Difference between revisions of "User:Tohline/Appendix/Ramblings/T3Integrals/QuadraticCase"
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\biggl[ \frac{\Lambda-1}{\Lambda} \biggr] | \biggl[ \frac{\Lambda^2-1}{\Lambda} \biggr] \frac{d\ln(\lambda_1/\lambda_2)}{dt} . | ||
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h_1^4\frac{d}{dt}\biggl[ \ln(\lambda_1/\lambda_2) | 2 h_1^4\frac{d}{dt}\biggl[ \ln(\lambda_1/\lambda_2) \biggr] | ||
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\frac{1}{ | \frac{1}{2} \biggl( \frac{\Lambda +1}{\Lambda} \biggr)^2 \biggl[ \frac{\Lambda^2-1}{\Lambda} \biggr]^{-1} \frac{d\Lambda}{dt} | ||
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\frac{1}{ | \frac{1}{2} \biggl( \frac{\Lambda +1}{\Lambda-1} \biggr) \frac{d\ln\Lambda}{dt} | ||
</math> | </math> | ||
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Revision as of 02:07, 10 June 2010
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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^2-1}{\Lambda} \biggr] \frac{d\ln(\lambda_1/\lambda_2)}{dt} . </math> |
Hence,
<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> |
© 2014 - 2021 by Joel E. Tohline |