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(→‎Special Case (Quadratic): Correct minor error in h_2 derivative)
(→‎Special Case (Quadratic): Subdivide into subsections and extend discussion)
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==Special Case (Quadratic)==
==Special Case (Quadratic)==
===Coordinate Relations===


When <math>q^2=2</math>, the two key coordinates are:
When <math>q^2=2</math>, the two key coordinates are:
Line 150: Line 152:
</table>
</table>


----


===Scale Factor Expressions===


We are now in a position to express the two key scale factors purely in terms of the two key T3 coordinates.  First, we note that,
We are now in a position to express the two key scale factors purely in terms of the two key T3 coordinates.  First, we note that,
<div align="center">
<div align="center">
<math>
<math>
\ell^{2} = [\varpi^2 + 4z^2]^{-1} = \frac{2}{\lambda_2^2(\Lambda - 1)\Lambda}.
\ell^{2} = [\varpi^2 + 4z^2]^{-1} = \frac{2}{\lambda_2^2(\Lambda - 1)\Lambda} ,
</math>
</math>
</div>
</div>
and,
<div align="center">
<math>
\biggl(\frac{\lambda_1}{\lambda_2}\biggr)^2 = \frac{1}{4}\Chi^2 = \frac{1}{4}(\Lambda^2 -1) = \frac{1}{4}(\Lambda -1)(\Lambda +1) .
</math>
</div>
Hence,
Hence,


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</table>
</table>


Hence,
Hence, starting from the [[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|''general'' expression for <math>d\ln h_2/dt</math> derived elsewhere]], we deduce that,
 


<table border="0" align="center" cellpadding="5">
<table border="0" align="center" cellpadding="5">
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</tr>
</tr>
</table>
</table>
(This identical expression also can be derived straightforwardly from the ''specific'' expression for <math>h_2</math> given above.)  From the expression given above for <math>h_1</math>, we also deduce that,
<div align="center">
<math>
\frac{d\ln h_1}{dt} = - \frac{1}{2(\Lambda + 1)} \frac{d\ln\Lambda}{dt} .
</math>
</div>
As a check, we note that the relationship between <math>d\ln h_1/dt</math> and <math>d\ln h_2/dt</math> in this ''specific'' case (<i>i.e.</i>, <math>q^2 = 2</math>) matches the ''general'' relationship between these two logarithmic time-derivatives that has been [[User:Tohline/Appendix/Ramblings/T3Integrals#Logarithmic_Derivatives_of_Scale_Factors|derived elsewhere]], namely,
<div align="center">
<math>
(h_1 \lambda_1)^2 \frac{d\ln h_1}{dt} + (h_2 \lambda_2)^2 \frac{d\ln h_2}{dt} = 0 .
</math>
</div>




&nbsp;<br />
&nbsp;<br />
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{{LSU_HBook_footer}}

Revision as of 13:50, 10 June 2010

Whitworth's (1981) Isothermal Free-Energy Surface
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T3 Coordinates (continued)

On one accompanying wiki page we have introduced T3 Coordinates and on another we have described how Jay Call's Characteristic Vector applies to T3 Coordinates. Here we investigate the properties of our T3 Coordinate system in the special case when <math>q^2 = 2</math>; Jay Call's independent analysis is recorded on a separate page.

Special Case (Quadratic)

Coordinate Relations

When <math>q^2=2</math>, the two key coordinates are:

<math> \lambda_1 </math>

<math> = </math>

<math> \varpi \cosh\Zeta </math>

<math> \lambda_2 </math>

<math> = </math>

<math> \frac{\varpi}{\sinh\Zeta} </math>

Note also:   

<math> \Chi \equiv 2\frac{\lambda_1}{\lambda_2} </math>

<math> = </math>

<math> 2 \sinh\Zeta \cosh\Zeta = \sinh(2\Zeta) , </math>

where, in this case,

<math> \Zeta \equiv \sinh^{-1} \biggl( \frac{\sqrt{2}z}{\varpi} \biggr) . </math>

For this special case, we can invert these coordinate relations to obtain analytic expressions for both <math>\varpi</math> and <math>z</math> in terms of <math>\lambda_1</math> and <math>\lambda_2</math>. Specifically, the relation,

<math> 1 = \cosh^2\Zeta - \sinh^2\Zeta = \biggl(\frac{\lambda_1}{\varpi}\biggr)^2 - \biggl(\frac{\varpi}{\lambda_2}\biggr)^2 </math>

implies that the function <math>\varpi(\lambda_1,\lambda_2)</math> can be obtained from the physically relevant root of the following equation:

<math> \varpi^4 \lambda_2^{-2} + \varpi^2 - \lambda_1^2 = 0. </math>

The relevant root gives,

<math> \varpi^2 = \frac{\lambda_2^{2}}{2} \biggl[ \sqrt{1 + (2\lambda_1/\lambda_2)^2} - 1 \biggr] = \frac{\lambda_2^{2}}{2} \biggl[ \cosh(2\Zeta) - 1 \biggr] </math>

