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> 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> 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>.

Related Wikepedia Links

Quartic Function
mathworld.wolfram.com

© 2014 - 2021 by Joel E. Tohline |   H_Book Home   |   YouTube   | Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS | Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation