SSC/Structure/StahlerMassRadius

Stahler's Mass-Radius Relationship for Embedded Polytropes

Review

In an accompanying chapter that discusses detailed force-balanced models of embedded (and pressure-truncated) polytropes, we have summarized the derivation by 📚 S. W. Stahler (1983, ApJ, Vol. 268, pp. 165 - 184) of the pair of parametric relations for the equilibrium mass and equilibrium radius for such systems, namely,

$M$	$=$	$M_{\mathrm{S}\mathrm{W}\mathrm{S}}\left(\frac{n^3}{4\pi }{)}^{1/2}\right\{{\theta }_n^{(n-3)/2}{\xi }^2\left|\frac{d{\theta }_n}{d\xi }\right|{\}}_{{\xi }_e}$
$R_{\mathrm{e}\mathrm{q}}$	$=$	$R_{\mathrm{S}\mathrm{W}\mathrm{S}}\left(\frac{n}{4\pi }{)}^{1/2}\right\{\xi {\theta }_n^{(n-1)/2}{\}}_{{\xi }_e}$

where,

$M_{\mathrm{S}\mathrm{W}\mathrm{S}}=(\frac{n+1}{nG}{)}^{3/2}{K}_n^{2n/(n+1)}{P}_{\mathrm{e}}^{(3-n)/[2(n+1)]},$

$R_{\mathrm{S}\mathrm{W}\mathrm{S}}=(\frac{n+1}{nG}{)}^{1/2}{K}_n^{n/(n+1)}{P}_{\mathrm{e}}^{(1-n)/[2(n+1)]}.$

As we also have reviewed, 📚 Stahler (1983) — hereafter, SWS — also explicitly states (see his equation B13) that the relevant mass-radius relationship for $n=5$ embedded polytropes is,

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

$=$

$0.$

In what was intended to be a complementary discussion, our free-energy analysis of embedded polytropes produced a virial equilibrium expression of the general form,

${𝒜}-{ℬ}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_g}+{𝒟}{\chi }_{\mathrm{e}\mathrm{q}}^4=0,$

where,

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

and,

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

and,

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

When we went back to compare the mass-radius relationship that results from our very general virial equilibrium expression to the one published by Stahler for pressure-truncated $n=5$ polytropes, they did not appear to agree. In what follows, we methodically plow through this comparison in considerable detail to uncover whatever discrepancies might exist.

Comparison

First, let's insert the definitions of the coefficients ${𝒜}$, ${ℬ}$, and ${𝒞}$ into the virial equilibrium expression, replacing, where necessary, the adiabatic exponent in favor of the polytropic index, using the relation, ${\gamma }_g=(n+1)/n$.

$0$	$=$	${𝒜}-{ℬ}{\chi }_{\mathrm{e}\mathrm{q}}^{(n-3)/n}+{𝒟}{\chi }_{\mathrm{e}\mathrm{q}}^4$
	$=$	$\frac{1}{5}\cdot \left[\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^2\cdot {\widetilde{\mathfrak{𝔣}}}_W-\frac{4\pi }{3}[\frac{3}{4\pi }\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}_{\mathrm{e}\mathrm{q}}^{(n+1)/n}\cdot {\widetilde{\mathfrak{𝔣}}}_A(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{(n-3)/n}+(\frac{4\pi }{3}\left)\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^4$

Next, explicitly spelling out as well the definitions of our adopted normalization radius and normalization pressure — recognizing that $P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^4=G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2$ — and multiply the expression through by $[3G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2/(4\pi )]$.

$0$	$=$	$\frac{3}{20\pi }\left(\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\right)G{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^2-G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2\left[\frac{3}{4\pi }\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left){]}_{\mathrm{e}\mathrm{q}}^{(n+1)/n}\right[\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}}\left]\right[\left(\frac{K}{G}\right)M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{(1-n)/n}]R_{\mathrm{e}\mathrm{q}}^{(n-3)/n}+P_e{R}_{\mathrm{e}\mathrm{q}}^4$
	$=$	$\frac{3}{20\pi }\left(\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\right)G{M}_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}^2-K[\frac{3}{4\pi }\cdot \frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{{\widetilde{\mathfrak{𝔣}}}_M}{]}_{\mathrm{e}\mathrm{q}}^{(n+1)/n}{\mathfrak{𝔣}}_A{R}_{\mathrm{e}\mathrm{q}}^{(n-3)/n}+P_e{R}_{\mathrm{e}\mathrm{q}}^4.$

As has been pointed out in our separate, more general discussion of the virial equilibrium of polytropes, if we multiply this expression through by $R_{\mathrm{e}\mathrm{q}}^{-4}$, set all three structural form factors, ${\mathfrak{𝔣}}_i$, equal to unity, and replace $M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}$ with the notation, $M_0$, the expression exactly matches the one presented as equation (5) of Whitworth, which reads:

Whitworth (1981, MNRAS, 195, 967)

But I like this last version of our derived expression as well because it shows some resemblance to the mass-radius relationship presented by Stahler and highlighted above: The first term on the left-hand-side is a constant times the square of the mass; the third term is a constant times the fourth power of the equilibrium radius; and the middle term shows a cross-product of the mass and radius (in our case, each is raised to a power other than unity). In an effort to make the comparison with Stahler even clearer, let's rewrite our expression in terms of the mass and equilibrium radius, normalized respectively to $M_{\mathrm{S}\mathrm{W}\mathrm{S}}$ and $R_{\mathrm{S}\mathrm{W}\mathrm{S}}$.

