SSC/Structure/Polytropes/Analytic

Polytropic Spheres

Lane (1870)

Part I:  Isolated Polytropes

Part II:  Known Analytic Solutions

Part III:  Numerical Solutions

Known Analytic Solutions

Known Analytic Solutions

While the Lane-Emden equation has been studied for over 100 years, to date, analytic solutions to the equation (subject to the above specified boundary conditions) have been found only for three values of the polytropic index, $n$. We will review these three solutions here.  
 
 
 
 

n = 0 Polytrope

When the polytropic index, $n$, is set equal to zero, the right-hand-side of the Lane-Emden equation becomes a constant $(-1)$, so the equation can be straightforwardly integrated, twice, to obtain the desired solution for ${\mathrm{\Theta }}_H(\xi )$. Specifically, the first integration along with enforcement of the boundary condition on $d{\mathrm{\Theta }}_H/d\xi$ at the center gives,

${\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }=-\frac{1}{3}{\xi }^3.$

Then the second integration along with enforcement of the boundary condition on ${\mathrm{\Theta }}_H$ at the center gives,

${\mathrm{\Theta }}_H=1-\frac{1}{6}{\xi }^2.$

This function varies smoothly from unity at $\xi =0$ (as required by one of the boundary conditions) to zero at $\xi ={\xi }_1=\sqrt{6}$ (by tradition, the subscript "1" is used to indicate that it is the "first" zero of the Lane-Emden function), then becomes negative for values of $\xi >{\xi }_1$.

The astrophysically interesting surface of this spherical configuration is identified with the first zero of the function, that is, where the dimensionless enthalpy first goes to zero. In other words, the dimensionless radius ${\xi }_1$ should correspond with the dimensional radius of the configuration, $R$. From the definition of $\xi$, we therefore conclude that,

$a_{n=0}=\frac{R}{{\xi }_1}=\frac{R}{\sqrt{6}},$

and

$\xi =\sqrt{6}\left(\frac{r}{R}\right),$

Hence, the Lane-Emden function solution can also be written as,

${\mathrm{\Theta }}_H=\frac{H}{H_c}=1-(\frac{r}{R}{)}^2.$

Since,

$a_{n=0}^2=\frac{1}{4\pi G}\left(\frac{H_c}{{\rho }_c}\right)=\frac{R^2}{6},$

we also conclude that,

$H_c=\frac{2\pi G}{3}{\rho }_c{R}^2.$

This, combined with the Lane-Emden function solution, tells us that the run of enthalpy through the configuration is,

$H(r)=\frac{2\pi G}{3}{\rho }_c{R}^2[1-(\frac{r}{R}{)}^2].$

Now, it is always true for polytropic structures — see, for example, expressions at the top of this page of discussion — that $\rho$ can be related to $H$ through the expression,

$\left(\frac{\rho }{{\rho }_c}\right)=(\frac{H}{H_c}{)}^n={\mathrm{\Theta }}_H^n.$

Hence, for the specific case of an $n$ = 0 polytrope, we deduce that

$\frac{\rho }{{\rho }_c}=1.$

This means that an $n$ = 0 polytropic sphere is also a uniform-density sphere. It should come as no surprise to discover, therefore, that the functional behavior of $H$$(r)$ we have derived for the $n$ = 0 polytrope is identical to the $H$$(r)$ function that we have derived elsewhere for uniform-density spheres. All of the other summarized properties of uniform-density spheres can therefore also be assigned as properties of $n$ = 0 polytropes.

n = 0 Polytrope In particular, after integrating the hydrostatic-balance equation, $\frac{dP}{dr}=-\frac{G{M}_r\rho }{r^2}$ we find that the expression for the pressure is, $P(r)=P_c[1-(\frac{r}{R}{)}^2]$ , where, $P_c=\frac{2\pi G}{3}{\rho }_c^2{R}^2=\frac{3G}{8\pi }\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R^4}\right)$ .

n = 1 Polytrope

Primary E-Type Solution

When the polytropic index, $n$, is set equal to unity, the Lane-Emden equation takes the form of an inhomogeneous, $2^{\mathrm{n}\mathrm{d}}$-order ODE that is linear in the unknown function, ${\mathrm{\Theta }}_H$. Specifically, to derive the radial distribution of the Lane-Emden function ${\mathrm{\Theta }}_H(r)$ for an $n$ = 1 polytrope, we must solve,

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

subject to the above-specified boundary conditions. If we multiply this equation through by ${\xi }^2$ and move all the terms to the left-hand-side, we see that the governing ODE takes the form,

${\xi }^2\frac{d^2{\mathrm{\Theta }}_H}{d{\xi }^2}+2\xi \frac{d{\mathrm{\Theta }}_H}{d\xi }+{\xi }^2{\mathrm{\Theta }}_H=0,$

which is a relatively familiar $2^{\mathrm{n}\mathrm{d}}$-order ODE (the spherical Bessel differential equation) whose general solution involves a linear combination of the order zero spherical Bessel functions of the first and second kind, respectively,

$j_0(\xi )=\frac{\sin \xi }{\xi },$

and,

$y_0(\xi )=-\frac{\cos \xi }{\xi }.$

Given the boundary conditions that have been imposed on our astrophysical problem, we can rule out any contribution from the $y_0$ function. The desired solution is,

${\mathrm{\Theta }}_H(\xi )=j_0(\xi )=\frac{\sin \xi }{\xi }.$

This function is also referred to as the (unnormalized) sinc function.

