Editing Appendix/Ramblings/Dyson1893Part1

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=Dyson (1893a) Part I:  Some Details=
This chapter provides some derivation details relevant to our [[Apps/DysonPotential#Dyson_.281893.29|accompanying discussion]] of Dyson's analysis of the gravitational potential exterior to an ''anchor ring.''

==Overview==
In his pioneering work, [http://adsabs.harvard.edu/abs/1893RSPTA.184...43D F. W. Dyson (1893a, Philosophical Transactions of the Royal Society of London. A., 184, 43 - 95)] and [http://adsabs.harvard.edu/abs/1893RSPTA.184.1041D (1893b, Philosophical Transactions of the Royal Society of London. A., 184, 1041 - 1106)] used analytic techniques to determine the approximate equilibrium structure of axisymmetric, uniformly rotating, incompressible tori.  [http://adsabs.harvard.edu/abs/1974ApJ...190..675W C.-Y. Wong (1974, ApJ, 190, 675 - 694)] extended Dyson's work, using numerical techniques to obtain more accurate &#8212; but still approximate &#8212; equilibrium structures for incompressible tori having solid body rotation.  Since then, [http://adsabs.harvard.edu/abs/1981PThPh..65.1870E Y. Eriguchi &amp; D. Sugimoto (1981, Progress of Theoretical Physics, 65, 1870 - 1875)] and [http://adsabs.harvard.edu/abs/1988ApJS...66..315H I. Hachisu, J. E. Tohline &amp; Y. Eriguchi (1987, ApJ, 323, 592 - 613)] have mapped out the full sequence of Dyson-Wong tori, beginning from a bifurcation point on the Maclaurin spheroid sequence.

==External Potential in Terms of Angle &psi;==

===Step 1===
On p. 59, at the end of &sect;6 of [http://adsabs.harvard.edu/abs/1893RSPTA.184...43D Dyson (1893a)], we find the following expression for the potential at point "P", anywhere exterior to an [http://www.mathematicsdictionary.com/english/vmd/full/t/torusanchorring.htm anchor ring]:

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

<tr>
  <td align="right">
<math>~\frac{\pi V(r,\theta)}{M}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathfrak{I}(r,\theta,c)
~+~
\frac{a^2}{2^3} ~\frac{1}{c} \cdot \frac{d}{dc} \biggl[ \mathfrak{I}(r,\theta,c)\biggr]
~-~
\frac{a^4}{2^6\cdot 3} ~\frac{1}{c} \cdot \frac{d}{dc} \biggl\{ \frac{1}{c} \cdot \frac{d}{dc} \biggl[ \mathfrak{I}(r,\theta,c)\biggr]\biggr\}
~+~\cdots 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~+~(-1)^{n+1} \frac{2a^{2n}}{2n+2} \biggl[ \frac{1\cdot 3\cdot 5 \cdots (2n-3)}{2^2\cdot 4^2\cdot 6^2\cdots(2n)^2} \biggr]
\biggl( \frac{1}{c}\cdot \frac{d}{dc}\biggr)^n \biggl[ \mathfrak{I}(r,\theta,c)\biggr] ~+~ \cdots
</math>
  </td>
</tr>
</table>
where (see beginning of &sect;8 on p. 61),
<table border="0" cellpadding="10" align="right" width="35%"><tr><td align="center">
<table border="0" cellpadding="5" align="right" width="100%">
<tr>
<td align="center">
[[File:DysonTorusIllustration03.png|300px|center|Anchor Ring Schematic]]<br />
'''Caption:''' [http://www.mathematicsdictionary.com/english/vmd/full/t/torusanchorring.htm Anchor ring] schematic, adapted from figure near the top of &sect;2 (on p. 47) of [http://adsabs.harvard.edu/abs/1893RSPTA.184...43D Dyson (1893a)]
</td>
</tr>
</table>
</td></tr></table>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{I}(r,\theta,c)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\int_0^\pi d\phi \biggl[r^2 - 2cr\sin\theta \cos\phi  +c^2\biggr]^{-1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\int_0^{\pi/2} d\phi \biggl[ R_1^2 - (R_1^2-R^2)\sin^2\phi \biggr]^{-1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{R_1}\int_0^{\pi/2} d\phi \biggl[ 1 - \biggl( \frac{R_1^2-R^2}{R_1^2}\biggr) \sin^2\phi \biggr]^{-1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2K(k)}{R_1} \, ,
</math>
  </td>
</tr>
</table>
and, where furthermore,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~K(k)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\pi/2} d\phi \biggl[1 - k^2\sin^2\phi  \bigg]^{-1 / 2}
</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>~k</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{R_1^2-R^2}{R_1^2} \biggr]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>

===Step 2===
Taking a queue from our [[2DStructure/ToroidalGreenFunction#Basic_Elements_of_a_Toroidal_Coordinate_System|accompanying discussion of toroidal coordinates]], if we adopt the variable notation,
<div align="center">
<math>~\eta \equiv \ln\biggl(\frac{R_1}{R}\biggr) \, ,</math>
</div>
then we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\cosh\eta = \frac{1}{2}\biggl[e^\eta + e^{-\eta}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{R^2 + R_1^2}{2RR_1} \, ,</math>
  </td>
</tr>
</table>
which implies that,
<!--
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tanh^2\biggl(\frac{\eta}{2}\biggr) = \frac{\cosh\eta - 1}{\cosh\eta+1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{R_1 - R}{R_1 + R}\biggr]^2 \, ,</math>
  </td>
</tr>
</table>
and,
-->
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{2}{\coth\eta +1} \biggr]^{1 / 2} = [1 - e^{-2\eta}]^{1 / 2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 1 - \biggl(\frac{R}{R_1}\biggr)^2 \biggr]^{1 / 2} = k \, .</math>
  </td>
</tr>
</table>
Now, if we employ the [https://dlmf.nist.gov/19.8#ii ''Descending Landen Transformation'' for the complete elliptic integral of the first kind], we can make the substitution,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~K(k)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
(1 + \mu)K(\mu) \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; where, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\mu</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ 
\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}} \, .
</math>
  </td>
</tr>
</table>
But notice that, <math>~\sqrt{1-k^2} = e^{-\eta}</math>, in which case,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mu </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1-e^{-\eta}}{1+e^{-\eta}}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1-R/R_1}{1+R/R_1}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{R_1-R}{R_1+R} \, .
</math>
  </td>
</tr>
</table>
Hence, we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{I}(r,\theta,c) = \frac{2K(k)}{R_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2}{R_1} \biggl[(1+\mu)K(\mu) \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2K(\mu)}{R_1} \biggl[1+\frac{R_1-R}{R_1+R}  \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4K(\mu)}{R_1+R} \, .</math>
  </td>
</tr>
</table>

This is the expression for <math>~\mathfrak{I}(r,\theta,c) </math> that was adopted by Dyson at the beginning of his &sect;8.  

