Editing SSC/Structure/Polytropes/Analytic (section)

==Known Analytic Solutions==
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<font size="-1">[[H_BookTiledMenu#Equilibrium_Structures|<b>Known<br />Analytic<br />Solutions</b>]]</font>
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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, {{Math/MP_PolytropicIndex}}.  We will review these three solutions here.
&nbsp;<br />
&nbsp;<br />
&nbsp;<br />
&nbsp;<br />
&nbsp;<br />

===n = 0 Polytrope===
When the polytropic index, {{Math/MP_PolytropicIndex}}, is set equal to zero, the right-hand-side of the Lane-Emden equation becomes a constant <math>~(-1)</math>, so the equation can be straightforwardly integrated, twice, to obtain the desired solution for <math>~\Theta_H(\xi)</math>.  Specifically, the first integration along with enforcement of the boundary condition on <math>~d\Theta_H/d\xi</math> at the center gives,

<div align="center">
<math>
\xi^2 \frac{d\Theta_H}{d\xi} = - \frac{1}{3}\xi^3 .
</math>
</div>

Then the second integration along with enforcement of the boundary condition on <math>~\Theta_H</math> at the center gives,

<div align="center">
<math>
~\Theta_H = 1 - \frac{1}{6}\xi^2 .
</math>
</div>
This function varies smoothly from unity at <math>~\xi = 0</math> (as required by one of the boundary conditions) to zero at <math>~\xi = \xi_1 = \sqrt{6}</math> (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 <math>~\xi > \xi_1</math>.  

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 <math>~\xi_1</math> should correspond with the dimensional radius of the configuration, <math>~R</math>.  From the definition of <math>~\xi</math>, we therefore conclude that,

<div align="center">
<math>~
a_{n=0} = \frac{R}{\xi_1} = \frac{R}{\sqrt{6}} ,
</math>
</div>
and

<div align="center">
<math>
\xi = \sqrt{6} \biggl(\frac{r}{R} \biggr) ,
</math>
</div>
Hence, the Lane-Emden function solution can also be written as,

<div align="center">
<math>
\Theta_H = \frac{H}{H_c} = 1 - \biggl(\frac{r}{R}\biggr)^2 .
</math>
</div>
Since,
<div align="center">
<math>
a_{n=0}^2 = \frac{1}{4\pi G} \biggl(\frac{H_c}{\rho_c}\biggr) = \frac{R^2}{6} , 
</math>
</div>

we also conclude that,
<div align="center">
<math>~
H_c = \frac{2\pi G}{3} \rho_c R^2 .
</math>
</div>
This, combined with the Lane-Emden function solution, tells us that the run of enthalpy through the configuration is,
<div align="center">
<math>~
H(r) = \frac{2\pi G}{3} \rho_c R^2 \biggl[ 1 - \biggl(\frac{r}{R}\biggr)^2 \biggr].
</math>
</div>

Now, it is always true for polytropic structures &#8212; see, for example, expressions at the top of this page of discussion &#8212; that {{Math/VAR_Density01}} can be related to {{Math/VAR_Enthalpy01}} through the expression,
<div align="center">
<math>~
\biggl( \frac{\rho}{\rho_c} \biggr) = \biggl( \frac{H}{H_c} \biggr)^n = \Theta_H^n .
</math>
</div>
Hence, for the specific case of an {{Math/MP_PolytropicIndex}} = 0 polytrope, we deduce that
<div align="center">
<math>~
\frac{\rho}{\rho_c} = 1 .
</math>
</div>
This means that an {{Math/MP_PolytropicIndex}} = 0 polytropic sphere is also a uniform-density sphere.  It should come as no surprise to discover, therefore, that the functional behavior of {{Math/VAR_Enthalpy01}}<math>(r)~</math> we have derived for the {{Math/MP_PolytropicIndex}} = 0 polytrope is identical to the {{Math/VAR_Enthalpy01}}<math>(r)~</math> function that we have [[SSC/Structure/UniformDensity|derived elsewhere for uniform-density spheres]].  All of the other [[SSC/Structure/UniformDensity#Summary|summarized properties of uniform-density spheres]] can therefore also be assigned as properties of {{Math/MP_PolytropicIndex}} = 0 polytropes.

<table border="1" align="center" width="80%"  cellpadding="5"><tr><td align="left" bgcolor="lightgreen">
<div align="center">'''n = 0 Polytrope'''</div>
[[SSC/Structure/UniformDensity#Solution_Technique_1|In particular]], after integrating the hydrostatic-balance equation, 
<div align="center">
{{Math/EQ_SShydrostaticBalance01}} 
</div>
we find that the expression for the pressure is,

<div align="center">
<math>P(r) = P_c\biggl[1 - \biggl(\frac{r}{R}\biggr)^2 \biggr] </math> ,
</div>
where,

<div align="center">
<math>P_c = \frac{2\pi G}{3} \rho_c^2 R^2 = \frac{3G}{8\pi}\biggl( \frac{M^2_\mathrm{tot}}{R^4} \biggr)</math> .
</div>
</td></tr></table>

===n = 1 Polytrope===
====Primary E-Type Solution====
When the polytropic index, {{Math/MP_PolytropicIndex}}, is set equal to unity, the Lane-Emden equation takes the form of an inhomogeneous, <math>2^\mathrm{nd}</math>-order ODE that is linear in the unknown function, <math>~\Theta_H</math>.  Specifically, to derive the radial distribution of the Lane-Emden function <math>~\Theta_H(r)</math> for an {{Math/MP_PolytropicIndex}} = 1 polytrope, we must solve,
<div align="center">
<math>\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr) = - \Theta_H</math> ,
</div>
subject to the above-specified [[SSC/Structure/Polytropes#Boundary_Conditions|boundary conditions]].  If we multiply this equation through by <math>~\xi^2</math> and move all the terms to the left-hand-side, we see that the governing ODE takes the form,
<div align="center">
<math>~\xi^2 \frac{d^2\Theta_H}{d\xi^2} + 2\xi \frac{d\Theta_H}{d\xi}  + \xi^2 \Theta_H = 0 \, ,</math>
</div>
which is a relatively familiar <math>2^\mathrm{nd}</math>-order ODE (the [http://mathworld.wolfram.com/SphericalBesselFunction.html ''spherical Bessel differential equation'']) whose general solution involves a linear combination of the [http://en.wikipedia.org/wiki/Bessel_function order zero spherical Bessel functions] of the first and second kind, respectively,
<div align="center">
<math>
~j_0(\xi) = \frac{\sin\xi}{\xi} ,
</math>
</div>
and,
<div align="center">
<math>
~y_0(\xi) = - \frac{\cos\xi}{\xi} .
</math>
</div>
Given the boundary conditions that have been imposed on our astrophysical problem, we can rule out any contribution from the <math>~y_0</math> function.  The desired solution is,
<div align="center">
<math>
\Theta_H(\xi) = j_0(\xi) = \frac{\sin\xi}{\xi} .
</math>
</div>
<span id="NIST">This function</span> is also referred to as the (unnormalized) [http://en.wikipedia.org/wiki/Sinc_function sinc function].

