SSC/Structure/PolytropesEmbedded/n5

Embedded Polytropic Spheres

Part I:   General Properties

Part II:  Truncated Configurations with n = 1

Part III:  Truncated Configurations with n = 5

Part IV:  Other Considerations

Truncated Configurations with n = 5

Drawing from the earlier discussion of isolated polytropes, we will reference various radial locations within a spherical $n$ = 5 polytrope by the dimensionless radius,

$\xi \equiv \frac{r}{a_{\mathrm{n}\mathrm{=}\mathrm{5}}},$

where,

$a_{n=5}=[\frac{(n+1)K}{4\pi G}{\rho }_c^{(1/n-1)}{]}_{n=5}^{1/2}=[\frac{3K}{2\pi G}{]}^{1/2}{\rho }_c^{-2/5}.$

The solution to the Lane-Emden equation for $n=5$ is,

${\theta }_5$

$=$

$(1+\frac{{\xi }^2}{3}{)}^{-1/2},$

hence,

$\frac{d{\theta }_5}{d\xi }$

$=$

$-\frac{\xi }{3}(1+\frac{{\xi }^2}{3}{)}^{-3/2}.$

Review

Again, from the earlier discussion, we can describe the properties of an isolated, spherical $n$ = 5 polytrope as follows:

• Mass:

In terms of the central density, ${\rho }_c$, and $K_{\mathrm{n}}$, the total mass is,

$M=[\frac{2\cdot 3^4{K}^3}{\pi G^3}{]}^{1/2}{\rho }_c^{-1/5}$ ;

and, expressed as a function of $M$, the mass that lies interior to the dimensionless radius $\xi$ is,

$\frac{M_{\xi }}{M}={\xi }^3(3+{\xi }^2{)}^{-3/2}.$

Hence,

$M_{\xi }=\left[\frac{2\cdot 3^4{K}^3}{\pi G^3}{]}^{1/2}{\rho }_c^{-1/5}\right[{\xi }^3(3+{\xi }^2{)}^{-3/2}].$

• Pressure:

The central pressure of the configuration is,

$P_c=[\frac{\pi M^2{G}^3}{2\cdot 3^4}{]}^{1/3}{\rho }_c^{4/3}=[\frac{\pi G^3}{2\cdot 3^4}\left(\frac{2\cdot 3^4{K}^3}{\pi G^3}\right){\rho }_c^{-2/5}{]}^{1/3}{\rho }_c^{4/3}=K{\rho }_c^{6/5}$ ;

and, expressed in terms of the central pressure $P_c$, the variation with radius of the pressure is,

$P_{\xi }=P_c[1+\frac{1}{3}{\xi }^2{]}^{-3}$ .

Hence,

$P_{\xi }=K{\rho }_c^{6/5}[1+\frac{1}{3}{\xi }^2{]}^{-3}=3^{3}K{\rho }_c^{6/5}[3+{\xi }^2{]}^{-3}$ .

Extension to Bounded Sphere

Eliminating ${\rho }_c$ between the last expression for $M_{\xi }$ and the last expression for $P_{\xi }$ gives,

$P_{\xi }$	$=$	$3^{3}K[3+{\xi }^2{]}^{-3}[\frac{2\cdot 3^4{K}^3}{\pi G^3}{]}^3{M}_{\xi }^{-6}[{\xi }^3(3+{\xi }^2{)}^{-3/2}{]}^6$
	$=$	$\left(\frac{2^3\cdot 3^{15}{K}^{10}}{{\pi }^3{M}_{\xi }^6{G}^9}\right)\frac{{\xi }^{18}}{(3+{\xi }^2{)}^{12}}.$

