Editing Apps/Wong1973Potential (section)

====Phase 0D====

We therefore have,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1 - \epsilon^2) \biggl[ \frac{2^2}{\pi^2} \biggr] C_0(\epsilon)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~  
\biggl[ \frac{2^2}{\pi^2} \biggr]\biggl\{ (K - E)^2 \biggr\}
~-~ \epsilon \biggl[ \frac{2^2}{\pi^2} \biggr]\biggl\{ E\cdot E -K\cdot K\biggr\}
~+~2\epsilon^{2}  \biggl[ \frac{2^2}{\pi^2} \biggr]\biggl\{ K \cdot E \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~  
\biggl\{ \biggl( \frac{1}{2^2} \biggr)k^4 
+ \biggl( \frac{3}{2^4} \biggr) k^6 
+ \biggl( \frac{3 \cdot 13}{2^8}  \biggr) k^8 + \mathcal{O}(k^{10})
\biggr\}
~+~ \epsilon \biggl\{ k^2 ~+~ \frac{3}{2^3} ~ k^4 ~+~\frac{3^2}{2^5} ~ k^6 + \mathcal{O}(k^{8})\biggr\}
~+~2\epsilon^{2} \biggl\{ 1 ~+~\frac{1}{2^5} ~k^4 
~+~\frac{1}{2^5} ~ k^6 + \mathcal{O}(k^{8}) \biggr\} \, .
</math>
  </td>
</tr>
</table>

Now, given that,
<div align="center">
<math>~k_0^2 = 2\epsilon (1+\epsilon)^{-1} \, ,</math>
</div>
we furthermore have,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1 - \epsilon^2) \biggl[ \frac{2^2}{\pi^2} \biggr] C_0(\epsilon)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~  
\biggl\{ \biggl( \frac{1}{2^2} \biggr)[2\epsilon (1+\epsilon)^{-1}]^2 
+ \biggl( \frac{3}{2^4} \biggr) [2\epsilon (1+\epsilon)^{-1}]^3 
+ \biggl( \frac{3 \cdot 13}{2^8}  \biggr) [2\epsilon (1+\epsilon)^{-1}]^4 + \mathcal{O}(\epsilon^{5})
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~+~ \epsilon \biggl\{ [2\epsilon (1+\epsilon)^{-1}] ~+~ \frac{3}{2^3} ~ [2\epsilon (1+\epsilon)^{-1}]^2 ~+~\frac{3^2}{2^5} ~ [2\epsilon (1+\epsilon)^{-1}]^3 + \mathcal{O}(\epsilon^{4})\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math> 
~+~2\epsilon^{2} \biggl\{ 1 ~+~\frac{1}{2^5} ~[2\epsilon (1+\epsilon)^{-1}]^2 ~+~\frac{1}{2^5} ~ [2\epsilon (1+\epsilon)^{-1}]^3 + \mathcal{O}(\epsilon^{4}) \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~  \biggl\{
\epsilon^2 (1+\epsilon)^{-2} 
+ \biggl( \frac{3}{2} \biggr) \epsilon^3 (1+\epsilon)^{-3} 
+ \biggl( \frac{3 \cdot 13}{2^4}  \biggr) \epsilon^4 (1+\epsilon)^{-4} 
\biggr\}
~+~ \biggl\{ 2\epsilon^2 (1+\epsilon)^{-1} ~+~ \frac{3}{2} ~ \epsilon^3 (1+\epsilon)^{-2} ~+~\frac{3^2}{2^2} ~ \epsilon^4 (1+\epsilon)^{-3} \biggr\}
~+~ \biggl\{ 2\epsilon^{2} ~+~\frac{1}{2^2} ~\epsilon^4 (1+\epsilon)^{-2}  \biggr\} 
+ \mathcal{O}(\epsilon^{5}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\epsilon^2 \biggl\{ 2 + 2(1+\epsilon)^{-1} - (1+\epsilon)^{-2} \biggr\}
~+~ \epsilon^3 \biggl\{ \frac{3}{2} (1+\epsilon)^{-2} - \biggl( \frac{3}{2} \biggr) (1+\epsilon)^{-3}\biggr\}
~+~ \epsilon^4 \biggl\{ \frac{1}{2^2} (1+\epsilon)^{-2} + \frac{3^2}{2^2} (1+\epsilon)^{-3} - \biggl( \frac{3 \cdot 13}{2^4}  \biggr) (1+\epsilon)^{-4}\biggr\}
+ \mathcal{O}(\epsilon^{5}) 
</math>
  </td>
</tr>
</table>