===Step 3===

Subsequently, Dyson was able to obtain analytic expressions for successive derivatives of the function, <math>~\mathfrak{I}(r,\theta,c) </math>, by first demonstrating that
[[File:CommentButton02.png|right|100px|Comment by J. E. Tohline on 17 September 2018:  In the middle of p. 61 of Dyson(1893a), there appears to be a typographical error in the expression for the derivative of R<sub>1</sub> with respect to ''c''; as we have indicated here, the numerator should be 4cR<sub>1</sub> instead of 4cR.]]
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dR}{dc}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4c^2 + R^2 - R_1^2}{4cR} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{dR_1}{dc}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4c^2 + R_1^2 - R^2}{4cR_1} \, ,</math> &nbsp; &nbsp; &nbsp; and,
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{d\mu}{dc}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\mu}{c} \cos\psi \, ,</math>
  </td>
</tr>
</table>
where &#8212; as shown above in the [http://www.mathematicsdictionary.com/english/vmd/full/t/torusanchorring.htm Anchor ring] schematic &#8212; <math>~\psi</math> is the angle between <math>~R</math> and <math>~R_1</math> for which (according to the law of cosines),
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\cos\psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{R^2 + R_1^2 - 4c^2}{2RR_1} \, .</math>
  </td>
</tr>
</table>
It will be useful for us to note that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d(\cos\psi)}{dc}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{dc}\biggl[
\frac{R^2 + R_1^2 - 4c^2}{2RR_1}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2RR_1}\frac{d}{dc}\biggl[
R^2 + R_1^2 - 4c^2
\biggr]
~+~(R^2 + R_1^2 - 4c^2)
\frac{d}{dc}\biggl[
\frac{1}{2RR_1}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2RR_1}\biggl[2R\frac{dR}{dc} + 2R_1\frac{dR_1}{dc} - 8c
\biggr]
~+~(R^2 + R_1^2 - 4c^2)\biggl[
- \frac{1}{2R^2R_1} \frac{dR}{dc} - \frac{1}{2RR_1^2}\frac{dR_1}{dc}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4cRR_1}\biggl[ 4c^2 + R^2 - R_1^2 + 4c^2 + R_1^2 - R^2 - 2^4c^2
\biggr]
~-~(R^2 + R_1^2 - 4c^2)\biggl\{
\biggl[ \frac{4c^2 + R^2 - R_1^2}{8cR^3R_1}\biggr] + \biggl[ \frac{4c^2 + R_1^2 - R^2}{8cR_1^3 R} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{2c}{RR_1}
~-~\frac{\cos\psi}{4cR^2 R_1^2} \biggl[4c^2(R_1^2 + R^2) + 2R_1^2 R^2 - R_1^4 - R^4 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\cos\psi}{4cR^2 R_1^2} \biggl[
R_1^4 + R^4- 4c^2(R_1^2 + R^2) - 2R_1^2 R^2 
\biggr]
-~\frac{2c}{RR_1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4cR^2 R_1^2}\biggl\{
\cos\psi 
[R_1^4 + R^4 - 2R_1^2 R^2 - 4c^2(R_1^2 + R^2) ]
-~8c^2R R_1 \biggr\}
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~[R_1^4 + R^4 - 2R_1^2 R^2 - 4c^2(R_1^2 + R^2)]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(R_1^2 - R^2)^2 
- 4c^2(R_1^2 + R^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[(R_1+R)(R_1-R)]^2 
- 4c^2(R_1^2 + R^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(R_1^2 + 2R_1 R +R^2)(R_1-R)^2 
- 4c^2(R_1^2 + R^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2R_1 R(R_1-R)^2 
+(R_1^2 + R^2)(R_1-R)^2 
- 4c^2(R_1^2 + R^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2R_1 R(R_1-R)^2 
+(R_1^2 + R^2)[R_1^2 -2R_1 R + R^2 - 4c^2]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2R_1 R[ (R_1-R)^2 
+(R_1^2 + R^2)(\cos\psi - 1)]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2R_1 R[  (R_1^2 + R^2)\cos\psi - 2R_1 R]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2R_1 R)^2 \biggl[ \cos^2\psi  + \biggl(\frac{2c^2 }{R_1 R}\biggr)\cos\psi - 1 \biggr]
</math>
  </td>
</tr>
</table>
Hence,

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

<tr>
  <td align="right">
<math>~
\frac{d(\cos\psi)}{dc}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(2R_1 R)^2 }{4cR^2 R_1^2}\biggl\{
\cos\psi 
\biggl[ \cos^2\psi  + \biggl(\frac{2c^2 }{R_1 R}\biggr)\cos\psi - 1 \biggr]
-~\frac{8c^2R R_1}{(2R_1 R)^2 } \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{c}\biggl\{
\cos\psi 
\biggl[ \cos^2\psi  + \biggl(\frac{2c^2 }{R_1 R}\biggr)\cos\psi - 1 \biggr]
-~\frac{2c^2}{R_1 R } \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~
\Rightarrow ~~~ 2c\cdot \frac{d(\cos\psi)}{dc}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\cos\psi \biggl[ \cos^2\psi  + \biggl(\frac{2c^2 }{R_1 R}\biggr)\cos\psi - 1 \biggr]
-~\frac{4c^2}{R_1 R } 
</math>
  </td>
</tr>
</table>

===Step 4===

Then, drawing upon known expressions for the derivatives of elliptic integrals, as are now tidily catalogued online in [http://dlmf.nist.gov/ NIST's ''Digital Library of Mathematical Functions''], Dyson showed that,


<table border="1" align="center" cellpadding="8" width="70%">
<tr>
  <th align="center" bgcolor="yellow">
LaTeX mathematical expressions cut-and-pasted directly from
<br />
NIST's ''Digital Library of Mathematical Functions''
  </th>
</tr>
<tr>
  <td align="left">
According to [http://dlmf.nist.gov/19.4.i &sect;19.4 of NIST's ''Digital Library of Mathematical Functions''],
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\mathrm{d}K\left(k\right)}{\mathrm{d}k}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{E\left(k\right)-{k^{\prime}}^{2}K\left(k\right)}{k{k^{\prime}}^{2}},
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\mathrm{d}(E\left(k\right)-{k^{\prime}}^{2}K\left(k\right))}{\mathrm{d}k}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
kK\left(k\right),
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\mathrm{d}E\left(k\right)}{\mathrm{d}k}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{E\left(k\right)-K\left(k\right)}{k},
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\mathrm{d}(E\left(k\right)-K\left(k\right))}{\mathrm{d}k}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{kE\left(k\right)}{{k^{\prime}}^{2}},
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{{\mathrm{d}}^{2}E\left(k\right)}{{\mathrm{d}k}^{2}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{1}{k}\frac{\mathrm{d}K\left(k\right)}{\mathrm{d}k}=\frac{{k^{\prime}}^{2}K\left(k\right)-E\left(k\right)}{k^{2}{k^{\prime}}^{2}} \, ,
</math>
  </td>
</tr>
</table>
where,
<div align="center">
<math>~k^{\prime} \equiv \sqrt{1 - k^2} \, .</math>
</div>
  </td>
</tr>
</table>