Now, if we rip off an outer layer of the star down to some dimensionless radius ${\xi }_e<\mathrm{\infty }$, the interior of the configuration that remains — containing mass $M_{{\xi }_e}$ — should remain in equilibrium if we impose the appropriate amount of externally applied pressure $P_e=P_{{\xi }_e}$ at that radius. (This will work only for spherically symmetric configurations, as the gravitation acceleration at any location only depends on the mass contained inside that radius.) If we rescale our solution such that the mass enclosed within ${\xi }_e$ is the original total mass $M$, then the pressure that must be imposed by the external medium in which the configuration is embedded is,

$P_e=\left(\frac{2^3\cdot 3^{15}{K}^{10}}{{\pi }^3{M}^6{G}^9}\right)\frac{{\xi }_e^{18}}{(3+{\xi }_e^2{)}^{12}}$ .

The associated equilibrium radius of this pressure-confined configuration is,

$R_{\mathrm{e}\mathrm{q}}={\xi }_e{a}_{\mathrm{n}\mathrm{=}\mathrm{5}}=[\frac{3K}{2\pi G}{]}^{1/2}{\rho }_c^{-2/5}{\xi }_e=[\frac{\pi M^4{G}^5}{2^3\cdot 3^7{K}^5}{]}^{1/2}\frac{(3+{\xi }_e^2{)}^3}{{\xi }_e^5}.$

Overlap with Whitworth's Presentation

The curve labeled $n=5$ in the top two panels of Figure 1 shows how $R_{\mathrm{e}\mathrm{q}}$ varies with the applied external pressure $P_e$; as shown, the curve has two segments — configurations that are stable (blue diamonds) and configurations that are unstable (red squares). Following the lead of Whitworth (1981, MNRAS, 195, 967) — for clarification, read the accompanying ASIDE — these two quantities have been respectively normalized (or, "referenced") to,

$R_{\mathrm{r}\mathrm{f}}{|}_{\mathrm{n}\mathrm{=}\mathrm{5}}\equiv \frac{2^6}{3^3}\left(\frac{\pi }{5^5}{)}^{1/2}\right[\frac{G^5{M}^4}{K^5}{]}^{1/2}\Rightarrow \frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{r}\mathrm{f}}}=(\frac{5^5}{2^{15}\cdot 3}{)}^{1/2}\frac{(3+{\xi }_e^2{)}^3}{{\xi }_e^5};$

and,

$P_{\mathrm{r}\mathrm{f}}{|}_{\mathrm{n}\mathrm{=}\mathrm{5}}\equiv \frac{3^{12}{5}^9}{2^{26}{\pi }^3}\left(\frac{K^{10}}{G^9{M}^6}\right)\Rightarrow \frac{P_e}{P_{\mathrm{r}\mathrm{f}}}=\left(\frac{2^{29}\cdot 3^3}{5^9}\right)\frac{{\xi }_e^{18}}{(3+{\xi }_e^2{)}^{12}}.$

We see that this $n=5$ model sequence bends back on itself. That is to say, for this polytropic index there is an externally applied pressure above which no equilibrium configuration exists. This limiting pressure arises along the curve where,

$\frac{d{P}_e}{d{R}_{\mathrm{e}\mathrm{q}}}=\left(\frac{d{P}_e}{d{\xi }_e}\right)(\frac{d{R}_{\mathrm{e}\mathrm{q}}}{d{\xi }_e}{)}^{-1}=0.$

Evaluation of this expression shows that the limiting pressure occurs precisely at ${\xi }_e=3$, that is,

$(\frac{P_e}{P_{\mathrm{r}\mathrm{f}}}{)}_{\mathrm{m}\mathrm{a}\mathrm{x}}=(\frac{2^{29}\cdot 3^3}{5^9})\frac{3^{18}}{12^{12}}=\frac{2^5\cdot 3^9}{5^9},$

and the radius of this limiting configuration is,

$\left(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{r}\mathrm{f}}}\right)=(\frac{5^5}{2^{15}\cdot 3}{)}^{1/2}\frac{12^3}{3^5}=(\frac{5^5}{2^3\cdot 3^5}{)}^{1/2}.$