LaTeX mathematical expressions cut-and-pasted directly from NIST's Digital Library of Mathematical Functions

As an additional point of reference, note that according to §10.47 of NIST's Digital Library of Mathematical Functions, a Spherical Bessel Function is the solution to the 2nd-order ODE, $z^2\frac{{\mathrm{d}}^{2}w}{{\mathrm{d}z}^2}+2z\frac{\mathrm{d}w}{\mathrm{d}z}+(z^2-m(m+1))w$ $=$ $0.$ This is our governing ODE if we set the parameter, $m\to 0$, in which case, according to §10.49 of NIST's Digital Library of Mathematical Functions, the solutions are, ${\mathsf{j}}_0\left(z\right)$ $=$ $\frac{\sin z}{z},$ ${\mathsf{y}}_0\left(z\right)$ $=$ $-\frac{\cos z}{z}.$

Because, by definition, $H/H_c={\mathrm{\Theta }}_H$, and for an $n$ = 1 polytrope $\rho /{\rho }_c=H/H_c$, we can immediately conclude from this Lane-Emden function solution that,

$\frac{\rho (\xi )}{{\rho }_c}=\frac{H(\xi )}{H_c}=\frac{\sin \xi }{\xi }.$

Furthermore, because the relation ($n$ + 1)$P$ = $H$$\rho$ holds for all polytropic gases, we conclude that the pressure distribution inside an $n$ = 1 polytrope is,

$\frac{P(\xi )}{P_c}=(\frac{\sin \xi }{\xi }{)}^2.$

The functions $P$$(\xi )$, $H$$(\xi )$, and $\rho$$(\xi )$ all first drop to zero when $\xi =\pi$. Hence, for an $n$ = 1 polytrope, ${\xi }_1=\pi$ and, in terms of the configuration's radius, $R$, the polytropic scale length is,

$a_{n=1}=\frac{R}{{\xi }_1}=\frac{R}{\pi }.$

So, throughout the configuration, we can relate $\xi$ to the dimensional spherical coordinate $r$ through the relation,

$\xi =\pi \left(\frac{r}{R}\right);$

and, from the general definition of $a_n$, the central value of $H$ can be expressed in terms of $R$ and ${\rho }_c$ via the relation,

$H_c=\frac{4G}{\pi }{\rho }_c{R}^2.$

Again because the relation ($n$ + 1)$P$ = $H$$\rho$ must hold everywhere inside a polytrope, this means that the central pressure is given by the expression,

$P_c=\frac{2G}{\pi }{\rho }_c^2{R}^2.$

Given the radial distribution of $\rho$, we can determine the functional behavior of the integrated mass. Specifically,

$M_r(\xi )$	$=$	$\underset{0}{\overset{r}{\int }}4\pi r^2\rho dr$
	$=$	$4\pi {\rho }_c(\frac{R}{\pi }{)}^3\underset{0}{\overset{\xi }{\int }}\xi \sin \xi d\xi$
	$=$	$\frac{4}{{\pi }^2}{\rho }_c{R}^3[\sin \xi -\xi \cos \xi ].$

Because $\xi =\pi$ at the surface of this spherical configuration — in which case the term inside the square brackets is $\pi$ — we conclude as well that the total mass of the configuration is,

$M=\frac{4}{\pi }{\rho }_c{R}^3.$

n = 1 Polytrope Let's verify the expression for the pressure by integrating the hydrostatic-balance equation, $\frac{dP}{dr}=-\frac{G{M}_r\rho }{r^2}$ From our introductory discussion of the Lane-Emden Equation $\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }\right)=-{\mathrm{\Theta }}_H^n$ we appreciate that, for a $n=1$ polytrope, $\rho ={\rho }_c{\mathrm{\Theta }}_H={\rho }_c\left(\frac{\sin \xi }{\xi }\right),$ and, $r=[\frac{K_1}{2\pi G}{]}^{1/2}\xi$,            in which case, $[\frac{K_1}{2\pi G}{]}^{1/2}=\frac{R}{\pi }.$ Combining these expressions with our above-derived expression for $M_r$, namely, $M_r(\xi )$ $=$ $\underset{0}{\overset{r}{\int }}4\pi r^2\rho dr$   $=$ $4\pi {\rho }_c[\frac{K_1}{2\pi G}{]}^{3/2}\underset{0}{\overset{\xi }{\int }}\xi \sin \xi d\xi$   $=$ $4\pi {\rho }_c(\frac{R}{\pi }{)}^3[\sin \xi -\xi \cos \xi ],$ the RHS of the hydrostatic-balance relation can be written as, $\mathrm{R}\mathrm{H}\mathrm{S}=-G\left[M_r\right]\left[\rho \right][r{]}^{-2}$ $=$ $-G[\frac{4}{{\pi }^2}{\rho }_c{R}^3[\sin \xi -\xi \cos \xi ]]\left[{\rho }_c\right(\frac{\sin \xi }{\xi }\left)\right][\frac{R\xi }{\pi }{]}^{-2}$   $=$ $-G{\rho }_c^2{R}^3\cdot \frac{{\pi }^2}{R^2}\left(\frac{4}{{\pi }^2}\right)[\sin \xi -\xi \cos \xi ]\left(\frac{\sin \xi }{{\xi }^3}\right)$   $=$ $-4G{\rho }_c^{2}R\{\frac{{\sin }^2\xi }{{\xi }^3}-\frac{\sin \xi \cos \xi }{{\xi }^2}\}.$ Now, let's integrate the hydrostatic-balance equation: $\underset{P_c}{\overset{P}{\int }}dP$ $=$ $\frac{4G{\rho }_c^2{R}^2}{\pi }\underset{0}{\overset{\xi }{\int }}\{\frac{\sin \xi \cos \xi }{{\xi }^2}-\frac{{\sin }^2\xi }{{\xi }^3}\}d\xi$ $\Rightarrow P-P_c$ $=$ $\frac{4G{\rho }_c^2{R}^2}{\pi }[\frac{{\sin }^2\xi }{2{\xi }^2}{]}_0^{\xi }$ $\Rightarrow P$ $=$ $P_c+\frac{2G{\rho }_c^2{R}^2}{\pi }[\frac{{\sin }^2\xi }{{\xi }^2}-1]$ $\Rightarrow \frac{P}{P_c}$ $=$ $(\frac{\sin \xi }{\xi }{)}^2,$ where, in order to make the last step, we have set the central pressure to, $P_c=2G{\rho }_c^2{R}^2/\pi$. This agrees with the above derivation.

Summary

From the above derivations, we can describe the properties of a spherical $n$ = 1 polytrope as follows:

• Mass:

Given the density, ${\rho }_c$, and the radius, $R$, of the configuration, the total mass is,

$M=\frac{4}{\pi }{\rho }_c{R}^3$ ;

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

$\frac{M_r}{M}=\frac{1}{\pi }[\sin (\frac{\pi r}{R})-(\frac{\pi r}{R})\cos (\frac{\pi r}{R}\left)\right]$ .