<table border="1" align="center" cellpadding="8" width="70%">
<tr>
  <th align="center" bgcolor="yellow">
LaTeX mathematical expressions cut-and-pasted directly from
<br />
NIST's ''Digital Library of Mathematical Functions''
  </th>
</tr>
<tr>
  <td align="left">
As an additional point of reference, note that according to [http://dlmf.nist.gov/10.47 &sect;10.47 of NIST's ''Digital Library of Mathematical Functions''], a Spherical Bessel Function is the solution to the 2<sup>nd</sup>-order ODE,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
z^{2}\frac{{\mathrm{d}}^{2}w}{{\mathrm{d}z}^{2}}+2z\frac{\mathrm{d}w}{\mathrm{d}z}+\left(z^{2}-m(m+1)\right)w
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0.</math>
  </td>
</tr>
</table>

This is our governing ODE if we set the parameter, <math>~m\rightarrow 0</math>, in which case, according to [http://dlmf.nist.gov/10.49 &sect;10.49 of NIST's ''Digital Library of Mathematical Functions''], the solutions are,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\mathsf{j}_{0}\left(z\right)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\sin z}{z},</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~
\mathsf{y}_{0}\left(z\right)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{\cos z}{z}.</math>
  </td>
</tr>
</table>

  </td>
</tr>
</table>


Because, by definition, <math>~H/H_c = \Theta_H</math>, and for an {{Math/MP_PolytropicIndex}} = 1 polytrope <math>~\rho/\rho_c = H/H_c</math>, we can immediately conclude from this Lane-Emden function solution that,
<div align="center">
<math>
~\frac{\rho(\xi)}{\rho_c} = \frac{H(\xi)}{H_c} = \frac{\sin\xi}{\xi} .
</math>
</div>
Furthermore, because the relation ({{Math/MP_PolytropicIndex}} + 1){{Math/VAR_Pressure01}} = {{Math/VAR_Enthalpy01}}{{Math/VAR_Density01}} holds for all polytropic gases, we conclude that the pressure distribution inside an {{Math/MP_PolytropicIndex}} = 1 polytrope is,
<div align="center">
<math>
~\frac{P(\xi)}{P_c} = \biggl( \frac{\sin\xi}{\xi} \biggr)^2 .
</math>
</div>
The functions {{Math/VAR_Pressure01}}<math>(\xi)~</math>, {{Math/VAR_Enthalpy01}}<math>(\xi)~</math>, and {{Math/VAR_Density01}}<math>(\xi)~</math> all first drop to zero when <math>~\xi = \pi</math>.  Hence, for an {{Math/MP_PolytropicIndex}} = 1 polytrope, <math>~\xi_1 = \pi</math> and, in terms of the configuration's radius, <math>~R</math>, the polytropic scale length is,
<div align="center">
<math>
~a_{n=1} = \frac{R}{\xi_1} = \frac{R}{\pi} .
</math>
</div>
So, throughout the configuration, we can relate <math>~\xi</math> to the dimensional spherical coordinate <math>~r</math> through the relation,
<div align="center">
<math>
~\xi = \pi \biggl(\frac{r}{R}\biggr) ;
</math>
</div>
and, from the general definition of <math>~a_n</math>, the central value of {{Math/VAR_Enthalpy01}} can be expressed in terms of <math>~R</math> and <math>~\rho_c</math> via the relation,
<div align="center">
<math>
~H_c = \frac{4G}{\pi}\rho_c R^2 .
</math>
</div>
Again because the relation ({{Math/MP_PolytropicIndex}} + 1){{Math/VAR_Pressure01}} = {{Math/VAR_Enthalpy01}}{{Math/VAR_Density01}} must hold everywhere inside a polytrope, this means that the central pressure is given by the expression,
<div align="center">
<math>
~P_c = \frac{2G}{\pi}\rho_c^2 R^2 .
</math>
</div>

Given the radial distribution of {{Math/VAR_Density01}}, we can determine the functional behavior of the integrated mass.  Specifically, 
<div align="center">
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
~M_r(\xi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_0^r 4\pi r^2 \rho~ dr  
</math>
  </td>  <td rowspan="3">
[[Image:WolframN1polytropeMass.jpg|border|240px|right]] 
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>  
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~4\pi \rho_c \biggl(\frac{R}{\pi}\biggr)^3 \int_0^\xi \xi\sin\xi ~d\xi
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>  
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4}{\pi^2} \rho_c R^3 [ \sin\xi - \xi\cos\xi  ]  \, .
</math>
  </td>
</tr>
</table>
</div>
Because <math>~\xi = \pi</math> at the surface of this spherical configuration &#8212; in which case the term inside the square brackets is <math>~\pi</math> &#8212; we conclude as well that the total mass of the configuration is,
<div align="center">
<math>
~M = \frac{4}{\pi}\rho_c R^3 .
</math>
</div>


<table border="1" align="center" width="80%"  cellpadding="5"><tr><td align="left" bgcolor="lightgreen">
<div align="center">'''n = 1 Polytrope'''</div>
Let's verify the expression for the pressure by integrating the hydrostatic-balance equation, 
<div align="center">
{{Math/EQ_SShydrostaticBalance01}} 
</div>
From our [[SSC/Structure/Polytropes|introductory discussion]] of the 
<div align="center">
<span id="LaneEmdenEquation"><font color="#770000">'''Lane-Emden Equation'''</font></span>
<br />
{{Math/EQ_SSLaneEmden01}}
</div>
we appreciate that, for a <math>n=1</math> polytrope,

<div align="center">
<math>
\rho = \rho_c \Theta_H = \rho_c \biggl(\frac{\sin\xi}{\xi} \biggr) \, ,
</math>
</div>
and,
<div align="center">
<math>
r = \biggl[ \frac{K_1}{2\pi G}\biggr]^{1 / 2} \xi</math>,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; in which case, 
<math>\biggl[ \frac{K_1}{2\pi G}\biggr]^{1 / 2} = \frac{R}{\pi} \,  .</math>
</div>
Combining these expressions with our  above-derived expression for <math>M_r</math>, namely,
<table align="center" border="0" cellpadding="5">

<tr>
  <td align="right">
<math>
~M_r(\xi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_0^r 4\pi r^2 \rho~ dr  
</math>
  </td>  
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>  
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~4\pi \rho_c \biggl[ \frac{K_1}{2\pi G}\biggr]^{3 / 2} \int_0^\xi \xi\sin\xi ~d\xi
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>  
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\pi \rho_c \biggl(\frac{R}{\pi}\biggr)^3 [ \sin\xi - \xi\cos\xi  ]  \, ,
</math>
  </td>
</tr>
</table>

the RHS of the hydrostatic-balance relation can be written as,
<table align="center" cellpadding="8">

<tr>
  <td align="right"><math>\mathrm{RHS} = - G \biggl[M_r\biggr]~\biggl[\rho\biggr]~\biggl[ r \biggr]^{-2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- G \biggl[\frac{4}{\pi^2} \rho_c R^3 [ \sin\xi - \xi\cos\xi  ] \biggr]
~\biggl[\rho_c \biggl(\frac{\sin\xi}{\xi} \biggr)\biggr]
~\biggl[ \frac{R\xi}{\pi} \biggr]^{-2}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- G \rho_c^2 R^3 \cdot \frac{\pi^2}{R^2} \biggl(\frac{4}{\pi^2}\biggr)[ \sin\xi - \xi\cos\xi  ] 
~\biggl(\frac{\sin\xi}{\xi^3} \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- 4G \rho_c^2 R 
\biggl\{
\frac{\sin^2\xi}{\xi^3} - \frac{\sin\xi \cos\xi}{\xi^2}
\biggr\}\, .
</math>
  </td>
</tr>

</table>

Now, let's integrate the hydrostatic-balance equation:

<table align="center" cellpadding="8">
<tr>
  <td align="right"><math>\int_{P_c}^{P}dP</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{4G \rho_c^2 R^2}{\pi} 
~\int_0^\xi\biggl\{
\frac{\sin\xi \cos\xi}{\xi^2} - \frac{\sin^2\xi}{\xi^3}
\biggr\}d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ P - P_c</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{4G \rho_c^2 R^2}{\pi} 
~\biggl[
\frac{\sin^2\xi}{2\xi^2}
\biggr]_0^\xi
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ P </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
P_c + 
\frac{2G \rho_c^2 R^2}{\pi} 
~
\biggl[
\frac{\sin^2\xi}{\xi^2} - 1
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~\frac{P}{P_c}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(
\frac{\sin\xi}{\xi} 
\biggr)^2 \, ,
</math>
  </td>
</tr>

</table>
where, in order to make the last step, we have set the central pressure to, <math>P_c = 2G\rho_c^2 R^2/\pi</math>.  This agrees with the above derivation.

