Editing SSC/Structure/Polytropes/Analytic (section)

===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>