Editing Appendix/Ramblings/T3Integrals/QuadraticCase

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=T3 Coordinates (continued)=
On one accompanying wiki page we have [[Appendix/Ramblings/T3Integrals#Integras_of_Motion_in_T3_Coordinates|introduced T3 Coordinates]] and on another we have described how [[Appendix/Ramblings/T3CharacteristicVector#Characteristic_Vector_for_T3_Coordinates|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 [[User:Jaycall/T3_Coordinates/Special_Case|separate page]].

==Special Case (Quadratic)==

===Coordinate Relations===

When <math>q^2=2</math>, the two key coordinates are:
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right" colspan="2">
<math>
\lambda_1
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\varpi \cosh\Zeta
</math>
  </td>
</tr>

<tr>
  <td align="right" colspan="2">
<math>
\lambda_2
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{\varpi}{\sinh\Zeta}
</math>
  </td>
</tr>

<tr>
  <td align="left">
Note also:&nbsp;&nbsp;&nbsp;
  </td>
  <td align="right" colspan="1">
<math>
\Chi \equiv 2\frac{\lambda_1}{\lambda_2}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
2 \sinh\Zeta \cosh\Zeta = \sinh(2\Zeta) ,
</math>
  </td>
</tr>
</table>
where, in this case, 
<div align="center">
<math>
\Zeta \equiv \sinh^{-1} \biggl( \frac{\sqrt{2}z}{\varpi} \biggr) .
</math>
</div>

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,
<div align="center">
<math>
1 = \cosh^2\Zeta - \sinh^2\Zeta = \biggl(\frac{\lambda_1}{\varpi}\biggr)^2 - \biggl(\frac{\varpi}{\lambda_2}\biggr)^2
</math>
</div>
implies that the function <math>\varpi(\lambda_1,\lambda_2)</math> can be obtained from the physically relevant root of the following equation:
<div align="center">
<math>
\varpi^4 \lambda_2^{-2} + \varpi^2 - \lambda_1^2 = 0.
</math>
</div>
The relevant root gives,
<div align="center">
<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><br />

<math>
\Rightarrow ~~~~~\varpi =  \frac{\lambda_2}{\sqrt{2}}\biggl[ \cosh(2\Zeta) - 1 \biggr]^{1/2} .
</math>
</div>
The desired function <math>z(\lambda_1,\lambda_2)</math> is therefore,
<div align="center">
<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><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>


===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>

===Equation of Motion===

According to [[Appendix/Ramblings/T3Integrals#EOM.01|'''Equation EOM.01''']], as derived elsewhere in the context of T3 coordinates, a general expression for the second component of the equation of motion is,
<div align="center">
<math>
\frac{d(h_2 \dot{\lambda}_2)}{dt} =  \biggl(\frac{\lambda_2 \dot{\lambda}_1}{\lambda_1}\biggr) \frac{dh_2}{dt} 
</math>
<br /><br />
<math>
\Rightarrow ~~~~~ \frac{\ddot{\lambda}_2}{\lambda_2} =  \biggl[\frac{\dot{\lambda}_1}{\lambda_1} - \frac{\dot{\lambda}_2}{\lambda_2} \biggr]\frac{d\ln h_2 }{dt} = 
\biggl[\frac{d\ln(\lambda_1/\lambda_2)}{dt} \biggr]\frac{d\ln h_2 }{dt} .
</math>
</div>
Based on the derivations provided above, both factors that make up the term on the RHS of this expression can be written entirely in terms of the variable, <math>\Lambda</math>.  This allows us to rewrite [[Appendix/Ramblings/T3Integrals#EOM.01|'''Equation EOM.01''']] as,
<div align="center">
<math>
\frac{\ddot{\lambda}_2}{\lambda_2} =  \biggl[\frac{\Lambda}{\Lambda^2-1} \frac{d\Lambda}{dt}\biggr]\biggl[ \frac{1}{2\Lambda} \biggl( \frac{\Lambda + 1}{\Lambda - 1} \biggr) \frac{d\Lambda}{dt} \biggr] = \frac{1}{2(\Lambda - 1)^2}  \biggl[\frac{d\Lambda}{dt} \biggr]^2
</math><br /><br />
</div>

<span id="T3Q.01"><table align="right" border="1" cellpadding="10" width="10%">
<tr><th><font color="darkblue">T3Q.01</font></th></tr>
</table></span>
<div align="center">
<math>
\Rightarrow ~~~~~ 2\frac{\ddot{\lambda}_2}{\lambda_2} =  \biggl[\frac{d\ln(\Lambda-1)}{dt} \biggr]^2.
</math>
</div>

===Solution Strategy===
I'm not sure whether the following strategy is fully legitimate, but let's explore it anyway.  Because the LHS of [[Appendix/Ramblings/T3Integrals/QuadraticCase#T3Q.01|'''Equation T3Q.01''']] displays an explicit dependence only on the coordinate <math>\lambda_2</math> while the RHS displays an explicit dependence only on <math>\Lambda</math> &#8212; that is, only on the ''ratio'' of the two coordinates <math>\lambda_1/\lambda_2</math> &#8212; perhaps we can use a ''separation of variables'' technique to derive a solution.  Specifically, suppose the LHS and the RHS separately are set equal to the same value, call it <math>n</math>. 

