figure 3, below. It is almost — but definitely not — coincident with the configuration along the sequence (marked by the yellow circular dot) that is associated with the minimum-radius turning point.

Figure 3:   Case M Equilibrium Sequence
Case M pressure-truncated n=5 polytrope

Now for a movie!

Case P Equilibrium Conditions

The equilibrium radius and corresponding configuration mass from a "Case P" analysis are,

$\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}$	$=$	$(\frac{n}{4\pi }{)}^{1/2}\tilde{\xi }{\tilde{\theta }}_n^{(n-1)/2},$
$\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}$	$=$	$(\frac{n^3}{4\pi }{)}^{1/2}{\tilde{\theta }}_n^{(n-3)/2}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\text{'}}).$

According to our review of, especially, Kimura's work, the turning point associated with $M_{\mathrm{m}\mathrm{a}\mathrm{x}}$ occurs where,

$\frac{{\tilde{\theta }}^{n+1}}{({\tilde{\theta }}^{\text{'}}{)}^2}$

$=$

$\frac{(n-3)}{2}$

…

For n = 5, this occurs when $\tilde{\xi }=3.$

And a turning point associated with $R_{\mathrm{m}\mathrm{a}\mathrm{x}}$ occurs where,

$\frac{\tilde{\xi }(-{\tilde{\theta }}^{\text{'}})}{\tilde{\theta }}$

$=$

$\frac{2}{(n-1)}.$

For $n=5$ configurations, this means,

$\frac{1}{2}$	$=$	$\tilde{\xi }(1+\frac{{\tilde{\xi }}^2}{3}{)}^{1/2}\frac{\tilde{\xi }}{3}(1+\frac{{\tilde{\xi }}^2}{3}{)}^{-3/2}$
	$=$	$\frac{{\ell }^2}{(1+{\ell }^2)}$
$\Rightarrow (1+{\ell }^2)$	$=$	$2{\ell }^2$
$\Rightarrow {\ell }^2$	$=$	$1$
$\Rightarrow \tilde{\xi }$	$=$	$\sqrt{3}.$

Five-One Bipolytropes

Basic Properties

For bipolytropes, in general, let $n=n_c$ and $j=n_e$. The statement of virial equilibrium is, then,

$\frac{b}{n_{c}c}\cdot x_{\mathrm{e}\mathrm{q}}^{(n_c-3)/n_c}-\frac{a}{3c}+\frac{1}{n_e}\cdot x_{\mathrm{e}\mathrm{q}}^{(n_e-3)/n_e}$

$=$

$0.$

And the critical equilibrium configuration has,

$[x_{\mathrm{e}\mathrm{q}}^{(n_e-3)/n_e}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$

$=$

$\frac{a}{3^{2}c}\left[\frac{n_e^2(n_c-3)}{n_c-n_e}\right].$

Here we choose to set $n_c=5$ and $n_e=1$. Hence, these two conditions become, respectively,

$\frac{b}{5c}\cdot x_{\mathrm{e}\mathrm{q}}^{2/5}+x_{\mathrm{e}\mathrm{q}}^{-2}$

$=$

$\frac{a}{3c},$

and,

$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$

$=$

$[\frac{2\cdot 3^{2}c}{a}{]}^{1/2}.$

Comparing the general free-energy expression at the beginning of this chapter with the free-energy expression provided via our accompanying summary discussion of five-one bipolytropes, we find that,

$a$	$=$	$(3{\mathfrak{𝔏}}_i+12{\mathfrak{𝔎}}_i)x_{\mathrm{e}\mathrm{q}}\left[\right(\frac{3^6}{2^5\pi }{)}^{1/2}\frac{{\ell }_i^3}{(1+{\ell }_i^2{)}^3}]$
$b$	$=$	$(5{\mathfrak{𝔏}}_i)x_{\mathrm{e}\mathrm{q}}^{3/5}\left[\right(\frac{3^6}{2^5\pi }{)}^{1/2}\frac{{\ell }_i^3}{(1+{\ell }_i^2{)}^3}]$
$c$	$=$	$(4{\mathfrak{𝔎}}_i)x_{\mathrm{e}\mathrm{q}}^3\left[\right(\frac{3^6}{2^5\pi }{)}^{1/2}\frac{{\ell }_i^3}{(1+{\ell }_i^2{)}^3}]$

