User:Tohline/SR/Ptot QuarticSolution
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Determining Temperature from Density and Pressure
As has been derived elsewhere, the normalized total pressure can be written as,
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<math>~p_\mathrm{total} = \biggl(\frac{\mu_e m_p}{\bar{\mu} m_u} \biggr) 8 \chi^3 \frac{T}{T_e} + F(\chi) + \frac{8\pi^4}{15} \biggl( \frac{T}{T_e} \biggr)^4</math> |
To solve this algebraic equation for the normalized temperature <math>T/T_e</math>, given values of the normalized total pressure <math>p_\mathrm{total}</math> and the normalized density <math>\chi</math>, we first realize that the equation can be written in the form,
<math> a_4z^4 + a_1 z - a_0 = 0 , </math>
where,
<math> z \equiv \frac{T}{T_e} , </math>
and the coefficients,
<math>
a_4 \equiv \frac{8\pi^4}{15} ,
</math>
<math>
a_1 \equiv 8\biggl(\frac{\mu_e m_p}{\bar{\mu} m_u} \biggr) \chi^3 ,
</math>
<math>
a_0 \equiv \biggl[p_\mathrm{total} - F(\chi) \biggr] .
</math>
Mathematical Manipulation
Quartic Equation Solution
Following the 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 — for example, real and non-negative — is,
<math> 2z = \biggl[\frac{2a_1}{a_4 W}-W^2\biggr]^{1/2} - W , </math>
where,
<math>
\frac{1}{2}W^2 \equiv R^{-1/3}\biggl[R^{2/3} - \frac{a_0}{3a_4} \biggr] ,
</math>
<math>
R \equiv \biggl( \frac{a_1}{4a_4} \biggr)^2 \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr] ,
</math>
<math>
\lambda^3 \equiv \frac{2^8 a_0^3 a_4}{3^3 a_1^4} .
</math>
Defining,
<math> \phi \equiv \frac{2a_1}{a_4 W^3} ~~~~~\Rightarrow ~~~~~ W = 2\biggl(\frac{a_1}{4a_4 \phi} \biggr)^{1/3} , </math>
and realizing that, from one of the above expressions,
<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>
we can rewrite the desired root of our quartic equation in the form,
<math> z = \biggl(\frac{a_1}{4a_4}\biggr)^{1/3} \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr] , </math>
with,
<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>
Limiting Regimes
We can immediately see that this solution makes sense in the present context. In order for the temperature — that is, <math>z</math> — to be real and nonnegative, the function <math>\phi(\lambda)</math> must be greater than or equal to 2. 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>,
<math>
\phi \approx 2 + \frac{3}{2^{2/3}}\lambda
</math>
<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>
We see, as well, that when <math>\lambda \gg 1</math>,
<math>
\phi \approx \biggl(\frac{3\lambda}{2^2} \biggr)^{3/2}
</math>
<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>
Physical Implications
Clearly, a key dimensionless physical parameter for this problem is,
<math> \mathcal{K} \equiv \biggl( \frac{8a_0}{3a_1^2} \biggr)^2 = \biggl\{ \frac{1}{5} \biggl[ \frac{\pi^2}{3} \biggl(\frac{\bar{\mu} m_u}{\mu_e m_p} \biggr) \biggr]^2 \biggl[ \frac{p_\mathrm{total} - F(\chi)}{\chi^6} \biggr] \biggr\}^2 . </math>
And, since <math>z_3 \propto T</math> and <math>a_1 \propto \rho</math>, the above solution tells us that the product <math>T \rho^{-1/3}</math> can be expressed as a function of this single parameter, <math>\mathcal{K}</math>.
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