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

<tr>
  <td align="right">
<math>~\frac{d\mathfrak{I}(r,\theta,c)}{dc} = \frac{d}{dc}\biggl[\frac{4K(\mu)}{R_1+R}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
4K(\mu) \frac{d}{dc}\biggl[ R_1 + R \biggr]^{-1}
+
\frac{4}{(R_1+R)} \biggl[ \frac{E\left(\mu\right)-{\mu^{\prime}}^{2}K\left(\mu\right)}{\mu{\mu^{\prime}}^{2}} \biggr]\frac{d\mu}{dc}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
-~\frac{4K(\mu)}{(R_1+R)^2} \biggl[ \frac{dR_1}{dc} + \frac{dR}{dc} \biggr]
~-~
\frac{4}{(R_1+R)} \biggl[ \frac{K\left(\mu\right)}{\mu} \biggr]\frac{\mu}{c}\cos\psi
+
\frac{4}{(R_1+R)} \biggl[ \frac{E\left(\mu\right)}{\mu{\mu^{\prime}}^{2}} \biggr]\frac{\mu}{c}\cos\psi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
-~\frac{4K(\mu)}{(R_1+R)^2} \biggl[ \frac{4c^2 + R_1^2 - R^2}{4cR_1} + \frac{4c^2 + R^2 - R_1^2}{4cR} \biggr]
~-~
\biggl[ \frac{4K\left(\mu\right)}{c(R_1+R)} \biggr] \cos\psi
+
\frac{4E\left(\mu\right)}{c(R_1+R)} \biggl[ \frac{(R_1+R)^2}{4RR_1} \biggr]\cos\psi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{cRR_1}\biggr] \cos\psi
~-~
\biggl[ \frac{4K\left(\mu\right)}{c(R_1+R)}\biggr] \cos\psi
~-~
\frac{4K(\mu)}{4cR_1 R (R_1+R)^2} \biggl[ R(4c^2 + R_1^2 - R^2) + R_1(4c^2 + R^2 - R_1^2) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{cRR_1}\biggr] \cos\psi
~-~
\biggl[ \frac{4K\left(\mu\right)}{c(R_1+R)}\biggr] \cos\psi
~-~
\frac{K(\mu)}{cR_1 R (R_1+R)^2}\biggl[(4c^2 + R_1R)(R_1+R) - (R_1^3 + R^3) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{cRR_1}\biggr] \cos\psi
~-~
\biggl[ \frac{4K\left(\mu\right)}{c(R_1+R)}\biggr] \cos\psi
~-~
\frac{K(\mu)}{cR_1 R (R_1+R)}\biggl[(4c^2 + R_1R) - (R_1^2 + R^2 - R_1R) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{cRR_1}\biggr] \cos\psi
~-~
\biggl[ \frac{4K\left(\mu\right)}{c(R_1+R)}\biggr] \cos\psi
~-~
\frac{K(\mu)}{cR_1 R (R_1+R)}\biggl[4c^2  - (R_1 - R)^2 \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{cRR_1}\biggr] \cos\psi
~-~
\biggl[ \frac{4K\left(\mu\right)}{c(R_1+R)}\biggr] \cos\psi
~-~
\frac{K(\mu)}{c (R_1+R)}\biggl[2(1-\cos\psi) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{cRR_1}\biggr] \cos\psi~-~\frac{2K(\mu)}{c (R_1+R)}(1+\cos\psi) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{1}{c}\cdot \frac{d\mathfrak{I}(r,\theta,c)}{dc} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{c^2}\biggl\{\biggl[ \frac{E(\mu)(R+R_1)}{RR_1} \biggr] \cos\psi ~-~\biggl[ \frac{4K(\mu)}{R+R_1} \biggr] \cos^2\frac{\psi}{2} \biggr\} \, .
</math>
  </td>
</tr>
</table>
This expression appears at the top of Dyson's p. 62.

