User:Tohline/SR/Ptot QuarticSolution - Revision history

Tohline: /* Determining Temperature from Density and Pressure */ Clean up presentation

2015-07-13T03:00:28Z

Determining Temperature from Density and Pressure: Clean up presentation

← Older revision		Revision as of 03:00, 13 July 2015
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	=Determining Temperature from Density and Pressure=		=Determining Temperature from Density and Pressure=
	{{LSU_HBook_header}}		{{LSU_HBook_header}}
	As has been [~~http://www.vistrails.org/index.php/~~User:Tohline/SR/PressureCombinations#Derivation derived elsewhere], the normalized total pressure can be written as,		As has been [[User:Tohline/SR/PressureCombinations#Derivation\|derived elsewhere]], the normalized total pressure can be written as,

	<div align="center">		<div align="center">
	{{User:Tohline/Math/EQ_PressureTotal01}}		{{User:Tohline/Math/EQ_PressureTotal01}}
	</div>		</div>
	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,		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,
	<div align="center">		<div align="center">
	<math>		<math>~
	a_4z^4 + a_1 z - a_0 = 0 ,		a_4z^4 + a_1 z - a_0 = 0 \, ,
	</math>		</math>
	</div>		</div>
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	<div align="center">		<div align="center">
	<math>		<math>
	z \equiv \frac{T}{T_e} ,		~z \equiv \frac{T}{T_e} \, ,
	</math>		</math>
	</div>		</div>
	and the coefficients,		and the coefficients,
	<div align="center">		<div align="center">
	<math>		<math>~
	a_4 \equiv \frac{8\pi^4}{15} ,		a_4 \equiv \frac{8\pi^4}{15} \, ,
	</math><br/>		</math><br/>
	<math>		<math>~
	a_1 \equiv 8\biggl(\frac{\mu_e m_p}{\bar{\mu} m_u} \biggr) \chi^3 ,		a_1 \equiv 8\biggl(\frac{\mu_e m_p}{\bar{\mu} m_u} \biggr) \chi^3 \, ,
	</math><br/>		</math><br/>
	<math>		<math>~
	a_0 \equiv \biggl[p_\mathrm{total} - F(\chi) \biggr] .		a_0 \equiv \biggl[p_\mathrm{total} - F(\chi) \biggr] \, .
	</math>		</math>
	</div>		</div>
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	From the above solution, there appear to be two key dimensionless parameters that can be formed from a strategic combination of the coefficients of the original quartic equation. They are,		From the above solution, there appear to be two key dimensionless parameters that can be formed from a strategic combination of the coefficients of the original quartic equation. They are,
	<div align="center">		<div align="center">
	<math>		<math>~
	\lambda^3 \equiv \frac{2^8 a_0^3 a_4}{3^3 a_1^4} , ~~~~~\mathrm{and}~~~~~		\lambda^3 \equiv \frac{2^8 a_0^3 a_4}{3^3 a_1^4} , ~~~~~\mathrm{and}~~~~~
	\theta^3 \equiv \frac{a_1}{2^2 a_4} .		\theta^3 \equiv \frac{a_1}{2^2 a_4} .
	</math>		</math>
	</div>		</div>
	[Note, as well, that the product <math>\lambda\theta = (4a_0)/(3a_1)</math>.]		[Note, as well, that the product <math>~\lambda\theta = (4a_0)/(3a_1)</math>.]
	The desired solution of our quartic equation is a product of <math>\theta</math> and an expression that is only a function of <math>\lambda</math>. Specifically, we can write,		The desired solution of our quartic equation is a product of <math>~\theta</math> and an expression that is only a function of <math>~\lambda</math>. Specifically, we can write,
	<div align="center">		<div align="center">
	<math>		<math>~
	\frac{z}{\theta} = \mathcal{K}(\lambda) ,		\frac{z}{\theta} = \mathcal{K}(\lambda) ,
	</math>		</math>
	</div>		</div>
	where, in terms of the above-defined function <math>\phi(\lambda)</math>,		where, in terms of the above-defined function <math>~\phi(\lambda)</math>,
	<div align="center">		<div align="center">
	<math>		<math>~
	\mathcal{K}(\phi(\lambda)) \equiv \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr] .		\mathcal{K}(\phi(\lambda)) \equiv \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr] .
	</math>		</math>
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	===Limiting Regimes===		===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.		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 <math>~2</math>. 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>,		Looking at the limiting functional behavior of our solution, we see that when <math>~0 \le \lambda \ll 1</math>,
	<div align="center">		<div align="center">
	<math>		<math>~
	\phi \approx 2 + \frac{3}{2^{2/3}}\lambda		\phi \approx 2 + \frac{3}{2^{2/3}}\lambda
	</math><br />		</math><br />