• Pressure:

Given values for the pair of model parameters $({\rho }_c,R)$, or $(M,R)$, or $({\rho }_c,M)$, the central pressure of the configuration is,

$P_c=\frac{2G}{\pi }{\rho }_c^2{R}^2=\frac{\pi G}{8}\left(\frac{M^2}{R^4}\right)=[\frac{1}{2\pi }{G}^3{\rho }_c^4{M}^2{]}^{1/3}$ ;

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

$P(r)=P_c[\frac{R}{\pi r}\sin (\frac{\pi r}{R}){]}^2$ .

• Enthalpy:

Throughout the configuration, the enthalpy is given by the relation,

$H(r)=\frac{2P(r)}{\rho (r)}=\frac{GM}{R}[\frac{R}{\pi r}\sin (\frac{\pi r}{R}\left)\right]$ .

• Gravitational potential:

Throughout the configuration — that is, for all $r\le R$ — the gravitational potential is given by the relation,

${\mathrm{\Phi }}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}-\mathrm{\Phi }(r)=H(r)=\frac{GM}{R}[\frac{R}{\pi r}\sin (\frac{\pi r}{R}\left)\right]$ .

Outside of this spherical configuration— that is, for all $r\ge R$ — the potential should behave like a point mass potential, that is,

$\mathrm{\Phi }(r)=-\frac{GM}{r}$ .

Matching these two expressions at the surface of the configuration, that is, setting ${\mathrm{\Phi }}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}=-GM/R$, we have what is generally considered the properly normalized prescription for the gravitational potential inside a spherically symmetric, $n$ = 1 polytropic configuration:

$\mathrm{\Phi }(r)=-\frac{GM}{R}\{1+[\frac{R}{\pi r}\sin \left(\frac{\pi r}{R}\right)\left]\right\}$ .

• Mass-Radius relationship:

We see that, for a given value of ${\rho }_c$, the relationship between the configuration's total mass and radius is,

$M\propto R^3\mathrm{o}\mathrm{r}R\propto M^{1/3}$ .

• Central- to Mean-Density Ratio:

The ratio of the configuration's central density to its mean density is,

$\frac{{\rho }_c}{\overline{\rho }}=\left(\frac{\pi M}{4R^3}\right)\left(\frac{3M}{4\pi R^3}\right)=\frac{{\pi }^2}{3}$ .

file = Dropbox/WorkFolder/Wiki edits/EmbeddedPolytropes/n1.xlsx --- worksheet = Sheet1 Figure 1:   Mass vs. Radius for n = 1 polytrope

For the purposes of comparing the internal structure of configurations having different polytropic indexes — see, for example Figure 4 in an accompanying chapter — we have found it useful in each case to graphically illustrate how the normalized mass, $M/M_{\mathrm{S}\mathrm{W}\mathrm{S}}$, varies with the normalized radius, $R/R_{\mathrm{S}\mathrm{W}\mathrm{S}}$, where the definition of these two functions is drawn from an accompanying discussion of pressure-truncated polytropic configurations. In the case of an $n=1$ polytrope, both functions are expressible analytically; specifically, we have,

$\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{|}_{n=1}$	$\equiv$	$(\frac{1}{4\pi }{)}^{1/2}\xi ;$
$\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{|}_{n=1}$	$=$	$\left(\frac{1}{4\pi }{)}^{1/2}\right[\frac{{\xi }^2}{{\theta }_n}\left|\frac{d{\theta }_n}{d\xi }\right|{]}_{n=1}$
	$=$	$\left(\frac{1}{4\pi }{)}^{1/2}\frac{{\xi }^3}{\sin \xi }\right[\frac{\sin \xi -\xi \cos \xi }{{\xi }^2}]$
	$=$	$\left(\frac{1}{4\pi }{)}^{1/2}\xi \right[1-\xi \cot \xi ].$

As Figure 1 illustrates, this normalized mass increases monotonically with radius. Given that the surface of the configuration is associated with the parameter value, $\xi =\pi$, we recognize that, at the surface, $R/R_{\mathrm{S}\mathrm{W}\mathrm{S}}=\sqrt{\pi /4}\approx 0.8862269$ and $M/M_{\mathrm{S}\mathrm{W}\mathrm{S}}$ formally climbs to infinity.

Published n = 1 Tabulations

Published Tabulations of n = 1 Polytropic Structure (Primary E-Type Solution)
Copied from p. 75 of Emden (1907)				Copied from p. 73 of Horedt (2004)
${\mathfrak{𝔯}}_1$	$u_1$	$-\frac{d{u}_1}{d{\mathfrak{𝔯}}_1}$		$\xi$	${\mathrm{\Theta }}_H$	$-\frac{d{\mathrm{\Theta }}_H}{d\xi }$
$0$	$1$	$0$		$0$	$1$	$0$
$$	$$	$$		$\frac{1}{10}$	$9.983342\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$	$3.330001\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{2}$
$\frac{1}{4}$	$0.98960$	$0.08280$		$$	$$	$$
$\frac{1}{2}$	$0.95882$	$0.16250$		$\frac{1}{2}$	$9.588511\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$	$1.625370\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$
$\frac{3}{4}$	$0.90886$	$0.23623$		$$	$$	$$
$1$	$0.84148$	$0.30117$		$1$	$8.414710\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$	$3.011687\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$
$1\frac{1}{4}$	$0.75918$	$0.35511$		$$	$$	$$
$1\frac{1}{2}$	$0.66500$	$0.39622$		$$	$$	$$
$2$	$0.45464$	$0.43541$		$2$	$4.546487\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$	$4.353978\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$
$2\frac{1}{2}$	$0.23938$	$0.41621$		$$	$$	$$
$3$	$0.04703$	$0.34569$		$3$	$4.704000\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{2}$	$3.3456775\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$
$$	$$	$$		$3.140$	$5.072143\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{4}$	$3.186325\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$
$\pi$	$0$	$0.31831$		$\pi$	$0$	$3.183099\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$
$3\frac{1}{4}$	$-0.03330$	$0.29564$		$$	$$	$$

n = 5 Polytrope

Primary E-Type Solution

To derive the radial distribution of the Lane-Emden function ${\mathrm{\Theta }}_H(r)$ for an $n$ = 5 polytrope, we must solve,