===Step 5===
Differentiating a second time gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{dc}\biggl[\frac{1}{c}\cdot \frac{d\mathfrak{I}(r,\theta,c)}{dc}\biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{d}{dc}\biggl\{
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{c^2RR_1}\biggr] \cos\psi~-~\frac{2K(\mu)}{c^2 (R_1+R)}(1+\cos\psi) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
~-~\frac{2}{c^3}\biggl\{
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \cos\psi~-~\frac{2K(\mu)}{ (R_1+R)}(1+\cos\psi) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\frac{1}{c^2}\cdot \frac{d}{dc}\biggl\{
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \cos\psi~-~\frac{2K(\mu)}{ (R_1+R)}(1+\cos\psi) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
~-~\frac{2}{c^3}\biggl\{
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \cos\psi~-~\frac{2K(\mu)}{ (R_1+R)}(1+\cos\psi) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\frac{1}{c^2}\cdot \frac{d}{dc}\biggl\{
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}~-~\frac{2K(\mu)}{ (R_1+R)}\biggr] 
\cos\psi\biggr\}
~-~\frac{1}{c^2}\cdot \frac{d}{dc}\biggl[ \frac{2K(\mu)}{ (R_1+R)} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
~-~\frac{2}{c^3}\biggl\{
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \cos\psi~-~\frac{2K(\mu)}{ (R_1+R)}(1+\cos\psi) 
\biggr\}
~+~\frac{1}{c^2}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}~-~\frac{2K(\mu)}{ (R_1+R)}\biggr] \frac{d(\cos\psi)}{dc}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\frac{\cos\psi}{c^2}\cdot \frac{d}{dc}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] 
~-~\frac{(1+\cos\psi)}{2c^2}\cdot \frac{d}{dc}\biggl[\frac{4K(\mu)}{ (R_1+R)}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{c^3}\biggl[ \frac{4K(\mu)}{ (R_1+R)}\biggr]\biggl[(1+\cos\psi) -  \biggl(\frac{c}{2}\biggr) \frac{d(\cos\psi)}{dc}\biggr]
~-~\frac{2}{c^3}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl[ \cos\psi -  \biggl(\frac{c}{2}\biggr) \frac{d(\cos\psi)}{dc}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\frac{\cos\psi}{c^2}\cdot \frac{d}{dc}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] 
~-~\frac{(1+\cos\psi)}{2c^3} \biggl\{\biggl[ \frac{E(\mu)(R+R_1)}{RR_1} \biggr] \cos\psi ~-~\biggl[ \frac{4K(\mu)}{R+R_1} \biggr] \cos^2\frac{\psi}{2} \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{c^3}\biggl[ \frac{4K(\mu)}{ (R_1+R)}\biggr]\biggl[\frac{(1+\cos\psi)^2}{2^2}  + (1+\cos\psi) -  \biggl(\frac{c}{2}\biggr) \frac{d(\cos\psi)}{dc}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~\frac{2}{c^3}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl[\frac{(\cos\psi+\cos^2\psi)}{4} ~+~ \cos\psi ~-~  \biggl(\frac{c}{2}\biggr) \frac{d(\cos\psi)}{dc}\biggr]
~+~\frac{\cos\psi}{c^2}\cdot \frac{d}{dc}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{4c^3}\biggl[ \frac{4K(\mu)}{ (R_1+R)}\biggr]\biggl[5+6\cos\psi + \cos^2\psi  -  2c \cdot \frac{d(\cos\psi)}{dc}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~\frac{1}{2c^3}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl[5\cos\psi+\cos^2\psi ~-~  2c \cdot \frac{d(\cos\psi)}{dc}\biggr]
~+~\frac{\cos\psi}{c^2}\cdot \frac{d}{dc}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~
\frac{d}{dc}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{(R_1+R)}{RR_1}\biggr]\frac{dE\left(\mu\right)}{d\mu} \cdot \frac{d\mu}{dc}
~+~
E\left(\mu\right)\frac{d}{dc}\biggl[ \frac{(R_1+R)}{RR_1}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{(R_1+R)}{RR_1}\biggr]\biggl[\frac{E(\mu) - K(\mu)}{c} \biggr] \cos\psi
~+~
E\left(\mu\right)\biggl\{ 
\frac{1}{RR_1}\biggl[\frac{dR_1}{dc} + \frac{dR}{dc}\biggr]
~-~ (R_1+R)\biggl[\frac{1}{R R_1^2} \frac{dR_1}{dc}  + \frac{1}{R^2 R_1} \frac{dR}{dc}\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{(R_1+R)}{RR_1}\biggr]\biggl[\frac{E(\mu) - K(\mu)}{c} \biggr] \cos\psi
~-~
E\left(\mu\right)\biggl\{ 
\biggl[\frac{1}{R^2} \frac{dR}{dc}\biggr]
~+~ \biggl[\frac{1}{R_1^2} \frac{dR_1}{dc}  \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~K(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
+E(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
~-~
E\left(\mu\right)\biggl\{ 
\biggl[\frac{4c^2 + R^2-R_1^2}{4cR^3}\biggr]
~+~ \biggl[ \frac{4c^2+R_1^2-R^2}{4cR_1^3}  \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~K(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
+E(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
~+~
E\left(\mu\right)\biggl\{ 
\biggl[\frac{R_1^2 + R^2-4c^2 }{4cR^3}\biggr]
~-~\biggl[\frac{R^2 }{2cR^3}\biggr]
~+~ \biggl[ \frac{R_1^2 + R^2-4c^2}{4cR_1^3}  \biggr]
~-~ \biggl[ \frac{R_1^2}{2cR_1^3}  \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~K(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
+E(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
~+~
\frac{E\left(\mu\right)}{2cR_1^2R^2} \biggl\{ 
R_1^3 \cos\psi ~-~R R_1^2 ~+~ R^3\cos\psi ~-~ R_1 R^2
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~K(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
+E(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
~+~
\frac{E\left(\mu\right)}{2cR_1^2R^2} \biggl\{ 
(R_1^3 + R^3) \cos\psi ~-~R R_1(R_1 + R)
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~K(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
+E(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
~+~
\frac{E\left(\mu\right)(R_1+R)}{2cR_1^2R^2} \biggl\{ 
(R_1^2 + R^2 - 4c^2) \cos\psi + 4c^2\cos\psi~-~R R_1(1+\cos\psi)
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~K(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
+E(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
~+~
\frac{E\left(\mu\right)(R_1+R)}{cR_1 R} \biggl\{ 
\cos^2\psi
~+~ \frac{2c^2\cos\psi }{R_1R}
~-~\frac{1}{2}(1+\cos\psi)
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~K(\mu)\biggl[ \frac{(R_1+R)}{cRR_1}\biggr] \cos\psi
~+~
\frac{E\left(\mu\right)(R_1+R)}{cR_1 R} \biggl[ 
\cos^2\psi
~+~\frac{1}{2}(\cos\psi - 1)
~+~ \frac{2c^2\cos\psi }{R_1R}
\biggr] \, .
</math>
  </td>
</tr>
</table>

We therefore have,

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

<tr>
  <td align="right">
<math>~\frac{1}{c} \cdot \frac{d}{dc}\biggl[\frac{1}{c}\cdot \frac{d\mathfrak{I}(r,\theta,c)}{dc}\biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{4c^4}\biggl[ \frac{4K(\mu)}{ (R_1+R)}\biggr]\biggl\{ 
5+6\cos\psi + \cos^2\psi  
~-~2\cos\psi \biggl[ \cos^2\psi  + \biggl(\frac{2c^2 }{R_1 R}\biggr)\cos\psi - 1 \biggr]~+~\frac{4c^2}{R_1 R }
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~\frac{1}{2c^4}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl\{
5\cos\psi+\cos^2\psi 
~-~2\cos\psi \biggl[ \cos^2\psi  + \biggl(\frac{2c^2 }{R_1 R}\biggr)\cos\psi - 1 \biggr]~+~\frac{4c^2}{R_1 R }
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\frac{\cos\psi}{c^4} \biggl\{ 
\frac{E\left(\mu\right)(R_1+R)}{R_1 R} \biggl[ 
\cos^2\psi
~+~\frac{1}{2}(\cos\psi - 1)
~+~ \frac{2c^2\cos\psi }{R_1R}
\biggr] 
~-~K(\mu)\biggl[ \frac{(R_1+R)}{RR_1}\biggr] \cos\psi
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{c^4}\biggl[ \frac{K(\mu)}{ (R_1+R)}\biggr]\biggl\{ 
5+6\cos\psi + \cos^2\psi  
~-~2\cos\psi \biggl[ \cos^2\psi  + \biggl(\frac{2c^2 }{R_1 R}\biggr)\cos\psi - 1 \biggr]~+~\frac{4c^2}{R_1 R }
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~\frac{1}{2c^4}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl\{
5\cos\psi+\cos^2\psi 
~-~2\cos\psi \biggl[ \cos^2\psi  + \biggl(\frac{2c^2 }{R_1 R}\biggr)\cos\psi - 1 \biggr]~+~\frac{4c^2}{R_1 R }
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~\frac{1}{2c^4} \biggl[ 
\frac{E\left(\mu\right)(R_1+R)}{R_1 R}\biggr] \biggl[ 
~-~2\cos^3\psi
~-~\cos^2\psi ~+~ \cos\psi
~-~ \frac{4c^2\cos^2\psi }{R_1R}
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\frac{1}{c^4}\biggl[\frac{K(\mu)}{(R_1 + R)}\biggr] \biggl[~-~ \frac{(R_1+R)^2\cos^2\psi}{RR_1}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{c^4}\biggl[ \frac{K(\mu)}{ (R_1+R)}\biggr]\biggl\{ 
5 ~+~ 8\cos\psi ~+~ \cos^2\psi  
~-~2\cos^3\psi  
~+~\frac{4c^2}{R_1 R }
~-~ \biggl[ \frac{R_1^2 + R^2 - 4c^2 }{R_1 R} \biggr]\cos^2\psi
~-~ \biggl[ \frac{2R_1R }{R_1 R} \biggr]\cos^2\psi
~-~ \biggl[ \frac{8c^2 }{R_1 R} \biggr]\cos^2\psi
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~\frac{1}{2c^4}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl\{
8\cos\psi
~-~4\cos^3\psi  ~+~ \biggl(\frac{4c^2 }{R_1 R}\biggr)(1-2\cos^2\psi) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{c^4}\biggl[ \frac{K(\mu)}{ (R_1+R)}\biggr]\biggl\{ 
5 ~+~ 8\cos\psi ~-~ \cos^2\psi  
~-~4\cos^3\psi  
~+~\frac{4c^2}{R_1 R }(1-2\cos^2\psi)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~\frac{1}{c^4}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl\{
4\cos\psi
~-~2\cos^3\psi  ~+~ \biggl(\frac{2c^2 }{R_1 R}\biggr)(1-2\cos^2\psi) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{c^4}\biggl[ \frac{K(\mu)}{ (R_1+R)}\biggr]\biggl\{ 
5 ~+~ 8\cos\psi ~-~ \cos^2\psi  
~-~4\cos^3\psi  
~-~\biggl( \frac{4c^2}{R_1 R } \biggr) \cos 2\psi
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\frac{1}{c^4}\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl\{
~-~4\cos\psi
~+~2\cos^3\psi  ~+~ \biggl(\frac{2c^2 }{R_1 R}\biggr)\cos 2\psi 
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
which exactly matches the second equation from the top of p. 62 in [http://adsabs.harvard.edu/abs/1893RSPTA.184...43D Dyson (1893a)].

