SSCpt1/Virial/FormFactors/Pt3

Structural Form Factors (Pt 3)

Part I:  Synopsis

Part II:  n = 5 Polytrope

Part III:  n = 1 Polytrope

Second Detailed Example (n = 1)

Foundation (n = 1)

We use the following normalizations, as drawn from our more general introductory discussion:

Adopted Normalizations $(n=1;\gamma =2)$
$R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ $\equiv$ $(\frac{K}{G}{)}^{1/2}$ $P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ $\equiv$ $\left(\frac{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{K^2}\right)$ $E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ $\equiv$ $P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3=(\frac{G^3}{K}{)}^{1/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2$ ${\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ $\equiv$ $\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}=\frac{3}{4\pi }(\frac{G}{K}{)}^{3/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}$ $c_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^2$ $\equiv$ $\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\frac{4\pi }{3}(\frac{G^3}{K}{)}^{1/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}$
Note that the following relations also hold: $E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}=P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3=\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\left(\frac{3}{4\pi }\right)M_{\mathrm{t}\mathrm{o}\mathrm{t}}{c}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^2$

As is detailed in our discussion of the properties of isolated polytropes, in terms of the dimensionless Lane-Emden coordinate, $\xi \equiv r/a_1$, where,

$a_1=(\frac{K}{2\pi G}{)}^{1/2},$

the radial profile of various physical variables is as follows:

$\frac{r}{(K/G{)}^{1/2}}$	$=$	$(\frac{1}{2\pi }{)}^{1/2}\xi ,$
$\frac{\rho }{{\rho }_0}$	$=$	$\frac{\sin \xi }{\xi },$
$\frac{P}{K{\rho }_0^2}$	$=$	$(\frac{\sin \xi }{\xi }{)}^2,$
$\frac{M_r}{(K/G{)}^{3/2}{\rho }_0}$	$=$	$(\frac{2}{\pi }{)}^{1/2}(\sin \xi -\xi \cos \xi ).$

Notice that, in these expressions, the central density, ${\rho }_0$, has been used instead of $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ to normalize the relevant physical variables. We can switch from one normalization to the other by realizing that — see, again, our accompanying discussion — in isolated $n=1$ polytropes, the total mass is given by the expression,

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}=[\frac{2\pi K^3}{G^3}{]}^{1/2}{\rho }_0\Rightarrow {\rho }_0=[\frac{G^3}{2\pi K^3}{]}^{1/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}.$

Employing this mapping to switch to our "preferred" adopted normalizations, as defined in the above boxed-in table, the four radial profiles become,

$r^{†}\equiv \frac{r}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$(\frac{1}{2\pi }{)}^{1/2}\xi ,$
${\rho }^{†}\equiv \frac{\rho }{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$(\frac{2^3\pi }{3^2}{)}^{1/2}\frac{\sin \xi }{\xi },$
$P^{†}\equiv \frac{P}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\frac{1}{2\pi }(\frac{\sin \xi }{\xi }{)}^2,$
$\frac{M_r}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	$=$	$\frac{1}{\pi }(\sin \xi -\xi \cos \xi ).$

Mass1 (n = 1)

While we already know the expression for the $M_r$ profile, having copied it from our discussion of detailed force-balanced models of isolated polytropes, let's show how that profile can be derived by integrating over the density profile. After employing the norm-subscripted quantities, as defined above, to normalize the radial coordinate and the mass density in our introductory discussion of the virial theorem, we obtained the following integral defining the,

Normalized Mass:

$M_r(r^{†})$

$=$

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\underset{0}{\overset{r^{†}}{\int }}3(r^{†}{)}^2{\rho }^{†}d{r}^{†}.$

Plugging in the profiles for $r^{†}$ and ${\rho }^{†}$ gives, with the help of Mathematica's Online Integrator,

$\frac{M_r(\xi )}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	$=$	$3\underset{0}{\overset{\xi }{\int }}\frac{{\xi }^2}{2\pi }(\frac{2^3\pi }{3^2}{)}^{1/2}\frac{\sin \xi }{\xi }\cdot \frac{d\xi }{(2\pi {)}^{1/2}}$
	$=$	$3(2\pi {)}^{-3/2}(\frac{2^3\pi }{3^2}{)}^{1/2}\underset{0}{\overset{\xi }{\int }}\xi \sin \xi d\xi$
	$=$	$\frac{1}{\pi }(\sin \xi -\xi \cos \xi ).$