</td></tr></table>

====Summary====

From the above derivations, we can describe the properties of a spherical {{Math/MP_PolytropicIndex}} = 1 polytrope as follows:
* <font color="red">Mass</font>:  
: Given the density, <math>\rho_c</math>, and the radius, <math>R</math>, of the configuration, the total mass is,
<div align="center">
<math>M = \frac{4}{\pi} \rho_c R^3 </math> ;
</div>

: and, expressed as a function of <math>M</math>, the mass that lies interior to radius <math>r</math> is,
<div align="center">
<math>\frac{M_r}{M} = \frac{1}{\pi} \biggl[ \sin\biggl(\frac{\pi r}{R} \biggr) - \biggl(\frac{\pi r}{R} \biggr)\cos\biggl(\frac{\pi r}{R} \biggr) \biggr]</math> .
</div>

* <font color="red">Pressure</font>: 
: Given values for the pair of model parameters <math>( \rho_c , R )</math>, or <math>( M , R )</math>, or <math>( \rho_c , M )</math>, the central pressure of the configuration is,
<div align="center">
<math>P_c = \frac{2 G}{\pi} \rho_c^2 R^2 = \frac{\pi G}{8}\biggl( \frac{M^2}{R^4} \biggr) = \biggl[ \frac{1}{2\pi} G^3 \rho_c^4 M^2 \biggr]^{1/3}</math> ; 
</div>

: and, expressed in terms of the central pressure <math>P_c</math>, the variation with radius of the pressure is,
<div align="center">
<math>P(r)= P_c \biggl[\frac{R}{\pi r} \sin\biggl(\frac{\pi r}{R}\biggr)  \biggr]^2</math> .
</div>

* <font color="red">Enthalpy</font>: 
: Throughout the configuration, the enthalpy is given by the relation,
<div align="center">
<math>H(r) = \frac{2 P(r)}{ \rho(r)} = \frac{GM}{R} \biggl[\frac{R}{\pi r} \sin\biggl(\frac{\pi r}{R}\biggr)  \biggr]</math> .
</div>

* <font color="red">Gravitational potential</font>: 
: Throughout the configuration &#8212; that is, for all <math>r \leq R</math> &#8212; the gravitational potential is given by the relation,
<div align="center">
<math>\Phi_\mathrm{surf} - \Phi(r) = H(r) = \frac{GM}{R} \biggl[\frac{R}{\pi r} \sin\biggl(\frac{\pi r}{R}\biggr)  \biggr] </math> .
</div>
: Outside of this spherical configuration&#8212; that is, for all <math>r \geq R</math> &#8212;  the potential should behave like a point mass potential, that is,
<div align="center">
<math>\Phi(r) = - \frac{GM}{r} </math> .
</div>
: Matching these two expressions at the surface of the configuration, that is, setting <math>\Phi_\mathrm{surf} = - GM/R</math>, we have what is generally considered the properly normalized prescription for the gravitational potential inside a spherically symmetric, {{Math/MP_PolytropicIndex}} = 1 polytropic configuration:
<div align="center">
<math>\Phi(r) = - \frac{G M}{R} \biggl\{ 1 + \biggl[\frac{R}{\pi r} \sin\biggl(\frac{\pi r}{R}\biggr)  \biggr] \biggr\} </math> .
</div>

* <font color="red">Mass-Radius relationship</font>:
: We see that, for a given value of <math>\rho_c</math>, the relationship between the configuration's total mass and radius is,
<div align="center">
<math>M \propto R^3  ~~~~~\mathrm{or}~~~~~R \propto M^{1/3} </math> .
</div>

* <font color="red">Central- to Mean-Density Ratio</font>:
: The ratio of the configuration's central density to its mean density is,
<div align="center">
<math>\frac{\rho_c}{\bar{\rho}} = \biggl(\frac{\pi M}{4 R^3}  \biggr)\biggl(\frac{3 M}{4 \pi R^3}  \biggr) = \frac{\pi^2}{3} </math> .
</div>

<table border="1" cellpadding="8" align="right">
<tr>
  <th align="center">[[File:DataFileButton02.png|right|60px|file = Dropbox/WorkFolder/Wiki edits/EmbeddedPolytropes/n1.xlsx --- worksheet = Sheet1]]Figure 1: &nbsp; Mass vs. Radius <br />for n = 1 polytrope</th>
</tr>
<tr>
  <td align="center">[[File:N100SequenceA.png|300px|n = 1 mass vs. radius diagram]]</td>
</tr>
</table>
For the purposes of comparing the internal structure of configurations having different polytropic indexes &#8212; see, for example [[SSC/Structure/Polytropes/Numerical#Fig4|Figure 4 in an accompanying chapter]] &#8212; we have found it useful in each case to graphically illustrate how the normalized mass, <math>~M/M_\mathrm{SWS}</math>, varies with the normalized radius, <math>~R/R_\mathrm{SWS}</math>, where the definition of these two functions is drawn from an [[SSC/Structure/PolytropesEmbedded#Stahler.27s_Presentation|accompanying discussion of pressure-truncated polytropic configurations]]. In the case of an <math>~n=1</math> polytrope, both functions are expressible analytically; specifically, we have, 
<div align="center">
<table border="0" cellpadding="3">

<tr>
  <td align="right">
<math>
~\frac{R}{R_\mathrm{SWS}}\biggr|_{n=1}
</math>
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{1}{4\pi} \biggr)^{1/2} \xi   \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
~\frac{M}{M_\mathrm{SWS}}\biggr|_{n=1}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{1}{4\pi} \biggr)^{1/2} \biggl[ \frac{\xi^2}{\theta_n} 
\biggl| \frac{d\theta_n}{d\xi} \biggr|  ~\biggr]_{n=1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{1}{4\pi} \biggr)^{1/2} \frac{\xi^3}{\sin\xi} 
\biggl[\frac{\sin\xi - \xi\cos\xi}{\xi^2} \biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{1}{4\pi} \biggr)^{1/2} \xi 
\biggl[1 - \xi\cot\xi \biggr]  \, .
</math>
  </td>
</tr>
</table>
</div>

