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> |
Relevant Quartic Equation
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>
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, non-imaginary values that are positive — 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>
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© 2014 - 2021 by Joel E. Tohline |