where,

${\mathfrak{𝔏}}_i$	$\equiv$	$\frac{({\ell }_i^4-1)}{{\ell }_i^2}+\frac{(1+{\ell }_i^2{)}^3}{{\ell }_i^3}\cdot {\tan }^{-1}{\ell }_i,$
${\mathfrak{𝔎}}_i$	$\equiv$	$\frac{{\mathrm{\Lambda }}_i}{{\eta }_i}+\frac{(1+{\mathrm{\Lambda }}_i^2)}{{\eta }_i}[\frac{\pi }{2}+{\tan }^{-1}{\mathrm{\Lambda }}_i],$
${\mathrm{\Lambda }}_i$	$\equiv$	$\frac{1}{{\eta }_i}-{\ell }_i,$
${\eta }_i$	$=$	$3\left(\frac{{\mu }_e}{{\mu }_c}\right)\left[\frac{{\ell }_i}{(1+{\ell }_i^2)}\right],$

and,

$x_{\mathrm{e}\mathrm{q}}=\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$(\frac{\pi }{2^3}{)}^{1/2}\frac{{\nu }^2}{q}\cdot \frac{(1+{\ell }_i^2{)}^3}{3^3{\ell }_i^5}$
	$=$	$\left(\frac{\pi }{2^3}{)}^{1/2}\frac{(1+{\ell }_i^2{)}^3}{3^3{\ell }_i^3}\right[\frac{1}{(1+{\mathrm{\Lambda }}_i^2)}\left]\right\{1+\frac{1}{{\eta }_i}[\frac{\pi }{2}+{\tan }^{-1}{\mathrm{\Lambda }}_i]{\}}^{-1}$

In rewriting this last expression, we have made use of the two relations,

$q\equiv \frac{r_i}{R}$	$=$	$\{1+\frac{1}{{\eta }_i}[\frac{\pi }{2}+{\tan }^{-1}{\mathrm{\Lambda }}_i]{\}}^{-1},$
$\nu \equiv \frac{M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	$=$	$\frac{{\ell }_{i}q}{(1+{\mathrm{\Lambda }}_i^2{)}^{1/2}}.$

Consistent with our generic discussion of the stability of bipolytropes and the specific discussion of the stability of bipolytropes having $(n_c,n_e)=(5,1)$, it can straightforwardly be shown that $\partial \mathfrak{𝔊}/\partial \chi =0$ is satisfied by setting $\mathrm{X}=1$; that is, the equilibrium condition is,

Furthermore, the equilibrium configuration is unstable whenever ${\partial }^2\mathfrak{𝔊}/\partial {\chi }^2<0$, that is, the transition from stable to unstable configurations whenever,

$\frac{{\mathfrak{𝔏}}_i}{{\mathfrak{𝔎}}_i}$

$=$

$20.$

Table 1 of an accompanying chapter — and the red-dashed curve in the figure adjacent to that table — identifies some key properties of the model that marks the transition from stable to unstable configurations along equilibrium sequences that have various values of the mean-molecular weight ratio, ${\mu }_e/{\mu }_c$.

Coincidence Between Points of Secular and Dynamical Instability

From the accompanying graphical display of equilibrium sequences, it seems that a ${\nu }_{\mathrm{m}\mathrm{a}\mathrm{x}}$ turning point will only exist in five-one bipolytropes for ${\mu }_e/{\mu }_c$ less than some value — call it, $({\mu }_e/{\mu }_c{)}_{\mathrm{b}\mathrm{e}\mathrm{g}\mathrm{i}\mathrm{n}}$ — which is less than but approximately equal to $\frac{1}{3}$. As we move along any sequence for which ${\mu }_e/{\mu }_c<({\mu }_e/{\mu }_c{)}_{\mathrm{b}\mathrm{e}\mathrm{g}\mathrm{i}\mathrm{n}}$, in the direction of increasing ${\ell }_i$, it is fair to ask whether the system becomes dynamically unstable (at $[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$) before or after it encounters the point of secular instability marked by ${\nu }_{\mathrm{m}\mathrm{a}\mathrm{x}}$.