As it should, this expression exactly matches the normalized $M_r$ profile shown above. Notice that if we decide to truncate an $n=1$ polytrope at some radius, $\tilde{\xi }<{\xi }_1$ — as in the discussion that follows — the mass of this truncated configuration will be, simply,

$\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}=\frac{M_r(\tilde{\xi })}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$

$=$

$\frac{1}{\pi }(\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi }).$

Mass2 (n = 1)

Alternatively, as has been laid out in our accompanying summary of normalized expressions that are relevant to free-energy calculations,

$\frac{M_r(x)}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$

$=$

$\left(\frac{{\rho }_c}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{q}}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left)\underset{0}{\overset{x}{\int }}3x^2\right[\frac{\rho (x)}{{\rho }_0}]dx,$

where, $M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}$ is the "total" mass of the polytropic configuration that is truncated at $R_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}$; keep in mind that, here,

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}=[\frac{2\pi K^3}{G^3}{]}^{1/2}{\rho }_0,$

is the total mass of the isolated $n=1$ polytrope, that is, a polytrope whose Lane-Emden radius extends all the way to ${\xi }_1=\pi$. In our discussions of truncated polytropes, we often will use $\tilde{\xi }\le {\xi }_1$ to specify the truncated radius in terms of the familiar, dimensionless Lane-Emden radial coordinate, so here we will set,

$R_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}=a_1\tilde{\xi }\Rightarrow x=\frac{r}{R_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}=\frac{a_1\xi }{a_1\tilde{\xi }}=\frac{\xi }{\tilde{\xi }}.$

Hence, in terms of the desired integration coordinate, $x$, the density profile provided above becomes,

$\frac{\rho (x)}{{\rho }_0}$ $=$ $\frac{\sin (\tilde{\xi }x)}{\tilde{\xi }x},$

and the integral defining $M_r(x)$ becomes,

$\frac{M_r(x)}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	$=$	$\left(\frac{{\rho }_c}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{q}}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{3}{\tilde{\xi }}\underset{0}{\overset{x}{\int }}x\sin (\tilde{\xi }x)dx$
	$=$	$\left(\frac{{\rho }_c}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{q}}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{3}{{\tilde{\xi }}^3}[\sin (\tilde{\xi }x)-(\tilde{\xi }x)\cos (\tilde{\xi }x)].$

In this case, integrating "all the way out to the surface" means setting $r=R_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}$ and, hence, $x=1$; by definition, it also means $M_r(x)=M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}$. Therefore we have,

$\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	$=$	$\left(\frac{{\rho }_c}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{q}}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{3}{{\tilde{\xi }}^3}[\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi }]$
$\Rightarrow (\frac{\overline{\rho }}{{\rho }_c}{)}_{\mathrm{e}\mathrm{q}}$	$=$	$\frac{3}{{\tilde{\xi }}^3}[\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi }].$

Using this expression for the mean-to-central density ratio along with the expression for the ratio, $M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}/M_{\mathrm{t}\mathrm{o}\mathrm{t}}$, derived in the preceding subsection, we also can state that for truncated $n=1$ polytropes,

$\frac{M_r(x)}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	$=$	$\frac{1}{\pi }(\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi })\left\{\frac{[\sin (\tilde{\xi }x)-(\tilde{\xi }x)\cos (\tilde{\xi }x)]}{(\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi })}\right\}$
	$=$	$\frac{1}{\pi }[\sin (\tilde{\xi }x)-(\tilde{\xi }x)\cos (\tilde{\xi }x)]$

By making the substitution, $x\to \xi /\tilde{\xi }$, this expression becomes identical to the $M_r/M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ profile presented just before the "Mass1" subsection, above. In summary, then, we have the following two equally valid expressions for the $M_r$ profile — one expressed as a function of $\xi$ and the other expressed as a function of $x$:

$\frac{M_r(\xi )}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$ $=$ $\frac{1}{\pi }(\sin \xi -\xi \cos \xi );$ $\frac{M_r(x)}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$ $=$ $\frac{1}{\pi }[\sin (\tilde{\xi }x)-(\tilde{\xi }x)\cos (\tilde{\xi }x)].$

