Editing Jaycall/KillingVectorApproach (section)

===Strategic Choice of Characteristic Vector===
So we're looking for a vector <math>\vec{C}</math> such that

<div align="center">
<math>
{h_i}^2 \dot{\lambda}_i \dot{C}_i + {h_k}^2 \Gamma^k_{ij} \dot{\lambda}_j \dot{\lambda}_k C_i - C_i \ \partial_i \Phi = 0 .
</math>
</div>

The good news is that this is a single scalar equation constraining the three independent components of <math>\vec{C}</math>, so several families of solutions should exist.  Consequently, we can afford to be choosey!  

In the problem we're exploring using [[Appendix/Ramblings/T3Integrals|T3 coordinates]], we already know a conserved quantity associated with the azimuthal coordinate &#8212; angular momentum.  We're looking for an additional (independent) conserved quantity associated with the two radial coordinates.  So it makes sense to look for one of the solutions that has <math>C_3 = 0</math>.  Furthermore, we'd like to avoid dealing with the unknown potential function (which varies only with <math>\lambda_1</math>) as much as possible, so as long as we're being choosey, let's look for a solution that has <math>C_1 = 0</math>.  We now have three conditions on three components.  The condition constraining <math>C_2</math> is

<div align="center">
<math>
{h_2}^2 \dot{\lambda}_2 \dot{C}_2 + {h_k}^2 \Gamma^k_{2j} \dot{\lambda}_j \dot{\lambda}_k C_2 = 0 .
</math>
</div>

This can be rewritten more simply as

<div align="center">
<math>
\dot{C}_2 = - \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) C_2 .
</math>
</div>

Perhaps the best approach to solving this condition is separation of variables.

<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
<math>
\int \frac{dC_2}{C_2} 
</math>
  </td>
  <td align="center">
<math>
= 
</math>
  </td>
  <td align="left">
<math>
- \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow 
</math>
  </td>
  <td align="right">
<math>
\ln C_2 
</math>
  </td>
  <td align="center">
<math>
= 
</math>
  </td>
  <td align="left">
<math>
- \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow 
</math>
  </td>
  <td align="right">
<math>
C_2
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\exp \left\{ - \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt \right\} .
</math>
  </td>
</tr>

</table>

The final necessary step will be to write the quantity in parentheses as the exact derivative of some other quantity, call it <math>\Xi</math>.  The long-sought conserved quantity will then be <math> m {h_2}^2 \dot{\lambda}_2 \exp \left( - \Xi \right) </math> , where

<div align="center">
<math>
\Xi \equiv \int \left( \frac{{h_k}^2}{{h_2}^2} \Gamma^k_{2j} \frac{\dot{\lambda}_j \dot{\lambda}_k}{\dot{\lambda}_2} \right) dt .
</math>
</div>


====Question from Joel====
Is the following logic correct?  

Suppose we examine the term in which <math>j=2</math> and <math>k=1</math>.  The integral becomes,

<div align="center">
<math>
\Xi\biggr|_{j=2,k=1} = \int \left( \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_1}{\dot{\lambda}_2} \right) dt = \int \left( \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{22}  \right) d\lambda_1
</math>
</div>

Given that the scale factors and the Christoffel symbols are expressible entirely in terms of <math>\lambda_1</math> and <math>\lambda_2</math>, one could imagine a situation &#8212; for example in the quadratic case of <math>q^2=2</math> &#8212; in which this integral could be completed analytically.  (I presume that wherever <math>\lambda_2</math> appears inside this integral, it can be treated as a constant because the two coordinates are independent of one another.)

====Response from Jay====

Yes, I believe you've worked this term out correctly, but as you can see in what follows, it cancels out with the other cross term.  Writing out each of the terms in <math>\Xi</math> gives,

<div align="center">
<math>
\frac{d\Xi}{dt} = \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{21} \frac{\dot{\lambda}_1 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_1}^2}{{h_2}^2} \Gamma^1_{23} \frac{\dot{\lambda}_3 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \Gamma^2_{21} \frac{\dot{\lambda}_1 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \Gamma^2_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \Gamma^2_{23} \frac{\dot{\lambda}_3 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \Gamma^3_{21} \frac{\dot{\lambda}_1 \dot{\lambda}_3}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \Gamma^3_{22} \frac{\dot{\lambda}_2 \dot{\lambda}_3}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \Gamma^3_{23} \frac{\dot{\lambda}_3 \dot{\lambda}_3}{\dot{\lambda}_2} .
</math>
</div>

Plugging in values for the Christoffel symbols leads to the expression

<div align="center">
<math>
\frac{d\Xi}{dt} = \frac{{h_1}^2}{{h_2}^2} \frac{\partial_2 h_1}{h_1} \frac{\dot{\lambda}_1 \dot{\lambda}_1}{\dot{\lambda}_2} - \frac{{h_1}^2}{{h_2}^2} \frac{h_2}{h_1} \frac{\partial_1 h_2}{h_1} \frac{\dot{\lambda}_2 \dot{\lambda}_1}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \frac{\partial_1 h_2}{h_2} \frac{\dot{\lambda}_1 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_2}^2}{{h_2}^2} \frac{\partial_2 h_2}{h_2} \frac{\dot{\lambda}_2 \dot{\lambda}_2}{\dot{\lambda}_2} + \frac{{h_3}^2}{{h_2}^2} \frac{\partial_2 h_3}{h_3} \frac{\dot{\lambda}_3 \dot{\lambda}_3}{\dot{\lambda}_2} .
</math>
</div>

And limiting our interest to motion within the meridional plane (setting <math>\dot{\lambda}_3 = 0</math>) and simplifying

<span id="CV.02"><table align="right" border="1" cellpadding="10" width="10%">
<tr><th><font color="darkblue">CV.02</font></th></tr>
</table></span>
<table align="center" border="1" cellpadding="5">
<tr>
  <td align="right">
<math>
\frac{d\Xi}{dt} \equiv - \frac{d\ln C_2}{dt}
</math>
  </td>
  <td align="center">
<math>
= 
</math>
  </td>
  <td align="left">
<math>
\frac{h_1}{h_2} \frac{\partial_2 h_1}{h_2} \frac{{\dot{\lambda}_1}^2}{\dot{\lambda}_2} - \cancel{\frac{\partial_1 h_2}{h_2} \dot{\lambda}_1} + \cancel{\frac{\partial_1 h_2}{h_2} \dot{\lambda}_1} + \frac{\partial_2 h_2}{h_2} \dot{\lambda}_2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
= 
</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl(\frac{h_1 \dot{\lambda}_1}{h_2 \dot{\lambda}_2}\biggr)^2 \frac{\partial \ln h_1}{\partial\ln\lambda_2}  + \frac{\partial \ln h_2}{\partial \ln\lambda_2}\biggr] \frac{d \ln{\lambda}_2}{dt} 
</math>
  </td>
</tr>

<tr>
  <td align="left" colspan="3">
<font color="red"><b>NOTE:</b> Sign error fixed on 15 July 2010</font>.  Specifically, <math>d\Xi/dt \equiv d\ln C_2/dt</math> changed to <math>d\Xi/dt \equiv - d\ln C_2/dt</math>.
  </td>
</tr>
</table>