SSC/Structure/BiPolytropes/Analytic1.53/Pt2

BiPolytrope with (n_c, n_e) = (3/2, 3)

Part I:  Milne's (1930) EOS

Part II:  Point-Source Model

Part III:  Our Derivation

The Point-Source Model

According to Chapter IX.3 (p. 332) of [C67], in the so-called "point-source" model, "… it is assumed that the entire source of energy is liberated at the center of the star; analytically, the assumption is that $L_r=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}=L$."

Handling Radiation Transport

Here we begin with the familiar expression for the radiation flux,

${\overrightarrow{F}}_{\mathrm{r}\mathrm{a}\mathrm{d}}$	$=$	$-\frac{c}{3\rho {\kappa }_R}\mathrm{\nabla }(a_{\mathrm{r}\mathrm{a}\mathrm{d}}{T}^4)$	$=$	$-{\chi }_{\mathrm{r}\mathrm{a}\mathrm{d}}\mathrm{\nabla }T,$
[Shu92], Vol. I, §2, p. 17, Eq. (2.17)			and    [T78], §3.4, p. 57, Eq. (67)

where [T78] refers to

${\chi }_{\mathrm{r}\mathrm{a}\mathrm{d}}$

$\equiv$

$\frac{4c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}{T}^3}{3\kappa \rho },$

[T78], §3.4, p. 57, Eq. (68)

as the coefficient of radiative conductivity. When modeling spherically symmetric configurations, the radiation flux has only a radial component, that is, ${\overrightarrow{F}}_{\mathrm{r}\mathrm{a}\mathrm{d}}={\hat{e}}_r(F_r)$. And, as pointed out in the context of Eq. (170) on p. 214 of [C67] … the quantity $L_r\equiv 4\pi r^2{F}_r$, which is the net amount of energy crossing a spherical surface of radius $r$, is generally introduced instead of $F_r$. We therefore have,

$\frac{L_r}{4\pi r^2}$	$=$	$-\frac{c}{3\rho {\kappa }_R}\frac{d}{dr}(a_{\mathrm{r}\mathrm{a}\mathrm{d}}{T}^4)$
$\Rightarrow \frac{dT}{dr}$	$=$	$-\frac{3}{4c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}\frac{\rho {\kappa }_R}{T^3}\frac{L_r}{4\pi r^2}$

[C67], Chapter V, Eq. (171)
[Clayton68], §6, Eq. (6-4a)
[KW94], §9.1, Eq. (9.6)
[HK94], §7.1, Eq. (7.8)
[BLRY07], §5.2, Eq. (5.15)

Dimensional Analysis: $\frac{{\kappa }_{R}L}{c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}$ $\sim$ $\frac{r{T}^4}{\rho }\sim m^{-1}{\ell }^4({}^{\circ }K{)}^4.$ NOTE: This is consistent with the opacity, ${\kappa }_R\sim ({\ell }^2{m}^{-1})$.

Harrison's Approach

Following 📚 M. Hall Harrison (1946, ApJ, Vol. 103, pp. 193 - 206), we seek to solve this last expression in concert with solutions to a pair of additional key governing relations for spherically symmetric equilibrium configurations, namely,

$\frac{dP}{dr}=-\frac{G{M}_r\rho }{r^2}$	;	$\frac{d{M}_r}{dr}=4\pi r^2\rho$
📚 Hall Harrison (1946), p. 196, Eq. (20)

while adopting (see related discussion)

Form A
of the Ideal Gas Equation of State,

$P_{\mathrm{g}\mathrm{a}\mathrm{s}}=\frac{\mathrm{\Re }}{\overline{\mu }}\rho T$

and adopting Kramers' opacity law, that is,

${\kappa }_R\to {\kappa }_{\mathrm{K}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{r}\mathrm{s}}$

$=$

${\kappa }_0\rho T^{-7/2}.$

Dimensional Analysis: $\left[\frac{{\kappa }_{0}L}{c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}\right]$ $\sim$ $\frac{r{T}^{15/2}}{{\rho }^2}\sim m^{-2}{\ell }^7({}^{\circ }K{)}^{15/2}$ $\frac{\mathfrak{\Re }}{\mu }$ $\sim$ $t^{-2}{\ell }^2({}^{\circ }K{)}^{-1}$ We note as well that the leading coefficient in Kramers' opacity is, ${\kappa }_0$ $\approx$ $5\times 10^{22}{\mathrm{c}\mathrm{m}}^5{\mathrm{g}}^{-2}({}^{\circ }K{)}^{7/2}.$ [Clayton68], §3.3, Eq. (3-170) [KW94], §17.2, Eq. (17.5) [HK94], §4.4.2, Eq. (4.35)

Does a Polytropic Relation Work?

