SSC/Virial/Polytropes

Virial Equilibrium of Adiabatic Spheres

Highlights of the rather detailed discussion presented below have been summarized in an accompanying chapter of this H_Book.

Review

Adopted Normalizations

In an introductory discussion of the virial equilibrium structure of spherically symmetric configurations, we adopted the following physical parameter normalizations for adiabatic systems.

Adopted Normalizations for Adiabatic Systems

$R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ $\equiv$ $\left[\right(\frac{G}{K})M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{2-{\gamma }_g}{]}^{1/(4-3{\gamma }_g)}$ $P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ $\equiv$ $[\frac{K^4}{G^{3{\gamma }_g}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{2{\gamma }_g}}{]}^{1/(4-3{\gamma }_g)}$ $E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ $\equiv$ $P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3=[K{G}^{3(1-{\gamma }_g)}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{6-5{\gamma }_g}{]}^{1/(4-3{\gamma }_g)}$ ${\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ $\equiv$ $\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}=\frac{3}{4\pi }[\frac{K^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}^{1/(4-3{\gamma }_g)}$ $c_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^2$ $\equiv$ $\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\frac{4\pi }{3}[\frac{K}{(G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2{)}^{{\gamma }_g-1}}{]}^{1/(4-3{\gamma }_g)}$

Virial Equilibrium

Also in our introductory discussion — see especially the section titled, Energy Extrema — we deduced that an adiabatic system's dimensionless equilibrium radius,

${\chi }_{\mathrm{e}\mathrm{q}}\equiv \frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}},$

is given by the root(s) of the following equation:

$2C{\chi }_{\mathrm{e}\mathrm{q}}^{-2}+3B{\chi }_{\mathrm{e}\mathrm{q}}^{3-3{\gamma }_g}-3A{\chi }_{\mathrm{e}\mathrm{q}}^{-1}-3D{\chi }_{\mathrm{e}\mathrm{q}}^3=0,$

where the definitions of the various coefficients are,

$A$	$\equiv$	$\frac{1}{5}\cdot \left[\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\mathfrak{𝔣}}_M}{]}^2\cdot {\mathfrak{𝔣}}_W,$
$B$	$\equiv$	$\frac{4\pi }{3}\left[\frac{3}{4\pi }\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\mathfrak{𝔣}}_M}{]}_{\mathrm{e}\mathrm{q}}^{\gamma }\cdot {\mathfrak{𝔣}}_A$
	$=$	$\frac{4\pi }{3}\left[\right(\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}){\chi }^{3\gamma }{]}_{\mathrm{e}\mathrm{q}}\cdot {\mathfrak{𝔣}}_A,$
$C$	$\equiv$	$\frac{3\cdot 5}{2^4\pi }\left[\frac{J^2{c}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^2}{G^2{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^4}\right]\cdot \frac{{\mathfrak{𝔣}}_T}{{\mathfrak{𝔣}}_M},$
$D$	$\equiv$	$\left(\frac{4\pi }{3}\right)\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}.$

(The dimensionless structural form factors, ${\mathfrak{𝔣}}_i,$ that appear in these expressions are defined for isolated polytropes in our accompanying introductory discussion and are discussed further, below.) Once the pressure exerted by the external medium ($P_e$), and the configuration's mass ($M_{\mathrm{t}\mathrm{o}\mathrm{t}}$), angular momentum ($J$), and specific entropy (via $K$) have been specified, the values of all of the coefficients are known and ${\chi }_{\mathrm{e}\mathrm{q}}$ can be determined.

Isolated Nonrotating Adiabatic Configuration

For a nonrotating configuration $(C=J=0)$ that is not influenced by the effects of a bounding external medium $(D=P_e=0)$, the statement of virial equilibrium is,

$3B{\chi }_{\mathrm{e}\mathrm{q}}^{3-3{\gamma }_g}-3A{\chi }_{\mathrm{e}\mathrm{q}}^{-1}=0.$

Hence, one equilibrium state exists for each value of ${\gamma }_g$ and it occurs where,

${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_g}=(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{4-3{\gamma }_g}$

$=$

$\frac{A}{B}.$

Two Points of View
In terms of $K$ and $M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}(=M_{\mathrm{t}\mathrm{o}\mathrm{t}})$	In terms of $P_c$ and $M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}(=M_{\mathrm{t}\mathrm{o}\mathrm{t}})$
${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_g}$ $=$ $\{\frac{1}{5}\cdot [\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{{\mathfrak{𝔣}}_M}{]}^2\cdot {\mathfrak{𝔣}}_W\}$     $\times \left\{\frac{3}{4\pi }\right[\frac{3}{4\pi }\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{{\mathfrak{𝔣}}_M}{]}_{\mathrm{e}\mathrm{q}}^{-{\gamma }_g}\cdot \frac{1}{{\mathfrak{𝔣}}_A}\}$ $\Rightarrow R_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_g}$ $=$ $\frac{4\pi }{3\cdot 5}\left[\frac{3}{4\pi }\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\mathfrak{𝔣}}_M}{]}_{\mathrm{e}\mathrm{q}}^{2-{\gamma }_g}\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_A}[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{2-{\gamma }_g}}{K}]$ $\Rightarrow \frac{K{R}_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_g}}{G{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^{2-{\gamma }_g}}$ $=$ $\frac{1}{5}(\frac{4\pi }{3}{)}^{{\gamma }_g-1}\frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_A{\mathfrak{𝔣}}_M^{2-{\gamma }_g}}$ — — — — — —     or, inverted and setting ${\gamma }_g=1+1/n$     — — — — — — $4\pi (\frac{G}{K}{)}^n{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^{n-1}{R}_{\mathrm{e}\mathrm{q}}^{3-n}=(\frac{5{\mathfrak{𝔣}}_A{\mathfrak{𝔣}}_M}{{\mathfrak{𝔣}}_W}{)}^n\frac{3}{{\mathfrak{𝔣}}_M}$	${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_g}$ $=$ $\{\frac{1}{5}\cdot [\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{{\mathfrak{𝔣}}_M}{]}^2\cdot {\mathfrak{𝔣}}_W\}$     $\times \left\{\frac{3}{4\pi }\right[\left(\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{P_c}\right){\chi }^{-3\gamma }{]}_{\mathrm{e}\mathrm{q}}\cdot \frac{1}{{\mathfrak{𝔣}}_A}\}$ $\Rightarrow {\chi }_{\mathrm{e}\mathrm{q}}^4$ $=$ $\frac{3}{20\pi }\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2\right(\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{P_c})\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_A\cdot {\mathfrak{𝔣}}_M^2}$ $\Rightarrow \frac{P_c{R}_{\mathrm{e}\mathrm{q}}^4}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^4}(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{)}^2$ $=$ $\frac{3}{20\pi }\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_A\cdot {\mathfrak{𝔣}}_M^2}$ $\Rightarrow \frac{P_c{R}_{\mathrm{e}\mathrm{q}}^4}{G{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^2}$ $=$ $\frac{3}{20\pi }\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_A\cdot {\mathfrak{𝔣}}_M^2}$

According to the solution shown in the left-hand column, for fluid with a given specify entropy content, the equilibrium mass-radius relationship for adiabatic configurations is,

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{({\gamma }_g-2)}\propto R_{\mathrm{e}\mathrm{q}}^{(3{\gamma }_g-4)}.$

We see that, for ${\gamma }_g=2$, the equilibrium radius depends only on the specific entropy of the gas and is independent of the configuration's mass. Conversely, for ${\gamma }_g=4/3$, the mass of the configuration is independent of the radius. For ${\gamma }_g>2$ or ${\gamma }_g<4/3$, configurations with larger mass (but the same specific entropy) have larger equilibrium radii. However, for ${\gamma }_g$ in the range, $2>{\gamma }_g>4/3$, configurations with larger mass have smaller equilibrium radii. (Note that the related result for isothermal configurations can be obtained by setting ${\gamma }_g=1$ in this adiabatic solution, because $K=c_s^2$ when ${\gamma }_g=1$.)