Mean-to-Central Density (n = 1)

Following the line of reasoning provided above, we can use the just-derived central-to-mean density ratio to specify one of the structural form factors. Specifically,

${\mathfrak{𝔣}}_M{|}_{n=1}=\frac{\overline{\rho }}{{\rho }_c}=\frac{3}{{\tilde{\xi }}^3}[\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi }].$

Gravitational Potential Energy (n = 1)

As presented at the top of this page, the structural form factor associated with determination of the gravitational potential energy is,

${\mathfrak{𝔣}}_W$

$\equiv$

$3\cdot 5\underset{0}{\overset{1}{\int }}\left\{\underset{0}{\overset{x}{\int }}\right[\frac{\rho (x)}{{\rho }_0}\left]x^{2}dx\right\}\left[\frac{\rho (x)}{{\rho }_0}\right]xdx.$

From the derivations already presented, above, for $n=1$ polytropic configurations, we know all of the functions under this integral. We know, for example, that,

$\left\{\underset{0}{\overset{x}{\int }}\right[\frac{\rho (x)}{{\rho }_0}\left]x^{2}dx\right\}$

$=$

$\frac{1}{{\tilde{\xi }}^3}[\sin (\tilde{\xi }x)-(\tilde{\xi }x)\cos (\tilde{\xi }x)].$

Hence, with the help of Mathematica's Online Integrator, we have,

${\mathfrak{𝔣}}_W$	$=$	$\frac{3\cdot 5}{{\tilde{\xi }}^4}\underset{0}{\overset{1}{\int }}\{[\sin (\tilde{\xi }x)-(\tilde{\xi }x)\cos (\tilde{\xi }x)]\}\sin (\tilde{\xi }x)dx$
	$=$	$\frac{3\cdot 5}{{\tilde{\xi }}^4}\{\frac{x}{2}-\frac{\sin (2\tilde{\xi }x)}{4\tilde{\xi }}-\tilde{\xi }[\frac{\sin (2\tilde{\xi }x)-2\tilde{\xi }x\cos (2\tilde{\xi }x)}{8{\tilde{\xi }}^2}]{\}}_0^1$
	$=$	$\frac{3\cdot 5}{2^3{\tilde{\xi }}^6}[4{\tilde{\xi }}^2-3\tilde{\xi }\sin (2\tilde{\xi })+2{\tilde{\xi }}^2\cos (2\tilde{\xi })].$

ASIDE: Now that we have expressions for, both, ${\mathfrak{𝔣}}_M$ and ${\mathfrak{𝔣}}_W$, we can determine an analytic expression for the normalized gravitational potential energy for truncated, $n=1$ polytropes. As is shown in a companion discussion, $\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$ $=$ $-3{𝒜}{\chi }^{-1},$ where, ${𝒜}$ $\equiv$ $\frac{1}{5}\cdot \left[\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\mathfrak{𝔣}}_M}{]}^2\cdot {\mathfrak{𝔣}}_W$ $\chi$ $\equiv$ $\frac{R_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=(\frac{1}{2\pi }{)}^{1/2}\tilde{\xi }.$ A summary of derived expressions, from above, gives, ${\mathfrak{𝔣}}_M$ $=$ $\frac{3}{{\tilde{\xi }}^3}[\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi }];$ ${\mathfrak{𝔣}}_W$ $=$ $\frac{3\cdot 5}{2^3{\tilde{\xi }}^6}[4{\tilde{\xi }}^2-3\tilde{\xi }\sin (2\tilde{\xi })+2{\tilde{\xi }}^2\cos (2\tilde{\xi })];$ $\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$ $=$ $\frac{1}{\pi }(\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi }).$ Hence, ${𝒜}$ $=$ $\frac{1}{5}[\frac{{\tilde{\xi }}^3}{3\pi }{]}^2{\mathfrak{𝔣}}_W$   $=$ $\frac{1}{2^3\cdot 3{\pi }^2}[4{\tilde{\xi }}^2-3\tilde{\xi }\sin (2\tilde{\xi })+2{\tilde{\xi }}^2\cos (2\tilde{\xi })]$ $\Rightarrow \frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$ $=$ $-\left(\frac{1}{2^5{\pi }^3}{)}^{1/2}\right[4\tilde{\xi }-3\sin (2\tilde{\xi })+2\tilde{\xi }\cos (2\tilde{\xi })].$