Recalling that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~D_0 \equiv \frac{2^{3 / 2}}{3\pi^2} \biggl[\frac{\sinh^3\eta_0}{\cosh\eta_0}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2^{3 / 2}}{3\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr] \, ,</math>
  </td>
</tr>
</table>
to lowest order we see that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2^{3 / 2} D_0 \cdot C_0</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{2^{3}}{3\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr] \epsilon^2 \biggl\{ 2 + 2(1+\epsilon)^{-1} - (1+\epsilon)^{-2} \biggr\} (1-\epsilon^2)^{-1} \biggl[ \frac{\pi^2}{2^2}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{2}{3} \biggl\{ 3 \biggr\} = 2 \, ,</math>
  </td>
</tr>
</table>
as [[#Speculation|speculated above]].

What about the next order?  Using the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial expansion]] we have,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1 - \epsilon^2) \biggl[ \frac{2^2}{\pi^2} \biggr] C_0(\epsilon)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\epsilon^2 \biggl\{ 2 + 2(1-\epsilon + \epsilon^2) - (1-2\epsilon +3\epsilon^2) \biggr\}
~+~ \epsilon^3 \biggl\{ \frac{3}{2} (1-2\epsilon) - \biggl( \frac{3}{2} \biggr) (1-3\epsilon)\biggr\}
~+~ \epsilon^4 \biggl\{ \frac{1}{2^2}  + \frac{3^2}{2^2}  - \biggl( \frac{3 \cdot 13}{2^4}  \biggr) \biggr\}
+ \mathcal{O}(\epsilon^{5}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\epsilon^2 \biggl\{ 3 + 2(-\epsilon + \epsilon^2) - (-2\epsilon +3\epsilon^2) \biggr\}
~+~ \epsilon^3 \biggl\{ \frac{3}{2} (-2\epsilon) - \biggl( \frac{3}{2} \biggr) (-3\epsilon)\biggr\}
~+~ \frac{1}{2^4}\epsilon^4 \biggl\{ 40  - 3 \cdot 13  \biggr\}
+ \mathcal{O}(\epsilon^{5}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\epsilon^2 \biggl\{ 3  -\epsilon^2 \biggr\}
~+~ \epsilon^3 \biggl\{ \frac{3}{2} (\epsilon) \biggr\}
~+~ \frac{1}{2^4}\epsilon^4 
+ \mathcal{O}(\epsilon^{5}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3\epsilon^2   -\epsilon^4 
~+~ \frac{3}{2} \epsilon^4 
~+~ \frac{1}{2^4}\epsilon^4 
+ \mathcal{O}(\epsilon^{5}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3\epsilon^2   \biggl[ 1  
~+~ \frac{3}{2^4}  \epsilon^2\biggr]
+ \mathcal{O}(\epsilon^{5}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~2^{3 / 2} D_0 \cdot C_0(\epsilon)</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~2^{3 / 2}D_0
\cdot (3\epsilon^2 )  \biggl[ 1  
~+~ \frac{3}{2^4}  \epsilon^2\biggr](1 - \epsilon^2)^{-1} \biggl[ \frac{\pi^2}{2^2} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~2^{3 / 2}\biggl\{ \frac{2^{3 / 2}}{3\pi^2} \biggl[ \epsilon^{-2} (1 + \epsilon^2)^{3 / 2} \biggr]\biggr\}
\cdot (3\epsilon^2 )  \biggl[ 1  
~+~ \frac{3}{2^4}  \epsilon^2\biggr](1 - \epsilon^2)^{-1} \biggl[ \frac{\pi^2}{2^2} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2 \biggl[ 1  ~+~ \frac{3}{2^4}  \epsilon^2\biggr](1 - \epsilon^2)^{-1}  (1 + \epsilon^2)^{3 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2 \biggl[ 1  ~+~ \frac{3}{2^4}  \epsilon^2\biggr](1 + \epsilon^2)  (1 + \tfrac{3}{2}\epsilon^2) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2 \biggl[ 1  ~+~ \frac{3}{2^4}  \epsilon^2\biggr]\biggl[1 + \frac{5}{2}\epsilon^2\biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
2  ~+~ \frac{3 + 2^3\cdot 5}{2^3}  \epsilon^2 \, .
</math>
  </td>
</tr>
</table>

This does ''not'' match the <math>~\epsilon^2</math> term, as [[#Speculation|speculated above]].  But keep in mind that Wong's coordinate system is slightly shifted from the one used by Dyson, so perhaps this difference can be entirely reconciled via the proper coordinate-system mapping.