Editing Appendix/Ramblings/Dyson1893Part1 (section)

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

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[[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)]
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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,
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<math>~\frac{R_1}{\sin\chi}</math>
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<math>~=</math>
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<math>~\frac{2c}{\sin\psi}</math>
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<math>~=</math>
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<math>~\frac{R}{\sin(\pi - \psi - \chi)} \, .</math>
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This means that the following associations may be used as well:
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<math>~\sin\psi</math>
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<math>~=</math>
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<math>~
\biggl(\frac{2c}{R_1} \biggr)\sin\chi \, ;
</math>
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<math>~\cos^2\psi = 1 - \sin^2\psi</math>
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<math>~=</math>
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<math>~
1 - \biggl[\biggl(\frac{2c}{R_1} \biggr)\sin\chi \biggr]^2\, .
</math>
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From an [[Apps/DysonPotential#gammaInverse|accompanying discussion]] that builds upon the law of cosines, we also may write,

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<math>~2\biggl(\frac{R_1}{c}\biggr)^{-1} </math>
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<math>~=</math>
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<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>
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&nbsp;
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&nbsp;
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<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>
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