===Step 6 (Summary)===

In summary, then, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\pi V(r,\theta)}{M}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathfrak{I}(r,\theta,c)
~+~
\frac{a^2}{2^3} ~\frac{1}{c} \cdot \frac{d}{dc} \biggl[ \mathfrak{I}(r,\theta,c)\biggr]
~-~
\frac{a^4}{2^6\cdot 3} ~\frac{1}{c} \cdot \frac{d}{dc} \biggl\{ \frac{1}{c} \cdot \frac{d}{dc} \biggl[ \mathfrak{I}(r,\theta,c)\biggr]\biggr\}
~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4K(\mu)}{R_1+R} 
~+~
\frac{1}{2^3} ~\frac{a^2}{c^2}\biggl\{\biggl[ \frac{E(\mu)(R+R_1)}{RR_1} \biggr] \cos\psi ~-~\biggl[ \frac{4K(\mu)}{R+R_1} \biggr] \cos^2\frac{\psi}{2} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~ \frac{1}{2^6\cdot 3} \frac{a^4}{c^4} \biggl\{
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl[
~-~4\cos\psi
~+~2\cos^3\psi  ~+~ \biggl(\frac{2c^2 }{R_1 R}\biggr)\cos 2\psi 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~
\biggl[ \frac{K(\mu)}{ (R_1+R)}\biggr]\biggl[ 
5 ~+~ 8\cos\psi ~-~ \cos^2\psi  
~-~4\cos^3\psi  
~-~\biggl( \frac{4c^2}{R_1 R } \biggr) \cos 2\psi
\biggr]
\biggr\}
~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \, .
</math>
  </td>
</tr>
</table>

In preparation for a [[Apps/Wong1973Potential#Compare_With_Dyson_.281893.29|comparison between Dyson's expression and the expression derived by Wong (1973)]], let's regroup terms according to the various trigonometric functions.  Keep in mind that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\cos^2 \frac{\psi}{2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2}(1 + \cos\psi) \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\cos^2 \psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2}(1 + \cos2\psi) \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\cos^3 \psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4}(3\cos\psi + \cos3\psi) \, .</math>
  </td>
</tr>
</table>

Hence, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\pi V(r,\theta)}{M}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4K(\mu)}{R_1+R} 
~+~
\frac{1}{2^3} ~\frac{a^2}{c^2}\biggl\{\biggl[ \frac{E(\mu)(R+R_1)}{RR_1} \biggr] \cos\psi 
~-~\biggl[ \frac{2K(\mu)}{R+R_1} \biggr] (1+\cos\psi) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~ \frac{1}{2^6\cdot 3} \frac{a^4}{c^4} \biggl\{
\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl[
~-~4\cos\psi
~+~\frac{1}{2}(3\cos\psi + \cos3\psi)  ~+~ \biggl(\frac{2c^2 }{R_1 R}\biggr)\cos 2\psi 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~
\biggl[ \frac{K(\mu)}{ (R_1+R)}\biggr]\biggl[ 
5 ~+~ 8\cos\psi ~-~ \frac{1}{2}(1 + \cos2\psi)  
~-~(3\cos\psi + \cos3\psi)  
~-~\biggl( \frac{4c^2}{R_1 R } \biggr) \cos 2\psi
\biggr]
\biggr\}
~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4K(\mu)}{R_1+R} 
~-~\frac{1}{2^3} ~\frac{a^2}{c^2}\biggl[ \frac{2K(\mu)}{R+R_1} \biggr] 
~+~
\frac{1}{2^3} ~\frac{a^2}{c^2}\biggl[ \frac{E(\mu)(R+R_1)}{RR_1} \biggr] \cos\psi 
~-~\frac{1}{2^3} ~\frac{a^2}{c^2}\biggl[ \frac{2K(\mu)}{R+R_1} \biggr] \cos\psi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~ \frac{1}{2^7\cdot 3} \frac{a^4}{c^4} \biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl[
~-~5\cos\psi ~+~ \biggl(\frac{4c^2 }{R_1 R}\biggr)\cos 2\psi ~+~ \cos3\psi  
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~ \frac{1}{2^7\cdot 3} \frac{a^4}{c^4} \biggl[ \frac{K(\mu)}{ (R_1+R)}\biggr]\biggl[ 
9 ~+~ 10\cos\psi ~-~ \cos2\psi  
~-~\biggl( \frac{8c^2}{R_1 R } \biggr) \cos 2\psi
~-~2 \cos3\psi  
\biggr]
~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2K(\mu)}{R_1+R} \biggl[ 2
~-~\frac{1}{2^3} ~\frac{a^2}{c^2} 
~-~ \frac{3}{2^8} \frac{a^4}{c^4}~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\biggl\{
~-~ \frac{5}{2^6\cdot 3} \frac{a^4}{c^4} \biggl[ \frac{K(\mu)}{ (R_1+R)}\biggr]
~+~\frac{1}{2^3} ~\frac{a^2}{c^2}\biggl[ \frac{E(\mu)(R+R_1)}{RR_1} \biggr] 
~-~\frac{1}{2^3} ~\frac{a^2}{c^2}\biggl[ \frac{2K(\mu)}{R+R_1} \biggr] 
~+~ \frac{5}{2^7\cdot 3} \frac{a^4}{c^4} \biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] 
\biggr\}\cos\psi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~ \biggl\{
\frac{1}{2^7\cdot 3} \frac{a^4}{c^4} \biggl[ \frac{K(\mu)}{ (R_1+R)}\biggr]\biggl[ 1  ~+~\biggl( \frac{8c^2}{R_1 R } \biggr) \biggr]
~-~ \frac{1}{2^7\cdot 3} \frac{a^4}{c^4} \biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl(\frac{4c^2 }{R_1 R}\biggr)
\biggr\} \cos 2\psi   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\biggl\{
\frac{1}{2^6\cdot 3} \frac{a^4}{c^4} \biggl[ \frac{K(\mu)}{ (R_1+R)}\biggr] 
~-~ \frac{1}{2^7\cdot 3} \frac{a^4}{c^4} \biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggr\}\cos3\psi  
~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2K(\mu)}{R_1+R} \biggl[ 2
~-~\frac{1}{2^3} ~\frac{a^2}{c^2} 
~-~ \frac{3}{2^8} \frac{a^4}{c^4}~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~\biggl[ \frac{2K(\mu)}{R+R_1} ~-~ \frac{E(\mu)(R+R_1)}{RR_1} \biggr] 
\biggl[ \frac{1}{2^3} ~\frac{a^2}{c^2} ~+~ \frac{5}{2^7\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr)  \biggr]\cos\psi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~ \biggl\{
\biggl[ \frac{ 2K(\mu)}{ (R_1+R)}\biggr]\biggl[ \frac{1}{2}  ~+~\biggl( \frac{4c^2}{R_1 R } \biggr) \biggr]
~-~\biggl[ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl(\frac{4c^2 }{R_1 R}\biggr)
\biggr\} \biggl[ \frac{1}{2^7\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \biggr]\cos 2\psi   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~
\biggl[ \frac{2K(\mu)}{ (R_1+R)} ~-~ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] 
\biggl[ \frac{1}{2^7\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr)\biggr] \cos3\psi  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2K(\mu)}{R_1+R} \biggl[ 2
~-~\frac{1}{2^3} ~\frac{a^2}{c^2} 
~-~ \frac{3}{2^8} \frac{a^4}{c^4}~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~\biggl[ \frac{2K(\mu)}{R+R_1} ~-~ \frac{E(\mu)(R+R_1)}{RR_1} \biggr] 
\biggl[ \frac{1}{2^3} ~\frac{a^2}{c^2} ~+~ \frac{5}{2^7\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr)  \biggr]\cos\psi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~ 
\biggl[ \frac{ 2K(\mu)}{ (R_1+R)}~-~\frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] \biggl(\frac{4c^2 }{R_1 R}\biggr)
\biggl[ \frac{1}{2^7\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \biggr]\cos 2\psi   
~+~
\biggl[ \frac{ 2K(\mu)}{ (R_1+R)}\biggr] 
\biggl[ \frac{1}{2^8\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \biggr]\cos 2\psi   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\biggl[ \frac{2K(\mu)}{ (R_1+R)} ~-~ \frac{(R_1+R)E\left(\mu\right)}{RR_1}\biggr] 
\biggl[ \frac{1}{2^7\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr)\biggr] \cos3\psi  
</math>
  </td>
</tr>
</table>