$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }\right)=-({\mathrm{\Theta }}_H{)}^5$ ,

subject to the above-specified boundary conditions. Following Emden (1907), [C67] (pp. 93-94) shows that by making the substitutions,

$\xi =\frac{1}{x}=e^{-t};{\mathrm{\Theta }}_H=(\frac{x}{2}{)}^{1/2}z=(\frac{1}{2}{e}^t{)}^{1/2}z,$

the differential equation can be rewritten as,

$\frac{d^{2}z}{d{t}^2}=\frac{1}{4}z(1-z^4).$

This equation has the solution,

$z=\pm [\frac{12C{e}^{-2t}}{(1+C{e}^{-2t}{)}^2}{]}^{1/4},$

that is,

${\mathrm{\Theta }}_H=[\frac{3C}{(1+C{\xi }^2{)}^2}{]}^{1/4}.$

where $C$ is an integration constant. Because ${\mathrm{\Theta }}_H$ must go to unity when $\xi =0$, we see that $C=1/3$. Hence,

${\mathrm{\Theta }}_H=[1+\frac{1}{3}{\xi }^2{]}^{-1/2}.$

From this Lane-Emden function solution, we obtain,

$\frac{\rho }{{\rho }_c}={\mathrm{\Theta }}_H^5=[1+\frac{1}{3}{\xi }^2{]}^{-5/2},$

and,

$\frac{P}{P_c}=(\frac{\rho }{{\rho }_c}{)}^{6/5}=[1+\frac{1}{3}{\xi }^2{]}^{-3}.$

Notice that, for this polytropic structure, the density and pressure don't go to zero until $\xi \to \mathrm{\infty }$. Hence, ${\xi }_1=\mathrm{\infty }$. However, the radial scale length,

$a_5=\left[\frac{1}{4\pi G}\right(\frac{H_c}{{\rho }_c}){]}^{1/2}=[\frac{(n+1)K}{4\pi G}{\rho }_c^{(1/n-1)}{]}^{1/2}=[\frac{3K}{2\pi G}{]}^{1/2}{\rho }_c^{-2/5}.$

Hence,

$M_r(\xi )$	$=$	$4\pi {\rho }_c{a}_5^3\underset{0}{\overset{\xi }{\int }}{\xi }^2[1+\frac{1}{3}{\xi }^2{]}^{-5/2}d\xi$
	$=$	$4\pi \left[\frac{3K}{2\pi G}{]}^{3/2}{\rho }_c^{-1/5}\right\{\frac{\sqrt{3}{\xi }^3}{(3+{\xi }^2{)}^{3/2}}\}$
	$=$	$[\frac{2\cdot 3^4{K}^3}{\pi G^3}{]}^{1/2}{\rho }_c^{-1/5}\{{\xi }^3(3+{\xi }^2{)}^{-3/2}\}.$

n = 5 Polytrope Let's verify the expression for the pressure by integrating the hydrostatic-balance equation, $\frac{dP}{dr}=-\frac{G{M}_r\rho }{r^2}$ From our introductory discussion of the Lane-Emden Equation $\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }\right)=-{\mathrm{\Theta }}_H^n$ we appreciate that, for a $n=5$ polytrope, $\rho ={\rho }_c{\mathrm{\Theta }}_H^5={\rho }_c(1+\frac{1}{3}{\xi }^2{)}^{-5/2},$ and, $r=(\frac{3K_5}{2\pi G}{)}^{1/2}{\rho }_c^{-2/5}\xi$. Combining these expressions with our above-derived expression for $M_r$, namely, $M_r(\xi )$ $=$ $(\frac{2\cdot 3^4{K}^3}{\pi G^3}{)}^{1/2}{\rho }_c^{-1/5}\{{\xi }^3(3+{\xi }^2{)}^{-3/2}\},$ the RHS of the hydrostatic-balance relation can be written as, $\mathrm{R}\mathrm{H}\mathrm{S}=-G\left[M_r\right]\left[\rho \right][r{]}^{-2}$ $=$ $-G\left[\right(\frac{2\cdot 3^4{K}^3}{\pi G^3}{)}^{1/2}{\rho }_c^{-1/5}\{{\xi }^3(3+{\xi }^2{)}^{-3/2}\}]\left[{\rho }_c\right(1+\frac{1}{3}{\xi }^2{)}^{-5/2}\left]\right[(\frac{3K_5}{2\pi G}{)}^{1/2}{\rho }_c^{-2/5}\xi {]}^{-2}$   $=$ $-G{3}^{5/2}\left(\frac{2\cdot 3^4{K}^3}{\pi G^3}{)}^{1/2}{\rho }_c^{-1/5}{\rho }_c\right[\left(\frac{2\pi G}{3K_5}\right){\rho }_c^{4/5}\left]\right[\xi (3+{\xi }^2{)}^{-4}]$   $=$ $-(2^3\cdot 3^7\pi \cdot GK{)}^{1/2}{\rho }_c^{8/5}[\xi (3+{\xi }^2{)}^{-4}]$ Now, let's integrate the hydrostatic-balance equation: $\underset{P_c}{\overset{P}{\int }}dP$ $=$ $-(2^3\cdot 3^7\pi \cdot GK{)}^{1/2}{\rho }_c^{8/5}(\frac{3K_5}{2\pi G}{)}^{1/2}{\rho }_c^{-2/5}\underset{0}{\overset{\xi }{\int }}[\xi (3+{\xi }^2{)}^{-4}]d\xi$ $\Rightarrow P-P_c$ $=$ $2\cdot 3^4\cdot K{\rho }_c^{6/5}\{\frac{1}{6(3+{\xi }^2{)}^3}{\}}_0^{\xi }$ $\Rightarrow P$ $=$ $P_c+3^3\cdot K{\rho }_c^{6/5}[\frac{1}{(3+{\xi }^2{)}^3}-\frac{1}{(3{)}^3}]$   $=$ $P_c+K{\rho }_c^{6/5}[(1+{\xi }^2/3{)}^{-3}-1]$   $=$ $P_c(1+\frac{{\xi }^2}{3}{)}^{-3},$ where, $P_c=K{\rho }_c^{6/5}$.

