Editing Jaycall/KillingVectorApproach (section)

====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,

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<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

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<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">
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<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>

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&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>

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  <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>.
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