$0$	$=$	$\frac{3G}{20\pi }\left(\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\right)\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2\right[(\frac{n+1}{nG}{)}^{3/2}{K}_n^{2n/(n+1)}{P}_{\mathrm{e}}^{(3-n)/[2(n+1)]}{]}^2+P_e(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4\left[\right(\frac{n+1}{nG}{)}^{1/2}{K}_n^{n/(n+1)}{P}_{\mathrm{e}}^{(1-n)/[2(n+1)]}{]}^4$
		$-K{\mathfrak{𝔣}}_A[\frac{3}{4\pi }\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{(n+1)/n}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}\left[\right(\frac{n+1}{nG}{)}^{3/2}{K}_n^{2n/(n+1)}{P}_{\mathrm{e}}^{(3-n)/[2(n+1)]}{]}^{(n+1)/n}\left(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-3)/n}\right[(\frac{n+1}{nG}{)}^{1/2}{K}_n^{n/(n+1)}{P}_{\mathrm{e}}^{(1-n)/[2(n+1)]}{]}^{(n-3)/n}$
	$=$	$\frac{3}{20\pi }\left(\frac{n+1}{n}{)}^3\right(\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\left)\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2[K_n^{4n}{P}_{\mathrm{e}}^{(3-n)}{]}^{1/(n+1)}{G}^{-2}+(\frac{n+1}{n}{)}^2\left(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4\right[K_n^{4n}{P}_{\mathrm{e}}^{[(n+1)+2(1-n)]}{]}^{1/(n+1)}{G}^{-2}$
		$-{\mathfrak{𝔣}}_A[\frac{3}{4\pi }\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{(n+1)/n}(\frac{n+1}{nG}{)}^{[3(n+1)+(n-3)]/2n}\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-3)/n}\left[K^{1+2+(n-3)/(n+1)}\right][P_{\mathrm{e}}^{(3-n)+(1-n)(n-3)/(n+1)}{]}^{1/(2n)}$
	$=$	$\frac{3}{20\pi }\left(\frac{n+1}{n}{)}^3\right(\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\left)\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2[K_n^{4n}{P}_{\mathrm{e}}^{(3-n)}{]}^{1/(n+1)}{G}^{-2}+(\frac{n+1}{n}{)}^2\left(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4\right[K_n^{4n}{P}_{\mathrm{e}}^{(3-n)}{]}^{1/(n+1)}{G}^{-2}$
		$-{\mathfrak{𝔣}}_A[\frac{3}{4\pi }\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{(n+1)/n}(\frac{n+1}{n}{)}^2\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-3)/n}\left[K^{4n/(n+1)}\right]\left[P_{\mathrm{e}}^{(3-n)/(n+1)}\right]G^{-2}$
	$=$	$\frac{3}{20\pi }\left(\frac{n+1}{n}\right)\left(\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\right)(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2-{\mathfrak{𝔣}}_A[\frac{3}{4\pi }\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{(n+1)/n}\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-3)/n}+(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4$

On 30 September 2014, J.E. Tohline showed that this expression is perfectly satisfied using Stahler's definitions of the normalized mass and normalized equilibrium radius along with Tohline's definitions of the structural form factors. The middle term on the right-hand side includes the structural form factor, ${\mathfrak{𝔣}}_A$, which in turn is the sum of two pieces; the first piece of this form factor supplies the term that cancels the first term on the right-hand side of the equilibrium expression and the second piece cancels the third term on the right-hand side. We have noticed that the first term on the right-hand side (via the factor, ${\mathfrak{𝔣}}_W$) and the first portion of the middle term (via the first term in the expression for ${\mathfrak{𝔣}}_A$), contain factors of $(5-n{)}^{-1}$, which will cause these terms to blow up when considering truncated polytropes of index, $n=5$. But this is precisely the case for which Stahler provides an analytic mass-radius relationship. What happens to the virial expression if we multiply through by $(5-n)$?

Experimentation

First Finding

Let's break the form factor, ${\mathfrak{𝔣}}_A$, into two distinct pieces, namely,

${\widetilde{\mathfrak{𝔣}}}_{A1}$	$\equiv$	$\frac{3(n+1)}{(5-n)}[{\mathrm{\Theta }}^{\text{'}}{]}_{\tilde{\xi }}^2,$
${\widetilde{\mathfrak{𝔣}}}_{A2}$	$\equiv$	${\tilde{\mathrm{\Theta }}}^{n+1},$

then multiply the virial equilibrium expression through by $(5-n)$.