Now, drawing from our [[2DStructure/ToroidalGreenFunction#Appendix_B:_Elliptic_Integrals|accompanying discussion of relationships between complete elliptic integrals and relationships between their parameter arguments]] and, in particular, associating Dyson's parameter, <math>~\mu = (R_1-R)/(R_1+R)</math>, with <math>~k_1</math>, we understand that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{2E(\mu)}{1+\mu} - (1-\mu)K(\mu)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~E(k) \, ,</math>
  </td>
</tr>
</table>
where the alternate parameter, 
<div align="center">
<math>~k = \frac{2\sqrt{\mu}}{1+\mu} = \biggl[1 - \biggr(\frac{R}{R_1}\biggr)^2\biggr]^{1 / 2} 
= \biggl[ \frac{2}{\coth\eta+1}\biggr]^{1 / 2} = \sqrt{1-e^{-2\eta}} \, .</math>
</div>
<span id="EKrelation">Hence, we appreciate that,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~E(k)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2E(\mu) \biggl[ 1 + \frac{R_1-R}{R_1+R} \biggr]^{-1} - \biggl[ 1 - \frac{R_1-R}{R_1+R} \biggr]K(\mu)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2E(\mu) \biggl[ \frac{2R_1}{R_1+R} \biggr]^{-1} - \biggl[ \frac{2R}{R_1+R} \biggr]K(\mu)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
R \biggl[ \frac{E(\mu)(R_1+R)}{RR_1}  - \frac{2K(\mu)}{R_1+R} \biggr]
</math>
  </td>
</tr>
</table>

<span id="Comparison">As a result, we can rewrite Dyson's expression for the external potential as</span>,

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

<tr>
  <td align="right">
<math>~\frac{\pi V(r,\theta)}{M}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2K(\mu)}{R_1+R} \biggl[ 2
~-~\frac{1}{2^3} ~\frac{a^2}{c^2} 
~-~ \frac{3}{2^8} \frac{a^4}{c^4}~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \biggr]
~+~\frac{E(k)}{R} 
\biggl[ \frac{1}{2^3} ~\frac{a^2}{c^2} ~+~ \frac{5}{2^7\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr)  \biggr]\cos\psi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~ 
\frac{E(k)}{R}\biggl(\frac{4c^2 }{R_1 R}\biggr)
\biggl[ \frac{1}{2^7\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \biggr]\cos 2\psi   
~+~
\biggl[ \frac{ 2K(\mu)}{ (R_1+R)}\biggr] 
\biggl[ \frac{1}{2^8\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \biggr]\cos 2\psi   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~-~\frac{E(k)}{R} 
\biggl[ \frac{1}{2^7\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr)\biggr] \cos3\psi  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2K(\mu)}{R_1+R} \biggl\{\biggl[ 2
~-~\frac{1}{2^3} ~\frac{a^2}{c^2} 
~-~ \frac{3}{2^8} \frac{a^4}{c^4}~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \biggr]
~+~\biggl[ \frac{1}{2^8\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \biggr]\cos 2\psi   \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
~+~\frac{E(k)}{R} \biggl\{
\biggl[ \frac{1}{2^3} ~\frac{a^2}{c^2} ~+~ \frac{5}{2^7\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr)  \biggr]\cos\psi 
~-~ 
\biggl(\frac{4c^2 }{R_1 R}\biggr)
\biggl[ \frac{1}{2^7\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \biggr]\cos 2\psi   
~-~ 
\biggl[ \frac{1}{2^7\cdot 3} \frac{a^4}{c^4} ~+~\mathcal{O}\biggl(\frac{a^5}{c^5}\biggr)\biggr] \cos3\psi  \biggr\} \, .
</math>
  </td>
</tr>
</table>

