Editing Appendix/Ramblings/T3Integrals/QuadraticCase (section)

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

<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="0" 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^2-1}{\Lambda} \biggr] \frac{d\ln(\lambda_1/\lambda_2)}{dt} .
</math>
  </td>
</tr>
</table>

Hence, starting from the [[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">
<tr>
  <td align="right">
<math>
\frac{d\ln h_2}{dt}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
2 h_1^4\frac{d}{dt}\biggl[ \ln(\lambda_1/\lambda_2) \biggr]
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{2} \biggl( \frac{\Lambda +1}{\Lambda} \biggr)^2 \biggl[ \frac{\Lambda^2-1}{\Lambda} \biggr]^{-1} \frac{d\Lambda}{dt}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{2} \biggl( \frac{\Lambda +1}{\Lambda-1} \biggr) \frac{d\ln\Lambda}{dt} .
</math>
  </td>
</tr>
</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 [[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>