Editing SSC/Structure/Polytropes/Analytic (section)

====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>