Thermal Energy (n = 1)

As presented at the top of this page, the structural form factor associated with determination of the configuration's thermal energy is,

${\mathfrak{𝔣}}_A$

$\equiv$

$\underset{0}{\overset{1}{\int }}3\left[\frac{P(x)}{P_0}\right]x^{2}dx,$

Given that an expression for the normalized pressure profile, $P/P_0$, has already been provided, above, we can carry out the integral immediately. Specifically, we have,

$\frac{P(\xi )}{P_0}$	$=$	$(\frac{\sin \xi }{\xi }{)}^2$
$\Rightarrow \frac{P(x)}{P_0}$	$=$	$[\frac{\sin (\tilde{\xi }x)}{(\tilde{\xi }x)}{]}^2.$

Hence, with the aid of Mathematica's Online Integrator, the relevant integral gives,

${\mathfrak{𝔣}}_A$	$=$	$\frac{3}{{\tilde{\xi }}^2}\underset{0}{\overset{1}{\int }}{\sin }^2(\tilde{\xi }x)dx$
	$=$	$\frac{3}{{\tilde{\xi }}^2}[\frac{x}{2}-\frac{\sin (2\tilde{\xi }x)}{4\tilde{\xi }}{]}_0^1=\frac{3}{2^2{\tilde{\xi }}^3}[2\tilde{\xi }-\sin (2\tilde{\xi })].$

ASIDE: Having this expression for ${\mathfrak{𝔣}}_A$ allows us to determine an analytic expression for the coefficient, ${ℬ}$, that appears in our general expression for the free energy, and that can be straightforwardly used to obtain an expression for the thermal energy content of $n=1(\gamma =2)$ polytropic configurations. From our accompanying introductory discussion, we have, ${ℬ}$ $=$ $\left(\frac{3}{2^2\pi }\right)\left[\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\mathfrak{𝔣}}_M}{]}_{\mathrm{e}\mathrm{q}}^2\cdot {\mathfrak{𝔣}}_A.$ The various factors in the definition of ${ℬ}$ and $S_{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}}$ are (see above), $\chi$ $=$ $(\frac{1}{2\pi }{)}^{1/2}\tilde{\xi };$ $\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{{\mathfrak{𝔣}}_M}$ $=$ $\frac{{\tilde{\xi }}^3}{3\pi };$ ${\mathfrak{𝔣}}_A$ $=$ $\frac{3}{2^2{\tilde{\xi }}^3}[2\tilde{\xi }-\sin (2\tilde{\xi })].$ Hence, ${ℬ}$ $=$ $\left(\frac{3}{2^2\pi }\right)\left[\frac{{\tilde{\xi }}^3}{3\pi }{]}^2\frac{3}{2^2{\tilde{\xi }}^3}\right[2\tilde{\xi }-\sin (2\tilde{\xi })]$   $=$ $\left(\frac{{\tilde{\xi }}^3}{2^4{\pi }^3}\right)[2\tilde{\xi }-\sin (2\tilde{\xi })]$ and (see here and here), $\frac{S_{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$ $=$ $\frac{3}{2}(\gamma -1)\left[\frac{{\mathfrak{𝔖}}_{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]=\frac{3}{2}\cdot {\chi }^{3(1-\gamma )}{ℬ}=\frac{3}{2}\cdot {\chi }^{-3}{ℬ}$   $=$ $\left(\frac{3{\tilde{\xi }}^3}{2^5{\pi }^3}\right)[2\tilde{\xi }-\sin (2\tilde{\xi })][(2\pi {)}^{3/2}{\tilde{\xi }}^{-3}]$   $=$ $\left(\frac{3^2}{2^7{\pi }^3}{)}^{1/2}\right[2\tilde{\xi }-\sin (2\tilde{\xi })].$

Summary (n = 1)