	<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} .		\Rightarrow ~~~~~ z \approx \frac{3\lambda}{2^2} \biggl( \frac{a_1}{2^2a_4} \biggr)^{1/3} = \frac{a_0}{a_1} .
	</math>		</math>
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	We see, as well, that when <math>\lambda \gg 1</math>,		We see, as well, that when <math>~\lambda \gg 1</math>,
	<div align="center">		<div align="center">
	<math>		<math>~
	\phi \approx \biggl(\frac{3\lambda}{2^2} \biggr)^{3/2}		\phi \approx \biggl(\frac{3\lambda}{2^2} \biggr)^{3/2}
	</math><br />		</math><br />

	<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} .		\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>		</math>
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	Clearly, a key dimensionless physical parameter for this problem is,		Clearly, a key dimensionless physical parameter for this problem is,
	<div align="center">		<div align="center">
	<math>		<math>~
	\mathcal{K} \equiv \biggl( \frac{8a_0}{3a_1^2} \biggr)^2 =		\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{1}{5} \biggl[ \frac{\pi^2}{3} \biggl(\frac{\bar{\mu} m_u}{\mu_e m_p} \biggr) \biggr]^2
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	</math>		</math>
	</div>		</div>
	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>.		And, because <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=		=Related Wikepedia Links=

Tohline: Move title image

2012-11-06T17:48:46Z

Move title image

← Older revision		Revision as of 17:48, 6 November 2012
Line 1:		Line 1:
			=Determining Temperature from Density and Pressure=
	{{LSU_HBook_header}}		{{LSU_HBook_header}}

	~~=Determining Temperature from Density and Pressure=~~
	As has been [http://www.vistrails.org/index.php/User:Tohline/SR/PressureCombinations#Derivation derived elsewhere], the normalized total pressure can be written as,		As has been [http://www.vistrails.org/index.php/User:Tohline/SR/PressureCombinations#Derivation derived elsewhere], the normalized total pressure can be written as,

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	</div>		</div>
	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>.		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=		=Related Wikepedia Links=

Tohline: /* Quartic Equation Solution */ correct error in earlier link address

2010-03-13T20:58:46Z

Quartic Equation Solution: correct error in earlier link address

← Older revision		Revision as of 20:58, 13 March 2010
Line 36:		Line 36:
	===Quartic Equation Solution===		===Quartic Equation Solution===

	Following the [http://en.wikipedia.org/wiki/~~Quartic_function#~~Quartic_function#Summary_of_Ferrari.27s_method 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,		Following the [http://en.wikipedia.org/wiki/Quartic_function#Summary_of_Ferrari.27s_method 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,
	<div align="center">		<div align="center">
	<math>		<math>

Tohline: /* Quartic Equation Solution */ insert more specific link address to Ferrari's method

2010-03-13T20:57:50Z

Quartic Equation Solution: insert more specific link address to Ferrari's method

← Older revision		Revision as of 20:57, 13 March 2010
Line 36:		Line 36:
	===Quartic Equation Solution===		===Quartic Equation Solution===

	Following the [http://en.wikipedia.org/wiki/Quartic_function#Summary_of_Ferrari~~'s_method~~ 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,		Following the [http://en.wikipedia.org/wiki/Quartic_function#Quartic_function#Summary_of_Ferrari.27s_method 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,
	<div align="center">		<div align="center">
	<math>		<math>
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	</math>		</math>
	</div>		</div>


	===Key Dimensionless Parameters===		===Key Dimensionless Parameters===

Tohline: /* Mathematical Manipulation */ Discuss key dimensionless parameters

2010-03-12T00:44:34Z

Mathematical Manipulation: Discuss key dimensionless parameters

← Older revision		Revision as of 00:44, 12 March 2010
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	\phi = 2^{3/2} \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{1/2}		\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} .		\biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\}^{-3/2} .
			</math>
			</div>