Role of Structural Form Factors

When employing a virial analysis to determine the radius of an equilibrium configuration, it is customary to set the structural form factors, ${\mathfrak{𝔣}}_M$, ${\mathfrak{𝔣}}_W$ and ${\mathfrak{𝔣}}_A$, to unity and accept that the expression derived for $R_{\mathrm{e}\mathrm{q}}$ is an estimate of the configuration's radius that is good to within a factor of order unity. As has been demonstrated in our related discussion of the equilibrium of uniform-density spheres, these form factors can be evaluated if/when the internal structural profile of an equilibrium configuration is known from a complementary detailed force-balance analysis. In the case being discussed here of isolated, spherical polytropes, solutions to the,

Lane-Emden Equation

$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }\right)=-{\mathrm{\Theta }}_H^n$

can provide the desired internal structural information. Here we draw on Chandrasekhar's [C67] discussion of the structure of spherical polytropes to show precisely how our structural form factors can be expressed in terms of the Lane-Emden function, ${\mathrm{\Theta }}_H$, dimensionless radial coordinate, $\xi$, and the function derivative, ${\mathrm{\Theta }}^{\text{'}}=d{\mathrm{\Theta }}_H/d\xi$.

Mass

We note, first, that Chandrasekhar [C67] — see his Equation (78) on p. 99 — presents the following expression for the mean-to-central density ratio:

$\frac{\overline{\rho }}{{\rho }_c}$

$=$

$[-\frac{3{\mathrm{\Theta }}^{\text{'}}}{\xi }{]}_{{\xi }_1},$

where the notation at the bottom of the closing square bracket means that everything inside the square brackets should be, "evaluated at the surface of the configuration," that is, at the radial location, ${\xi }_1$, where the Lane-Emden function, ${\mathrm{\Theta }}_H(\xi )$, first goes to zero. But, as we pointed out when defining the structural form factors, the form factor associated with the configuration mass, ${\mathfrak{𝔣}}_M$, is equivalent to the mean-to-central density ratio. We conclude, therefore, that,

${\mathfrak{𝔣}}_M$

$=$

$[-\frac{3{\mathrm{\Theta }}^{\text{'}}}{\xi }{]}_{{\xi }_1}.$

Gravitational Potential Energy

Second, we note that Chandrasekhar's [C67] expression for the gravitational potential energy — see his Equation (90), p. 101 — is,

$-W$

$=$

$\frac{3}{5-n}\left(\frac{G{M}^2}{R}\right),$

whereas our analogous expression is,

$-W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$

$=$

$\frac{3}{5}\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}}\right)\frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}.$

We conclude, therefore, that,

$\frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}$	$=$	$\frac{5}{5-n}$
$\Rightarrow {\mathfrak{𝔣}}_W$	$=$	$\frac{3^2\cdot 5}{5-n}[\frac{{\mathrm{\Theta }}^{\text{'}}}{\xi }{]}_{{\xi }_1}^2.$

Mass-Radius Relationship

Third, Chandrasekhar [C67] shows — see his Equation (72), p. 98 — that the general mass-radius relationship for isolated spherical polytropes is,

$G{M}^{(n-1)/n}{R}^{(3-n)/n}$

$=$

$\frac{(n+1)K}{(4\pi {)}^{1/n}}[-{\xi }^{(n+1)/(n-1)}\frac{d{\mathrm{\Theta }}_H}{d\xi }{]}_{\xi ={\xi }_1}^{(n-1)/n},$

which we choose to rewrite as,

$4\pi (\frac{G}{K}{)}^n{M}^{(n-1)}{R}^{(3-n)}$	$=$	$(n+1{)}^n[{\xi }^{(n+1)}(-{\mathrm{\Theta }}^{\text{'}}{)}^{(n-1)}{]}_{\xi ={\xi }_1}$
	$=$	$-\left(\frac{\xi }{{\mathrm{\Theta }}^{\text{'}}}{)}_{\xi ={\xi }_1}\right[(n+1)\xi (-{\mathrm{\Theta }}^{\text{'}}){]}_{\xi ={\xi }_1}^n.$

By comparison, the expression for the equilibrium radius that has been derived, above, from an analysis of extrema in the free energy function — specifically, see the last expression in the left-hand column of the table titled "Two Points of View" — we obtain,

$4\pi (\frac{G}{K}{)}^n{M}^{(n-1)}{R}_{\mathrm{e}\mathrm{q}}^{(3-n)}$

$=$

$\frac{3}{{\mathfrak{𝔣}}_M}(\frac{5{\mathfrak{𝔣}}_A{\mathfrak{𝔣}}_M}{{\mathfrak{𝔣}}_W}{)}^n.$

Hence, it appears as though, quite generally,

$\frac{1}{{\mathfrak{𝔣}}_M}(\frac{5{\mathfrak{𝔣}}_A{\mathfrak{𝔣}}_M}{{\mathfrak{𝔣}}_W}{)}^n$

$=$

$-\left(\frac{\xi }{3{\mathrm{\Theta }}^{\text{'}}}{)}_{\xi ={\xi }_1}\right[(n+1)\xi (-{\mathrm{\Theta }}^{\text{'}}){]}_{\xi ={\xi }_1}^n.$

Or, taking into account the expressions for ${\mathfrak{𝔣}}_M$ and ${\mathfrak{𝔣}}_W$ that have just been uncovered, we conclude that,

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

Central Pressure

It is also worth pointing out that Chandrasekhar [C67] — see his Equations (80) & (81), p. 99 — introduces a dimensionless structural form factor, $W_n$, for the central pressure via the expression,

$P_c$

$=$

$W_n\left(\frac{G{M}^2}{R^4}\right),$

and demonstrates that,

$\frac{1}{W_n}$

$\equiv$

$4\pi (n+1)[{\mathrm{\Theta }}^{\text{'}}{]}_{{\xi }_1}^2.$

It is therefore clear that a spherical polytrope's central pressure is expressible in terms of our structural form factor, ${\mathfrak{𝔣}}_A$, as,

$P_c$

$=$

$\frac{3}{4\pi (5-n)}\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4}\right)\frac{1}{{\mathfrak{𝔣}}_A}.$

Alternate Derivation of Gravitational Potential Energy

As has been discussed elsewhere, we have learned from Chandrasekhar's discussion of polytropic spheres [C67] — see his Equation (16), p. 64 — that if a spherically symmetric system is in hydrostatic balance, the total gravitational potential energy can be obtained from the following integral:

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$

$=$

$+\frac{1}{2}\underset{0}{\overset{R}{\int }}\mathrm{\Phi }(r)dm.$

Using "technique #3" to solve the differential equation that governs the statement of hydrostatic balance, we know that in any polytropic sphere, $\mathrm{\Phi }(r)$ is related to the configuration's radial enthalpy profile, $H(r)$, via the algebraic expression,

$\mathrm{\Phi }(r)+H(r)$

$=$

$C_B,$

where, $C_B$, is an integration constant. At the surface of the equilibrium configuration, $H=0$ and $\mathrm{\Phi }=-G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}/R_{\mathrm{e}\mathrm{q}}$, so the integration constant is,

$C_B$

$=$

$-\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{R_{\mathrm{e}\mathrm{q}}},$

which implies,

$\mathrm{\Phi }(r)$

$=$

$-H(r)-\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{R_{\mathrm{e}\mathrm{q}}}.$

Now, from our general discussion of barotropic relations, we can write,

$H(r)$

$=$

$(n+1)\frac{P(r)}{\rho (r)}.$

Hence,

$-\mathrm{\Phi }(r)$

$=$

$(n+1)\frac{P(r)}{\rho (r)}+\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{R_{\mathrm{e}\mathrm{q}}},$

and,

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$	$=$	$-\frac{1}{2}\underset{0}{\overset{R}{\int }}[(n+1)\frac{P(r)}{\rho (r)}+\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{R_{\mathrm{e}\mathrm{q}}}]4\pi \rho (r)r^{2}dr$
	$=$	$-2\pi \{(n+1)\underset{0}{\overset{R}{\int }}P(r)r^{2}dr+\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{R_{\mathrm{e}\mathrm{q}}}\underset{0}{\overset{R}{\int }}\rho (r)r^{2}dr\}$
	$=$	$-2\pi \{\frac{1}{3}(n+1)P_c{R}_{\mathrm{e}\mathrm{q}}^3\underset{0}{\overset{1}{\int }}3[\frac{P(x)}{P_c}]x^{2}dx+\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{3R_{\mathrm{e}\mathrm{q}}}({\rho }_c{R}_{\mathrm{e}\mathrm{q}}^3\left)\underset{0}{\overset{1}{\int }}3\right[\frac{\rho (x)}{{\rho }_c}\left]x^{2}dx\right\}$
	$=$	$-2\pi \{\frac{1}{3}(n+1)P_c{R}_{\mathrm{e}\mathrm{q}}^3{\mathfrak{𝔣}}_A+\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{3R_{\mathrm{e}\mathrm{q}}}({\rho }_c{R}_{\mathrm{e}\mathrm{q}}^3\left){\mathfrak{𝔣}}_M\right\}$
	$=$	$-\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}}\{\frac{2\pi }{3}(n+1)[\frac{P_c{R}_{\mathrm{e}\mathrm{q}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}]{\mathfrak{𝔣}}_A+\frac{1}{2}[\frac{4\pi {\rho }_c{R}_{\mathrm{e}\mathrm{q}}^3}{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left]{\mathfrak{𝔣}}_M\right\}$
	$=$	$-\frac{1}{2}\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}}\{\frac{4\pi }{3}(n+1)[\frac{P_c{R}_{\mathrm{e}\mathrm{q}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}]{\mathfrak{𝔣}}_A+1\}.$