In summary, for $n=1$ structures we have,

Structural Form Factors (n = 1)
${\mathfrak{𝔣}}_M$ $=$ $\frac{3}{{\tilde{\xi }}^3}[\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi }]$ ${\mathfrak{𝔣}}_W$ $=$ $\frac{3\cdot 5}{2^3{\tilde{\xi }}^6}[4{\tilde{\xi }}^2-3\tilde{\xi }\sin (2\tilde{\xi })+2{\tilde{\xi }}^2\cos (2\tilde{\xi })]$ ${\mathfrak{𝔣}}_A$ $=$ $\frac{3}{2^2{\tilde{\xi }}^3}[2\tilde{\xi }-\sin (2\tilde{\xi })]$
Free-Energy Coefficients (n = 1)
${𝒜}$ $=$ $\frac{1}{2^3\cdot 3{\pi }^2}[4{\tilde{\xi }}^2-3\tilde{\xi }\sin (2\tilde{\xi })+2{\tilde{\xi }}^2\cos (2\tilde{\xi })]$ ${ℬ}$ $=$ $\left(\frac{{\tilde{\xi }}^3}{2^4{\pi }^3}\right)[2\tilde{\xi }-\sin (2\tilde{\xi })]$
Normalized Energies (n = 1)
$\frac{S_{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$ $=$ $\left(\frac{3^2}{2^7{\pi }^3}{)}^{1/2}\right[2\tilde{\xi }-\sin (2\tilde{\xi })]$ $\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$ $=$ $-\left(\frac{1}{2^5{\pi }^3}{)}^{1/2}\right[4\tilde{\xi }-3\sin (2\tilde{\xi })+2\tilde{\xi }\cos (2\tilde{\xi })]$

Reality Checks (n = 1)

Expectation from Stahler's Equilibrium Models

If we add twice the thermal energy to the gravitational potential energy, we obtain,

$2\left(\frac{S_{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)+\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\left(\frac{1}{2^5{\pi }^3}{)}^{1/2}\right\{[6\tilde{\xi }-3\sin (2\tilde{\xi })]-[4\tilde{\xi }-3\sin (2\tilde{\xi })+2\tilde{\xi }\cos (2\tilde{\xi })]\}$
	$=$	$\left(\frac{1}{2^5{\pi }^3}{)}^{1/2}2\tilde{\xi }\right\{1-\cos (2\tilde{\xi })\}=(\frac{1}{2{\pi }^3}{)}^{1/2}\tilde{\xi }{\sin }^{}(\tilde{\xi }).$

For embedded polytropes, this should be compared against the expectation (prediction) provided by Stahler's equilibrium models, as detailed above. Given that, for $n=1$ polytropes — see the "Mass1" discussion above and our accompanying tabular summary of relevant properties,

$\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}=\frac{1}{\pi }(\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi })$

;

${\theta }_1=\frac{\sin \tilde{\xi }}{\tilde{\xi }}$

and

$-\frac{d{\theta }_1}{d\xi }{|}_{\tilde{\xi }}=\frac{1}{{\tilde{\xi }}^2}(\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi }),$

the expectation is that,

$\frac{4\pi P_e{R}_{\mathrm{e}\mathrm{q}}^3}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\left[\frac{(n+1{)}^3}{4\pi }{]}^{1/(n-3)}\right[\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\frac{1}{(-\theta {\text{'}}_n{)}_{\tilde{\xi }}}{]}^{(n-5)/(n-3)}({\theta }_n{)}_{\tilde{\xi }}^{(n+1)}{\tilde{\xi }}^{(n+1)/(n-3)}$
	$=$	$\left[\frac{2}{\pi }{]}^{-1/2}\right[\frac{1}{\pi }(\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi }){\tilde{\xi }}^2(\sin \tilde{\xi }-\tilde{\xi }\cos \tilde{\xi }{)}^{-1}{]}^2(\frac{\sin \tilde{\xi }}{\tilde{\xi }}{)}^2{\tilde{\xi }}^{-1}$
	$=$	$(\frac{1}{2{\pi }^3}{)}^{1/2}\tilde{\xi }{\sin }^2\tilde{\xi }.$

This precisely matches our sum of the thermal and gravitational potential energies, as just determined using our expressions for the structural form factors, giving us additional confidence that our form-factor expressions are correct.