			===Key Dimensionless Parameters===
			From the above solution, there appear to be two key dimensionless parameters that can be formed from a strategic combination of the coefficients of the original quartic equation. They are,
			<div align="center">
			<math>
			\lambda^3 \equiv \frac{2^8 a_0^3 a_4}{3^3 a_1^4} , ~~~~~\mathrm{and}~~~~~
			\theta^3 \equiv \frac{a_1}{2^2 a_4} .
			</math>
			</div>
			[Note, as well, that the product <math>\lambda\theta = (4a_0)/(3a_1)</math>.]
			The desired solution of our quartic equation is a product of <math>\theta</math> and an expression that is only a function of <math>\lambda</math>. Specifically, we can write,
			<div align="center">
			<math>
			\frac{z}{\theta} = \mathcal{K}(\lambda) ,
			</math>
			</div>
			where, in terms of the above-defined function <math>\phi(\lambda)</math>,
			<div align="center">
			<math>
			\mathcal{K}(\phi(\lambda)) \equiv \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr] .
	</math>		</math>
	</div>		</div>
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	</math>		</math>
	</div>		</div>



	==Physical Implications==		==Physical Implications==

Tohline: Discuss limits

2010-03-12T00:03:05Z

Discuss limits

← Older revision		Revision as of 00:03, 12 March 2010
Line 81:		Line 81:
	</div>		</div>

	===~~Relevant Cubic Formula~~===
	~~The relevant cubic equation~~ is,		===Limiting Regimes===
	<~~div align="center"~~>		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.
	<math>
	~~y^3 +b_1 y +b_0 = 0~~ ,		Looking at the limiting functional behavior of our solution, we see that when <math>0 \le \lambda \ll 1</math>,
	</math>
	</~~div~~>
	~~where~~,
	~~<div align="center">~~
	<math>
	~~b_1~~ \~~equiv -4a_0 ~~~~~ \mathrm{and} ~~~~~ b_0 \equiv -a_1^2 .~~
	</math><~~br/~~>
	</~~div~~>
	~~According to [http://mathworld.wolfram.com/CubicFormula.html mathworld.wolfram.com],~~ the ~~roots~~ of ~~a cubic~~ equation ~~having this form include a~~ real ~~root given by~~ the ~~expression~~,
	<div align="center">		<div align="center">
	<math>		<math>
	~~y_1 =~~ \~~mathcal{S} + \mathcal{T} ,~~		\phi \approx 2 + \frac{3}{2^{2/3}}\lambda
	~~</math>~~
	~~</div>~~
	~~where,~~
	~~<div align="center">~~
	~~<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><br />		</math><br />

	<math>		<math>
	\~~mathcal{S} \equiv~~ \~~biggl[ -~~\frac{~~b_0}{2} + \mathcal{D}^{1/2} \biggr]^{1/~~3}		\Rightarrow ~~~~~ z \approx \frac{3\lambda}{2^2} \biggl( \frac{a_1}{2^2a_4} \biggr)^{1/3} = \frac{a_0}{a_1} .
	~~= \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><br/>~~
	~~<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>		</math>
	</div>		</div>

	~~===Summary===~~
	~~Hence~~, ~~defining,~~		We see, as well, that when <math>\lambda \gg 1</math>,
	~~<div align="center">~~
	~~<math>~~
	~~\mathcal{K} \equiv \biggl( \frac{8a_0}{3a_1^2} \biggr)^2 ,~~
	~~</math>~~
	~~</div>~~
	~~and~~
	~~<div align="center">~~
	~~<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>~~
	~~</div>~~
	~~we can write the desired solution of the quartic equation~~ as,
	~~<div align="center">~~
	<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>~~
	~~</div>~~
	~~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~~,
	<div align="center">		<div align="center">
	<math>		<math>
	8^~~{1/~~2} ~~[f_Q(~~\~~mathcal{K}~~)]^{-3/2} ~~\ge 2~~		\phi \approx \biggl(\frac{3\lambda}{2^2} \biggr)^{3/2}
	</math><br />		</math><br />