The function of $\xi$ inside the curly brackets of this last expression goes to unity as $\xi \to \mathrm{\infty }$, so the integrated mass is finite even though the configuration extends to infinity. Specifically, the total mass is,

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

We can invert this formula to obtain an expression for $K$ in terms of $M$ and ${\rho }_c$, namely,

$K=[\frac{\pi M^2{G}^3}{2\cdot 3^4}{]}^{1/3}{\rho }_c^{2/15}.$

This, in turn, means that the central pressure,

$P_c=K{\rho }_c^{6/5}=[\frac{\pi M^2{G}^3}{2\cdot 3^4}{]}^{1/3}{\rho }_c^{4/3},$

and,

$H_c=\frac{6P_c}{{\rho }_c}=[\frac{2^2\pi M^2{G}^3}{3}{]}^{1/3}{\rho }_c^{1/3}.$

file = Dropbox/WorkFolder/Wiki edits/EmbeddedPolytropes/NewN5.xlsx --- worksheet = AnalyticMR Figure 2:   Mass vs. Radius for n = 5 polytrope

For the purposes of comparing the internal structure of configurations having different polytropic indexes — see, for example Figure 4, below — we have found it useful in each case to graphically illustrate how the normalized mass, $M/M_{\mathrm{S}\mathrm{W}\mathrm{S}}$, varies with the normalized radius, $R/R_{\mathrm{S}\mathrm{W}\mathrm{S}}$, where the definition of these two functions is drawn from an accompanying discussion of pressure-truncated polytropic configurations. In the case of an $n=5$ polytrope, both functions are expressible analytically; specifically, we have,

$\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}$	$=$	$\left(\frac{5}{4\pi }{)}^{1/2}\right[\xi {\theta }^2{]}_{n=5}$
	$=$	$\left(\frac{5}{4\pi }{)}^{1/2}\xi \right[1+\frac{{\xi }^2}{3}{]}^{-1}$
	$=$	$\left(\frac{5}{4\pi }{)}^{1/2}\right[\frac{3\xi }{3+{\xi }^2}];$
$\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}$	$\equiv$	$\left(\frac{5^3}{4\pi }{)}^{1/2}\right[\theta {\xi }^2\left|\frac{d\theta }{d\xi }\right|{]}_{n=5}$
	$\equiv$	$\left(\frac{5^3}{4\pi }{)}^{1/2}\right\{{\xi }^2[1+\frac{{\xi }^2}{3}{]}^{-1/2}\frac{\xi }{3}[1+\frac{{\xi }^2}{3}{]}^{-3/2}\}$
	$\equiv$	$(\frac{5^3}{4\pi }{)}^{1/2}\frac{3{\xi }^3}{(3+{\xi }^2{)}^2}.$

As Stahler has pointed out, for an $n=5$ polytrope, this mass-radius relation can also be precisely couched in the form of a quadratic equation, namely,

$0$	$=$	$(\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2-5(\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}\left)\right(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}})+\frac{2^2\cdot 5\pi }{3}(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4$
$\Rightarrow \frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}$	$=$	$\frac{5}{2}\left(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}\right)[1\pm \sqrt{1-\frac{2^4\pi }{3\cdot 5}(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2}].$

As Figure 2 illustrates, this mass-radius relationship exhibits two turning points:   The maximum radius occurs at coordinate location,

$[\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}},\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{]}_{\mathrm{R}\mathrm{\_}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}}=[(\frac{3\cdot 5}{2^4\pi }{)}^{1/2},(\frac{3\cdot 5^3}{2^6\pi }{)}^{1/2}]\approx [0.5462742,1.3656855];$

and the maximum mass occurs at coordinate location,

$[\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}},\frac{M}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{]}_{\mathrm{M}\mathrm{\_}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}}=[(\frac{3^2\cdot 5}{2^6\pi }{)}^{1/2},(\frac{3^4\cdot 5^3}{2^{10}\pi }{)}^{1/2}]\approx [0.4730873,1.7740776].$

Published n = 5 Tabulations

Published Tabulations of n = 5 Polytropic Structure (Primary E-Type Solution)
Copied from p. 76 of Emden (1907)				Copied from p. 75 of Horedt (2004)
${\mathfrak{𝔯}}_1$	$u_1$	$-\frac{d{u}_1}{d{\mathfrak{𝔯}}_1}$		$\xi$	${\mathrm{\Theta }}_H$	$-\frac{d{\mathrm{\Theta }}_H}{d\xi }$
$0$	$1$	$0$		$0$	$1$	$0$
$$	$$	$$		$\frac{1}{10}$	$9.983375\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$	$3.316736\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{2}$
$\frac{1}{4}$	$0.98974$	$0.08079$		$$	$$	$$
$\frac{2}{4}$	$0.96078$	$0.14781$		$\frac{1}{2}$	$9.607689\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$	$1.478106\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$
$\frac{3}{4}$	$0.91768$	$0.19320$		$$	$$	$$
$1$	$0.86602$	$0.21650$		$1$	$8.660254\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$	$2.165064\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$
$\frac{3}{2}$	$0.75593$	$0.21598$		$$	$$	$$
$2$	$0.65465$	$0.18704$		$$	$$	$$
$\frac{5}{2}$	$0.56950$	$0.15392$		$$	$$	$$
$3$	$0.50000$	$0.12500$		$$	$$	$$
$\frac{7}{2}$	$0.44353$	$0.10180$		$$	$$	$$
$4$	$0.39736$	$0.08365$		$$	$$	$$
$5$	$0.32733$	$0.05845$		$5$	$3.273268\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$	$5.845122\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{2}$
$6$	$0.27735$	$0.04267$		$$	$$	$$
$7$	$0.24020$	$0.03233$		$$	$$	$$
$8$	$0.21160$	$0.02527$		$$	$$	$$
$10$	$0.17066$	$0.01657$		$10$	$1.706640\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{1}$	$1.656932\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{2}$
$12$	$0.14286$	$0.01166$		$$	$$	$$
$16$	$0.10763$	$0.00665$		$$	$$	$$
$20$	$0.08628$	$0.00428$		$$	$$	$$
$30$	$0.05764$	$0.00192$		$$	$$	$$
$$	$$	$$		$50$	$3.462025\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{2}$	$6.915751\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{4}$
$$	$$	$$		$100$	$1.731791\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{2}$	$1.731272\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{4}$
$$	$$	$$		$500$	$3.464081\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{3}$	$6.928079\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{6}$
$$	$$	$$		$1000$	$1.732048\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{3}$	$1.732043\mathrm{E}\mathrm{-}\mathrm{0}\mathrm{6}$
$$	$$	$$		$\mathrm{\infty }$	$0.000000\mathrm{E}\mathrm{+}\mathrm{0}\mathrm{0}$	$0.000000\mathrm{E}\mathrm{+}\mathrm{0}\mathrm{0}$

