User:Tohline/SR/Ptot QuarticSolution

Determining Temperature from Density and Pressure

As has been derived elsewhere, the normalized total pressure can be written as,

<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> z^4 + a_1 z + a_0 = 0 , </math>

where,

<math> z \equiv \frac{T}{T_e} , </math>
<math> a_1 \equiv \frac{15}{\pi^4} \biggl(\frac{\mu_e m_p}{\bar{\mu} m_u} \biggr) \chi^3 , </math>
<math> a_0 \equiv \frac{15}{8\pi^4}\biggl[p_\mathrm{total} - F(\chi) \biggr] . </math>

Mathematical Manipulation

Quartic Equation Solution

Following the outline and notation used by mathworld.wolfram.com to identify the roots of an arbitrary quartic equation, we can set <math>a_3 = a_2 = 0</math> and deduce that the only root that will give physically relevant temperatures — for example, real and non-negative — is,

<math> z_3 = \frac{1}{2}\biggl[E-R\biggr] , </math>

where,

<math> R \equiv y_1^{1/2} , </math>
<math> E \equiv \biggl[ \frac{2a_1}{R} - R^2 \biggr]^{1/2} = y_1^{1/2}\biggl[ 2a_1 y_1^{-3/2} - 1 \biggr]^{1/2}, </math>

and <math>y_1</math> is the real root of the following cubic equation:

<math> y^3 -4a_0 y -a_1^2 = 0 . </math>

So, fully in terms of the real root of this cubic equation, the desired solution of our quartic equation is,

<math> z_3 = \frac{1}{2}y_1^{1/2} \biggl\{ \biggl[ 2a_1 y_1^{-3/2} - 1 \biggr]^{1/2} -1 \biggr\} . </math>

Relevant Cubic Formula

The relevant cubic equation is,

<math> y^3 +b_1 y +b_0 = 0 , </math>

where,

<math> b_1 \equiv -4a_0 ~~~~~ \mathrm{and} ~~~~~ b_0 \equiv -a_1^2 . </math>

According to mathworld.wolfram.com, the roots of a cubic equation having this form include a real root given by the expression,

<math> y_1 = \mathcal{S} + \mathcal{T} , </math>

where,

<math> \mathcal{D} \equiv \biggl( \frac{b_1}{3} \biggr)^2 + \biggl(\frac{b_0}{2}\biggr)^2 = \biggl( \frac{4a_0}{3} \biggr)^2 + \biggl(\frac{a_1^2}{2}\biggr)^2 = \biggl(\frac{a_1^2}{2}\biggr)^2 \biggl[ 1 + \biggl( \frac{8a_0}{3a_1^2} \biggr)^2 \biggr] , </math>
<math> \mathcal{S} \equiv \biggl[ -\frac{b_0}{2} + \mathcal{D}^{1/2} \biggr]^{1/3} = \biggl[ \frac{a_1^2}{2} + \mathcal{D}^{1/2} \biggr]^{1/3} = \biggl[ \frac{a_1^2}{2} \biggr]^{1/3} \biggl\{1 + \biggl[ 1 + \biggl( \frac{8a_0}{3a_1^2} \biggr)^2 \biggr]^{1/2} \biggr\}^{1/3} , </math>
<math> \mathcal{T} \equiv \biggl[ -\frac{b_0}{2} - \mathcal{D}^{1/2} \biggr]^{1/3} = \biggl[ \frac{a_1^2}{2} - \mathcal{D}^{1/2} \biggr]^{1/3} = \biggl[ \frac{a_1^2}{2} \biggr]^{1/3} \biggl\{1 - \biggl[ 1 + \biggl( \frac{8a_0}{3a_1^2} \biggr)^2 \biggr]^{1/2} \biggr\}^{1/3} . </math>

Summary

Hence, defining,

<math> \mathcal{K} \equiv \biggl( \frac{8a_0}{3a_1^2} \biggr)^2 , </math>

and

<math> f_Q(\mathcal{K}) \equiv y_1 \biggl[ \frac{2}{a_1^2} \biggr]^{1/3} = \biggl[1 + \biggl( 1 + \mathcal{K} \biggr)^{1/2} \biggr]^{1/3} + \biggl[1 - \biggl( 1 + \mathcal{K} \biggr)^{1/2} \biggr]^{1/3} , </math>

we can write the desired solution of the quartic equation as,

<math> z_3 = 2^{-7/6} a_1^{1/3} [f_Q(\mathcal{K})]^{1/2} \biggl\{ \biggl[ 8^{1/2} [f_Q(\mathcal{K})]^{-3/2} - 1 \biggr]^{1/2} -1 \biggr\} . </math>

Note that, in order for the solution, <math>z_3</math>, to be real and non-negative, the function <math>f_Q(\mathcal{K})</math> must be limited to a range of values given by the expression,

<math> 8^{1/2} [f_Q(\mathcal{K})]^{-3/2} \ge 2 </math>
<math> \Rightarrow ~~ f_Q(\mathcal{K}) \le 2^{1/3} . </math>

This, in turn, implies that the dimensionless parameter, <math>\mathcal{K}</math>, must be limited to values,

<math> \mathcal{K} \ge 0 . </math>

Our solution takes on a somewhat cleaner form if we define a function,

<math> g_Q(\mathcal{K}) \equiv 2^{1/2} f_Q^{-3/2} = 2^{1/2} \biggl\{ \biggl[1 + \biggl( 1 + \mathcal{K} \biggr)^{1/2} \biggr]^{1/3} + \biggl[1 - \biggl( 1 + \mathcal{K} \biggr)^{1/2} \biggr]^{1/3} \biggr\}^{-3/2} , </math>

which is limited to values,

<math> g_Q(\mathcal{K}) \ge 1 . </math>

Then we can write,

<math> z_3 a_1^{-1/3} = \frac{1}{2}\biggl[\frac{1}{g_Q(\mathcal{K})}\biggr]^{1/3} \biggl\{ \biggl[ 2g_Q(\mathcal{K}) - 1 \biggr]^{1/2} -1 \biggr\} . </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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