Difference between revisions of "User:Tohline/Appendix/Ramblings/T3Integrals/QuadraticCase"

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(→‎T3 Coordinates (continued): Invert coordinate relations)
(→‎Special Case (Quadratic): Finish inversion expressions)
Line 87: Line 87:
<div align="center">
<div align="center">
<math>
<math>
\varpi^2 =  \frac{\lambda_2^{2}}{2}  \biggl[ \sqrt{1 + (2\lambda_1/\lambda_2)^2} - 1 \biggr].
\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><br />
 
<math>
\Rightarrow ~~~~~\varpi =  \frac{\lambda_2}{\sqrt{2}}\biggl[ \cosh(2\Zeta) - 1 \biggr]^{1/2} .
</math>
</math>
</div>
</div>
Line 93: Line 97:
<div align="center">
<div align="center">
<math>
<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].
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><br />
 
<math>
\Rightarrow ~~~~~ z = \frac{1}{2}~\frac{\lambda_2 }{\sqrt{2}} \biggl[ \cosh(2\Zeta) - 1 \biggr] .
</math>
</div>
 
In an effort to simplify the appearance of these and future expressions, we will henceforth adopt the notation,
<div align="center">
<math>
\Lambda \equiv \cosh(2\Zeta) ~~~~~ \Rightarrow ~~~~~ \Lambda = \sqrt{1 + \Chi^2} .
</math>
</div>
In terms of <math>\Lambda</math>, then, we have,
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\varpi
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{\lambda_2 }{\sqrt{2}} \biggl[ \Lambda - 1 \biggr]^{1/2} ;
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>
z
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{2}~\frac{\lambda_2 }{\sqrt{2}} \biggl[ \Lambda - 1 \biggr] .
</math>
  </td>
</tr>
</table>
 
----
 
 
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">
<math>
\ell^{2} = [\varpi^2 + 4z^2]^{-1} = \frac{2}{\lambda_2^2(\Lambda - 1)\Lambda}.
</math>
</math>
</div>
</div>
Hence,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right">
<math>
h_1^2
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\lambda_1^2 \ell^2
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{2}\biggl[ \frac{\Lambda + 1}{\Lambda} \biggr] ;
</math>
  </td>
</tr>


<tr>
  <td align="right">
<math>
h_2^2
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
(q^2-1)\biggl( \frac{\varpi z \ell}{\lambda_2} \biggr)^2
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{8}~ \frac{(\Lambda - 1)^2}{\Lambda} .
</math>
  </td>
</tr>
</table>
We note also that,
<table border="1" align="center" cellpadding="5">
<tr>
  <td align="right">
<math>
\frac{d\Lambda}{dt}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{d}{dt}\biggl[ 1+(2\lambda_1/\lambda_2)^2 \biggr]^{1/2}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\Lambda-1}{\Lambda} \biggr]^2 \frac{d\ln(\lambda_1/\lambda_2)}{dt} .
</math>
  </td>
</tr>
</table>
Hence,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right">
<math>
\frac{d\ln h_2}{dt}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
h_1^4\frac{d}{dt}\biggl[ \ln(\lambda_1/\lambda_2)^2 \biggr]
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{4} \biggl( \frac{\Lambda +1}{\Lambda} \biggr)^2 \biggl[ \frac{\Lambda-1}{\Lambda} \biggr]^{-2} \frac{d\Lambda}{dt}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{4} \biggl( \frac{\Lambda +1}{\Lambda-1} \biggr)^2 \frac{d\Lambda}{dt}
</math>
  </td>
</tr>
</table>




&nbsp;<br />
&nbsp;<br />
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Revision as of 01:41, 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)

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>



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>

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

Hence,


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

<math> = </math>

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

<math> = </math>

<math> \frac{1}{4} \biggl( \frac{\Lambda +1}{\Lambda} \biggr)^2 \biggl[ \frac{\Lambda-1}{\Lambda} \biggr]^{-2} \frac{d\Lambda}{dt} </math>

<math> = </math>

<math> \frac{1}{4} \biggl( \frac{\Lambda +1}{\Lambda-1} \biggr)^2 \frac{d\Lambda}{dt} </math>


 

Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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