Srivastava's F-Type Solution

Demonstration of Function's Validity

In a short paper, S. Srivastava (1968, ApJ, 136, 680) presents another, analytically prescribable solution to the Lane-Emden equation of index $n=5$ that we will call upon in our discussion of one category of bipolytropic configurations. Rather than repeat Srivastava's derivation here, we will simply specify his functional solution then demonstrate that it satisfies the Lane-Emden equation. Srivastiva's Lane-Emden function is (see his equations 12 & 13),

${\theta }_{5F}$

$=$

$\frac{\sin [\ln (A\xi {)}^{1/2})]}{{\xi }^{1/2}\{3-2{\sin }^2[\ln (A\xi {)}^{1/2}]{\}}^{1/2}},$

where, $A$ is an arbitrary (positive) constant. Adopting the shorthand notation,

$\mathrm{\Delta }\equiv \ln (A\xi {)}^{1/2},$

and, recognizing that,

$\frac{d}{d\ln (A\xi )}[\ln (A\xi {)}^{1/2}]=\frac{1}{2}$

$\Rightarrow$

$\frac{d\mathrm{\Delta }}{d\xi }=\frac{1}{2\xi },$

the first derivative of Srivastava's Lane-Emden function is,

$\frac{d{\theta }_{5F}}{d\xi }$	$=$	$\frac{\cos \mathrm{\Delta }}{2{\xi }^{3/2}(3-2{\sin }^2\mathrm{\Delta }{)}^{1/2}}-\frac{\sin \mathrm{\Delta }}{2{\xi }^{3/2}(3-2{\sin }^2\mathrm{\Delta }{)}^{1/2}}+\frac{{\sin }^2\mathrm{\Delta }\cos \mathrm{\Delta }}{{\xi }^{3/2}(3-2{\sin }^2\mathrm{\Delta }{)}^{3/2}}$
	$=$	$\frac{1}{2{\xi }^{3/2}(3-2{\sin }^2\mathrm{\Delta }{)}^{3/2}}[(\cos \mathrm{\Delta }-\sin \mathrm{\Delta })(3-2{\sin }^{}\mathrm{\Delta })+2{\sin }^{}\mathrm{\Delta }\cos \mathrm{\Delta }]$
	$=$	$\frac{3\cos \mathrm{\Delta }-3\sin \mathrm{\Delta }+2{\sin }^{}\mathrm{\Delta }}{2{\xi }^{3/2}(3-2{\sin }^2\mathrm{\Delta }{)}^{3/2}}.$

Hence, the left-hand-side of the,

Lane-Emden Equation

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

is,

LHS	$=$	$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left[\frac{{\xi }^{1/2}(3\cos \mathrm{\Delta }-3\sin \mathrm{\Delta }+2{\sin }^{}\mathrm{\Delta })}{2(3-2{\sin }^2\mathrm{\Delta }{)}^{3/2}}\right]$
	$=$	$\frac{1}{{\xi }^2}[\frac{(3\cos \mathrm{\Delta }-3\sin \mathrm{\Delta }+2{\sin }^{}\mathrm{\Delta })}{4{\xi }^{1/2}(3-2{\sin }^2\mathrm{\Delta }{)}^{3/2}}+\frac{(-3\sin \mathrm{\Delta }-3\cos \mathrm{\Delta }+6{\sin }^{}\mathrm{\Delta }\cos \mathrm{\Delta })}{4{\xi }^{1/2}(3-2{\sin }^2\mathrm{\Delta }{)}^{3/2}}$
		$+\frac{3(3\cos \mathrm{\Delta }-3\sin \mathrm{\Delta }+2{\sin }^{}\mathrm{\Delta })\sin \mathrm{\Delta }\cos \mathrm{\Delta }}{2{\xi }^{1/2}(3-2{\sin }^2\mathrm{\Delta }{)}^{5/2}}]$
	$=$	$2^{-2}{\xi }^{-5/2}(3-2{\sin }^2\mathrm{\Delta }{)}^{-5/2}[(3-2{\sin }^2\mathrm{\Delta })(3\cos \mathrm{\Delta }-3\sin \mathrm{\Delta }+2{\sin }^{}\mathrm{\Delta })$
		$+(3-2{\sin }^2\mathrm{\Delta })(-3\sin \mathrm{\Delta }-3\cos \mathrm{\Delta }+6{\sin }^{}\mathrm{\Delta }\cos \mathrm{\Delta })+6(3\cos \mathrm{\Delta }-3\sin \mathrm{\Delta }+2{\sin }^{}\mathrm{\Delta })\sin \mathrm{\Delta }\cos \mathrm{\Delta }]$
	$=$	$2^{-2}{\xi }^{-5/2}(3-2{\sin }^2\mathrm{\Delta }{)}^{-5/2}[(3-2{\sin }^2\mathrm{\Delta })(-6\sin \mathrm{\Delta }+2{\sin }^{}\mathrm{\Delta }+6{\sin }^{}\cos \mathrm{\Delta })$
		$+6(3\cos \mathrm{\Delta }-3\sin \mathrm{\Delta }+2{\sin }^{}\mathrm{\Delta })\sin \mathrm{\Delta }\cos \mathrm{\Delta }]$
	$=$	$2^{-2}{\xi }^{-5/2}(3-2{\sin }^2\mathrm{\Delta }{)}^{-5/2}[-18\sin \mathrm{\Delta }+6{\sin }^{}\mathrm{\Delta }+18{\sin }^{}\cos \mathrm{\Delta }+12{\sin }^{}\mathrm{\Delta }-4{\sin }^{}\mathrm{\Delta }-12{\sin }^{}\cos \mathrm{\Delta }$
		$+18\sin \mathrm{\Delta }{\cos }^{}\mathrm{\Delta }-18{\sin }^{}\mathrm{\Delta }\cos \mathrm{\Delta }+12{\sin }^{}\mathrm{\Delta }\cos \mathrm{\Delta }]$
	$=$	$2^{-2}{\xi }^{-5/2}(3-2{\sin }^2\mathrm{\Delta }{)}^{-5/2}[-4{\sin }^5\mathrm{\Delta }]$
	$=$	$-{\theta }_{5F}^5.$