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

====Published n = 1 Tabulations====

<table border="1" cellpadding="8" align="center">
<tr>
  <th align="center" colspan="7">Published Tabulations of n = 1 Polytropic Structure (Primary E-Type Solution)</th>
</tr>
<tr>
  <td align="center" colspan="3">
Copied from p. 75 of [https://books.google.com/books?id=MiDQAAAAMAAJ&printsec=frontcover#v=onepage&q&f=true Emden (1907)]
  </td>
  <td align="center" width="10%">&nbsp;</td>
  <td align="center" colspan="3">
Copied from p. 73 of [http://adsabs.harvard.edu/abs/2004ASSL..306.....H Horedt (2004)]
  </td>
</tr>
<tr>
  <td align="center"><math>~\mathfrak{r}_1</math></td>
  <td align="center"><math>~u_1</math></td>
  <td align="center"><math>~- \frac{du_1}{d\mathfrak{r}_1}</math></td>
  <td align="center">&nbsp;</td>
  <td align="center"><math>~\xi</math></td>
  <td align="center"><math>~\Theta_H</math></td>
  <td align="center"><math>~- \frac{d\Theta_H}{d\xi}</math></td>
</tr>
<tr>
  <td align="right"><math>~0</math></td>
  <td align="center"><math>~1</math></td>
  <td align="center"><math>~0</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~0</math></td>
  <td align="center"><math>~1</math></td>
  <td align="center"><math>~0</math></td>
</tr>
<tr>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~\tfrac{1}{10}</math></td>
  <td align="center"><math>~9.983342\mathrm{E-01}</math></td>
  <td align="center"><math>~3.330001\mathrm{E-02}</math></td>
</tr>
<tr>
  <td align="right"><math>~\tfrac{1}{4}</math></td>
  <td align="right"><math>~0.98960</math></td>
  <td align="right"><math>~0.08280</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~\tfrac{1}{2}</math></td>
  <td align="right"><math>~0.95882</math></td>
  <td align="right"><math>~0.16250</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~\tfrac{1}{2}</math></td>
  <td align="center"><math>~9.588511\mathrm{E-01}</math></td>
  <td align="center"><math>~1.625370\mathrm{E-01}</math></td>
</tr>
<tr>
  <td align="right"><math>~\tfrac{3}{4}</math></td>
  <td align="right"><math>~0.90886</math></td>
  <td align="right"><math>~0.23623</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~1</math></td>
  <td align="right"><math>~0.84148</math></td>
  <td align="right"><math>~0.30117</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~1</math></td>
  <td align="center"><math>~8.414710\mathrm{E-01}</math></td>
  <td align="center"><math>~3.011687\mathrm{E-01}</math></td>
</tr>
<tr>
  <td align="right"><math>~1 \tfrac{1}{4}</math></td>
  <td align="right"><math>~0.75918</math></td>
  <td align="right"><math>~0.35511</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~1\tfrac{1}{2}</math></td>
  <td align="right"><math>~0.66500</math></td>
  <td align="right"><math>~0.39622</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~2</math></td>
  <td align="right"><math>~0.45464</math></td>
  <td align="right"><math>~0.43541</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>2</math></td>
  <td align="center"><math>~4.546487\mathrm{E-01}</math></td>
  <td align="center"><math>~4.353978\mathrm{E-01}</math></td>
</tr>
<tr>
  <td align="right"><math>~2\tfrac{1}{2}</math></td>
  <td align="right"><math>~0.23938</math></td>
  <td align="right"><math>~0.41621</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~3</math></td>
  <td align="right"><math>~0.04703</math></td>
  <td align="right"><math>~0.34569</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~3</math></td>
  <td align="center"><math>~4.704000\mathrm{E-02}</math></td>
  <td align="center"><math>~3.3456775\mathrm{E-01}</math></td>
</tr>
<tr>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~3.140</math></td>
  <td align="center"><math>~5.072143\mathrm{E-04}</math></td>
  <td align="center"><math>~3.186325\mathrm{E-01}</math></td>
</tr>
<tr>
  <td align="right"><math>~\pi</math></td>
  <td align="right"><math>~0</math></td>
  <td align="right"><math>~0.31831</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~\pi</math></td>
  <td align="center"><math>~0</math></td>
  <td align="center"><math>~3.183099\mathrm{E-01}</math></td>
</tr>
<tr>
  <td align="right"><math>~3\tfrac{1}{4}</math></td>
  <td align="right"><math>~-0.03330</math></td>
  <td align="right"><math>~0.29564</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
</table>

===n = 5 Polytrope===
====Primary E-Type Solution====
To derive the radial distribution of the Lane-Emden function <math>\Theta_H(r)</math> for an {{Math/MP_PolytropicIndex}} = 5 polytrope, we must solve,
<div align="center">
<math>\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr) = - (\Theta_H)^5</math> ,
</div>
subject to the above-specified [[SSC/Structure/Polytropes#Boundary_Conditions|boundary conditions]].  Following [https://books.google.com/books?id=MiDQAAAAMAAJ&printsec=frontcover#v=onepage&q&f=true Emden (1907)], [[Appendix/References#C67|[<font color="red">C67</font>] ]] (pp. 93-94) shows that by making the substitutions,
<div align="center">
<math>
\xi = \frac{1}{x} = e^{-t} \, ; ~~~~~\Theta_H = \biggl(\frac{x}{2}\biggr)^{1/2} z = \biggl(\frac{1}{2}e^t\biggr)^{1/2}z \, ,
</math>
</div>
the differential equation can be rewritten as,
<div align="center">
<math>
\frac{d^2 z}{dt^2} = \frac{1}{4}z (1 - z^4) \, .
</math>
</div>
This equation has the solution,
<div align="center">
<math>
z = \pm \biggl[ \frac{12 C e^{-2t}}{(1 + C e^{-2t})^2} \biggr]^{1/4} \, ,
</math>
</div>
that is,
<div align="center">
<math>
\Theta_H = \biggl[ \frac{3 C }{(1 + C \xi^2)^2} \biggr]^{1/4} \, .
</math>
</div>
where <math>C</math> is an integration constant.  Because <math>\Theta_H</math> must go to unity when <math>\xi = 0</math>, we see that <math>C=1/3</math>.  Hence,
<div align="center">
<math>
\Theta_H = \biggl[ 1 + \frac{1}{3}\xi^2 \biggr]^{-1/2} \, .
</math>
</div>

From this Lane-Emden function solution, we obtain,
<div align="center">
<math>
\frac{\rho}{\rho_c} = \Theta_H^5 = \biggl[ 1 + \frac{1}{3}\xi^2 \biggr]^{-5/2} \, ,
</math>
</div>
and,
<div align="center">
<math>
\frac{P}{P_c} = \biggr(\frac{\rho}{\rho_c}\biggr)^{6/5} = \biggl[ 1 + \frac{1}{3}\xi^2 \biggr]^{-3} \, .
</math>
</div>
Notice that, for this polytropic structure, the density and pressure don't go to zero until <math>\xi \rightarrow \infty</math>.  Hence, <math>\xi_1 = \infty</math>. However, the radial scale length,
<div align="center">
<math>
a_5 = \biggr[ \frac{1}{4\pi G} \biggl( \frac{H_c}{\rho_c} \biggr) \biggr]^{1/2} = 
\biggr[ \frac{(n+1)K}{4\pi G} \rho_c^{(1/n - 1)} \biggr]^{1/2} =
\biggr[ \frac{3K}{2\pi G} \biggr]^{1/2}  \rho_c^{-2/5}  \, .
</math>
</div>
Hence, 
<div align="center">
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
M_r(\xi)
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4\pi \rho_c a_5^3 \int_0^\xi \xi^2 \biggl[ 1 + \frac{1}{3}\xi^2 \biggr]^{-5/2} d\xi  
</math>
  </td>  <td rowspan="3">
[[Image:WolframN5polytropeMass.jpg|border|240px|right]] 
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>  
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4\pi \biggr[ \frac{3K}{2\pi G} \biggr]^{3/2}  \rho_c^{-1/5} ~\biggl\{ \frac{\sqrt{3} \xi^3}{(3 + \xi^2)^{3/2}} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>  
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggr[ \frac{2\cdot 3^4 K^3}{\pi G^3} \biggr]^{1/2}  \rho_c^{-1/5} ~\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \} \, .
</math>
  </td>
</tr>
</table>
</div>


<table border="1" align="center" width="80%"  cellpadding="5"><tr><td align="left" bgcolor="lightgreen">
<div align="center">'''n = 5 Polytrope'''</div>
Let's verify the expression for the pressure by integrating the hydrostatic-balance equation, 
<div align="center">
{{Math/EQ_SShydrostaticBalance01}} 
</div>
From our [[SSC/Structure/Polytropes|introductory discussion]] of the 
<div align="center">
<span id="LaneEmdenEquation"><font color="#770000">'''Lane-Emden Equation'''</font></span>
<br />
{{Math/EQ_SSLaneEmden01}}
</div>
we appreciate that, for a <math>n=5</math> polytrope,

<div align="center">
<math>
\rho = \rho_c \Theta_H^5 = \rho_c \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} \, ,
</math>
</div>
and,
<div align="center">
<math>
r = \biggl( \frac{3K_5}{2\pi G}\biggr)^{1 / 2}\rho_c^{-2/5} \xi
</math>.
</div>
Combining these expressions with our  above-derived expression for <math>M_r</math>, namely,
<table align="center" border="0" cellpadding="5">