Compare With General Expressions Based on VH74 Work

Based on the general expressions derived above in the context of VH74, for the specific case of $n=1$ polytropic configurations, the three structural form factor should be,

${\widetilde{\mathfrak{𝔣}}}_M$	$=$	$(-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }}),$
${\widetilde{\mathfrak{𝔣}}}_W$	$=$	$\frac{3\cdot 5}{4{\tilde{\xi }}^2}[{\tilde{\theta }}^2+3({\tilde{\theta }}^{\text{'}}{)}^2-{\widetilde{\mathfrak{𝔣}}}_M\tilde{\theta }],$
${\widetilde{\mathfrak{𝔣}}}_A$	$=$	$\frac{1}{2}[3{\tilde{\theta }}^2+3({\tilde{\theta }}^{\text{'}}{)}^2-{\widetilde{\mathfrak{𝔣}}}_M\tilde{\theta }].$

Also, remember that,

$\theta$	$=$	$\frac{\sin \xi }{\xi }$
$\Rightarrow {\theta }^{\text{'}}\equiv \frac{d\theta }{d\xi }$	$=$	$\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}$
$\Rightarrow ({\theta }^{\text{'}}{)}^2$	$=$	$\frac{1}{{\xi }^4}[\xi \cos \xi -\sin \xi {]}^2=\frac{1}{{\xi }^4}[{\xi }^2{\cos }^{}\xi -2\xi \sin \xi \cos \xi +{\sin }^{}\xi ].$

Now, let's look at the structural form factors, one at a time. First, we have,

${\mathfrak{𝔣}}_M$

$=$

$\frac{3}{{\xi }^3}[\sin \xi -\xi \cos \xi ]$

which matches the expression presented in the summary table, above. Next,

${\mathfrak{𝔣}}_W$	$=$	$\frac{3\cdot 5}{4{\xi }^2}[\frac{{\sin }^2\xi }{{\xi }^2}+\frac{3}{{\xi }^4}({\xi }^2{\cos }^2\xi -2\xi \sin \xi \cos \xi +{\sin }^{}\xi )-\frac{3\sin \xi }{{\xi }^4}(\sin \xi -\xi \cos \xi \left)\right]$
	$=$	$\frac{3\cdot 5}{4{\xi }^6}[{\xi }^2{\sin }^2\xi +3({\xi }^2{\cos }^2\xi -2\xi \sin \xi \cos \xi +{\sin }^{}\xi )-3\sin \xi (\sin \xi -\xi \cos \xi \left)\right]$
	$=$	$\frac{3\cdot 5}{4{\xi }^6}[{\xi }^2+2{\xi }^2{\cos }^2\xi -3\xi \sin \xi \cos \xi ]$
	$=$	$\frac{3\cdot 5}{4{\xi }^6}\{{\xi }^2+{\xi }^2[1+\cos (2\xi )]-\frac{3}{2}\xi \sin (2\xi )\}$
	$=$	$\frac{3\cdot 5}{8{\xi }^6}[4{\xi }^2+2{\xi }^2\cos (2\xi )-3\xi \sin (2\xi )],$

which also matches the expression presented in the summary table, above. Finally,

${\mathfrak{𝔣}}_A$	$=$	$\frac{1}{2}[\frac{3{\sin }^2\xi }{{\xi }^2}+\frac{3}{{\xi }^4}({\xi }^2{\cos }^2\xi -2\xi \sin \xi \cos \xi +{\sin }^{}\xi )-\frac{3\sin \xi }{{\xi }^4}(\sin \xi -\xi \cos \xi \left)\right]$
	$=$	$\frac{1}{2{\xi }^4}[3{\xi }^2{\sin }^2\xi +3({\xi }^2{\cos }^2\xi -2\xi \sin \xi \cos \xi +{\sin }^{}\xi )-3\sin \xi (\sin \xi -\xi \cos \xi \left)\right]$
	$=$	$\frac{1}{2{\xi }^4}[3{\xi }^2{\sin }^2\xi +3({\xi }^2{\cos }^2\xi -2\xi \sin \xi \cos \xi )+3\xi \sin \xi \cos \xi ]$
	$=$	$\frac{3}{2{\xi }^4}[{\xi }^2-\xi \sin \xi \cos \xi ]$
	$=$	$\frac{3}{2^2{\xi }^3}[2\xi -\sin (2\xi )],$

which also matches the expression presented in the summary table, above. So this adds support to the deduction, above, that VH74 have provided us with the information necessary to develop general expressions for the three structural form factors.