This demonstrates that Srivastava's function satisfies the Lane-Emden equation of index $n=5$.

Function Properties

The function, ${\theta }_{5F}$, looks like a damped oscillator with the following specific properties:

• As $\xi$ increases from zero, the function oscillates with an ever increasing period; the function goes through zero when $\mathrm{\Delta }=\pm \pi m$ (m is an integer), that is, when $(A\xi )=e^{\pm 2\pi m}$.
• The amplitude of the oscillation drops approximately as ${\xi }^{-1/2}$.
• In an astrophysical context, the function can be used as a physically realistic representation of a spherical shell inside of a self-gravitating configuration only over the interval of a single oscillation for which ${\theta }_{5F}$ is positive (ensuring that the mass density is everywhere positive) and, at the same time, $d{\theta }_{5F}/d\xi$ is negative (ensuring that the density and pressure are a decreasing function of the radial coordinate). In the following example, the astrophysically relevant segment of the function is identified with the parameter interval, ${\xi }_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}\le (A\xi )\le e^{2\pi }$.

Example Interval

As an example, let's set $A=1$ and examine the oscillation interval between $m=0$ and $m=1$, that is, over the range, $0\le \mathrm{\Delta }\le \pi$ which corresponds to the parameter interval $\xi =[1,e^{2\pi }]$. The denominator of ${\theta }_{5F}$ is positive for all values of $\xi$ and, over this specified interval, the numerator of ${\theta }_{5F}$ is also always positive. The blue curve in the following figure presents a plot of ${\theta }_{5F}(x)$ and the green curve presents a plot of the first derivative (the slope) of the function $d{\theta }_{5F}(x)/d\xi$ over the desired interval, where $x\equiv \xi /e^{2\pi }$; note that the horizontal axis is shown in logarithmic units.

Figure 3:   Our Determination and Presentation of a Segment of the ${\theta }_{5F}$ Function as originally derived by 📚 Srivastava (1962)

Srivastava's Lane-Emden function for n = 5

At both ends of the chosen parameter interval — that is, at $\mathrm{\Delta }=0$ and at $\mathrm{\Delta }=\pi$ — the function ${\theta }_{5F}=0$ and, correspondingly as depicted in the figure, the blue curve touches the horizontal axis. At the beginning of the interval ($\mathrm{\Delta }=0$), the slope of the function and, correspondingly, the green curve, has the (positive) value,

$\frac{d{\theta }_{5F}}{d\xi }$

$=$

$\frac{3\cos (0)-3\sin (0)+2{\sin }^{}(0)}{2{\xi }^{3/2}[3-2{\sin }^2(0){]}^{3/2}}=\frac{3}{2(3^{3/2})}=(2^2\cdot 3{)}^{-1/2}\approx 0.28868.$

At the end of the interval ($\mathrm{\Delta }=\pi$), the slope of the function as well as the green curve, has the (negative) value,

$\frac{d{\theta }_{5F}}{d\xi }$

$=$

$\frac{3\cos (\pi )-3\sin (\pi )+2{\sin }^{}(\pi )}{2{\xi }^{3/2}[3-2{\sin }^2(\pi ){]}^{3/2}}=\frac{-3}{2e^{3\pi }(3^{3/2})}=-e^{-3\pi }(2^2\cdot 3{)}^{-1/2}\approx -2.3296\times 10^{-5}.$

Over this interval, ${\theta }_{5F}$ reaches its maximum when the slope of the function is zero, that is, at the value of $\mathrm{\Delta }$ where,

$0$	$=$	$3\cos \mathrm{\Delta }-3\sin \mathrm{\Delta }+2(1-{\cos }^{}\mathrm{\Delta })\sin \mathrm{\Delta }$
	$=$	$3\cos \mathrm{\Delta }-\sin \mathrm{\Delta }-2{\cos }^{}\mathrm{\Delta }\sin \mathrm{\Delta }$
$\Rightarrow 1$	$=$	$3\cot \mathrm{\Delta }-2{\cos }^{}\mathrm{\Delta }.$

Rewriting both of these trigonometric functions in terms of the tangent function and adopting the shorthand notation,

$y\equiv \tan \mathrm{\Delta },$

this condition becomes,

$1$	$=$	$\frac{3}{y}-\frac{2}{1+y^2}$
$\Rightarrow y(y^2+1)$	$=$	$3(y^2+1)-2y$
$\Rightarrow y^3-3y^2+3y-3$	$=$	$0.$