<tr>
  <td align="right">
<math>
~M_r(\xi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggr( \frac{2\cdot 3^4 K^3}{\pi G^3} \biggr)^{1/2}  \rho_c^{-1/5} ~\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \} \, ,
</math>
  </td>
</tr>
</table>

the RHS of the hydrostatic-balance relation can be written as,
<table align="center" cellpadding="8">

<tr>
  <td align="right"><math>\mathrm{RHS} = - G \biggl[M_r\biggr]~\biggl[\rho\biggr]~\biggl[ r \biggr]^{-2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- G \biggl[\biggr( \frac{2\cdot 3^4 K^3}{\pi G^3} \biggr)^{1/2}  \rho_c^{-1/5} ~\{ \xi^3 ( 3 + \xi^2 )^{-3/2} \}\biggr]
~\biggl[ \rho_c \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} \biggr]
~\biggl[ \biggl( \frac{3K_5}{2\pi G}\biggr)^{1 / 2}\rho_c^{-2/5} \xi \biggr]^{-2}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- G ~3^{5/2}\biggr( \frac{2\cdot 3^4 K^3}{\pi G^3} \biggr)^{1/2}  \rho_c^{-1/5} \rho_c \biggl[ \biggl( \frac{2\pi G}{3K_5}\biggr)\rho_c^{4/5} \biggr]
\biggl[~ \xi ( 3 + \xi^2 )^{-4} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- ~\biggr( 2^3\cdot 3^7\pi \cdot G K  \biggr)^{1/2}  \rho_c^{8/5} 
\biggl[~ \xi ( 3 + \xi^2 )^{-4} \biggr]
</math>
  </td>
</tr>

</table>

Now, let's integrate the hydrostatic-balance equation:

<table align="center" cellpadding="8">
<tr>
  <td align="right"><math>\int_{P_c}^{P}dP</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- ~\biggr( 2^3\cdot 3^7\pi \cdot G K  \biggr)^{1/2}  \rho_c^{8/5} \biggl( \frac{3K_5}{2\pi G}\biggr)^{1 / 2}\rho_c^{-2/5}
\int_0^\xi \biggl[~ \xi ( 3 + \xi^2 )^{-4} \biggr] d\xi
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ P - P_c</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
2\cdot 3^4 \cdot K   \rho_c^{6/5} 
\biggl\{
\frac{1}{ 6(3+\xi^2)^3 }
\biggr\}_0^\xi
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ P </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
P_c + 
3^3 \cdot K   \rho_c^{6/5} 
\biggl[
\frac{1}{ (3+\xi^2)^3 }
-\frac{1}{ (3)^3 }
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
P_c + 
K   \rho_c^{6/5} 
\biggl[(1+\xi^2/3)^{-3} -1 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
P_c \biggl(1+\frac{\xi^2}{3} \biggr)^{-3} \, ,
</math>
  </td>
</tr>

</table>
where, <math>P_c = K\rho_c^{6/5}</math>.
</td></tr></table>
The function of <math>\xi</math> inside the curly brackets of this last expression goes to unity as <math>\xi \rightarrow \infty</math>, so the integrated mass is finite even though the configuration extends to infinity.  Specifically, the total mass is,
<div align="center">
<math>M = \biggr[ \frac{2\cdot 3^4 K^3}{\pi G^3} \biggr]^{1/2}  \rho_c^{-1/5}  \, .</math>
</div>

We can invert this formula to obtain an expression for <math>K</math> in terms of <math>M</math> and <math>\rho_c</math>, namely,
<div align="center">
<math>
K = \biggr[ \frac{\pi M^2 G^3}{2\cdot 3^4} \biggr]^{1/3}  \rho_c^{2/15}  \, .
</math>
</div>
This, in turn, means that the central pressure,
<div align="center">
<math>
P_c = K\rho_c^{6/5} = \biggr[ \frac{\pi M^2 G^3}{2\cdot 3^4} \biggr]^{1/3}  \rho_c^{4/3}  \, ,
</math>
</div>
and,
<div align="center">
<math>
H_c = \frac{6P_c}{\rho_c} = \biggr[ \frac{2^2 \pi M^2 G^3}{3} \biggr]^{1/3}  \rho_c^{1/3}  \, .
</math>
</div>


<table border="1" cellpadding="8" align="right">
<tr>
  <th align="center">[[File:DataFileButton02.png|right|60px|file = Dropbox/WorkFolder/Wiki edits/EmbeddedPolytropes/NewN5.xlsx --- worksheet = AnalyticMR]]Figure 2: &nbsp; Mass vs. Radius <br />for n = 5 polytrope</th>
</tr>
<tr>
  <td align="center">[[File:N500SequenceB.png|300px|n = 5 mass vs. radius diagram]]</td>
</tr>
</table>
For the purposes of comparing the internal structure of configurations having different polytropic indexes &#8212; see, for example [[#Fig4|Figure 4, below]] &#8212; we have found it useful in each case to graphically illustrate how the normalized mass, <math>~M/M_\mathrm{SWS}</math>, varies with the normalized radius, <math>~R/R_\mathrm{SWS}</math>, where the definition of these two functions is drawn from an [[SSC/Structure/PolytropesEmbedded#Stahler.27s_Presentation|accompanying discussion of pressure-truncated polytropic configurations]]. In the case of an <math>~n=5</math> polytrope, both functions are expressible analytically; specifically, we have, 
<div align="center">
<table border="0" cellpadding="3">

<tr>
  <td align="right">
<math>
~\frac{R}{R_\mathrm{SWS}}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{5}{4\pi} \biggr)^{1/2} \biggl[ \xi \theta^{2}  \biggr]_{n=5}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{5}{4\pi} \biggr)^{1/2} \xi \biggl[1 + \frac{\xi^2}{3}  \biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{5}{4\pi} \biggr)^{1/2} \biggl[\frac{3\xi}{3 + \xi^2}  \biggr] \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
~\frac{M}{M_\mathrm{SWS}}
</math>
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{5^3}{4\pi} \biggr)^{1/2} \biggl[\theta \xi^2 
\biggl| \frac{d\theta}{d\xi} \biggr| ~\biggr]_{n=5}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{5^3}{4\pi} \biggr)^{1/2} \biggl\{\xi^2 \biggl[1 + \frac{\xi^2}{3}\biggr]^{-1 / 2} 
\frac{\xi}{3} \biggl[1 + \frac{\xi^2}{3}\biggr]^{-3 / 2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{5^3}{4\pi} \biggr)^{1/2} \frac{3\xi^3}{(3 + \xi^2)^2}\, .
</math>
  </td>
</tr>
</table>
</div>
As [[SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|Stahler has pointed out]], for an <math>~n = 5</math> polytrope, this mass-radius relation can also be precisely couched in the form of a quadratic equation, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{M}{M_\mathrm{SWS}} \biggr)^2 - 5 \biggl( \frac{M}{M_\mathrm{SWS}} \biggr)\biggl( \frac{R}{R_\mathrm{SWS}} \biggr)
+ \frac{2^2 \cdot 5 \pi}{3} \biggl( \frac{R}{R_\mathrm{SWS}} \biggr)^4
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{M}{M_\mathrm{SWS}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{5}{2} \biggl( \frac{R}{R_\mathrm{SWS}} \biggr) \biggl[ 1 \pm \sqrt{1-  \frac{2^4 \pi}{3\cdot 5}\biggl( \frac{R}{R_\mathrm{SWS}} \biggr)^2} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