We now recall two earlier expressions that show the role that our structural form factors play in the evaluation of $W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$ and $P_c$, namely,

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$

$=$

$-\frac{3}{5}\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}}\right)\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}$

and,

$P_c$

$=$

$\frac{3}{20\pi }\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4}\right)\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_A{\mathfrak{𝔣}}_M^2}.$

Plugging these into our newly derived expression for the gravitational potential energy gives,

$-\frac{3}{5}\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}$	$=$	$-\frac{1}{2}\{\frac{4\pi }{3}(n+1)[\frac{3}{20\pi }\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_A{\mathfrak{𝔣}}_M^2}]{\mathfrak{𝔣}}_A+1\}$
$\Rightarrow (2\cdot 3)\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}$	$=$	$(n+1)\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}+5$
$\Rightarrow (5-n)\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}$	$=$	$5$
$\Rightarrow \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}$	$=$	$\frac{5}{5-n}.$

As it should, this agrees with the expression for the ratio, ${\mathfrak{𝔣}}_W/{\mathfrak{𝔣}}_M^2$, that was derived in our above discussion of the gravitational potential energy.

Summary

In summary, expressions for the three structural form factors associated with isolated, spherically symmetric polytropes are as follows:

Structural Form Factors for Isolated Polytropes
${\mathfrak{𝔣}}_M$ $=$ $[-\frac{3{\mathrm{\Theta }}^{\text{'}}}{\xi }{]}_{{\xi }_1}$ ${\mathfrak{𝔣}}_W$ $=$ $\frac{3^2\cdot 5}{5-n}[\frac{{\mathrm{\Theta }}^{\text{'}}}{\xi }{]}_{{\xi }_1}^2$ ${\mathfrak{𝔣}}_A$ $=$ $\frac{3(n+1)}{(5-n)}[{\mathrm{\Theta }}^{\text{'}}{]}_{{\xi }_1}^2$

Nonrotating Adiabatic Configuration Embedded in an External Medium

For a nonrotating configuration $(C=J=0)$ that is embedded in, and is influenced by the pressure $P_e$ of, an external medium, the statement of virial equilibrium is,

$3B{\chi }_{\mathrm{e}\mathrm{q}}^{3-3{\gamma }_g}-3A{\chi }_{\mathrm{e}\mathrm{q}}^{-1}-3D{\chi }_{\mathrm{e}\mathrm{q}}^3=0.$

Solution Expressed in Terms of K and M (Whitworth's 1981 Relation)

This is precisely the same condition that derives from setting equation (3) to zero in the 📚 A. Whitworth (1981, MNRAS, Vol. 195, pp. 967 - 977) discussion of the Global Gravitational Stability for One-dimensional Polytropes. The overlap with Whitworth's narative is clearer after introducing the algebraic expressions for the coefficients $A$, $B$, and $D$, to obtain,

$4\pi \left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right){\chi }_{\mathrm{e}\mathrm{q}}^3$

$=$

$3\left(\frac{4\pi }{3}{)}^{1-{\gamma }_g}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{{\gamma }_g}\frac{{\mathfrak{𝔣}}_A}{{\mathfrak{𝔣}}_M^{{\gamma }_g}}\cdot {\chi }_{\mathrm{e}\mathrm{q}}^{3-3{\gamma }_g}-\frac{3}{5}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2\frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}\cdot {\chi }_{\mathrm{e}\mathrm{q}}^{-1};$

dividing the equation through by $(4\pi {\chi }_{\mathrm{e}\mathrm{q}}^3/P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}})$,

$P_e$	$=$	$P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\left[\right(\frac{3}{4\pi }{)}^{{\gamma }_g}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{{\gamma }_g}\frac{{\mathfrak{𝔣}}_A}{{\mathfrak{𝔣}}_M^{{\gamma }_g}}\cdot {\chi }_{\mathrm{e}\mathrm{q}}^{-3{\gamma }_g}-(\frac{3}{20\pi }\left)\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2\frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}\cdot {\chi }_{\mathrm{e}\mathrm{q}}^{-4}]$
	$=$	$P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^4\left[\right(\frac{3}{4\pi R_{\mathrm{e}\mathrm{q}}^3}{)}^{{\gamma }_g}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{{\gamma }_g}\frac{{\mathfrak{𝔣}}_A}{{\mathfrak{𝔣}}_M^{{\gamma }_g}}\cdot R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{3{\gamma }_g-4}-(\frac{3}{20\pi R_{\mathrm{e}\mathrm{q}}^4}\left)\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2\frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}];$

and inserting expressions for the parameter normalizations as defined in our accompanying introductory discussion to obtain,

$P_e$	$=$	$G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2\left[\right(\frac{3}{4\pi R_{\mathrm{e}\mathrm{q}}^3}{)}^{{\gamma }_g}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{{\gamma }_g}\frac{{\mathfrak{𝔣}}_A}{{\mathfrak{𝔣}}_M^{{\gamma }_g}}\cdot \frac{K{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{{\gamma }_g-2}}{G}-(\frac{3}{20\pi R_{\mathrm{e}\mathrm{q}}^4}\left)\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2\frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}]$
	$=$	$K(\frac{3M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{4\pi R_{\mathrm{e}\mathrm{q}}^3}{)}^{{\gamma }_g}\frac{{\mathfrak{𝔣}}_A}{{\mathfrak{𝔣}}_M^{{\gamma }_g}}-(\frac{3G{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^2}{20\pi R_{\mathrm{e}\mathrm{q}}^4})\frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}.$

If the structural form factors are set equal to unity, this exactly matches equation (5) of 📚 Whitworth (1981), which reads:

Equation and accompanying sentence drawn directly from p. 970 of A. Whitworth (1981) Global Gravitational Stability for One-Dimensional Polytropes Monthly Notices of the Royal Astronomical Society, Vol. 195, pp. 967 - 977 © Royal Astronomical Society

The general equilibrium condition, $(d{𝒰}/dR{)}_{R_0}=0$, reduces to $R_0\to R_{\mathrm{e}\mathrm{q}},$   $P_{\mathrm{e}\mathrm{x}}=K(3M_0/4\pi R_{\mathrm{e}\mathrm{q}}^3{)}^{\eta }-3G{M}_0^2/20\pi R_{\mathrm{e}\mathrm{q}}^4$ (5) (subscript 'eq' for equilibrium).

Notice that, when $P_e\to 0$, this expression reduces to the solution we obtained for an isolated polytrope, expressed in terms of $K$ and $M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}$ (see the left-hand column of our table titled "Two Points of View").

Solution Expressed in Terms of M and Central Pressure

Beginning again with the relevant statement of virial equilibrium, namely,

$A=B{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_g}-D{\chi }_{\mathrm{e}\mathrm{q}}^4,$

but adopting the alternate expression for the coefficient, $B$, given above, that is,

$B=\frac{4\pi }{3}\left[\right(\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}){\chi }^{3\gamma }{]}_{\mathrm{e}\mathrm{q}}\cdot {\mathfrak{𝔣}}_A,$

we can write,

$\frac{1}{5}\cdot \left[\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\mathfrak{𝔣}}_M}{]}^2\cdot {\mathfrak{𝔣}}_W$	$=$	$\frac{4\pi }{3}\left[\right(\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}){\chi }^{3\gamma }{]}_{\mathrm{e}\mathrm{q}}{\mathfrak{𝔣}}_A\cdot {\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_g}-(\frac{4\pi }{3})\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{\chi }_{\mathrm{e}\mathrm{q}}^4$
$\Rightarrow \frac{3}{20\pi }\cdot \left[\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\mathfrak{𝔣}}_M}{]}^2\cdot {\mathfrak{𝔣}}_W$	$=$	$\left[\right(\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}){\mathfrak{𝔣}}_A-\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}]{\chi }_{\mathrm{e}\mathrm{q}}^4$
	$=$	$[{\mathfrak{𝔣}}_A{P}_c-P_e]\frac{R_{\mathrm{e}\mathrm{q}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}$
$\Rightarrow \frac{3}{20\pi }\left(\frac{G{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4}\right)\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}$	$=$	${\mathfrak{𝔣}}_A{P}_c-P_e.$

Again notice that, when $P_e\to 0$, this expression reduces to the solution we obtained for an isolated polytrope, but this time expressed in terms of $P_c$ and $M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}$ (see the right-hand column of our table titled "Two Points of View").