Let's examine whether a point-source model can be represented by a polytropic relation.

Adopting Temperature (instead of enthalpy)

$P$	$=$	$K{\rho }^{1+1/n}\Rightarrow \rho =(\frac{P}{K}{)}^{n/(n+1)};$
$\left(\frac{\mathfrak{\Re }}{\mu }\right)\rho T$	$=$	$K{\rho }^{1+1/n}\Rightarrow T=\left(\frac{K}{\mathfrak{\Re }/\mu }\right){\rho }^{1/n};$
$T^n$	$=$	$(\frac{K}{\mathfrak{\Re }/\mu }{)}^n\rho \Rightarrow T^n=(\frac{K}{\mathfrak{\Re }/\mu }{)}^n(\frac{P}{K}{)}^{n/(n+1)}\Rightarrow T^{(n+1)}=(\frac{K}{\mathfrak{\Re }/\mu }{)}^{(n+1)}\left(\frac{P}{K}\right).$

Hydrostatic balance is governed by the single 2^nd order ODE,

$\frac{1}{r^2}\frac{d}{dr}\left[\frac{r^2}{\rho }\frac{dP}{dr}\right]$

$=$

$-4\pi G\rho .$

Normally in order to arrive at the Lane-Emden equation, $P$ is converted to $\rho$; here, let's convert both $P$ and $\rho$ to $T$. First, on the RHS we have,

$\rho$

$=$

$(\frac{\mathfrak{\Re }/\mu }{K}{)}^n{T}^n;$

and second, on the LHS we have,

$\frac{1}{\rho }\frac{dP}{dr}$	$=$	$(\frac{K}{\mathfrak{\Re }/\mu }{)}^n{T}^{-n}\cdot \frac{d}{dr}[K^{-n}(\mathfrak{\Re }/\mu {)}^{(n+1)}{T}^{(n+1)}]$
	$=$	$\left(\frac{\mathfrak{\Re }}{\mu }\right)T^{-n}\cdot \frac{d}{dr}\left[T^{(n+1)}\right]$
	$=$	$(n+1)\left(\frac{\mathfrak{\Re }}{\mu }\right)\frac{dT}{dr}$
	$=$	$(n+1)\left(\frac{\mathfrak{\Re }}{\mu }\right)\{-\frac{3}{4c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}\frac{\rho {\kappa }_R}{T^3}\frac{L}{4\pi r^2}\}$
	$=$	$-(n+1)\left(\frac{\mathfrak{\Re }}{\mu }\right)\left[\frac{3L}{16\pi c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}\right]\left[\frac{\rho }{r^2{T}^3}\right]{\kappa }_0\rho T^{-7/2}$
$\Rightarrow \frac{r^2}{\rho }\frac{dP}{dr}$	$=$	$-(n+1)\left(\frac{\mathfrak{\Re }}{\mu }\right)\left[\frac{3{\kappa }_{0}L}{16\pi c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}\right]{\rho }^2{T}^{-13/2}$
	$=$	$-(n+1)K^{-2n}\left(\frac{\mathfrak{\Re }}{\mu }{)}^{2n+1}\right[\frac{3{\kappa }_{0}L}{16\pi c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}]T^{(4n-13)/2}.$

So, the hydrostatic-balance condition becomes,

$-(n+1)K^{-2n}\left(\frac{\mathfrak{\Re }}{\mu }{)}^{2n+1}\right[\frac{3{\kappa }_{0}L}{16\pi c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}]\cdot \frac{1}{r^2}\frac{d{T}^{(4n-13)/2}}{dr}$	$=$	$-4\pi G(\frac{\mathfrak{\Re }/\mu }{K}{)}^n{T}^n$
$\Rightarrow \frac{(n+1)}{4\pi G{K}^{2n}}\left(\frac{\mathfrak{\Re }/\mu }{K}{)}^{-n}\right(\frac{\mathfrak{\Re }}{\mu }{)}^{2n+1}\left[\frac{3{\kappa }_{0}L}{16\pi c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}\right]\cdot \frac{1}{r^2}\frac{d{T}^{(4n-13)/2}}{dr}$	$=$	$T^n$
$\Rightarrow \frac{{𝒜}}{r^2}\frac{d{T}^{(4n-13)/2}}{dr}$	$=$	$T^n$
$\Rightarrow {{𝒜}}^{-1}{r}^{2}dr$	$=$	$T^{-n}\cdot d{T}^{(4n-13)/2},$

where,

${𝒜}$

$\equiv$

$\frac{(n+1)}{4\pi G{K}^n}\left(\frac{\mathfrak{\Re }}{\mu }{)}^{n+1}\right[\frac{3{\kappa }_{0}L}{16\pi c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}].$

Dimensional Analysis: $\left[\frac{{\kappa }_{0}L}{c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}\right]$ $\sim$ $m^{-2}{\ell }^7({}^{\circ }K{)}^{15/2}$ $\frac{\mathfrak{\Re }}{\mu }$ $\sim$ $t^{-2}{\ell }^2({}^{\circ }K{)}^{-1}$ polytropic $K$ $\sim$ $t^{-2}{\ell }^{2+3/n}{m}^{-1/n}$ Hence, ${𝒜}$ $\sim$ $G^{-1}{K}^{-n}\left(\frac{\mathfrak{\Re }}{\mu }{)}^{n+1}\right[m^{-2}{\ell }^7({}^{\circ }K{)}^{15/2}]$   $\sim$ $\left[{\ell }^{-3}m{t}^2\right]\left[t^{-2}{\ell }^{2+3/n}{m}^{-1/n}{]}^{-n}\right[t^{-2}{\ell }^2({}^{\circ }K{)}^{-1}{]}^{n+1}[m^{-2}{\ell }^7({}^{\circ }K{)}^{15/2}]$   $\sim$ $\left[{\ell }^{-3}\right]\left[{\ell }^{2+3/n}{]}^{-n}\right[{\ell }^2{]}^{n+1}\left[{\ell }^7\right]({}^{\circ }K{)}^{13/2-n}\sim {\ell }^3({}^{\circ }K{)}^{13/2-n}.$ Now, the characteristic length scale for polytropic configurations is given by the expression, $a_{\mathrm{n}}^2\equiv [\frac{(n+1)K}{4\pi G}\cdot {\rho }_c^{(1-n)/n}].$ If we divide ${𝒜}$ by $a_n^3$, the resulting expression should give us the characteristic temperature of the envelope. Specifically, we find that, $\frac{{𝒜}}{a_n^3}$ $=$ $\frac{(n+1)}{4\pi G{K}^n}\left(\frac{\mathfrak{\Re }}{\mu }{)}^{n+1}\right[\frac{3{\kappa }_{0}L}{16\pi c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}]\times [\frac{(n+1)K_n}{4\pi G}\cdot {\rho }_c^{(1-n)/n}{]}^{-3/2}$   $=$ $(n+1{)}^{-1/2}{\rho }_c^{3(n-1)/2n}(\frac{\mathfrak{\Re }}{\mu }{)}^{n+1}\left[\frac{3{\kappa }_{0}L}{16\pi c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}\right](4\pi G{)}^{1/2}{K}^{-n-3/2}$   $\sim$ $\left[m{\ell }^{-3}{]}^{3(n-1)/2n}\right[t^{-2}{\ell }^2({}^{\circ }K{)}^{-1}{]}^{n+1}[m^{-2}{\ell }^7({}^{\circ }K{)}^{15/2}]\left[{\ell }^3{m}^{-1}{t}^{-2}{]}^{1/2}\right[t^{-2}{\ell }^{2+3/n}{m}^{-1/n}{]}^{-n-3/2}$   $\sim$ $t^{-2n-2-1+2n+3}{m}^{3(n-1)/2n-2-1/2+1+3/2n}{\ell }^{-9(n-1)/2n+2n+2+7+3/2}[{\ell }^{(2n+3)/n}{]}^{-(2n+3)/2}({}^{\circ }K{)}^{13/2-n}$   $\sim$ ${\ell }^{(9+4n^2+12n)/2n}{\ell }^{-(2n+3{)}^2/2n}({}^{\circ }K{)}^{13/2-n}$   $\sim$ $({}^{\circ }K{)}^{(13-2n)/2}.$

Quite generally we can write,

$\frac{1}{\alpha }\frac{d{T}^{\alpha }}{dr}$

$=$

$T^{\alpha -1}\frac{dT}{dr}.$

Rewriting the hydrostatic-balance condition, we find that,

${{𝒜}}^{-1}{r}^2$

$=$

$T^{-n}\cdot \frac{d}{dr}\left[T^{(4n-13)/2}\right]=T^{-n}\left(\frac{4n-13}{2}\right)T^{(4n-15)/2}\cdot \frac{dT}{dr}=\left(\frac{4n-13}{2}\right)T^{(2n-15)/2}\cdot \frac{dT}{dr}.$