==External Potential in Terms of Angle &chi;==

<table border="0" cellpadding="10" align="right" width="35%"><tr><td align="center">
<table border="0" cellpadding="5" align="right" width="100%">
<tr>
<td align="center">
[[File:DysonTorusIllustration03.png|400px|center|Anchor Ring Schematic]]<br />
'''Caption:''' [http://www.mathematicsdictionary.com/english/vmd/full/t/torusanchorring.htm Anchor ring] schematic, adapted from figure near the top of &sect;2 (on p. 47) of [http://adsabs.harvard.edu/abs/1893RSPTA.184...43D Dyson (1893a)]
</td>
</tr>
</table>
</td></tr></table>
Is it relatively straightforward to develop a similar expression for the external potential that is written in terms of the angle, <math>~\chi</math>, instead of (as above) in terms of the angle, <math>~\psi</math> ?  This would make the transition to Dyson's Paper II smoother.  Initially I have in mind making the transformation via the law of sines, whereby,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{R_1}{\sin\chi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2c}{\sin\psi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{R}{\sin(\pi - \psi - \chi)} \, .</math>
  </td>
</tr>
</table>

This means that the following associations may be used as well:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sin\psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{2c}{R_1} \biggr)\sin\chi \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\cos^2\psi = 1 - \sin^2\psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 - \biggl[\biggl(\frac{2c}{R_1} \biggr)\sin\chi \biggr]^2\, .
</math>
  </td>
</tr>
</table>

From an [[Apps/DysonPotential#gammaInverse|accompanying discussion]] that builds upon the law of cosines, we also may write,

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

<tr>
  <td align="right">
<math>~2\biggl(\frac{R_1}{c}\biggr)^{-1} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 + \frac{1}{2}\biggl(\frac{a}{c}\biggr)\cos\chi 
+ \frac{1}{2^3}\biggl(\frac{a}{c}\biggr)^2 \biggl[ 3\cos^2\chi - 1 \biggr] 
+  \frac{1}{2^4}\biggl(\frac{a}{c}\biggr)^3\biggl[ 5\cos^3\chi 
~-~  3\cos\chi \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
~+~ \frac{1}{2^7} \biggl( \frac{a}{c}\biggr)^4 \biggl[ 3
~-~ 30 \cos^2\chi
~+~ 35 \cos^4\chi \biggr]
~+~ \mathcal{O}\biggl(\frac{a^5}{c^5}\biggr) \, .
</math>
  </td>
</tr>
</table>


=Compare Notations Used by Dyson and Wong=

<table border="1" cellpadding="5" align="center">
<tr>
<td align="center">
[[File:DysonWongCompareA0.png|500px|center|Dyson Anchor Ring Schematic]]<br />
'''Dyson's Notation''' (slightly modified)
</td>
<td align="center">
[[File:DysonWongCompareB0.png|500px|center|Wong Anchor Ring Schematic]]<br />
'''Wong's Notation''' (slightly modified)
</td>
</tr>
</table>


Key relationship, with understanding that, <math>~\epsilon \equiv d/c</math>:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~c^2 (1 - \epsilon^2)</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp; <math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~\frac{a}{c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1 - \epsilon^2)^{1 / 2} \, .</math>
  </td>
</tr>
</table>

==Relationship Between Lengths==

===Subtraction===
In both cases, lengths are referenced to the same "external" point whose meridional-plane coordinates are, <math>~(\varpi,z)</math>.  Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~R^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^2 + (c-\varpi)^2 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~r_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^2 + (a-\varpi)^2</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~R^2 - r_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(c-\varpi)^2 - (a-\varpi)^2</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(c^2 - 2c\varpi + \varpi^2) - (a^2 -2a\varpi +\varpi^2)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~c^2   - a^2 - 2(c-a)\varpi</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~\varpi </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{c^2   - a^2 + r_2^2 - R^2}{2(c-a)} \, .</math>
  </td>
</tr>
</table>

and,

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

<tr>
  <td align="right">
<math>~R_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^2 + (c+\varpi)^2</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~r_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^2 + (a+\varpi)^2</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ R_1^2 - r_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(c+\varpi)^2 -  (a+\varpi)^2</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(c^2 + 2c\varpi + \varpi^2) -  (a^2 + 2a\varpi +\varpi^2)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~c^2 -  a^2  + 2(c - a)\varpi </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\varpi  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \frac{R_1^2 - r_1^2 - (c^2 -  a^2)}{2(c-a)} \, .</math>
  </td>
</tr>
</table>
Put together, then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{c^2   - a^2 + r_2^2 - R^2}{2(c-a)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{R_1^2 - r_1^2 - (c^2 -  a^2)}{2(c-a)}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~(R^2 + R_1^2) - (r_1^2 + r_2^2) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2(c^2 -  a^2)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2c^2 -  2c^2(1-\epsilon^2)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2c^2\epsilon^2</math>
  </td>
</tr>
</table>

===Multiplication===

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

<tr>
  <td align="right">
<math>~R^2R_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[z^2 + (c-\varpi)^2][z^2 + (c+\varpi)^2]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^4 + z^2[(c-\varpi)^2 + (c+\varpi)^2] + [(c-\varpi)^2(c+\varpi)^2 - R^2R_1^2]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ 2z^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\pm \biggl[ [(c-\varpi)^2 + (c+\varpi)^2]^2 - 4[(c-\varpi)^2(c+\varpi)^2-R^2R_1^2]  \biggr]^{1 / 2} ~-~ [(c-\varpi)^2 + (c+\varpi)^2]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~r_1^2r_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[z^2 + (a-\varpi)^2][z^2 + (a+\varpi)^2]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^4 + z^2[(a-\varpi)^2 + (a+\varpi)^2  ] + [(a-\varpi)^2(a+\varpi)^2 - r_1^2r_2^2]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ 2z^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\pm \biggl[ [(a-\varpi)^2 + (a+\varpi)^2  ]^2 ~-~ 4[(a-\varpi)^2(a+\varpi)^2 - r_1^2r_2^2]  \biggr]^{1 / 2} ~-~  [(a-\varpi)^2 + (a+\varpi)^2  ]
</math>
  </td>
</tr>
</table>
Note:  In order to be physically relevant, the quantity, <math>~2z^2</math>, must be positive.  Given that the second term on the RHS is always negative, we conclude that only a solution with the superior (plus) sign is physically relevant.

