Difference between revisions of "User:Tohline/SR/Ptot QuarticSolution"

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(More revamping of derivation)
(→‎Quartic Equation Solution: Improve discussion of quartic solution)
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<div align="center">
<div align="center">
<math>
<math>
z_3 = \frac{1}{2}\biggl[E-R\biggr] ,
2z = \biggl[\frac{2a_1}{a_4 W}-W^2\biggr]^{1/2} - W ,
</math>
</math>
</div>
</div>
Line 45: Line 45:
<div align="center">
<div align="center">
<math>
<math>
R \equiv y_1^{1/2} ,
\frac{1}{2}W^2 \equiv R^{-1/3}\biggl[R^{2/3} - \frac{a_0}{3a_4} \biggr] ,
</math><br/>
</math><br/>
<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},
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>
</math>
</div>
</div>
and <math>y_1</math> is the <i>real</i> root of the following cubic equation:
Now, defining,
<div align="center">
<div align="center">
<math>
<math>
y^3 -4a_0 y -a_1^2 = 0 .
\phi \equiv \frac{2a_1}{a_4 W^3} ~~~~~\Rightarrow ~~~~~ W = 2\biggl(\frac{a_1}{4a_4 \phi} \biggr)^{1/3} ,
</math>
</math>
</div>
</div>
So, fully in terms of the real root of this cubic equation, the desired solution of our quartic equation is,
and realizing that, from one of the above expressions,
<div align="center">
<math>
\frac{1}{2}W^2  \equiv \biggl[ \frac{a_1}{2^8 a_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}{2^8 a_4 R}\biggr]^{1/3} \biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\} ,
</math><br/>
</div>
we can rewrite the quartic solution in the form,
<div align="center">
<div align="center">
<math>
<math>
z_3 = \frac{1}{2}y_1^{1/2} \biggl\{ \biggl[ 2a_1 y_1^{-3/2} - 1  \biggr]^{1/2} -1 \biggr\} .
z = \biggl(\frac{a_1}{4a_4}\biggr) \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr]
</math><br />
<math>
\phi^2 = \frac{2^2 a_1^2}{2^3 a_4^2} \biggl( \frac{2^8 a_4 R}{a_1} \biggr)
\biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\}^{-3} =
\biggl( \frac{2a_1}{a_4} \biggr)^3 \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]
\biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\}^{-3}
</math>
</math>
</div>
</div>

Revision as of 03:23, 11 March 2010

Whitworth's (1981) Isothermal Free-Energy Surface
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Determining Temperature from Density and Pressure

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

LSU Key.png

<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>

Now, 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 \equiv \biggl[ \frac{a_1}{2^8 a_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}{2^8 a_4 R}\biggr]^{1/3} \biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\} , </math>

we can rewrite the quartic solution in the form,

<math> z = \biggl(\frac{a_1}{4a_4}\biggr) \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr] </math>
<math> \phi^2 = \frac{2^2 a_1^2}{2^3 a_4^2} \biggl( \frac{2^8 a_4 R}{a_1} \biggr) \biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\}^{-3} = \biggl( \frac{2a_1}{a_4} \biggr)^3 \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr] \biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\}^{-3} </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

Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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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