Associating the exponents,

$\alpha -1$

$=$

$\frac{2n-15}{2}\Rightarrow \alpha =\frac{2n-13}{2},$

we can write,

${{𝒜}}^{-1}{r}^2$	$=$	$\left(\frac{4n-13}{2}\right)\left(\frac{2}{2n-13}\right)\frac{d}{dr}\left[T^{(2n-13)/2}\right]$
$\Rightarrow 3d(r^3)$	$=$	${𝒜}\left(\frac{4n-13}{2n-13}\right)d\left[T^{(2n-13)/2}\right]$
$\Rightarrow 0$	$=$	$d\left[{𝒜}\right(\frac{13-4n}{13-2n})T^{(2n-13)/2}-3r^3]$
$\Rightarrow$ constant	$=$	$\frac{{𝒜}}{a_n^3}\left(\frac{13-4n}{13-2n}\right)T^{(2n-13)/2}-3(\frac{r}{a_n}{)}^3.$

Adopting Enthalpy (instead of Temperature)

For polytropic configurations the enthalpy, $H$, can easily be adopted in place of temperature via the relation,

$\left(\frac{\mathfrak{\Re }}{\mu }\right)T$

$=$

$\frac{H}{(n+1)}.$

Hence,

$P$

$=$

$K{\rho }^{1+1/n};$

$H$

$=$

$K{\rho }^{1/n};$

$H^{n+1}$

$=$

$K^{n}P.$

And the radiation-transport equation can be rewritten in the form,

$\frac{L_r}{4\pi r^2}$	$=$	$-\frac{c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}{3\rho {\kappa }_R(n+1{)}^4}(\frac{\mathfrak{\Re }}{\mu }{)}^{-4}\frac{d{H}^4}{dr}$
	$=$	$-\frac{c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}{3{\kappa }_0(n+1{)}^4}\left(\frac{\mathfrak{\Re }}{\mu }{)}^{-4}\right[{\rho }^{-2}{T}^{7/2}]\frac{d{H}^4}{dr}$
	$=$	$-\frac{c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}{3{\kappa }_0(n+1{)}^4}\left(\frac{\mathfrak{\Re }}{\mu }{)}^{-4}\right\{\left(\frac{K}{H}{)}^{2n}\right[\left(\frac{\mathfrak{\Re }}{\mu }{)}^{-1}\frac{H}{(n+1)}{]}^{7/2}\right\}4H^3\frac{dH}{dr}$
	$=$	$-\frac{4c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}{K}^{2n}}{3{\kappa }_0(n+1{)}^{15/2}}(\frac{\mathfrak{\Re }}{\mu }{)}^{-15/2}{H}^{(13-4n)/2}\frac{dH}{dr}$
$\Rightarrow r^2\frac{dH}{dr}$	$=$	$-\left[\frac{3{\kappa }_{0}L(n+1{)}^{15/2}}{16\pi c{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}{K}^{2n}}\right(\frac{\mathfrak{\Re }}{\mu }{)}^{15/2}]H^{(4n-13)/2}.$

In terms of the enthalpy, the hydrostatic-balance expression becomes,

$-4\pi G\left[\frac{H^n}{K^n}\right]$	$=$	$\frac{1}{r^2}\frac{d}{dr}\left\{r^2\right[\frac{K^n}{H^n}\left]\frac{d}{dr}\right[K^{-n}{H}^{n+1}\left]\right\}$
$\Rightarrow -4\pi G{H}^n$	$=$	$\frac{K^n}{r^2}\frac{d}{dr}\left\{\right[\frac{r^2}{H^n}\left]\frac{d}{dr}\right[H^{n+1}\left]\right\}$
	$=$	$\frac{(n+1)K^n}{r^2}\frac{d}{dr}\left\{r^2\frac{dH}{dr}\right\}.$

Combining these two equations gives,

$r^2{H}^n$	$=$	${ℬ}\cdot \frac{d}{dr}\left\{H^{(4n-13)/2}\right\}$
	$=$	$\frac{(4n-13){ℬ}}{2}\cdot H^{(4n-15)/2}\frac{dH}{dr}$
$\Rightarrow \left[\frac{2}{(4n-13){ℬ}}\right]r^2$	$=$	$H^{(2n-15)/2}\frac{dH}{dr}$

where,

${ℬ}$

$\equiv$

$\left[\frac{3{\kappa }_{0}L(n+1{)}^{17/2}}{64{\pi }^{2}Gc{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}{K}^n}\right(\frac{\mathfrak{\Re }}{\mu }{)}^{15/2}].$