As Figure 2 illustrates, this mass-radius relationship exhibits two turning points: &nbsp; 
The maximum radius occurs at coordinate location,
<div align="center">
<math>~\biggl[ \frac{R}{R_\mathrm{SWS}}, \frac{M}{M_\mathrm{SWS}}\biggr]_\mathrm{R\_turn} = \biggl[ \biggl( \frac{3\cdot 5}{2^4 \pi} \biggr)^{1 / 2}, \biggl( \frac{3\cdot 5^3}{2^6 \pi} \biggr)^{1 / 2} \biggr] 
\approx \biggl[ 0.5462742,  1.3656855\biggr] \, ;</math>
</div>
and the maximum mass occurs at coordinate location,
<div align="center">
<math>~\biggl[ \frac{R}{R_\mathrm{SWS}}, \frac{M}{M_\mathrm{SWS}}\biggr]_\mathrm{M\_turn} = \biggl[ \biggl( \frac{3^2\cdot 5}{2^6 \pi} \biggr)^{1 / 2}, \biggl( \frac{3^4\cdot 5^3}{2^{10} \pi} \biggr)^{1 / 2} \biggr] 
\approx \biggl[ 0.4730873,  1.7740776\biggr] \, .</math>
</div>

====Published n = 5 Tabulations====

<table border="1" cellpadding="8" align="center">
<tr>
  <th align="center" colspan="7">Published Tabulations of n = 5 Polytropic Structure (Primary E-Type Solution)</th>
</tr>
<tr>
  <td align="center" colspan="3">
Copied from p. 76 of [https://books.google.com/books?id=MiDQAAAAMAAJ&printsec=frontcover#v=onepage&q&f=true Emden (1907)]
  </td>
  <td align="center" width="10%">&nbsp;</td>
  <td align="center" colspan="3">
Copied from p. 75 of [http://adsabs.harvard.edu/abs/2004ASSL..306.....H Horedt (2004)]
  </td>
</tr>
<tr>
  <td align="center"><math>~\mathfrak{r}_1</math></td>
  <td align="center"><math>~u_1</math></td>
  <td align="center"><math>~- \frac{du_1}{d\mathfrak{r}_1}</math></td>
  <td align="center">&nbsp;</td>
  <td align="center"><math>~\xi</math></td>
  <td align="center"><math>~\Theta_H</math></td>
  <td align="center"><math>~- \frac{d\Theta_H}{d\xi}</math></td>
</tr>
<tr>
  <td align="right"><math>~0</math></td>
  <td align="center"><math>~1</math></td>
  <td align="center"><math>~0</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~0</math></td>
  <td align="center"><math>~1</math></td>
  <td align="center"><math>~0</math></td>
</tr>
<tr>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~\tfrac{1}{10}</math></td>
  <td align="center"><math>~9.983375\mathrm{E-01}</math></td>
  <td align="center"><math>~3.316736\mathrm{E-02}</math></td>
</tr>
<tr>
  <td align="right"><math>~\tfrac{1}{4}</math></td>
  <td align="right"><math>~0.98974</math></td>
  <td align="right"><math>~0.08079</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~\tfrac{2}{4}</math></td>
  <td align="right"><math>~0.96078</math></td>
  <td align="right"><math>~0.14781</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~\tfrac{1}{2}</math></td>
  <td align="center"><math>~9.607689\mathrm{E-01}</math></td>
  <td align="center"><math>~1.478106\mathrm{E-01}</math></td>
</tr>
<tr>
  <td align="right"><math>~\tfrac{3}{4}</math></td>
  <td align="right"><math>~0.91768</math></td>
  <td align="right"><math>~0.19320</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~1</math></td>
  <td align="right"><math>~0.86602</math></td>
  <td align="right"><math>~0.21650</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~1</math></td>
  <td align="center"><math>~8.660254\mathrm{E-01}</math></td>
  <td align="center"><math>~2.165064\mathrm{E-01}</math></td>
</tr>
<tr>
  <td align="right"><math>~\tfrac{3}{2}</math></td>
  <td align="right"><math>~0.75593</math></td>
  <td align="right"><math>~0.21598</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~2</math></td>
  <td align="right"><math>~0.65465</math></td>
  <td align="right"><math>~0.18704</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~\tfrac{5}{2}</math></td>
  <td align="right"><math>~0.56950</math></td>
  <td align="right"><math>~0.15392</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~3</math></td>
  <td align="right"><math>~0.50000</math></td>
  <td align="right"><math>~0.12500</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~\tfrac{7}{2}</math></td>
  <td align="right"><math>~0.44353</math></td>
  <td align="right"><math>~0.10180</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~4</math></td>
  <td align="right"><math>~0.39736</math></td>
  <td align="right"><math>~0.08365</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~5</math></td>
  <td align="right"><math>~0.32733</math></td>
  <td align="right"><math>~0.05845</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~5</math></td>
  <td align="center"><math>~3.273268\mathrm{E-01}</math></td>
  <td align="center"><math>~5.845122\mathrm{E-02}</math></td>
</tr>
<tr>
  <td align="right"><math>~6</math></td>
  <td align="right"><math>~0.27735</math></td>
  <td align="right"><math>~0.04267</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~7</math></td>
  <td align="right"><math>~0.24020</math></td>
  <td align="right"><math>~0.03233</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~8</math></td>
  <td align="right"><math>~0.21160</math></td>
  <td align="right"><math>~0.02527</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~10</math></td>
  <td align="right"><math>~0.17066</math></td>
  <td align="right"><math>~0.01657</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~10</math></td>
  <td align="center"><math>~1.706640\mathrm{E-01}</math></td>
  <td align="center"><math>~1.656932\mathrm{E-02}</math></td>
</tr>
<tr>
  <td align="right"><math>~12</math></td>
  <td align="right"><math>~0.14286</math></td>
  <td align="right"><math>~0.01166</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~16</math></td>
  <td align="right"><math>~0.10763</math></td>
  <td align="right"><math>~0.00665</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~20</math></td>
  <td align="right"><math>~0.08628</math></td>
  <td align="right"><math>~0.00428</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~30</math></td>
  <td align="right"><math>~0.05764</math></td>
  <td align="right"><math>~0.00192</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~</math></td>
  <td align="center"><math>~</math></td>
  <td align="center"><math>~</math></td>
</tr>
<tr>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~50</math></td>
  <td align="center"><math>~3.462025\mathrm{E-02}</math></td>
  <td align="center"><math>~6.915751\mathrm{E-04}</math></td>
</tr>
<tr>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~100</math></td>
  <td align="center"><math>~1.731791\mathrm{E-02}</math></td>
  <td align="center"><math>~1.731272\mathrm{E-04}</math></td>
</tr>
<tr>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~500</math></td>
  <td align="center"><math>~3.464081\mathrm{E-03}</math></td>
  <td align="center"><math>~6.928079\mathrm{E-06}</math></td>
</tr>
<tr>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~1000</math></td>
  <td align="center"><math>~1.732048\mathrm{E-03}</math></td>
  <td align="center"><math>~1.732043\mathrm{E-06}</math></td>
</tr>
<tr>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="right"><math>~</math></td>
  <td align="center">&nbsp;</td>
  <td align="right"><math>~\infty</math></td>
  <td align="center"><math>~0.000000\mathrm{E+00}</math></td>
  <td align="center"><math>~0.000000\mathrm{E+00}</math></td>
</tr>
</table>

====Srivastava's F-Type Solution====
=====Demonstration of Function's Validity=====

In a short paper, [http://adsabs.harvard.edu/abs/1962ApJ...136..680S S. Srivastava (1968, ApJ, 136, 680)] presents another, analytically prescribable solution to the Lane-Emden equation of index <math>~n = 5</math> that we will call upon in our [[SSC/Structure/BiPolytropes/Analytic15#BiPolytrope_with_nc_.3D_1_and_ne_.3D_5|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 &amp; 13),

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\theta_{5F}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\sin[\ln(A\xi)^{1/2})]}{\xi^{1/2}\{3-2\sin^2[\ln(A\xi)^{1/2}]\}^{1/2}} \, ,</math>
  </td>
</tr>