Then, for the LHS:
<div align="center">
<math>
\ddot{\lambda}_2 = \frac{n}{2}\lambda_2 ;
</math>
</div>

And, for the RHS:
<div align="center">
<math>
\frac{d\ln(\Lambda-1)}{dt} = \sqrt{n} .
</math>
</div>
Now I suppose that, in general, <math>n</math> should be allowed to vary with time, but for exploratory purposes, let's assume that <math>n</math> is a constant.  The solution to the LHS's <math>2^\mathrm{nd}</math>-order ODE is,
<div align="center">
<math>
\lambda_2 = \lambda^0_2 \exp{[-\sqrt{n/2}~t]} ,
</math>
</div>
where, <math>\lambda^0_2</math> is the coordinate position <math>\lambda_2</math> at time <math>t=0</math>.  The solution to the RHS's <math>1^\mathrm{st}</math>-order ODE is,
<div align="center">
<math>
\sqrt{n} t = \ln\biggl( \frac{\Lambda-1}{\Lambda_0 -1} \biggr) ,
</math>
</div>
where, <math>\Lambda_0</math> is given by the coordinate ratio at time <math>t=0</math>, specifically, <math>\Lambda_0 \equiv \sqrt{1 + (2\lambda_1^0/\lambda_2^0)^2}</math>.

Now, if we replace "<math>\sqrt{n}~t</math>" in the first of these expressions by the second expression, we find,
<div align="center">
<math>
\frac{\lambda_2}{\lambda^0_2} = \exp{\biggl[ -\frac{1}{\sqrt{2}} \ln\biggl( \frac{\Lambda-1}{\Lambda_0 -1} \biggr) \biggr]} = \biggl( \frac{\Lambda_0 -1}{\Lambda-1} \biggr)^{1/\sqrt{2}} .
</math>
</div>
This would be a fantastically simple result, if it proved to be a proper solution to the governing equation of motion.  Unfortunately, if this relatively elementary equation is differentiated twice in an effort to reproduce [[Appendix/Ramblings/T3Integrals/QuadraticCase#T3Q.01|'''Equation T3Q.01''']], we find that an additional undesirable term appears that involves the second derivative of a function containing the variable <math>\Lambda</math>.  It can be shown that this undesirable term goes to zero if <math>\sqrt{n}</math> is assumed to be independent of time (as we did indeed assume, above).  Unfortunately, in reality, this does not seem to be a desirable assumption for the physical problem in which we have interest, so we must conclude that the derived elementary equation is not the desired solution of the equation of motion.

But can we learn something valuable from this failed separation of variables approach???

===Another Thought===

It can easily be shown that, in general,
<div align="center">
<math>
\frac{d}{dt}\biggl( \frac{\dot{F}}{F} \biggr) = \frac{\ddot{F}}{F} - \biggl[\frac{d\ln F}{dt}  \biggr]^2 .
</math>
</div>

Hence, [[Appendix/Ramblings/T3Integrals/QuadraticCase#T3Q.01|Equation T3Q.01]] can be rewritten as,
<div align="center">
<math>
\frac{d}{dt}\biggl( \frac{\dot{\lambda_2}}{\lambda_2} \biggr) + \biggl[\frac{d\ln \lambda_2}{dt}  \biggr]^2 = \biggl[\frac{d\ln (\Lambda-1)^{1/2}}{dt}  \biggr]^2 
</math><br /><br />

<math>
\Rightarrow ~~~~~ \frac{d}{dt}\biggl( \frac{d\ln\lambda_2}{dt} \biggr) = \biggl[\frac{d\ln (\Lambda-1)^{1/2}}{dt}  \biggr]^2 - \biggl[\frac{d\ln \lambda_2}{dt}  \biggr]^2 = \biggl[\frac{d\ln (\Lambda-1)^{1/2}}{dt} + \frac{d\ln \lambda_2}{dt}  \biggr] \biggl[\frac{d\ln (\Lambda-1)^{1/2}}{dt} - \frac{d\ln \lambda_2}{dt}  \biggr] 
</math><br /><br />

<math>
\Rightarrow ~~~~~ \frac{d}{dt}\biggl( \frac{d\ln\lambda_2}{dt} \biggr) = \frac{d\ln [(\Lambda-1)^{1/2}\lambda_2]}{dt} \cdot \frac{d\ln [(\Lambda-1)^{1/2} \lambda_2^{-1}]}{dt}  
</math>
</div>
Can we gain any insight from this form of the <math>2^\mathrm{nd}</math> component of the equation of motion?  For example, can the convert the RHS into the total time-derivative of some function?

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