As above, quite generally we can write,

$\frac{1}{\alpha }\frac{d{H}^{\alpha }}{dr}$

$=$

$H^{\alpha -1}\frac{dH}{dr}.$

So, associating the exponents, we appreciate that,

$\alpha -1$

$=$

$\frac{2n-15}{2}\Rightarrow \alpha =\frac{2n-13}{2}.$

Hence, we have,

$\left[\frac{2}{(4n-13){ℬ}}\right]r^2$	$=$	$\frac{2}{(2n-13)}\frac{d}{dr}\left[H^{(2n-13)/2}\right]$
$\Rightarrow r^2$	$=$	$\frac{(4n-13){ℬ}}{(2n-13)}\frac{d}{dr}\left[H^{(2n-13)/2}\right]$
$\Rightarrow d\left[\frac{r^3}{a_n^3}\right]$	$=$	$\frac{3(4n-13){ℬ}}{(2n-13)a_n^3}\cdot d\left[H^{(2n-13)/2}\right]$
$\Rightarrow d\left[{\xi }^3\right]$	$=$	$\left[\frac{3(4n-13)}{(2n-13)}\right][\frac{{ℬ}}{a_n^3}\cdot H_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{(2n-13)/2}]d[\frac{H}{H_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{(2n-13)/2};$

so, if we adopt the definition,

$H_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$

$\equiv$

$[\frac{(2n-13)}{3(4n-13)}\cdot \frac{a_n^3}{{ℬ}}{]}^{2/(2n-13)},$

the relation becomes,

$d\left[{\xi }^3\right]$

$=$

$d[\frac{H}{H_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{(2n-13)/2}.$

Power-Law Density Distribution

In an accompanying discussion, we have demonstrated that power-law density distributions can provide analytic solutions of the Lane-Emden equation, although the associated boundary conditions do not naturally conform to the boundary conditions that are suitable to astrophysical configurations. We have just shown that the point-source envelope configuration appears to admit a power-law temperature (alternatively, enthalpy) solution. Via the polytropic relation, $H=K{\rho }^{1/n}$, we can convert to the density-radius relation,

$d\left[{\xi }^3\right]$

$=$

$\left[\frac{K{\rho }_c^{1/n}}{H_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{(2n-13)/2}d\right[\frac{\rho }{{\rho }_c}{]}^{(2n-13)/2n}$

which, upon integration gives,

constant

$=$

$[\frac{\rho }{{\rho }_c}{]}^{(2n-13)/2n}-{\xi }^3,$

if we adopt the definition,

${\rho }_c$

$\equiv$

$(\frac{H_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{K}{)}^n.$

Setting the integration constant to zero, our result gives,

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

$\propto$

${\xi }^{6n/(2n-13)}.$

In astrophysically relevant configurations, the exponent on $\xi$ must be negative, which means that we are confined to models for which $n<\frac{7}{2}$.

Now, from our associated discussion of power-law density distributions in polytropes, we discovered that hydrostatic balance can be established at all radial positions within a spherically symmetric configuration for power-law density distributions of the form,

$\frac{\rho }{{\rho }_c}\propto {\xi }^{-2n/(n-1)}$

This matches our just-derived point-source model if,

$6n/(2n-13)$	$=$	$-2n/(n-1)$
$\Rightarrow 6(n-1)$	$=$	$2(13-2n)$
$\Rightarrow n$	$=$	$\frac{16}{5},$

which is less than $\frac{7}{2}$, so it is an astrophysically viable result.

See Also

• M. Hall Harrison (1944, ApJ, 100, 343 - 346), The Generalized Cowling Model — Bibliographic Code: 1944ApJ...100..343H
  (3^rd paragraph on p. 343): "We shall consider a composite model made up of a central core described by the Lane-Emden function of index $n=\frac{3}{2}$ and a point-source envelope."
• M. Hall Harrison (1946, ApJ, 103, 193 - 206), Stellar Models with Partially Degenerate Isothermal Cores and Point-Source Envelopes — Bibliographic Code: 1946ApJ...103..193H
• M. Hall Harrison (1947, ApJ, 105, 322 - 326), Stellar Models with Isothermal Cores and Point-Source Envelopes — Bibliographic Code: 1947ApJ...105..322H
• K. Suda & Z. Hitotuyanagi (1960, PASJapan, 12, 21 - 27), Stellar Models with Partially Degenerate Isothermal Cores