On the log-log plot displayed in the top-right panel of Figure 1, the location of this special point is $[\log (P_e/P_{\mathrm{r}\mathrm{f}}),\log (R_{\mathrm{e}\mathrm{q}}/R_{\mathrm{r}\mathrm{f}})]\approx [-0.49149,+0.10308].$

We note as well that a conversion from Whitworth's normalizations to the normalizations adopted by Horedt produce the following coordinates for the limiting model configuration:

$p_a{|}_{\mathrm{m}\mathrm{a}\mathrm{x}}$

$=$

$\frac{3^{12}}{2^{24}},$

and, at this bounding pressure, the model has an equilibrium radius,

$r_a$

$=$

$\frac{2^6}{3^3}.$

Overlap with Stahler's Presentation

We can invert the above expression for $P_e(K,M)$ to obtain the following expression for $M(K,P_e)$:

$M^6=\left(\frac{2^3\cdot 3^{15}{K}^{10}}{{\pi }^3{P}_e{G}^9}\right)\frac{{\xi }_e^{18}}{(3+{\xi }_e^2{)}^{12}}$ .

If, following Stahler's lead, we normalize this expression by $M_{\mathrm{S}\mathrm{W}\mathrm{S}}$ (evaluated for $n=5$) and we normalize the above expression for $R_{\mathrm{e}\mathrm{q}}$ by $R_{\mathrm{S}\mathrm{W}\mathrm{S}}$ (evaluated for $n=5$), we obtain,

$\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}$	$=$	$\left(\frac{2^3\cdot 3^{15}{K}^{10}}{{\pi }^3{P}_e{G}^9}{)}^{1/6}\frac{{\xi }_e^3}{(3+{\xi }_e^2{)}^2}\right[(\frac{2\cdot 3}{5G}{)}^{3/2}{K}^{5/3}{P}_{\mathrm{e}\mathrm{x}}^{-1/6}{]}^{-1}$
	$=$	$(\frac{3^2\cdot 5^3}{4\pi }{)}^{1/2}\frac{{\xi }_e^3}{(3+{\xi }_e^2{)}^2},$
$\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}$	$=$	$\left[\frac{\pi M^4{G}^5}{2^3\cdot 3^7{K}^5}{]}^{1/2}\frac{(3+{\xi }_e^2{)}^3}{{\xi }_e^5}\right[(\frac{2\cdot 3}{5G}{)}^{1/2}{K}^{5/6}{P}_{\mathrm{e}\mathrm{x}}^{-1/3}{]}^{-1}$
	$=$	$\left[\right(\frac{2^3\cdot 3^{15}{K}^{10}}{{\pi }^3{P}_e{G}^9}{)}^{1/3}\frac{{\xi }_e^6}{(3+{\xi }_e^2{)}^4}\left]\right[\frac{\pi G^5}{2^3\cdot 3^7{K}^5}{]}^{1/2}\frac{(3+{\xi }_e^2{)}^3}{{\xi }_e^5}\left(\frac{5G}{2\cdot 3}{)}^{1/2}\right[K^{-5/6}{P}_{\mathrm{e}\mathrm{x}}^{1/3}]$
	$=$	$(\frac{3^2\cdot 5}{2^2\pi }{)}^{1/2}\frac{{\xi }_e}{(3+{\xi }_e^2)}.$

This set of parametric relations that relate the mass of the truncated configuration to its radius via the parameter, ${\xi }_e$, has been recorded to the immediate right of Stahler's name in our $n=5$ summary table, below.