Contrast with Detailed Force-Balanced Solution

As has just been demonstrated, the virial theorem provides a mathematical expression that allows us to relate the equilibrium radius of a configuration to the applied external pressure, once the configuration's mass and either its specific entropy or central pressure are specified. In contrast to this, as has been discussed in detail in another chapter, 📚 Gp. Horedt (1970, MNRAS, Vol. 151, pp. 81 - 86), 📚 Whitworth (1981) and 📚 S. W. Stahler (1983, ApJ, Vol. 268, pp. 165 - 184) have each derived separate analytic expressions for $R_{\mathrm{e}\mathrm{q}}$ and $P_e$ — given in terms of the Lane-Emden function, $\mathrm{\Theta }$, and its radial derivative — without demonstrating how the equilibrium radius and external pressure directly relate to one another. That is to say, solution of the detailed force-balanced equations provides a pair of equilibrium expressions that are parametrically related to one another through the Lane-Emden function. For example — see our related discussion for more details — 📚 Horedt (1970) derives the following set of parametric equations relating the configuration's dimensionless radius, $r_a$, to a specified dimensionless bounding pressure, $p_a$:

$r_a\equiv \frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}$	$=$	$\tilde{\xi }(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{(1-n)/(n-3)},$
$p_a\equiv \frac{P_{\mathrm{e}}}{P_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}$	$=$	${\tilde{\theta }}_n^{n+1}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{2(n+1)/(n-3)},$

where,

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

It is important to appreciate that, in the expressions for $r_a$ and $p_a$, the tilde indicates that the Lane-Emden function and its derivative are to be evaluated, not at the radial coordinate, ${\xi }_1$, that is traditionally associated with the "first zero" of the Lane-Emden function and therefore with the surface of the isolated polytrope, but at the radial coordinate, $\tilde{\xi }$, where the internal pressure of the isolated polytrope equals $P_e$ and at which the embedded polytrope is to be truncated. The coordinate, $\tilde{\xi }$, therefore identifies the surface of the embedded — or, pressure-truncated — polytrope. We also have taken the liberty of attaching the subscript "limit" to $M$ in both defining relations because it is clear that 📚 Horedt (1970) intended for the normalization mass to be the mass of the pressure-truncated object, not the mass of the associated isolated (and untruncated) polytrope. In anticipation of further derivations, below, we note here the ratio of the 📚 Horedt (1970) normalization parameters to ours, assuming $\gamma =(n+1)/n$:

$(\frac{R_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{n-3}$	$=$	$\left[\frac{4\pi }{(n+1{)}^n}\right(\frac{G}{K}{)}^n{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^{n-1}\left]\right[\left(\frac{K}{G}{)}^n{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{1-n}\right]$
	$=$	$\frac{4\pi }{(n+1{)}^n}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{n-1},$
$(\frac{P_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{n-3}$	$=$	$K^{4n}\left[\frac{(n+1{)}^3}{4\pi G^3{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^2}{]}^{n+1}\right[\frac{G^{3(n+1)}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{2(n+1)}}{K^{4n}}]$
	$=$	$\left[\frac{(n+1{)}^3}{4\pi }\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{-2}{]}^{n+1}.$

Next, we demonstrate that this pair of parametric relations satisfies the virial theorem and, in so doing, demonstrate how $r_a$ and $p_a$ may be directly related to each other. Given that the normalization radius and normalization pressure chosen by 📚 Horedt (1970) are defined in terms of $K$ and $M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}$, we begin with the virial theorem derived above in terms of $K$ and $M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}$, setting ${\gamma }_g=(n+1)/n$.

$P_e$

$=$

$K(\frac{3M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{4\pi R_{\mathrm{e}\mathrm{q}}^3}{)}^{(n+1)/n}\frac{{\mathfrak{𝔣}}_A}{{\mathfrak{𝔣}}_M^{(n+1)/n}}-(\frac{3G{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^2}{20\pi R_{\mathrm{e}\mathrm{q}}^4})\frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}.$

After setting $R_{\mathrm{e}\mathrm{q}}=r_a{R}_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}$, a bit of algebraic manipulation shows that the first term on the right-hand side of the virial equilibrium expression becomes,

$K(\frac{3M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{4\pi R_{\mathrm{e}\mathrm{q}}^3}{)}^{(n+1)/n}\frac{{\mathfrak{𝔣}}_A}{{\mathfrak{𝔣}}_M^{(n+1)/n}}$

$=$

$r_a^{-3(n+1)/n}{\mathfrak{𝔣}}_A\left[\right(\frac{3}{{\mathfrak{𝔣}}_M}{)}^{n-3}\frac{(n+1{)}^{3n}}{(4\pi {)}^n}{]}^{(n+1)/[n(n-3)]}[K^{4n}{G}^{-3(n+1)}{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^{-2(n+1)}{]}^{1/(n-3)},$

while the second term on the right-hand side becomes,

$\left(\frac{3G{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^2}{20\pi R_{\mathrm{e}\mathrm{q}}^4}\right)\frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}$

$=$

$\frac{3}{5}\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}\cdot r_a^{-4}(4\pi {)}^{-(n+1)/(n-3)}(n+1{)}^{4n/(n-3)}[K^{4n}{G}^{-3(n+1)}{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^{-2(n+1)}{]}^{1/(n-3)}.$

But, using Horedt's expression for $P_e$, the left-hand side of the virial equilibrium equation becomes,

$P_e=p_a{P}_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}$

$=$

$p_a(4\pi {)}^{-(n+1)/(n-3)}(n+1{)}^{3(n+1)/(n-3)}[K^{4n}{G}^{-3(n+1)}{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^{-2(n+1)}{]}^{1/(n-3)}.$

Hence, the statement of virial equilibrium is,

$p_a$	$=$	$\left\{r_a^{-3(n+1)/n}{\mathfrak{𝔣}}_A\right[(\frac{3}{{\mathfrak{𝔣}}_M}{)}^{n-3}\frac{(n+1{)}^{3n}}{(4\pi {)}^n}{]}^{(n+1)/[n(n-3)]}$
		$-\frac{3}{5}\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}\cdot r_a^{-4}(4\pi {)}^{-(n+1)/(n-3)}(n+1{)}^{4n/(n-3)}\}(4\pi {)}^{(n+1)/(n-3)}(n+1{)}^{-3(n+1)/(n-3)}$
	$=$	${\mathfrak{𝔣}}_A(\frac{3}{{\mathfrak{𝔣}}_M\cdot r_a^3}{)}^{(n+1)/n}-\frac{3(n+1)}{5}\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}\cdot r_a^{-4};$

or, multiplying through by $r_a^4$ and rearranging terms,

${\mathfrak{𝔣}}_A(\frac{3}{{\mathfrak{𝔣}}_M}{)}^{(n+1)/n}{r}_a^{(n-3)/n}-p_a{r}_a^4$ $=$ $\frac{3(n+1)}{5}\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}.$

Now, 📚 Horedt (1970) has given analytic expressions for $r_a$ and $p_a$ in terms of the Lane-Emden function and its first derivative. The question is, what should the expressions for our structural form factors be in order for this virial expression to hold true for all pressure-truncated polytropic structures? As has been summarized above, in the case of an isolated polytrope, whose surface is located at ${\xi }_1$ and whose global properties are defined by evaluation of the Lane-Emden function at ${\xi }_1$, we know that (see the above summary),

Structural Form Factors for Isolated Polytropes
${\mathfrak{𝔣}}_M$ $=$ $[-\frac{3{\mathrm{\Theta }}^{\text{'}}}{\xi }{]}_{{\xi }_1}$ ${\mathfrak{𝔣}}_W$ $=$ $\frac{3^2\cdot 5}{5-n}[\frac{{\mathrm{\Theta }}^{\text{'}}}{\xi }{]}_{{\xi }_1}^2$ ${\mathfrak{𝔣}}_A$ $=$ $\frac{3(n+1)}{(5-n)}[{\mathrm{\Theta }}^{\text{'}}{]}_{{\xi }_1}^2$

These same expressions may or may not work for pressure-truncated polytropes, even if the evaluation radius is shifted from ${\xi }_1$ to $\tilde{\xi }$. Let's see …

January 13, 2015: As is noted in our accompanying outline of work, I no longer believe that ${\mathfrak{𝔣}}_W$ and ${\mathfrak{𝔣}}_A$ have the same expressions as in isolated polytropes. Hence, all of the material that follows is suspect and needs to be reworked.