$0$	$=$	$\frac{3}{20\pi }\left(\frac{n+1}{n}\right)\left[\frac{(5-n){\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\right](\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2-[(5-n){\mathfrak{𝔣}}_{A1}+(5-n){\mathfrak{𝔣}}_{A2}][\frac{3}{4\pi }\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{(n+1)/n}\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-3)/n}+(5-n)(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4$
	$=$	$\frac{3}{4\pi }\left(\frac{n+1}{n}\right)\left\{\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2-\frac{n}{(4\pi {)}^{1/n}}({\tilde{\theta }}^{\prime }{)}^2[\frac{3}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{(n+1)/n}\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-3)/n}\}$
		$-(5-n)\left[{\tilde{\theta }}^{n+1}\right(\frac{3}{4\pi }\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{)}^{(n+1)/n}\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-3)/n}+\left(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4\right]$

When $(n=5)$, the entire term that appears in the second line of this expression goes to zero. The remaining set of terms (that is, the first line of the expression) give,

$(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2$	$=$	$\frac{n({\tilde{\theta }}^{\prime }{)}^2}{(4\pi {)}^{1/n}}\left[\frac{3}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{(n+1)/n}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-3)/n}$
	$=$	$\frac{5({\tilde{\theta }}^{\prime }{)}^2}{(4\pi {)}^{1/5}}\left[\frac{3}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{6/5}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{6/5}(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{2/5}$
$\Rightarrow \left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-2}$	$=$	$\frac{5^5({\tilde{\theta }}^{\prime }{)}^{10}}{(4\pi )}[\frac{3}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^6$
	$=$	$\left(\frac{5^5}{4\pi }\right){\tilde{\xi }}^6({\tilde{\theta }}^{\prime }{)}^4$
$\Rightarrow \left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2\right(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-1}$	$=$	$(\frac{5^5}{4\pi }{)}^{1/2}{\tilde{\xi }}^3({\tilde{\theta }}^{\prime }{)}^2.$

It is straightforward to show that, when $n$ is set equal to $5$, Stahler's expressions for the normalized mass and normalized radius satisfy this relation.

Second Finding

To simplify our expressions, let's write Stahler's normalized mass and normalized radius as, respectively,

$M_n\equiv \frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}$

and

$R_n\equiv \frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}.$

Then insert the ratio, ${\widetilde{\mathfrak{𝔣}}}_W/{\widetilde{\mathfrak{𝔣}}}_M^2=5/(5-n)$, into the first term of the virial equilibrium expression and rearrange the terms:

$\frac{3}{20\pi }\left[\frac{5(n+1)}{n(5-n)}\right]M_n^2+R_n^4$

$=$

${\mathfrak{𝔣}}_A[\frac{3}{4\pi }\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{(n+1)/n}{M}_n^{(n+1)/n}{R}_n^{(n-3)/n}.$

Next, given Stahler's expressions for the normalized mass and normalized radius, can we develop an expression for the ratio, ${\widetilde{\mathfrak{𝔣}}}_A/{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}$, that is only in terms of Stahler's mass and radius? Well, first notice that,

${\tilde{\theta }}^{n+1}$

$=$

$4\pi n(-{\tilde{\theta }}^{\prime }{)}^2\cdot \frac{R_n^4}{M_n^2},$

so we can write,

${\widetilde{\mathfrak{𝔣}}}_A$

$=$

$({\tilde{\theta }}^{\prime }{)}^2[\frac{3(n+1)}{(5-n)}+4\pi n\cdot \frac{R_n^4}{M_n^2}].$

Then notice that,

${\tilde{\xi }}^{n+1}$

$=$

$(4\pi )n^{-n}(-{\tilde{\theta }}^{\prime }{)}^{1-n}{R}_n^{3-n}{M}_n^{n-1},$

so,

${\widetilde{\mathfrak{𝔣}}}_M^{-(n+1)/n}$	$=$	$({\tilde{\theta }}^{\prime }{)}^{-2}[(4\pi )3^{-(n+1)}{n}^{-n}{R}_n^{3-n}{M}_n^{n-1}{]}^{1/n}$      and,
${\widetilde{\mathfrak{𝔣}}}_A\left[\right(\frac{3}{4\pi })\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}^{(n+1)/n}$	$=$	$[\frac{3(n+1)}{4\pi n(5-n)}+\frac{R_n^4}{M_n^2}][R_n^{3-n}{M}_n^{n-1}{]}^{1/n}.$

Hence, the virial equilibrium expression becomes,

$\left[\frac{3(n+1)}{4\pi n(5-n)}\right]M_n^2+R_n^4$	$=$	$[\frac{3(n+1)}{4\pi n(5-n)}+\frac{R_n^4}{M_n^2}][R_n^{3-n}{M}_n^{n-1}{]}^{1/n}{M}_n^{(n+1)/n}{R}_n^{(n-3)/n}$
	$=$	$[\frac{3(n+1)}{4\pi n(5-n)}+\frac{R_n^4}{M_n^2}]M_n^2.$

Q.E.D. Just not sure what I've really accomplished here!