ASIDE: As is well known and documented — see, for example Wolfram MathWorld or Wikipedia's discussion of the topic — the roots of any cubic equation can be determined analytically. In order to evaluate the root(s) of our particular cubic equation, we have drawn from the utilitarian online summary provided by Eric Schechter at Vanderbilt University. For a cubic equation of the general form, $a{y}^3+b{y}^2+cy+d=0,$ a real root is given by the expression, $y=p+\{q+[q^2+(r-p^2{)}^3{]}^{1/2}{\}}^{1/3}+\{q-[q^2+(r-p^2{)}^3{]}^{1/2}{\}}^{1/3},$ where, $p\equiv -\frac{b}{3a},$      $q\equiv [p^3+\frac{bc-3ad}{6a^2}],$      and      $r=\frac{c}{3a}.$ In our particular case, $a=1,$      $b=-3,$      $c=+3,$      and      $d=-3.$ WolframAlpha Hence, interestingly enough, $p=q=r=+1,$ which implies that the real root is, $y$ $=$ $1+\{2{\}}^{1/3}+\{0{\}}^{1/3}.$ (There is also a pair of imaginary roots, but they are irrelevant in the context of our overarching astrophysical discussion.) Just for fun, we have also used WolframAlpha's online "cubic equation solver" widget to find the root(s) of our specific cubic equation. Clicking on the thumbnail image provided here, on the right, displays the key result that was returned by this WolframAlpha widget.

The single, real root of this cubic equation is,

$y=1+2^{1/3},$

which corresponds to,

$\mathrm{\Delta }={\tan }^{-1}(1+2^{1/3}).$

Comment by J. E. Tohline on 17 April 2015: As far as I have been able to determine, this analytic prescription of xi_crit has not previously been derived, although, as is made clear in what follows, Murphy (1983) has assessed its value numerically to six significant digits.

Hence, over this example interval, the maximum of Srivastava's

${\theta }_{5F}$

function — and, hence also, the location at which the function's slope transitions from positive to negative values (denoted by the vertical red line in the above figure) — occurs at,

${\xi }_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}\equiv e^{2{\tan }^{-1}(1+2^{1/3})}=10.05836783.$

The corresponding value of the function at this critical radial location is,

${\theta }_{5F}{|}_{\mathrm{m}\mathrm{a}\mathrm{x}}\equiv {\theta }_{5F}({\xi }_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}})=(1+2^{1/3})[3+(1+2^{1/3}{)}^2{]}^{-1/2}{e}^{-{\tan }^{-1}(1+2^{1/3})}=0.250260848.$

This agrees precisely with the determination made by 📚 J. O. Murphy (1983a, Proc. Astron. Soc. Australia, Vol. 5, no. 2, pp. 175 - 179) — see the excerpts from his paper displayed in the following boxed-in image — that the portion of the ${\theta }_{5F}$ function that falls in the interval $1\le (A\xi )<{\xi }_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$ (the segment of the blue curve that lies to the left of the vertical red line in the above figure) is unphysical because the slope of the function is positive throughout that interval.

Equation and text extracted† from p. 177 of … J. O. Murphy (1983) Composite Analytical Solutions of the Lane-Emden Equation with Polytropic Indices n = 1 and n = 5 Proceedings of the Astronomical Society of Australia, Vol. 5, no. 2, pp. 175 - 179

on the interval $[1,e^{2\pi }]$     …     $d{\theta }_{5F}/d\xi >0$ in the range [1, 10.0583] ${\theta }_{5F}(\zeta {)}_{\mathrm{M}\mathrm{A}\mathrm{X}}=0.2503\sqrt{A}$     at     $\zeta =10.0583/A$

†Equations and text displayed here, with presentation order & layout modified from the original publication.

On the other hand, the segment that falls in the interval, ${\xi }_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}\le (A\xi )\le e^{2\pi }$, whose function values lie in the range, ${\theta }_{5F}{|}_{\mathrm{m}\mathrm{a}\mathrm{x}}\ge (A^{-1/2}{\theta }_{5F})\ge 0$ — that is, the segment of the blue curve that lies to the right of the vertical red line in the above figure — can be used to describe the $n=5$ "envelope" of a bipolytropic configuration because the function value is positive while it's first derivative is negative.

Other (All) Solutions

In a very clearly written article titled, All Solutions of the n = 5 Lane-Emden Equation, Patryk Mach (2012, J. Math. Phys., 53, 062503) has pointed out that there are other families of solutions to the Lane-Emden equation of index, $n=5$, in addition to the two solutions that have just been detailed, which he includes as his equations (3) and (5):

Equations extracted† from pp. 062503-1 & -2 of Mach (2012) "All Solutions of the n = 5 Lane-Emden Equation" Journal of Mathematical Physics, vol. 53, pp. 062503-062503-6 © American Institute of Physics

$\theta (\xi )$ $=$ $\pm \frac{1}{\sqrt{1+{\xi }^2/3}}$       $(3)$ $\theta (\xi )$ $=$ $\pm \frac{\sin (\ln \sqrt{\xi })}{\sqrt{3\xi -2\xi {\sin }^2(\ln \sqrt{\xi })}}$       $(5)$

†Equations displayed here, with layout modified from the original publication.

For completeness, Mach mentions a well-known solution that works for all indexes, $n>3$, which we have discussed separately in the context of power-law density distributions, namely,

${\theta }^n(\xi )=\frac{\rho }{{\rho }_c}=[\frac{2(n-3)}{(n-1{)}^2}{]}^{n/(n-1)}{\xi }^{-2n/(n-1)}.$

In addition, Mach identifies the rarely referenced work of H. Goenner & P. Havas (2000, J. Math. Phys., 41, 7029), which presents a family of solutions that is expressed in terms of the Weierstrass elliptic function; and he derives a new family of solutions — see equation (10) in his §2.1 — that can be expressed entirely in terms of Jacobi elliptic functions. Mach's new solutions, in particular, are oscillatory (like Srivastava's solution) but have no zeros, so in isolation they are not likely to be useful for astrophysical models. But, as Mach suggests, they "can be used in composite stellar models on the same footing as Srivastava's solution" — see our accompanying description of a composite model using Srivastava's solution.

Other Analytically Definable, but Non-Polytropic Equilibrium Spheres

In an accompanying chapter, we summarize the results of published work in which analytic equilibrium structures have been constructed without adopting a polytropic pressure-density relation. In one case, the density is assumed to drop linearly from the center to the surface; in a second case, it is assumed that the configuration has a parabolic density distribution.