Hence,

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

<tr>
  <td align="right">
<math>~
\biggl[ [(c-\varpi)^2 + (c+\varpi)^2]^2 - 4[(c-\varpi)^2(c+\varpi)^2-R^2R_1^2]  \biggr]^{1 / 2} ~-~ [(c-\varpi)^2 + (c+\varpi)^2]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ [(a-\varpi)^2 + (a+\varpi)^2  ]^2 ~-~ 4[(a-\varpi)^2(a+\varpi)^2 - r_1^2r_2^2]  \biggr]^{1 / 2} ~-~  [(a-\varpi)^2 + (a+\varpi)^2  ]
</math>
  </td>
</tr>
</table>

===Ratios===

Let's work with the following (dimensionless) length ratios:

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

<tr>
  <td align="right">
<math>~\lambda_2^2 \equiv \frac{R^2}{r_2^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{z^2 + (c-\varpi)^2}{z^2 + (a-\varpi)^2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\lambda_1^2 \equiv \frac{R_1^2}{r_1^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{z^2 + (c+\varpi)^2}{z^2 + (a+\varpi)^2} \, .</math>
  </td>
</tr>
</table>

Solving for <math>~z^2</math> in the first case, we find,

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

<tr>
  <td align="right">
<math>~\lambda_2^2 [z^2 + (a-\varpi)^2]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^2 + (c-\varpi)^2</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~z^2( \lambda_2^2 - 1)   </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(c-\varpi)^2 - \lambda_2^2(a-\varpi)^2</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~z^2   </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(c-\varpi)^2 - \lambda_2^2(a-\varpi)^2}{( \lambda_2^2 - 1)} \, ;</math>
  </td>
</tr>
</table>

and in the second case, we find,

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

<tr>
  <td align="right">
<math>~\lambda_1^2 [z^2 + (a+\varpi)^2]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^2 + (c+\varpi)^2</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ z^2 (\lambda_1^2 - 1)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(c+\varpi)^2 - \lambda_1^2 (a+\varpi)^2</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ z^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(c+\varpi)^2 - \lambda_1^2(a+\varpi)^2}{(\lambda_1^2 - 1)} \, .</math>
  </td>
</tr>
</table>

Combined, this gives,

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

<tr>
  <td align="right">
<math>~\frac{(c-\varpi)^2 - \lambda_2^2(a-\varpi)^2}{( \lambda_2^2 - 1)}  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(c+\varpi)^2 - \lambda_1^2(a+\varpi)^2}{(\lambda_1^2 - 1)} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ (\lambda_1^2 - 1)[(c-\varpi)^2 - \lambda_2^2(a-\varpi)^2]  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~( \lambda_2^2 - 1) [ (c+\varpi)^2 - \lambda_1^2(a+\varpi)^2 ]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ (\lambda_1^2 - 1)[(c^2 - 2c\varpi + \varpi^2) - \lambda_2^2(a^2 - 2a\varpi +\varpi^2 )]  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~( \lambda_2^2 - 1) [ (c^2 + 2c\varpi + \varpi^2) - \lambda_1^2(a^2+ 2a \varpi + \varpi^2) ]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ (\lambda_1^2 - 1)[c^2 - 2c\varpi + \varpi^2 - a^2\lambda_2^2 + 2a\varpi\lambda_2^2 - \varpi^2\lambda_2^2 ]  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~( \lambda_2^2 - 1) [ c^2 + 2c\varpi + \varpi^2 -a^2\lambda_1^2 - 2a \varpi\lambda_1^2 - \varpi^2\lambda_1^2 ]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ (\lambda_1^2 - 1)[ ( c^2 - a^2\lambda_2^2 ) + ( 2a\lambda_2^2 - 2c )\varpi  + \varpi^2(1 - \lambda_2^2 ) ]  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~( \lambda_2^2 - 1) [ ( c^2   -a^2\lambda_1^2 ) + ( 2c - 2a \lambda_1^2)\varpi + \varpi^2( 1 - \lambda_1^2 ) ]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 
(\lambda_1^2 - 1)( c^2 - a^2\lambda_2^2 )  
+ \varpi^2(1 - \lambda_2^2 )(\lambda_1^2 - 1) + \varpi( 2a\lambda_2^2 - 2c )(\lambda_1^2 - 1) 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ ( \lambda_2^2 - 1)( c^2   -a^2\lambda_1^2 ) + \varpi( 2c - 2a \lambda_1^2)( \lambda_2^2 - 1) + \varpi^2( 1 - \lambda_1^2 )( \lambda_2^2 - 1) \, .</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 
\varpi^2 [(1 - \lambda_2^2 )(\lambda_1^2 - 1)  ~-~ ( 1 - \lambda_1^2 )( \lambda_2^2 - 1) ]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\varpi [ ( 2c - 2a \lambda_1^2)( \lambda_2^2 - 1) ~-~ ( 2a\lambda_2^2 - 2c )(\lambda_1^2 - 1) ]
~+~[ ( \lambda_2^2 - 1)( c^2   -a^2\lambda_1^2 ) ~-~(\lambda_1^2 - 1)( c^2 - a^2\lambda_2^2 )  ]
</math>
  </td>
</tr>
</table>

Now, because the coefficient of the <math>~\varpi^2</math> term (on the left-hand-side of this last expression) is zero, we find that,

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

<tr>
  <td align="right">
<math>~
2 \varpi
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{ (\lambda_1^2 - 1)( c^2 - a^2\lambda_2^2 ) ~-~( \lambda_2^2 - 1)( c^2   -a^2\lambda_1^2 ) }{ ( c - a \lambda_1^2)( \lambda_2^2 - 1) ~-~ ( a\lambda_2^2 - c )(\lambda_1^2 - 1) }
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ (\lambda_1^2 - \lambda_2^2)(c^2 - a^2) }{ (\lambda_1^2 + \lambda_2^2)(c+a) - 2c } \, .
</math>
  </td>
</tr>
</table>

==Relationship Between Key Angles==

Law of Cosines:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(2c)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
R_1^2 + R^2 - 2RR_1\cos\psi
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~(2a)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r_1^2 + r_2^2 - 2r_1 r_2 \cos\theta
</math>
  </td>
</tr>
</table>

==Try Again==
Let's transform from Wong's lengths and angles to the ones used by Dyson.  (This will permit us to compare Wong's analytic solution to the approximate one presented by Dyson.)  Given the coordinate-pair <math>~(\varpi, z)</math> and Dyson's length scale, <math>~c</math>, we have,

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

<tr>
  <td align="right">
<math>~\tan\chi_R</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{z}{c-\varpi} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~R^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^2 + (c-\varpi)^2 \, .</math>
  </td>
</tr>
</table>

From this pair of expressions we can transform from <math>~(R,\chi_R)</math> to <math>~(\varpi,z)</math> as follows:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~c-\varpi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z \cot\chi_R</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ R^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^2 (1 + \cot\chi^2_R) =\biggl[  \frac{z}{\sin\chi_R} \biggr]^{2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ z</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~R\sin\chi_R</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \varpi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~c - \cot\chi_R \cdot R\sin\chi_R = c - R\cos\chi_R \, .</math>
  </td>
</tr>
</table>

Next, given the coordinate-pair <math>~(\varpi, z)</math> and Wong's length scale, <math>~a</math>, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tan\chi_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{z}{a-\varpi} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~r_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^2 + (a-\varpi)^2 \, ;</math>
  </td>
</tr>
</table>

and, combining the two sets gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~r_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~R^2\sin^2\chi_R + [ R\cos\chi_R + (a - c) ]^2 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~R^2\sin^2\chi_R + [ R^2\cos^2\chi_R + 2(a-c)R\cos\chi_R + (a - c)^2 ] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~R^2 + (a - c)^2  - 2(c-a)R\cos\chi_R  \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\tan\chi_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{R\sin\chi_R}{R\cos\chi_R + (a - c)} \, .</math>
  </td>
</tr>
</table>

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