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Material that appears after this point in our presentation is under development and therefore
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Inserting the expressions for $r_a$ and $p_a$, as provided by 📚 Horedt (1970), into the virial equilibrium expression, we have,

$\frac{3(n+1)}{5}\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_M^2}$	$=$	${\mathfrak{𝔣}}_A(\frac{3}{{\mathfrak{𝔣}}_M}{)}^{(n+1)/n}[\tilde{\xi }(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{(1-n)/(n-3)}{]}^{(n-3)/n}-{\tilde{\theta }}_n^{n+1}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{2(n+1)/(n-3)}[\tilde{\xi }(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{(1-n)/(n-3)}{]}^4$
	$=$	${\mathfrak{𝔣}}_A(\frac{3}{{\mathfrak{𝔣}}_M}{)}^{(n+1)/n}[\tilde{\xi }(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{(1-n)/(n-3)}{]}^{(n-3)/n}-{\tilde{\theta }}_n^{n+1}{\tilde{\xi }}^4[(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{2(n+1)+4(1-n)}{]}^{1/(n-3)}$
	$=$	${\mathfrak{𝔣}}_A(\frac{3}{{\mathfrak{𝔣}}_M}{)}^{(n+1)/n}[{\tilde{\xi }}^{(n-3)}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{(1-n)}{]}^{1/n}-{\tilde{\theta }}_n^{n+1}{\tilde{\xi }}^4[(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{-2}]$
	$=$	${\mathfrak{𝔣}}_A(\frac{3}{{\mathfrak{𝔣}}_M}{)}^{(n+1)/n}{\tilde{\xi }}^{-(n+1)/n}(-{\tilde{\theta }}^{\prime }{)}^{(1-n)/n}-\frac{{\tilde{\theta }}_n^{n+1}}{(-{\tilde{\theta }}^{\prime }{)}^2}.$

If we assume that both of the structural form factors, ${\mathfrak{𝔣}}_W$ and ${\mathfrak{𝔣}}_M$, have the same functional expressions as in the case of isolated polytropes (but evaluated at $\tilde{\xi }$ instead of at ${\xi }_1$), the virial relation further reduces to the form,

$\frac{{\tilde{\theta }}_n^{n+1}}{(-{\tilde{\theta }}^{\prime }{)}^2}$	$=$	${\mathfrak{𝔣}}_A(\frac{\tilde{\xi }}{-{\tilde{\theta }}^{\prime }}{)}^{(n+1)/n}{\tilde{\xi }}^{-(n+1)/n}(-{\tilde{\theta }}^{\prime }{)}^{(1-n)/n}-\frac{3(n+1)}{5-n}$
	$=$	$\frac{{\mathfrak{𝔣}}_A}{(-{\tilde{\theta }}^{\prime }{)}^2}-\frac{3(n+1)}{5-n}$
$\Rightarrow {\mathfrak{𝔣}}_A$	$=$	$\frac{3(n+1)}{5-n}(-{\tilde{\theta }}^{\prime }{)}^2+{\tilde{\theta }}_n^{n+1}.$

This all seems to make a great deal of sense. Only the structural parameter that is derived from an integral over the pressure distribution, ${\mathfrak{𝔣}}_A$, gets modified when the polytropic configuration is truncated. Notice, as well, that the term that has been added to the definition of ${\mathfrak{𝔣}}_A$ naturally goes to zero in the limit of $\tilde{\xi }\to {\xi }_1$, that is, for an isolated polytrope. We should definitely go back to the original definitions of all three structural parameters and prove that this is the case. But, in the meantime, here is the summary:

WRONG!!  For the correct form-factor expressions, go here.
Structural Form Factors for Pressure-Truncated Polytropes
${\widetilde{\mathfrak{𝔣}}}_M$ $=$ $[-\frac{3{\mathrm{\Theta }}^{\text{'}}}{\xi }{]}_{\tilde{\xi }}$ ${\widetilde{\mathfrak{𝔣}}}_W$ $=$ $\frac{3^2\cdot 5}{5-n}[\frac{{\mathrm{\Theta }}^{\text{'}}}{\xi }{]}_{\tilde{\xi }}^2$ ${\widetilde{\mathfrak{𝔣}}}_A$ $=$ $\frac{3(n+1)}{(5-n)}[{\mathrm{\Theta }}^{\text{'}}{]}_{\tilde{\xi }}^2+{\tilde{\mathrm{\Theta }}}^{n+1}$

WRONG!!  For the correct form-factor expressions, go here.

Notice that, in an effort to differentiate them from their counterparts developed earlier for "isolated" polytropes, we have affixed a tilde to each of these three form-factors, ${\mathfrak{𝔣}}_i$.

Example

Outline

Let's identify an equilibrium configuration numerically, using the free-energy expression. From our introductory discussion, the relevant expression is,

${\mathfrak{𝔊}}^{*}=-3{𝒜}{\chi }^{-1}-\frac{1}{(1-{\gamma }_g)}{ℬ}{\chi }^{3-3{\gamma }_g}+{𝒟}{\chi }^3,$

where,

${𝒜}$	$\equiv$	$\frac{1}{5}\cdot \left[\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\mathfrak{𝔣}}_M}{]}^2\cdot {\mathfrak{𝔣}}_W,$
${ℬ}$	$\equiv$	$\frac{4\pi }{3}\left[\frac{3}{4\pi }\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\mathfrak{𝔣}}_M}{]}_{\mathrm{e}\mathrm{q}}^{\gamma }\cdot {\mathfrak{𝔣}}_A$
	$=$	$\frac{4\pi }{3}\left[\right(\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}){\chi }^{3\gamma }{]}_{\mathrm{e}\mathrm{q}}\cdot {\mathfrak{𝔣}}_A,$
${𝒟}$	$\equiv$	$\left(\frac{4\pi }{3}\right)\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}.$

For later use, note that, $(\frac{{𝒜}}{{ℬ}}{)}^n$ $=$ $\left\{\frac{3}{20\pi }\right[\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{{\mathfrak{𝔣}}_M}{]}^2\cdot {\mathfrak{𝔣}}_W\left[\frac{3}{4\pi }\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\mathfrak{𝔣}}_M}{]}^{-(n+1)/n}\cdot {\mathfrak{𝔣}}_A^{-1}{\}}^n$   $=$ $5^{-n}\left(\frac{3}{4\pi }{)}^n\right[\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{{\mathfrak{𝔣}}_M}{]}^{2n}\left[\frac{3}{4\pi }\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\mathfrak{𝔣}}_M}{]}^{-(n+1)}\cdot (\frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_A}{)}^n$   $=$ $\frac{1}{5^n}\left(\frac{4\pi }{3}\right)(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{n-1}\cdot \frac{{\mathfrak{𝔣}}_W^n}{{\mathfrak{𝔣}}_A^n{\mathfrak{𝔣}}_M^{1-n}};$ and, $\frac{{{ℬ}}^{4n}}{{{𝒜}}^{3(n+1)}}$ $=$ $\left\{\frac{1}{5}\right[\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{{\mathfrak{𝔣}}_M}{]}^2\cdot {\mathfrak{𝔣}}_W{\}}^{-3(n+1)}\left\{\frac{4\pi }{3}\right[\frac{3}{4\pi }\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{{\mathfrak{𝔣}}_M}{]}^{(n+1)/n}\cdot {\mathfrak{𝔣}}_A{\}}^{4n}$   $=$ $5^{3(n+1)}\left[\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left)\frac{1}{{\mathfrak{𝔣}}_M}{]}^{-6(n+1)}\right(\frac{4\pi }{3}{)}^{4n-4(n+1)}\left[\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\mathfrak{𝔣}}_M}{]}^{4(n+1)}\cdot \frac{{\mathfrak{𝔣}}_A^{4n}}{{\mathfrak{𝔣}}_W^{3(n+1)}}$   $=$ $5^{3(n+1)}\left(\frac{3}{4\pi }{)}^4\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{-2(n+1)}\cdot \frac{{\mathfrak{𝔣}}_A^{4n}{\mathfrak{𝔣}}_M^{2(n+1)}}{{\mathfrak{𝔣}}_W^{3(n+1)}}$