Stahler points out (see his equation B13) that, for this particular pressure-bounded polytropic sequence, ${\xi }_e$ can be eliminated between the expressions to obtain the following direct algebraic relationship between $M$ and $R_{\mathrm{e}\mathrm{q}}$:

$(\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2-5(\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}\left)\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}})+\frac{20\pi }{3}(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4$

$=$

$0.$

Viewed as a quadratic equation in the mass, the roots of this expression give,

$\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}$

$=$

$\frac{5}{2}\left(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}\right)\{1\pm [1-\frac{16\pi }{15}\left(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2{]}^{1/2}\right\}.$

[CORRECTION: Changed factor inside square root from $16\pi /3$ to $16\pi /15$ on 24 December 2014.] We have used this expression to generate the complete $n=5$ sequence shown here in the top panel of Figure 2 — the solid green segment of the curve shows the negative root and the solid red segment of the curve was generated using the positive root.

ASIDE: In his Appendix B, Stahler (1983) claims that the quadratic equation relating $M$ directly to $R_{\mathrm{e}\mathrm{q}}$ (his equation B13) can be obtained by analytically integrating the first-order ordinary differential equation presented as his equation B10. I don't think that this is possible without knowing ahead of time how $M$ relates to $R_{\mathrm{e}\mathrm{q}}$ through the above-derived parametric relations in ${\xi }_e$.

[29 September 2014 by J. E. Tohline] Now that (I think) I've finished deriving the properly defined virial equilibrium condition for embedded polytropes and have reconciled that equilibrium expression with Horedt's corresponding specification of the equilibrium radius and surface-pressure, it's time to revisit the concern that was expressed in this "ASIDE" regarding the mass-radius relationship for embedded, $n=5$ polytropes presented by Stahler.

Tabular Summary (n=5)

Table 2:  Properties of $n=5$ Polytropes Embedded in an External Medium of Pressure $P_e$ (and, accordingly, truncated at radius ${\xi }_e$)
${\theta }_5=(1+\frac{{\xi }_e^2}{3}{)}^{-1/2}$         and         $\frac{d{\theta }_5}{d\xi }{|}_{{\xi }_e}=-\frac{{\xi }_e}{3}(1+\frac{{\xi }_e^2}{3}{)}^{-3/2}$
📚 Horedt (1970) for fixed $(M,K_n)$	$r_a=\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}=\left\{3\right[\frac{({\xi }_e^2/3{)}^5}{(1+{\xi }_e^2/3{)}^6}]{\}}^{-1/2}$	$p_a=\frac{P_e}{P_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}=3^3[\frac{({\xi }_e^2/3{)}^3}{(1+{\xi }_e^2/3{)}^4}{]}^3$
📚 Whitworth (1981) for fixed $(M,K_n)$	$\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{r}\mathrm{f}}}=\left\{\frac{2^{15}}{5^5}\right[\frac{({\xi }_e^2/3{)}^5}{(1+{\xi }_e^2/3{)}^6}]{\}}^{-1/2}$	$\frac{P_e}{P_{\mathrm{r}\mathrm{f}}}=\frac{2^{29}}{5^9}[\frac{({\xi }_e^2/3{)}^3}{(1+{\xi }_e^2/3{)}^4}{]}^3$
📚 Stahler (1983) for fixed $(P_e,K_n)$	$\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}=\left\{\frac{3\cdot 5}{2^2\pi }\right[\frac{{\xi }_e^2/3}{(1+{\xi }_e^2/3{)}^2}]{\}}^{1/2}$	$\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}=\left[\right(\frac{3\cdot 5^3}{2^2\pi })\frac{({\xi }_e^2/3{)}^3}{(1+{\xi }_e^2/3{)}^4}{]}^{1/2}$
	$(\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2-5(\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}\left)\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}})+\frac{2^2\cdot 5\pi }{3}(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4=0$
NOTE: None of the analytic expressions for the dimensionless radius, pressure, or mass presented in this table explicitly appear in the referenced articles by Horedt, by Whitworth, or by Stahler but, as is discussed fully above, they are straightforwardly derivable from the more general relations that appear in these papers. The final polynomial relating the dimensionless mass to the dimensionless radius does explicitly appear as equation (B13) in 📚 Stahler (1983). Additional discussion of Stahler's analytic mass-radius relation is presented in an accompanying chapter.