	<math>		<math>
	\Rightarrow ~~ ~~f_Q(~~\~~mathcal{K})~~ \~~le 2^{1/3} .~~		\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>~~
	~~</div>~~
	~~This, in turn, implies that the dimensionless parameter, <math>\mathcal{K}</math>, must be limited to values,~~
	~~<div align="center">~~
	~~<math>~~
	~~\mathcal{K} \ge 0 .~~
	~~</math>~~
	~~</div>~~
	~~Our solution takes on a somewhat cleaner form if we define a function,~~
	~~<div align="center">~~
	~~<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>~~
	~~</div>~~
	~~which is limited to values,~~
	~~<div align="center">~~
	~~<math>~~
	~~g_Q(\mathcal{K}~~) ~~\ge 1 .~~
	~~</math>~~
	~~</div>~~
	~~Then we can write,~~
	~~<div align="center">~~
	~~<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>		</math>
	</div>		</div>
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	=Related Wikepedia Links=		=Related Wikepedia Links=
	[http://en.wikipedia.org/wiki/Quartic_function Quartic Function]~~</br>~~		* [http://en.wikipedia.org/wiki/Quartic_function Quartic Function]
	[http://mathworld.wolfram.com/QuarticEquation.html mathworld.wolfram.com]		* [http://mathworld.wolfram.com/QuarticEquation.html mathworld.wolfram.com]

	{{LSU_HBook_footer}}		{{LSU_HBook_footer}}

Tohline: /* Quartic Equation Solution */ Fine-tune presentation

2010-03-11T17:46:15Z

Quartic Equation Solution: Fine-tune presentation

← Older revision		Revision as of 17:46, 11 March 2010
Line 54:		Line 54:
	</math>		</math>
	</div>		</div>
	~~Now, defining~~,		Defining,
	<div align="center">		<div align="center">
	<math>		<math>
Line 63:		Line 63:
	<div align="center">		<div align="center">
	<math>		<math>
	\frac{1}{2}W^2 ~~\equiv~~ \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]		\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\} ,		= \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><br/>		</math><br/>
	</div>		</div>
	we can rewrite the quartic ~~solution~~ in the form,		we can rewrite the desired root of our quartic equation in the form,
	<div align="center">		<div align="center">
	<math>		<math>
	z = \biggl(\frac{a_1}{4a_4}\biggr)^{1/3} \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr]		z = \biggl(\frac{a_1}{4a_4}\biggr)^{1/3} \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr] ,
	</math><br />		</math>
			</div>
			with,
			<div align="center">
	<math>		<math>
	\phi = 2^{3/2} \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{1/2}		\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}		\biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\}^{-3/2} .
	</math>		</math>
	</div>		</div>

Tohline: /* Quartic Equation Solution */ Correct errors in math

2010-03-11T05:33:58Z

Quartic Equation Solution: Correct errors in math

← Older revision		Revision as of 05:33, 11 March 2010
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	<div align="center">		<div align="center">
	<math>		<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]		\frac{1}{2}W^2 \equiv \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}{2^8 a_4 R}\biggr]^{1/3} \biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \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><br/>		</math><br/>
	</div>		</div>
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	<div align="center">		<div align="center">
	<math>		<math>
	z = \biggl(\frac{a_1}{4a_4}\biggr) \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr]		z = \biggl(\frac{a_1}{4a_4}\biggr)^{1/3} \phi^{-1/3}\biggl[ (\phi - 1)^{1/2} - 1 \biggr]
	</math><br />		</math><br />
	<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)~~		\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} =~~		\biggl\{ \biggl[ 1 + (1 + \lambda^3)^{1/2} \biggr]^{2/3} - \lambda \biggr\}^{-3/2}
	~~\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>

Tohline: /* Quartic Equation Solution */ Improve discussion of quartic solution

2010-03-11T03:23:50Z

Quartic Equation Solution: Improve discussion of quartic solution

← Older revision		Revision as of 03:23, 11 March 2010
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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>
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	<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>

Tohline: More revamping of derivation

2010-03-11T02:16:48Z

More revamping of derivation

← Older revision		Revision as of 02:16, 11 March 2010
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	===Quartic Equation Solution===		===Quartic Equation Solution===

	Following the ~~outline and notation used by~~ [http://~~mathworld~~.~~wolfram~~.~~com~~/~~QuarticEquation.html 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,		Following the [http://en.wikipedia.org/wiki/Quartic_function#Summary_of_Ferrari's_method 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,
	<div align="center">		<div align="center">
	<math>		<math>
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	=Related Wikepedia Links=		=Related Wikepedia Links=
	[http://en.wikipedia.org/wiki/Quartic_function Quartic Function]		[http://en.wikipedia.org/wiki/Quartic_function Quartic Function]</br>
			[http://mathworld.wolfram.com/QuarticEquation.html mathworld.wolfram.com]
	{{LSU_HBook_footer}}		{{LSU_HBook_footer}}