Now, we could just blindly start setting values of the three leading coefficients, ${𝒜}$, ${ℬ}$, and ${𝒟}$, then plot ${\mathfrak{𝔊}}^{*}(\chi )$ to look for extrema. But let's accept a little guidance from this chapter's virial analysis before choosing the coefficient values. For embedded polytropes, we know that the structural form factors are,

${\widetilde{\mathfrak{𝔣}}}_M=\frac{\overline{\rho }}{{\rho }_c}$	$=$	$[-\frac{3{\mathrm{\Theta }}^{\text{'}}}{\xi }{]}_{\tilde{\xi }},$
${\widetilde{\mathfrak{𝔣}}}_W$	$=$	$\frac{3^2\cdot 5}{5-n}[\frac{{\mathrm{\Theta }}^{\text{'}}}{\xi }{]}_{\tilde{\xi }}^2=(\frac{5}{5-n}){\widetilde{\mathfrak{𝔣}}}_M^2,$
${\widetilde{\mathfrak{𝔣}}}_A=\frac{\overline{P}}{P_c}$	$=$	$\frac{3(n+1)}{(5-n)}({\mathrm{\Theta }}^{\text{'}}{)}_{\tilde{\xi }}^2+{\tilde{\mathrm{\Theta }}}^{n+1}.$

Hence, the coefficient expressions become,

${𝒜}$	$=$	$\frac{1}{(5-n)}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2,$
${ℬ}$	$=$	$\frac{4\pi }{3}\left(\frac{\overline{P}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right){\chi }_{\mathrm{e}\mathrm{q}}^{3(n+1)/n},$
${𝒟}$	$=$	$\left(\frac{4\pi }{3}\right)\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}.$

Strategy

Generic setup:

• Choose the polytropic index, $n$, which also sets the value of the adiabatic index, $\gamma =(n+1)/n$.
• Fix $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ and $K$, so that the radial and pressure normalizations are fixed; specifically,

$R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{n-3}=(\frac{G}{K}{)}^n{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{n-1}$

and

$P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{(n-3)}=K^{4n}(G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2{)}^{-(n+1)}.$

• Fix $M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}$, and let it be the normalization mass; that is, set $M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}=M_{\mathrm{t}\mathrm{o}\mathrm{t}}$.
• As a result of the above choices, the value of ${𝒜}$ is set, and fixed; specifically,

$A=\frac{1}{(5-n)}.$

Case I:

• Fix ${𝒟}$, which fixes the external pressure; specifically,

$P_e=\frac{3}{4\pi }\cdot {𝒟}{P}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}.$

• Choose a variety of values of the remaining coefficient, ${ℬ}$; then, for each value, plot ${\mathfrak{𝔊}}^{*}(\chi )$ and locate one or more extrema along with the value of ${\chi }_{\mathrm{e}\mathrm{q}}$ that is associated with each free energy extremum. This identifies the equilibrium value of the mean pressure inside the pressure-truncated polytrope via the expression,

$\overline{P}=\left(\frac{3}{4\pi }\right)P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{ℬ}{\chi }_{\mathrm{e}\mathrm{q}}^{-3(n+1)/n}.$

• In order to check whether we've identified the correct value of ${\chi }_{\mathrm{e}\mathrm{q}}$, we have to relate it to the radial coordinate, ${\xi }_e$, used in the analytic solution of the Lane-Emden equation. As has been explained in our discussion of detailed force-balanced models of polytropes, generically,

$R_{\mathrm{e}\mathrm{q}}$	$=$	$a_n{\xi }_e,$        where,
$a_{\mathrm{n}}$	$\equiv$	$\left[\frac{1}{4\pi G}\right(\frac{H_c}{{\rho }_c}){]}^{1/2}=[\frac{(n+1)K}{4\pi G}\cdot {\rho }_c^{(1-n)/n}{]}^{1/2}$

$\Rightarrow (\frac{a_n}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^2$	$=$	$\left[\frac{(n+1)K}{4\pi G}\right]{\rho }_c^{(1-n)/n}\cdot \frac{1}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^2}$
	$=$	$\left[\frac{(n+1)K}{4\pi G}\right]\left(\frac{{\rho }_c}{\overline{\rho }}{)}^{(1-n)/n}\right(\frac{3M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{4\pi R_{\mathrm{e}\mathrm{q}}^3}{)}^{(1-n)/n}\cdot \frac{1}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^2}$
	$=$	$\left[\frac{(n+1)K}{4\pi G}\right]\left(\frac{{\rho }_c}{\overline{\rho }}{)}^{(1-n)/n}\right(\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi }{)}^{(1-n)/n}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{(1-n)/n}{\chi }_{\mathrm{e}\mathrm{q}}^{3(n-1)/n}\cdot R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{(n-3)/n}$
	$=$	$\frac{(n+1)}{4\pi }\left[\frac{3}{4\pi }\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{(1-n)/n}{\chi }_{\mathrm{e}\mathrm{q}}^{3(n-1)/n}.$

Hence,

${\xi }_e^2$	$=$	$(\frac{R_{\mathrm{e}\mathrm{q}}}{a_n}{)}^2={\chi }_{\mathrm{e}\mathrm{q}}^2(\frac{a_n}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-2}$
	$=$	${\chi }_{\mathrm{e}\mathrm{q}}^2\left\{\frac{4\pi }{(n+1)}\right[\frac{3}{4\pi }\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{(n-1)/n}{\chi }_{\mathrm{e}\mathrm{q}}^{3(1-n)/n}\}$
	$=$	$\frac{4\pi }{(n+1)}\left[\frac{3}{4\pi }\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{(n-1)/n}{\chi }_{\mathrm{e}\mathrm{q}}^{(3-n)/n}.$

But, from above, we also know that,

$\frac{3}{4\pi }\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}$

$=$

$[\frac{3{ℬ}}{4\pi }\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_A}{]}^{n/(n+1)},$

where,

${\widetilde{\mathfrak{𝔣}}}_A=\frac{\overline{P}}{P_c}$

$=$

$\frac{3(n+1)}{(5-n)}({\mathrm{\Theta }}^{\text{'}}{)}_{\tilde{\xi }}^2+{\tilde{\mathrm{\Theta }}}^{n+1},$

Hence we can write,

${\xi }_e^2$

$=$

$\frac{4\pi }{(n+1)}[\frac{3{ℬ}}{4\pi }\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_A}{]}^{(n-1)/(n+1)}{\chi }_{\mathrm{e}\mathrm{q}}^{(3-n)/n},$

or,

${\xi }_e^2\left[\frac{3(n+1)}{(5-n)}\right({\mathrm{\Theta }}^{\text{'}}{)}_{\tilde{\xi }}^2+{\tilde{\mathrm{\Theta }}}^{n+1}{]}^{(n-1)/(n+1)}$

$=$

$\frac{4\pi }{(n+1)}[\frac{3{ℬ}}{4\pi }{]}^{(n-1)/(n+1)}{\chi }_{\mathrm{e}\mathrm{q}}^{(3-n)/n}.$

This last expression may be useful because the numerical value of the right-hand-side will be known once an extremum of a free-energy plot has been identified, while the function on the left-hand side can be evaluated separately, from knowledge of the internal structure of detailed force-balanced, isolated polytropes.