Equilibrium Sequences

Example Sequences

Pulling from, and setting n = 5 in, our above discussion of Chieze's presentation, we find the following … $\frac{P_e}{P_{\mathrm{C}\mathrm{h}}}$ $=$ ${\theta }^{n+1}=(1+\frac{{\xi }^2}{3}{)}^{-3}$ $\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{C}\mathrm{h}}}$ $=$ $[\frac{3}{2\pi }{]}^{1/2}\xi$ $\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{C}\mathrm{h}}}$ $=$ $[\frac{(n+1{)}^3}{4\pi }{]}^{1/2}(-{\xi }^2{\theta }^{\text{'}})=[\frac{2^3\cdot 3^3}{2^2\pi }{]}^{1/2}\left\{\frac{{\xi }^3}{3}\right(1+\frac{{\xi }^2}{3}{)}^{-3/2}\}=[\frac{2\cdot 3}{\pi }{]}^{1/2}\left\{{\xi }^3\right(1+\frac{{\xi }^2}{3}{)}^{-3/2}\}$ where, $P_{\mathrm{C}\mathrm{h}}$ $\equiv$ $K{\rho }_c^{(n+1)/n}=K{\rho }_c^{6/5}$ $R_{\mathrm{C}\mathrm{h}}$ $\equiv$ $\left[\right(\frac{K}{G}){\rho }_c^{1/n-1}{]}^{1/2}=[\left(\frac{K}{G}\right){\rho }_c^{-4/5}{]}^{1/2}$ $M_{\mathrm{C}\mathrm{h}}$ $\equiv$ $\left[\right(\frac{K}{G}{)}^3{\rho }_c^{-2/5}{]}^{1/2}.$ Hence, $P_e$ $=$ $(1+\frac{{\xi }^2}{3}{)}^{-3}K{\rho }_c^{6/5}$ $R_{\mathrm{e}\mathrm{q}}$ $=$ $\left[\frac{3}{2\pi }{]}^{1/2}\right(\frac{K}{G}{)}^{1/2}{\rho }_c^{-2/5}\xi$ $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ $=$ $\left[\frac{2\cdot 3}{\pi }{]}^{1/2}\right(\frac{K}{G}{)}^{3/2}{\rho }_c^{-1/5}{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-3/2}$

• External Pressure vs. Volume (fixed mass); displayed in panel "a" of Figure 3:

$P_e\left[G^9{K}^{-10}{M}^6\right]$	$=$	$\left(\frac{2\cdot 3}{\pi }{)}^3{\xi }^{18}\right(1+\frac{{\xi }^2}{3}{)}^{-12}$
$\left(\frac{4\pi R^3}{3}\right)\left[\right(\frac{K}{G}{)}^{15/2}{M}^{-6}]$	$=$	$\left(\frac{\pi }{2\cdot 3}{)}^{5/2}{\xi }^{-15}\right(1+\frac{{\xi }^2}{3}{)}^9$

• Mass vs. Radius (fixed external pressure); displayed in panel "b" of Figure 3:

$M\left[G^{3/2}{K}^{-5/3}{P}_e^{1/6}\right]$	$=$	$\left(\frac{2\cdot 3}{\pi }{)}^{1/2}{\xi }^3\right(1+\frac{{\xi }^2}{3}{)}^{-2}$
$R\left[G^{1/2}{K}^{-5/6}{P}_e^{1/3}\right]$	$=$	$\left(\frac{3}{2\pi }{)}^{1/2}\xi \right(1+\frac{{\xi }^2}{3}{)}^{-1}$

• Mass vs. Central Density (fixed external pressure); displayed in panel "c" of Figure 3:

$M\left[G^{3/2}{K}^{-5/3}{P}_e^{1/6}\right]$	$=$	$\left(\frac{2\cdot 3}{\pi }{)}^{1/2}{\xi }^3\right(1+\frac{{\xi }^2}{3}{)}^{-2}$
${\rho }_c[K{P}_e^{-1}{]}^{5/6}$	$=$	$(1+\frac{{\xi }^2}{3}{)}^{5/2}$

• Mass vs. Central Density (fixed radius); displayed in panel "d" of Figure 3:

$R_{\mathrm{e}\mathrm{q}}$	$=$	$\left[\frac{3}{2\pi }{]}^{1/2}\right(\frac{K}{G}{)}^{1/2}{\rho }_c^{-2/5}\xi$
$\Rightarrow {\rho }_c^{2/5}$	$=$	$\left[\frac{3}{2\pi }{]}^{1/2}\right(\frac{K}{G}{)}^{1/2}{R}^{-1}\xi$
$\Rightarrow {\rho }_c$	$=$	$\left[\frac{3}{2\pi }{]}^{5/4}\right[(\frac{K}{G}{)}^{1/2}{R}^{-1}{]}^{5/2}{\xi }^{5/2}$
$M$	$=$	$\left[\frac{2\cdot 3}{\pi }{]}^{1/2}\right(\frac{K}{G}{)}^{3/2}\left\{\right[\frac{3}{2\pi }{]}^{5/4}\left[\right(\frac{K}{G}{)}^{1/2}{R}^{-1}{]}^{5/2}{\xi }^{5/2}{\}}^{-1/5}{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-3/2}$
	$=$	$\left[\frac{2\cdot 3}{\pi }{]}^{1/2}\right(\frac{K}{G}{)}^{3/2}\left\{\right[\frac{3}{2\pi }{]}^{-1/4}\left[\right(\frac{K}{G}{)}^{1/2}{R}^{-1}{]}^{-1/2}{\xi }^{-1/2}\left\}{\xi }^3\right(1+\frac{{\xi }^2}{3}{)}^{-3/2}$
	$=$	$\left[\frac{2^2\cdot 3^2}{{\pi }^2}{]}^{1/4}\right[\frac{2\pi }{3}{]}^{1/4}\left(\frac{K}{G}{)}^{3/2}\right[\left(\frac{K}{G}{)}^{-1/4}{R}^{1/2}\right]{\xi }^{5/2}(1+\frac{{\xi }^2}{3}{)}^{-3/2}$
	$=$	$\left[\frac{2^3\cdot 3}{\pi }{]}^{1/4}\right[\left(\frac{K}{G}{)}^5{R}^2{]}^{1/4}{\xi }^{5/2}\right(1+\frac{{\xi }^2}{3}{)}^{-3/2}$

Figure 3:   Equilibrium Sequences of Pressure-Truncated, n = 5 Polytropic Spheres (viewed from several different astrophysical perspectives)
●	${\xi }_e$	†External Pressure vs. Volume (Fixed Mass)	Mass vs. Radius (Fixed External Pressure)	‡Mass vs. Central Density (Fixed External Pressure)	Mass vs. Central Density (Fixed Radius)
●	√3	(a) Pressure-Truncated Isothermal Equilibrium Sequence	(b) Pressure-Truncated Isothermal Equilibrium Sequence	(c) Pressure-Truncated Isothermal Equilibrium Sequence	(d) Pressure-Truncated Isothermal Equilibrium Sequence
●	3
●	√15
●	9.01
		$\left(\frac{2\cdot 3}{\pi }{)}^3\right[{\xi }^{18}(1+\frac{{\xi }^2}{3}{)}^{-12}{]}_{\tilde{\xi }}$ vs. $\left(\frac{\pi }{2\cdot 3}{)}^{5/2}\right[{\xi }^{-15}(1+\frac{{\xi }^2}{3}{)}^9{]}_{\tilde{\xi }}$	$\left(\frac{2\cdot 3}{\pi }{)}^{1/2}\right[{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-2}{]}_{\tilde{\xi }}$ vs. $\left(\frac{3}{2\pi }{)}^{1/2}\right[\xi (1+\frac{{\xi }^2}{3}{)}^{-1}{]}_{\tilde{\xi }}$	$\left(\frac{2\cdot 3}{\pi }{)}^{1/2}\right[{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-2}{]}_{\tilde{\xi }}$ vs. $\left[\right(1+\frac{{\xi }^2}{3}{)}^{5/2}{]}_{\tilde{\xi }}$	$\left[\frac{2^3\cdot 3}{\pi }{]}^{1/4}\right[{\xi }^{5/2}(1+\frac{{\xi }^2}{3}{)}^{-3/2}{]}_{\tilde{\xi }}$ vs. $[\frac{3}{2\pi }{]}^{5/4}{\tilde{\xi }}^{5/2}$