</table>
</div>
where, <math>~A</math> is an arbitrary (positive) constant.  Adopting the shorthand notation,
<div align="center">
<math>\Delta \equiv \ln(A\xi)^{1/2}\, ,</math>
</div>
and, recognizing that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>\frac{d}{d\ln(A\xi)}\biggl[ \ln(A\xi)^{1/2} \biggr] = \frac{1}{2}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; <math>~\Rightarrow~</math>&nbsp; &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{d\Delta}{d\xi} = \frac{1}{2\xi} \, ,
</math>
  </td>
</tr>

</table>
</div>
the first derivative of Srivastava's Lane-Emden function is,

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d\theta_{5F}}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\cos\Delta}{2\xi^{3/2}(3-2\sin^2\Delta)^{1/2}} 
- \frac{\sin\Delta}{2\xi^{3/2}(3-2\sin^2\Delta)^{1/2}} 
+ \frac{\sin^2\Delta \cos\Delta }{\xi^{3/2}(3-2\sin^2\Delta)^{3/2}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\xi^{3/2}(3-2\sin^2\Delta)^{3/2}} \biggl[ (\cos\Delta - \sin\Delta ) (3-2\sin^2\Delta) + 2\sin^2\Delta\cos\Delta \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3\cos\Delta-3\sin\Delta + 2\sin^3\Delta }{2\xi^{3/2}(3-2\sin^2\Delta)^{3/2}}  \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, the left-hand-side of the,
<div align="center">
<span id="LaneEmdenEquation"><font color="#770000">'''Lane-Emden Equation'''</font></span>
<br />
{{Math/EQ_SSLaneEmden01}}
</div>
is,

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\xi^2}\frac{d}{d\xi}\biggl[
\frac{\xi^{1/2}(3\cos\Delta-3\sin\Delta + 2\sin^3\Delta )}{2(3-2\sin^2\Delta)^{3/2}} 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\xi^2}\biggl[
\frac{(3\cos\Delta-3\sin\Delta + 2\sin^3\Delta )}{4\xi^{1/2}(3-2\sin^2\Delta)^{3/2}} 
+ \frac{(-3\sin\Delta-3\cos\Delta + 6\sin^2\Delta \cos\Delta )}{4\xi^{1/2}(3-2\sin^2\Delta)^{3/2}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{3(3\cos\Delta-3\sin\Delta + 2\sin^3\Delta )\sin\Delta \cos\Delta}{2\xi^{1/2}(3-2\sin^2\Delta)^{5/2}}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^{-2}\xi^{-5/2}(3-2\sin^2\Delta)^{-5/2}
[ (3-2\sin^2\Delta) (3\cos\Delta-3\sin\Delta + 2\sin^3\Delta ) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ (3-2\sin^2\Delta) (-3\sin\Delta-3\cos\Delta + 6\sin^2\Delta \cos\Delta ) 
+ 6(3\cos\Delta-3\sin\Delta + 2\sin^3\Delta )\sin\Delta \cos\Delta ]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^{-2}\xi^{-5/2}(3-2\sin^2\Delta)^{-5/2}
[ (3-2\sin^2\Delta) (-6\sin\Delta + 2\sin^3\Delta + 6\sin^2\cos\Delta) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ + 6(3\cos\Delta-3\sin\Delta + 2\sin^3\Delta )\sin\Delta \cos\Delta ]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^{-2}\xi^{-5/2}(3-2\sin^2\Delta)^{-5/2}
[ -18\sin\Delta + 6\sin^3\Delta + 18\sin^2\cos\Delta + 12\sin^3\Delta - 4\sin^5\Delta -12\sin^4\cos\Delta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ + 18\sin\Delta \cos^2\Delta-18\sin^2\Delta \cos\Delta + 12\sin^4\Delta \cos\Delta ]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^{-2}\xi^{-5/2}(3-2\sin^2\Delta)^{-5/2}[ -4\sin^5\Delta  ]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \theta_{5F}^5 \, .
</math>
  </td>
</tr>
</table>
</div>
This demonstrates that Srivastava's function satisfies the Lane-Emden equation of index <math>~n=5</math>.

=====Function Properties=====
The function, <math>~\theta_{5F}</math>, looks like a damped oscillator with the following specific properties:
* As <math>~\xi</math> increases from zero, the function oscillates with an ever increasing period; the function goes through zero when <math>~\Delta = \pm \pi m</math> (m is an integer), that is, when <math>~(A\xi) = e^{\pm 2\pi m}</math>.
* The amplitude of the oscillation drops approximately as <math>~\xi^{-1/2}</math>.
* 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 <math>~\theta_{5F}</math> is  positive (ensuring that the mass density is everywhere positive) and, at the same time,  <math>~d\theta_{5F}/d\xi</math> 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, <math>~\xi_\mathrm{crit} \le (A\xi) \le e^{2\pi}</math>.

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

<div align="center" id="Fig3">
<table border="2" cellpadding="10">
<tr>
  <td align="center">
Figure 3: &nbsp; Our Determination and Presentation of <br />a Segment of the <math>\theta_{5F}</math> Function as originally derived by<br />{{ Srivastava62 }} </td>
<tr>
  <td>
[[File:PlotTheta5F.png|450px|center|Srivastava's Lane-Emden function for n = 5]]
  </td>
</tr>
</table>
</div>

At both ends of the chosen parameter interval &#8212; that is, at <math>~\Delta = 0</math> and at <math>~\Delta = \pi</math> &#8212; the function <math>~\theta_{5F} = 0</math> and, correspondingly as depicted in the figure, the blue curve touches the horizontal axis.  At the beginning of the interval (<math>~\Delta =0</math>), the slope of the function and, correspondingly, the green curve, has the (positive) value,

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d\theta_{5F}}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3\cos(0)-3\sin(0) + 2\sin^3(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 
\, .
</math>
  </td>
</tr>
</table>
</div>
At the end of the interval (<math>~\Delta=\pi</math>), the slope of the function as well as the green curve, has the (negative) value,

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d\theta_{5F}}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3\cos(\pi)-3\sin(\pi) + 2\sin^3(\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} 
\, .
</math>
  </td>
</tr>
</table>
</div>

Over this interval, <math>~\theta_{5F}</math> reaches its maximum when the slope of the function is zero, that is, at the value of <math>~\Delta</math> where,

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
0
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\cos\Delta -3\sin\Delta +2(1-\cos^2\Delta)\sin\Delta</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\cos\Delta -\sin\Delta -2\cos^2\Delta \sin\Delta</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3\cot\Delta -2\cos^2\Delta \, .</math>
  </td>
</tr>
</table>
</div>
Rewriting both of these trigonometric functions in terms of the tangent function and adopting the shorthand notation,
<div align="center">
<math>~y \equiv \tan\Delta \, ,</math>
</div>
this condition becomes,

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3}{y} -\frac{2}{1+y^2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ y(y^2 + 1)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3(y^2+1) - 2y </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ y^3 - 3y^2 + 3y - 3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
</div>