Strategy2

• Pick the desired polytropic index, $n$, and a radial coordinate within the isolated polytropic model, $\tilde{\xi }\le {\xi }_1$, that will serve as the truncated edge of the embedded polytrope.
• Knowledge of the isolated polytrope's internal structure will give the value of the Lane-Emden function, $\tilde{\theta }$, and its radial derivative, ${\tilde{\theta }}^{\prime }$, at this truncated edge of the structure.
• According to 📚 Horedt (1970) — see our accompanying discussion of detailed force-balanced models — the physical radius and external pressure that corresponds to this choice of the truncated edge is given by the expressions,

$\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=r_a\left(\frac{R_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)$	$=$	$\tilde{\xi }(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{(1-n)/(n-3)}[\frac{4\pi }{(n+1{)}^n}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{n-1}{]}^{1/(n-3)},$
$\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=p_a\left(\frac{P_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)$	$=$	${\tilde{\theta }}_n^{n+1}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{2(n+1)/(n-3)}[\frac{(n+1{)}^3}{4\pi }(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{-2}{]}^{(n+1)/(n-3)}.$

• Using the chosen value of $\tilde{\xi }$ and its associated function values, $\tilde{\theta }$ and ${\tilde{\theta }}^{\text{'}}$, determine the values of the three relevant structural form factors via the following analytic relations:

${\widetilde{\mathfrak{𝔣}}}_M=\frac{\overline{\rho }}{{\rho }_c}$	$=$	$[-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }}],$
${\widetilde{\mathfrak{𝔣}}}_W$	$=$	$\frac{3^2\cdot 5}{5-n}[\frac{{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }}{]}^2=(\frac{5}{5-n}){\widetilde{\mathfrak{𝔣}}}_M^2,$
${\widetilde{\mathfrak{𝔣}}}_A=\frac{\overline{P}}{P_c}$	$=$	$\frac{3(n+1)}{(5-n)}({\tilde{\theta }}^{\text{'}}{)}^2+{\tilde{\theta }}^{n+1}.$

• Using these values of the structural form factors, determine the values of the three free-energy coefficients (set $M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}}/M_{\mathrm{t}\mathrm{o}\mathrm{t}}=1$ for the time being) via the expressions:

${{𝒜}}_{\mathrm{m}\mathrm{o}\mathrm{d}}\equiv (5-n){𝒜}$	$=$	$\frac{(5-n)}{5}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}=(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2,$
${{ℬ}}_{\mathrm{m}\mathrm{o}\mathrm{d}}\equiv (5-n){ℬ}$	$=$	$\left(\frac{3}{4\pi }{)}^{1/n}\right[\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}_{\mathrm{e}\mathrm{q}}^{(n+1)/n}\cdot [(5-n){\widetilde{\mathfrak{𝔣}}}_A]$
	$=$	$\frac{4\pi }{3}\left(\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right){\chi }_{\mathrm{e}\mathrm{q}}^{3(n+1)/n}\cdot [(5-n){\widetilde{\mathfrak{𝔣}}}_A]$
${{𝒟}}_{\mathrm{m}\mathrm{o}\mathrm{d}}\equiv (5-n){𝒟}$	$=$	$\left(\frac{4\pi }{3}\right)(5-n)\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}.$

• Plot the following free-energy function and see if the value of $\chi$ associated with the extremum is equal to the dimensionless equilibrium radius, $R_{\mathrm{e}\mathrm{q}}/R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$, as predicted by the 📚 Horedt (1970) expression, above:

$(5-n){\mathfrak{𝔊}}^{*}$	$=$	$-3{{𝒜}}_{\mathrm{m}\mathrm{o}\mathrm{d}}{\chi }^{-1}-\frac{1}{(1-{\gamma }_g)}{{ℬ}}_{\mathrm{m}\mathrm{o}\mathrm{d}}{\chi }^{3-3{\gamma }_g}+{{𝒟}}_{\mathrm{m}\mathrm{o}\mathrm{d}}{\chi }^3$
	$=$	$-3{{𝒜}}_{\mathrm{m}\mathrm{o}\mathrm{d}}{\chi }^{-1}+n{{ℬ}}_{\mathrm{m}\mathrm{o}\mathrm{d}}{\chi }^{-3/n}+{{𝒟}}_{\mathrm{m}\mathrm{o}\mathrm{d}}{\chi }^3.$

• Virial equilbrium — that is, an extremum in the free energy function — occurs when $\partial {\mathfrak{𝔊}}^{*}/\partial \chi =0$, that is, where,

${{ℬ}}_{\mathrm{m}\mathrm{o}\mathrm{d}}{\chi }_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-{{𝒟}}_{\mathrm{m}\mathrm{o}\mathrm{d}}{\chi }_{\mathrm{e}\mathrm{q}}^4$

$=$

${{𝒜}}_{\mathrm{m}\mathrm{o}\mathrm{d}}.$

Note that if the coefficient, ${ℬ}$, is written in terms of the normalized central pressure, the statement of virial equilibrium becomes,

${{𝒜}}_{\mathrm{m}\mathrm{o}\mathrm{d}}$	$=$	$\left[\frac{4\pi }{3}\right(\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}){\chi }_{\mathrm{e}\mathrm{q}}^{3(n+1)/n}\cdot [(5-n){\widetilde{\mathfrak{𝔣}}}_A]]{\chi }_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{4\pi }{3}(5-n)\left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right){\chi }_{\mathrm{e}\mathrm{q}}^4$
	$=$	$\frac{4\pi }{3}{\chi }_{\mathrm{e}\mathrm{q}}^4\left\{\right(\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}})\cdot (5-n){\widetilde{\mathfrak{𝔣}}}_A-(5-n)(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\left)\right\}$
	$=$	$\frac{4\pi }{3}{\chi }_{\mathrm{e}\mathrm{q}}^4\left\{\right(\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}})\cdot [3(n+1)({\tilde{\theta }}^{\text{'}}{)}^2+(5-n){\tilde{\theta }}^{n+1}]-(5-n)\left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)\}.$

But, in this situation, ${\tilde{\theta }}^{n+1}=P_e/P_c$, so virial equilibrium implies,

$\frac{3}{4\pi }{\chi }_{\mathrm{e}\mathrm{q}}^{-4}{{𝒜}}_{\mathrm{m}\mathrm{o}\mathrm{d}}$	$=$	$\left(\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)\cdot [3(n+1)({\tilde{\theta }}^{\text{'}}{)}^2+(5-n)\frac{P_e}{P_c}]-(5-n)(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}})$
	$=$	$3(n+1)({\tilde{\theta }}^{\text{'}}{)}^2(\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}})$
$\Rightarrow \frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\left(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^4\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{-2}$	$=$	$[4\pi (n+1)({\tilde{\theta }}^{\text{'}}{)}^2{]}^{-1}$
$\Rightarrow \frac{P_c{R}_{\mathrm{e}\mathrm{q}}^4}{G{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^2}$	$=$	$\frac{1}{4\pi (n+1)({\tilde{\theta }}^{\text{'}}{)}^2}.$

Compare With Detailed Force Balanced Solution

In a separate discussion, we presented the detailed force-balanced model of an $n=1$ polytrope that is embedded in an external medium. We showed that, for an applied external pressure given by,

$\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\frac{\pi }{2}[\frac{\sin {\xi }_e}{{\xi }_e(\sin {\xi }_e-{\xi }_e\cos {\xi }_e)}{]}^2,$

the associated equilibrium radius of the pressure-confined configuration is,

$R_{\mathrm{e}\mathrm{q}}={\xi }_e{a}_{\mathrm{n}\mathrm{=}\mathrm{1}}=[\frac{K}{2\pi G}{]}^{1/2}{\xi }_e.$

Flipping this around, after we use a plot of the free-energy expression to identify the equilibrium radius, ${\chi }_{\mathrm{e}\mathrm{q}}$, the corresponding dimensionless radius as used in the Lane-Emden equation should be,

${\xi }_e$	$=$	$[\frac{2\pi G}{K}{]}^{1/2}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }_{\mathrm{e}\mathrm{q}}$
	$=$	$(2\pi {)}^{1/2}{\chi }_{\mathrm{e}\mathrm{q}}.$

Keep in mind that, for an isolated $n=1$ polytrope, the (zero pressure) surface is identified by ${\xi }_1=\pi$. Hence we should expect free-energy extrema to occur at values of ${\chi }_{\mathrm{e}\mathrm{q}}\le \pi /2$.