Example Extrema

1. External Pressure vs. Volume (fixed mass):
   1. Maximum $P_e$ (green circular marker) …

      $0$	$=$	$\frac{d}{d\xi }\left\{{\xi }^{18}\right(1+\frac{{\xi }^2}{3}{)}^{-12}\}$
      	$=$	$18{\xi }^{17}(1+\frac{{\xi }^2}{3}{)}^{-12}-12{\xi }^{18}(1+\frac{{\xi }^2}{3}{)}^{-13}\frac{2\xi }{3}$
      	$=$	${\xi }^{17}(1+\frac{{\xi }^2}{3}{)}^{-13}[18(1+\frac{{\xi }^2}{3})-8{\xi }^2]$
      	$=$	${\xi }^{17}(1+\frac{{\xi }^2}{3}{)}^{-13}[18-2{\xi }^2]$
      $\Rightarrow {\xi }_{\mathrm{g}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{n}}^2$	$=$	$9.$

   2. Minimum Volume (purple circular marker) …

      $0$	$=$	$\frac{d}{d\xi }\left\{{\xi }^{-15}\right(1+\frac{{\xi }^2}{3}{)}^9\}$
      	$=$	$-15{\xi }^{-16}(1+\frac{{\xi }^2}{3}{)}^9+9{\xi }^{-15}(1+\frac{{\xi }^2}{3}{)}^8\frac{2\xi }{3}$
      	$=$	${\xi }^{-16}(1+\frac{{\xi }^2}{3}{)}^8[-15(1+\frac{{\xi }^2}{3})+6{\xi }^2]$
      	$=$	${\xi }^{-16}(1+\frac{{\xi }^2}{3}{)}^8[{\xi }^2-15]$
      $\Rightarrow {\xi }_{\mathrm{p}\mathrm{u}\mathrm{r}\mathrm{p}\mathrm{l}\mathrm{e}}^2$	$=$	$15.$

2. Mass vs. Central Density (fixed radius):
   1. Maximum mass (??? circular marker) …

      $0$	$=$	$\frac{d}{d\xi }\left\{{\xi }^{5/2}\right(1+\frac{{\xi }^2}{3}{)}^{-3/2}\}$
      	$=$	$\frac{5}{2}{\xi }^{3/2}(1+\frac{{\xi }^2}{3}{)}^{-3/2}-\frac{3}{2}{\xi }^{5/2}(1+\frac{{\xi }^2}{3}{)}^{-5/2}\frac{2\xi }{3}$
      	$=$	$\frac{{\xi }^{3/2}}{6}(1+\frac{{\xi }^2}{3}{)}^{-5/2}[5(3+{\xi }^2)-6{\xi }^2]$
      	$=$	$\frac{{\xi }^{3/2}}{6}(1+\frac{{\xi }^2}{3}{)}^{-5/2}[15-{\xi }^2]$
      $\Rightarrow {\xi }^2$	$=$	$15={\xi }_{\mathrm{p}\mathrm{u}\mathrm{r}\mathrm{p}\mathrm{l}\mathrm{e}}^2.$

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