Editing SR/Ptot QuarticSolution (section)

==Mathematical Manipulation==
===Quartic Equation Solution===

Following the [http://en.wikipedia.org/wiki/Quartic_function#Summary_of_Ferrari.27s_method Summary of Ferrari's method] that is presented in Wikipedia's discussion of the Quartic Function to identify the roots of an arbitrary quartic equation, we can set the coefficients of the cubic and quadratic terms both to zero and deduce that the only root that will give physically relevant temperatures &#8212; for example, real and non-negative &#8212; is,
<div align="center">
<math>
2z = \biggl[\frac{2a_1}{a_4 W}-W^2\biggr]^{1/2} - W ,
</math>
</div>
where,
<div align="center">
<math>
\frac{1}{2}W^2  \equiv R^{-1/3}\biggl[R^{2/3} - \frac{a_0}{3a_4}  \biggr] ,
</math><br/>
<math>
R \equiv \biggl( \frac{a_1}{4a_4} \biggr)^2 \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr] ,
</math><br />
<math>
\lambda^3 \equiv \frac{2^8 a_0^3 a_4}{3^3 a_1^4} .
</math>
</div>
Defining,
<div align="center">
<math>
\phi \equiv \frac{2a_1}{a_4 W^3} ~~~~~\Rightarrow ~~~~~ W = 2\biggl(\frac{a_1}{4a_4 \phi} \biggr)^{1/3} ,
</math>
</div>
and realizing that, from one of the above expressions, 
<div align="center">
<math>
\frac{1}{2}W^2  = \biggl[ \frac{a_1^4}{2^8 a_4^4 R}\biggr]^{1/3}\biggl[\biggl( \frac{2^4 a_4^2 R}{a_1^2}\biggr)^{2/3} - \biggl( \frac{2^8 a_0^3 a_4}{3^3 a_1^4} \biggr)^{1/3}  \biggr]
= \biggl[ \frac{a_1}{2^2 a_4}\biggr]^{2/3} \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{-1/3}\biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\} ,
</math><br/>
</div>
we can rewrite the desired root of our quartic equation in the form,
<div align="center">
<math>
z = \biggl(\frac{a_1}{4a_4}\biggr)^{1/3} \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr] ,
</math>
</div>
with,
<div align="center">
<math>
\phi = 2^{3/2} \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{1/2}
\biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\}^{-3/2} .
</math>
</div>

===Key Dimensionless Parameters===
From the above solution, there appear to be two key dimensionless parameters that can be formed from a strategic combination of the coefficients of the original quartic equation. They are,
<div align="center">
<math>
\lambda^3 \equiv \frac{2^8 a_0^3 a_4}{3^3 a_1^4} , ~~~~~\mathrm{and}~~~~~
\theta^3 \equiv \frac{a_1}{2^2 a_4} .
</math>
</div>
[Note, as well, that the product <math>\lambda\theta = (4a_0)/(3a_1)</math>.] 
The desired solution of our quartic equation is a product of <math>\theta</math> and an expression that is only a function of <math>\lambda</math>.  Specifically, we can write,
<div align="center">
<math>~
\frac{z}{\theta} = \mathcal{K}(\lambda) ,
</math>
</div>
where, in terms of the above-defined function <math>\phi(\lambda)</math>,
<div align="center">
<math>
\mathcal{K}(\phi(\lambda)) \equiv \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr] .
</math>
</div>

===Limiting Regimes===
We can immediately see that this solution makes sense in the present context.  In order for the temperature &#8212; that is, <math>z</math> &#8212; to be real and nonnegative, the function <math>\phi(\lambda)</math> must be greater than or equal to <math>2</math>.  This limiting value occurs when the dimensionless parameter, <math>\lambda = 0</math>.  The constraint <math>\lambda \ge 0</math> is satisfied as long as the three coefficients of the quartic equation are real and nonnegative, which is certainly true for our specific problem.

Looking at the limiting functional behavior of our solution, we see that when <math>~0 \le \lambda \ll 1</math>,
<div align="center">
<math>
\phi \approx 2 + \frac{3}{2^{2/3}}\lambda 
</math><br />

<math>~
\Rightarrow ~~~~~ z \approx \frac{3\lambda}{2^2} \biggl( \frac{a_1}{2^2a_4} \biggr)^{1/3} = \frac{a_0}{a_1} .
</math>
</div>


We see, as well, that when <math>\lambda \gg 1</math>,
<div align="center">
<math>
\phi \approx \biggl(\frac{3\lambda}{2^2} \biggr)^{3/2}
</math><br />

<math>
\Rightarrow ~~~~~ z \approx \biggl(\frac{3\lambda}{2^2} \biggr)^{1/4} \biggl( \frac{a_1}{2^2a_4} \biggr)^{1/3} = \biggl( \frac{a_0}{a_4} \biggr)^{1/4} .
</math>
</div>