Renormalization

Grunt Work

Returning to the dimensionless form of the virial expression and multiplying through by $[-{\chi }_{\mathrm{e}\mathrm{q}}/(3D)]$, we obtain,

${\chi }_{\mathrm{e}\mathrm{q}}^4=\frac{B}{D}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_g}-\frac{A}{D},$

or, after plugging in definitions of the coefficients, $A$, $B$, and $D$, and rewriting ${\chi }_{\mathrm{e}\mathrm{q}}$ explicitly as $R_{\mathrm{e}\mathrm{q}}/R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$,

$(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^4$

$=$

$\left(\frac{3}{4\pi }{)}^{{\gamma }_g}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{{\gamma }_g}\left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{{\gamma }_g}}\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{4-3{\gamma }_g}-\frac{3}{20\pi }\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2\right(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}.$

This relation can be written in a more physically concise form, as follows. First, normalize $P_e$ to a new pressure scale — call it $P_{\mathrm{a}\mathrm{d}}$ — and multiply through by $(R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}/R_{\mathrm{a}\mathrm{d}}{)}^4$ in order to normalizing $R_{\mathrm{e}\mathrm{q}}$ to a new length scale,$R_{\mathrm{a}\mathrm{d}}$:

$(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{a}\mathrm{d}}}{)}^4$	$=$	$\left(\frac{3}{4\pi }{)}^{{\gamma }_g}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{{\gamma }_g}\left(\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{P_{\mathrm{a}\mathrm{d}}}\right)\left(\frac{P_e}{P_{\mathrm{a}\mathrm{d}}}{)}^{-1}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{{\gamma }_g}}\right(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{a}\mathrm{d}}}{)}^{3{\gamma }_g}(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{a}\mathrm{d}}}{)}^{4-3{\gamma }_g}$
		$-\frac{3}{20\pi }\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2\right(\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{P_{\mathrm{a}\mathrm{d}}}\left)\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\right(\frac{P_e}{P_{\mathrm{a}\mathrm{d}}}{)}^{-1}(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{a}\mathrm{d}}}{)}^4,$

or,

${\chi }_{\mathrm{a}\mathrm{d}}^4$

$=$

$\left(\frac{3}{4\pi }{)}^{{\gamma }_g}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{{\gamma }_g}\left(\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{P_{\mathrm{a}\mathrm{d}}}\right)(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{a}\mathrm{d}}}{)}^{3{\gamma }_g}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{{\gamma }_g}}\cdot \frac{{\chi }_{\mathrm{a}\mathrm{d}}^{4-3{\gamma }_g}}{{\mathrm{\Pi }}_{\mathrm{a}\mathrm{d}}}-\frac{3}{20\pi }(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2\left(\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{P_{\mathrm{a}\mathrm{d}}}\right)(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{a}\mathrm{d}}}{)}^4\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\cdot \frac{1}{{\mathrm{\Pi }}_{\mathrm{a}\mathrm{d}}},$

where,

${\chi }_{\mathrm{a}\mathrm{d}}$	$\equiv$	$\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{a}\mathrm{d}}},$
${\mathrm{\Pi }}_{\mathrm{a}\mathrm{d}}$	$\equiv$	$\frac{P_{\mathrm{e}}}{P_{\mathrm{a}\mathrm{d}}}.$

By demanding that the leading coefficients of both terms on the right-hand-side of the expression are simultaneously unity — that is, by demanding that,

$\left(\frac{3}{4\pi }{)}^{{\gamma }_g}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{{\gamma }_g}\left(\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{P_{\mathrm{a}\mathrm{d}}}\right)(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{a}\mathrm{d}}}{)}^{3{\gamma }_g}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{{\gamma }_g}}$

$=$

$1,$

and,

$\frac{3}{20\pi }\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2\right(\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{P_{\mathrm{a}\mathrm{d}}}\left)\right(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{a}\mathrm{d}}}{)}^4\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}$

$=$

$1,$

we obtain the expressions for $R_{\mathrm{a}\mathrm{d}}/R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ and $P_{\mathrm{a}\mathrm{d}}/P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ as shown in the following table.

Renormalization for Adiabatic (ad) Systems

$\frac{R_{\mathrm{a}\mathrm{d}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$ $\equiv$ $\left[\frac{1}{5}\right(\frac{4\pi }{3}{)}^{{\gamma }_g-1}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{2-{\gamma }_g}\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_A\cdot {\widetilde{\mathfrak{𝔣}}}_M^{2-{\gamma }_g}}{]}^{1/(4-3{\gamma }_g)}$   $=$ $\left[\frac{1}{5^n}\right(\frac{4\pi }{3}\left)\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{n-1}\frac{{\widetilde{\mathfrak{𝔣}}}_W^n}{{\widetilde{\mathfrak{𝔣}}}_A^n\cdot {\widetilde{\mathfrak{𝔣}}}_M^{n-1}}{]}^{1/(n-3)}$ $\frac{P_{\mathrm{a}\mathrm{d}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$ $\equiv$ $[{\widetilde{\mathfrak{𝔣}}}_A^4\cdot (\frac{3\cdot 5^3}{4\pi }\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_M^2}{{\widetilde{\mathfrak{𝔣}}}_W^3}{)}^{{\gamma }_g}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{-2{\gamma }_g}{]}^{1/(4-3\gamma )}$   $=$ $\left[{\widetilde{\mathfrak{𝔣}}}_A^{4n}\right(\frac{3\cdot 5^3}{4\pi }{)}^{n+1}\left(\frac{{\widetilde{\mathfrak{𝔣}}}_M^2}{{\widetilde{\mathfrak{𝔣}}}_w^3}{)}^{n+1}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{-2(n+1)}{]}^{1/(n-3)}$

Using these new normalizations, we arrive at the desired, concise virial equilibrium relation, namely,

${\mathrm{\Pi }}_{\mathrm{a}\mathrm{d}}$

$=$

${\chi }_{\mathrm{a}\mathrm{d}}^{-3{\gamma }_g}-{\chi }_{\mathrm{a}\mathrm{d}}^{-4},$

or,

${\chi }_{\mathrm{a}\mathrm{d}}^{4-3{\gamma }_g}-{\mathrm{\Pi }}_{\mathrm{a}\mathrm{d}}{\chi }_{\mathrm{a}\mathrm{d}}^4$

$=$

$1.$

For the sake of completeness, we should develop expressions for both ${\chi }_{\mathrm{a}\mathrm{d}}$ and ${\mathrm{\Pi }}_{\mathrm{a}\mathrm{d}}$ that are entirely in terms of the Lane-Emden function, $\tilde{\theta }$, its derivative, ${\tilde{\theta }}^{\prime }$, and the associated dimensionless radial coordinate, $\tilde{\xi }$, at which the function and its derivative are to be evaluated. (Adopting a unified notation, we will set ${\gamma }_g\to (n+1)/n$.)

${\chi }_{\mathrm{a}\mathrm{d}}\equiv \frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{a}\mathrm{d}}}$	$=$	$\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}\cdot \frac{R_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\cdot \frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{a}\mathrm{d}}}$
	$=$	$r_a\cdot \left[\frac{4\pi }{(n+1{)}^n}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{n-1}{]}^{1/(n-3)}\cdot \left[5^n\right(\frac{3}{4\pi }\left)\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{1-n}\frac{{\widetilde{\mathfrak{𝔣}}}_A^n}{{\widetilde{\mathfrak{𝔣}}}_W^n\cdot {\widetilde{\mathfrak{𝔣}}}_M^{1-n}}{]}^{1/(n-3)}$
$\Rightarrow {\chi }_{\mathrm{a}\mathrm{d}}^{n-3}$	$=$	$3r_a^{n-3}\cdot [\frac{5}{(n+1)}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_W}{]}^n{\widetilde{\mathfrak{𝔣}}}_M^{n-1}.$

Inserting the functional expressions for $r_a$ from 📚 Horedt (1970), and our structural form factors, ${\widetilde{\mathfrak{𝔣}}}_i$, gives,

${\chi }_{\mathrm{a}\mathrm{d}}^{n-3}$	$=$	$3{\widetilde{\mathfrak{𝔣}}}_A^n[{\tilde{\xi }}^{n-3}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{1-n}\left]\right[\frac{5}{(n+1)}\cdot \frac{(5-n)}{5\cdot 3^2}(-\frac{\tilde{\xi }}{{\tilde{\theta }}^{\prime }}{)}^2{]}^n[-\frac{3{\tilde{\theta }}^{\prime }}{\tilde{\xi }}{]}^{n-1}$
	$=$	$[\frac{(5-n)}{3(n+1)}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_A}{(-{\tilde{\theta }}^{\prime }{)}^2}{]}^n$
$\Rightarrow {\chi }_{\mathrm{a}\mathrm{d}}^{(n-3)/n}$	$=$	$\frac{(5-n)}{3(n+1)}\cdot (-{\tilde{\theta }}^{\prime }{)}^{-2}[\frac{3(n+1)}{(5-n)}(-{\tilde{\theta }}^{\prime }{)}^2+{\theta }^{n+1}]$
	$=$	$1+\frac{(5-n)}{3(n+1)}\cdot \frac{{\tilde{\theta }}^{n+1}}{(-{\tilde{\theta }}^{\prime }{)}^2}.$

And,

${\mathrm{\Pi }}_{\mathrm{a}\mathrm{d}}\equiv \frac{P_e}{P_{\mathrm{a}\mathrm{d}}}$	$=$	$\frac{P_e}{P_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}\cdot \frac{P_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\cdot \frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{P_{\mathrm{a}\mathrm{d}}}$
	$=$	$p_a\cdot \left[\frac{(n+1{)}^3}{4\pi }\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{}}$