<div align="center" id="CubicRoot">
<table border="1" cellpadding="8" width="80%">
<tr>
  <td align="left">
<font color="red">'''ASIDE:'''</font>  As is well known and documented &#8212; see, for example [http://mathworld.wolfram.com/CubicFormula.html Wolfram MathWorld] or [http://en.wikipedia.org/wiki/Cubic_function Wikipedia's discussion] of the topic &#8212; 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 [http://www.math.vanderbilt.edu/~schectex/courses/cubic/ online summary provided by Eric Schechter at Vanderbilt University].  For a cubic equation of the general form,
<div align="center">
<math>~ay^3 + by^2 + cy + d = 0 \, ,</math>
</div> 
a real root is given by the expression,
<div align="center">
<math>~
y = p + \{q + [q^2 + (r-p^2)^3]^{1/2}\}^{1/3} + \{q - [q^2 + (r-p^2)^3]^{1/2}\}^{1/3}
\, ,</math>
</div> 
where,
<div align="center">
<math>~p \equiv -\frac{b}{3a} \, ,</math> &nbsp;&nbsp;&nbsp;&nbsp;
<math>~q \equiv \biggl[p^3 + \frac{bc-3ad}{6a^2} \biggr] \, ,</math>
&nbsp;&nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp; 
<math>~r=\frac{c}{3a} \, .</math>
</div>
In our particular case,
<div align="center">
<math>~a =1\, ,</math> &nbsp;&nbsp;&nbsp;&nbsp;
<math>~b =-3\, ,</math> &nbsp;&nbsp;&nbsp;&nbsp;
<math>~c = +3 \, ,</math>
&nbsp;&nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp; 
<math>~d = - 3 \, .</math>
</div>
[[File:WolframAlphaCubicSolver.png|thumbnail|right|100px|WolframAlpha]]Hence, interestingly enough,
<div align="center">
<math>~p = q = r = + 1 \, ,</math>
</div>
which implies that the real root is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~y</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 + \{2\}^{1/3} + \{0\}^{1/3} \, .</math>
  </td>
</tr>
</table>
</div>
(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.
  </td>
</tr>
</table>
</div>


The single, real root of this cubic equation is,
<div align="center">
<math>~y = 1 + 2^{1/3} \, ,</math>
</div>
which corresponds to,
<div align="center">
<math>~\Delta = \tan^{-1}(1 + 2^{1/3}) \, .</math>
</div>

[[File:CommentButton02.png|right|100px|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 <math>\theta_{5F}</math> function &#8212; 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) &#8212; occurs at,
<div align="center">
<math>\xi_\mathrm{crit} \equiv e^{2\tan^{-1}(1+2^{1/3})} = 10.05836783\, .</math>
</div>
The corresponding value of the function at this critical radial location is,
<div align="center">
<math>\theta_{5F}|_\mathrm{max} \equiv \theta_{5F}(\xi_\mathrm{crit}) = (1+2^{1/3})[3 + (1+2^{1/3})^2]^{-1/2} e^{-\tan^{-1}(1+2^{1/3})} = 0.250260848 \, .</math>
</div>
This agrees precisely with the determination made by {{ Murphy83afull }} &#8212; see the excerpts from his paper displayed in the following boxed-in image &#8212; that the portion of the <math>~\theta_{5F}</math> function that falls in the interval <math>~1 \le (A\xi) < \xi_\mathrm{crit}</math> (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.  
<!--
<table border="1" cellpadding="10" align="center">
<tr>
  <th align="center">Excerpts (edited) from 
[http://adsabs.harvard.edu/abs/1983PASAu...5..175M J. O. Murphy (1983, Proc. Astr. Soc. Australia, 5, 175)]
  </th>
<tr>
  <td>
[[File:Murphy1983Extremum02.png|600px|center|Murphy's (1983) examination of Srivastava's function]]
[[Image:AAAwaiting01.png|600px|center|Murphy's (1983) examination of Srivastava's function]]
  </td>
</tr>
</table>
-->

<div align="center">
<table border="1" cellpadding="5" width="80%">
<tr><td align="center">
Equation and text extracted<sup>&dagger;</sup> from p. 177 of &hellip;<br /> 
{{ Murphy83afigure }}
</td></tr>
<tr>
  <td align="center">
<!-- [[File:Murphy1983Extremum02.png|700px|center|Murphy's (1983) examination of Srivastava's function]] -->
<!-- [[Image:AAAwaiting01.png|400px|center|Norman &amp; Wilson (1978)]] -->
<table border="0" align="center" cellpadding="8" width="70%">
<tr>
  <td align="center">
on the interval <math>~[1, e^{2\pi} ]</math> &nbsp; &nbsp; &hellip; &nbsp; &nbsp; <math>~d\theta_{5F}/d\xi > 0</math> in the range [1, 10.0583]

<p></p>
----
<p></p>

<math>~\theta_{5F}(\zeta)_\mathrm{MAX} = 0.2503 \sqrt{A}</math> &nbsp; &nbsp; at &nbsp; &nbsp;  <math>~\zeta = 10.0583/A</math>
  </td>
</tr>
</table>
  </td>
</tr>
<tr><td align="left"><sup>&dagger;</sup>Equations and text displayed here,  with presentation order &amp; layout modified from the original publication.</td></tr>
</table>
</div>

On the other hand, the segment that falls in the interval, <math>~\xi_\mathrm{crit} \le (A\xi) \le e^{2\pi}</math>, whose function values lie in the range, <math>~\theta_{5F}|_\mathrm{max} \ge (A^{-1/2} \theta_{5F}) \ge 0</math> &#8212; that is, the segment of the blue curve that lies to the right of the vertical red line in the above figure &#8212; can be used to describe the <math>~n=5</math> "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'', [http://adsabs.harvard.edu/abs/2012JMP....53f2503M 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, <math>~n=5</math>, in addition to the two solutions that have just been detailed, which he includes as his equations (3) and (5):  

<!--
<div align="center">
<table border="2" cellpadding="10">
<tr>
  <th align="center">
Extracted (with minor editing) from [http://adsabs.harvard.edu/abs/2012JMP....53f2503M Mach (2012)]
  </th>
<tr>
  <td>
[[File:Mach2012Eqs.png|300px|center|Mach (2012)]]
[[Image:AAAwaiting01.png|300px|center|Mach (2012)]]
  </td>
</tr>
</table>
</div>
-->
<div align="center">
<table border="1" cellpadding="5" width="80%">
<tr><td align="center">
Equations extracted<sup>&dagger;</sup> from pp. 062503-1 &amp; -2 of [http://adsabs.harvard.edu/abs/2012JMP....53f2503M Mach (2012)]<p></p>
"''All Solutions of the n = 5 Lane-Emden Equation''"<p></p>
Journal of Mathematical Physics, vol. 53, pp. 062503-062503-6 &copy; American Institute of Physics
</td></tr>
<tr>
  <td align="center">
<!-- [[File:Mach2012Eqs.png|400px|center|Mach (2012)]] -->
<!-- [[Image:AAAwaiting01.png|400px|center|Norman &amp; Wilson (1978)]] -->
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\theta(\xi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pm \frac{1}{\sqrt{1 + \xi^2/3}}</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; <math>~(3)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\theta(\xi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pm \frac{\sin(\ln \sqrt{\xi})}{\sqrt{3\xi - 2\xi\sin^2(\ln\sqrt{\xi}) }}</math>
  </td>
  <td align="right">
&nbsp; &nbsp; &nbsp; <math>~(5)</math>
  </td>
</tr>
</table>
  </td>
</tr>
<tr><td align="left"><sup>&dagger;</sup>Equations displayed here, with layout modified from the original publication.</td></tr>
</table>
</div>


For completeness, Mach mentions a well-known solution that works for all indexes, <math>~n > 3</math>, which we have discussed separately in the context of  [[SSC/Structure/PowerLawDensity#Power-Law_Density_Distributions|power-law density distributions]], namely,
<div align="center">
<math>\theta^n(\xi) = 
\frac{\rho}{\rho_c} = \biggl[ \frac{2(n-3)}{(n-1)^2} \biggr]^{n/(n-1)} \xi^{- 2n/(n-1)} \, .
</math>
</div>

In addition, Mach identifies the rarely referenced work of [http://adsabs.harvard.edu/abs/2000JMP....41.7029G H. Goenner &amp; 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 &#8212; see equation (10) in ''his'' &sect;2.1 &#8212; 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" &#8212; see our [[SSC/Structure/BiPolytropes/Analytic15#Step_6:__Envelope_Solution|accompanying description of a composite model using Srivastava's solution]].