<math> \Rightarrow ~~~~~\varpi = \frac{\lambda_2}{\sqrt{2}}\biggl[ \cosh(2\Zeta) - 1 \biggr]^{1/2} . </math>

The desired function <math>z(\lambda_1,\lambda_2)</math> is therefore,

<math> z = \frac{\varpi}{\sqrt{2}} \sinh\Zeta = \frac{\varpi^2}{\sqrt{2}~\lambda_2} = \frac{\lambda_2 }{2\sqrt{2}} \biggl[ \sqrt{1 + (2\lambda_1/\lambda_2)^2} - 1 \biggr] </math>

<math> \Rightarrow ~~~~~ z = \frac{1}{2}~\frac{\lambda_2 }{\sqrt{2}} \biggl[ \cosh(2\Zeta) - 1 \biggr] . </math>

In an effort to simplify the appearance of these and future expressions, we will henceforth adopt the notation,

<math> \Lambda \equiv \cosh(2\Zeta) ~~~~~ \Rightarrow ~~~~~ \Lambda = \sqrt{1 + \Chi^2} . </math>

In terms of <math>\Lambda</math>, then, we have,

<math> \varpi </math>

<math> = </math>

<math> \frac{\lambda_2 }{\sqrt{2}} \biggl[ \Lambda - 1 \biggr]^{1/2} ; </math>

<math> z </math>

<math> = </math>

<math> \frac{1}{2}~\frac{\lambda_2 }{\sqrt{2}} \biggl[ \Lambda - 1 \biggr] . </math>


Scale Factor Expressions

We are now in a position to express the two key scale factors purely in terms of the two key T3 coordinates. First, we note that,

<math> \ell^{2} = [\varpi^2 + 4z^2]^{-1} = \frac{2}{\lambda_2^2(\Lambda - 1)\Lambda} , </math>

and,

<math> \biggl(\frac{\lambda_1}{\lambda_2}\biggr)^2 = \frac{1}{4}\Chi^2 = \frac{1}{4}(\Lambda^2 -1) = \frac{1}{4}(\Lambda -1)(\Lambda +1) . </math>

Hence,

<math> h_1^2 </math>

<math> = </math>

<math> \lambda_1^2 \ell^2 </math>

<math> = </math>

<math> \frac{1}{2}\biggl[ \frac{\Lambda + 1}{\Lambda} \biggr] ; </math>

<math> h_2^2 </math>

<math> = </math>

<math> (q^2-1)\biggl( \frac{\varpi z \ell}{\lambda_2} \biggr)^2 </math>

<math> = </math>

<math> \frac{1}{8}~ \frac{(\Lambda - 1)^2}{\Lambda} . </math>

We note also that,

<math> \frac{d\Lambda}{dt} </math>

<math> = </math>

<math> \frac{d}{dt}\biggl[ 1+(2\lambda_1/\lambda_2)^2 \biggr]^{1/2} </math>

<math> = </math>

<math> \biggl[ \frac{\Lambda^2-1}{\Lambda} \biggr] \frac{d\ln(\lambda_1/\lambda_2)}{dt} . </math>

Hence, starting from the general expression for <math>d\ln h_2/dt</math> derived elsewhere, we deduce that,

<math> \frac{d\ln h_2}{dt} </math>

<math> = </math>

<math> 2 h_1^4\frac{d}{dt}\biggl[ \ln(\lambda_1/\lambda_2) \biggr] </math>

<math> = </math>

<math> \frac{1}{2} \biggl( \frac{\Lambda +1}{\Lambda} \biggr)^2 \biggl[ \frac{\Lambda^2-1}{\Lambda} \biggr]^{-1} \frac{d\Lambda}{dt} </math>

<math> = </math>

<math> \frac{1}{2} \biggl( \frac{\Lambda +1}{\Lambda-1} \biggr) \frac{d\ln\Lambda}{dt} </math>

(This identical expression also can be derived straightforwardly from the specific expression for <math>h_2</math> given above.) From the expression given above for <math>h_1</math>, we also deduce that,

<math> \frac{d\ln h_1}{dt} = - \frac{1}{2(\Lambda + 1)} \frac{d\ln\Lambda}{dt} . </math>

As a check, we note that the relationship between <math>d\ln h_1/dt</math> and <math>d\ln h_2/dt</math> in this specific case (i.e., <math>q^2 = 2</math>) matches the general relationship between these two logarithmic time-derivatives that has been derived elsewhere, namely,

<math> (h_1 \lambda_1)^2 \frac{d\ln h_1}{dt} + (h_2 \lambda_2)^2 \frac{d\ln h_2}{dt} = 0 . </math>


 

Whitworth's (1981) Isothermal Free-Energy Surface

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