Fiddling Around

NOTE (from Tohline on 17 March 2015): Chronologically, this "Fiddling Around" subsection was developed before our discovery of the VH74 derivations. It put us on track toward the correct development of general expressions for the structural form factors that are applicable to pressure-truncated polytropic spheres. But this subsection's conclusions are superseded by the VH74 work.

In this subsection, for simplicity, we will omit the "tilde" over the variable $\xi$. In the case of $n=1$ structures,

${\theta }^{n+1}$	$=$	$(\frac{\sin \xi }{\xi }{)}^2=\frac{1}{2{\xi }^2}[1-\cos (2\xi )]=\frac{1}{2^2{\xi }^3}[2\xi -2\xi \cos (2\xi )]$
$\Rightarrow {\mathfrak{𝔣}}_A-{\theta }^{n+1}$	$=$	$\frac{1}{2^2{\xi }^3}[6\xi -3\sin (2\xi )]-\frac{1}{2^2{\xi }^3}[2\xi -2\xi \cos (2\xi )]$
	$=$	$\frac{1}{2^2{\xi }^3}[4\xi -3\sin (2\xi )+2\xi \cos (2\xi )].$

But, we also have shown that,

$\left(\frac{2^3{\xi }^5}{3\cdot 5}\right){\mathfrak{𝔣}}_W$

$=$

$[4\xi -3\sin (2\xi )+2\xi \cos (2\xi )].$

Hence, we see that,

$\left(\frac{2{\xi }^2}{3\cdot 5}\right){\mathfrak{𝔣}}_W$

$=$

${\mathfrak{𝔣}}_A-{\theta }^{n+1}.$

Similarly, in the case of $n=5$ structures,

${\theta }^{n+1}$	$=$	$(1+{\ell }^2{)}^{-3}$
$\Rightarrow \left(\frac{2^3}{3}{\ell }^3\right)[{\mathfrak{𝔣}}_A-{\theta }^{n+1}]$	$=$	$[{\tan }^{-1}(\ell )+\ell ({\ell }^4-1)(1+{\ell }^2{)}^{-3}]-(\frac{2^3}{3}{\ell }^3)(1+{\ell }^2{)}^{-3}$
	$=$	${\tan }^{-1}(\ell )+\ell ({\ell }^4-\frac{8}{3}{\ell }^2-1)(1+{\ell }^2{)}^{-3}.$

But, we also have shown that,

$(\frac{2^4}{5}\cdot {\ell }^5){\mathfrak{𝔣}}_W$

$=$

$\ell ({\ell }^4-\frac{8}{3}{\ell }^2-1)(1+{\ell }^2{)}^{-3}+{\tan }^{-1}(\ell ).$

Hence, we see that,

$(\frac{2\cdot 3}{5}\cdot {\ell }^2){\mathfrak{𝔣}}_W$	$=$	${\mathfrak{𝔣}}_A-{\theta }^{n+1}$
$\Rightarrow \left(\frac{2{\xi }^2}{5}\right){\mathfrak{𝔣}}_W$	$=$	${\mathfrak{𝔣}}_A-{\theta }^{n+1}.$

This is pretty amazing! Both examples produce almost exactly the same relationship between the two structural form factors, ${\mathfrak{𝔣}}_A$ and ${\mathfrak{𝔣}}_W$. I think that we are well on our way toward nailing down the generic, analytic relationship and, in turn, a generally applicable mass-radius relationship for pressure-truncated polytropic configurations.

Okay … here is the final piece of information. In the case of isolated polytropes, we know that the correct expressions for the structural form factors are as summarized in the following table:

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

We notice, from this, that the ratio,

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

Even in the case of the two pressure-truncated polytropes, analyzed above, this ratio proves to give the correct prefactor on ${\mathfrak{𝔣}}_W$. So we suspect that the universal relationship between the two form factors is,

$\left[\frac{(n+1){\xi }^2}{3\cdot 5}\right]{\mathfrak{𝔣}}_W$

$=$

${\mathfrak{𝔣}}_A-{\theta }^{n+1}.$

See Also