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====Attempt #2==== Now, drawing upon the Key Relation, <div align="center"> {{ Math/EQ_Toroidal02 }} </div> and setting n = m = 0 while adopting the association, <math>x \rightarrow \cosh\eta</math>, we obtain, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>Q_{-1 / 2}(\coth\eta)</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{\pi^{3/2}}{\sqrt{2\pi}}~ (\sinh\eta)^{1 / 2} P_{-1 / 2}(\cosh\eta) </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ P_{-1 / 2}(\cosh\eta)</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{\sqrt{2}}{\pi}~ (\sinh\eta)^{-1 / 2} Q_{-1 / 2}(\coth\eta) \, . </math> </td> </tr> </table> We therefore can write, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\biggl(\frac{a}{GM} \biggr) \Phi_{\mathrm{W}0} (\eta,\theta) \biggr|_\mathrm{Exterior} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -\frac{\sqrt{2} D_0\cdot C_0}{\pi} \biggl[ \frac{(\cosh\eta - \cos\theta)^{1 / 2}}{ (\sinh\eta)^{1 / 2}}\biggr]~Q_{- 1 / 2}(\coth\eta) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -\frac{\sqrt{2} D_0\cdot C_0}{\pi} \biggl[ \frac{(\cosh\eta - \cos\theta)^{1 / 2}}{ (\sinh\eta)^{1 / 2}}\biggr]~k K(k) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -\frac{2 D_0\cdot C_0}{\pi} \biggl[ \frac{\cosh\eta - \cos\theta}{ \sinh\eta + \cosh\eta}\biggr]^{1 / 2}~K(k) </math> </td> </tr> </table> where, <div align="center"> <math>k\equiv \biggl[ \frac{2}{1+\coth\eta } \biggr]^{1 / 2} = \biggl[ \frac{2\sinh\eta}{\sinh\eta+\cosh\eta } \biggr]^{1 / 2}\, .</math> </div> Now, converting back to cylindrical coordinates, we recognize that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>k^2</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{4a\varpi}{(\varpi + a)^2 + z^2 } \, , </math> </td> </tr> </table> and, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\frac{\cosh\eta - \cos\theta}{ \sinh\eta + \cosh\eta}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\frac{a\sinh\eta}{\varpi} \biggl[ \frac{ \{ [(\varpi - a)^2 + z^2]\cdot [(\varpi + a)^2 + z^2] \}^{1 / 2} }{ \varpi^2 + a^2 + z^2 +2a\varpi } \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ \frac{ 2a^2}{ \{ [(\varpi - a)^2 + z^2]\cdot [(\varpi + a)^2 + z^2] \}^{1 / 2} } \biggr] \biggl[ \frac{ \{ [(\varpi - a)^2 + z^2]\cdot [(\varpi + a)^2 + z^2] \}^{1 / 2} }{ (\varpi + a)^2 + z^2} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ \frac{ 2a^2}{ (\varpi + a)^2 + z^2 } \biggr] \, . </math> </td> </tr> </table> Hence, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\biggl(\frac{a}{GM} \biggr) \Phi_{\mathrm{W}0} (\varpi,z) \biggr|_\mathrm{Exterior} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -\frac{2 D_0\cdot C_0}{\pi} \biggl[ \frac{ 2a^2}{ (\varpi + a)^2 + z^2 } \biggr]^{1 / 2}~K(k) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - \biggl[ \frac{2^{3/2} D_0\cdot C_0}{\pi} \biggr] \frac{ aK(k) }{[ (\varpi + a)^2 + z^2]^{1 / 2} } </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>\approx</math> </td> <td align="left"> <math> - \biggl[ \frac{2^{3/2} }{\pi} \biggr] \biggl[ \frac{2^{3/2} }{3\pi^2} \cdot \frac{\sinh^3\eta_0}{\cosh\eta_0} \biggr] \biggl[\biggl( \frac{3 \pi^2}{2^2} \biggr) \frac{1}{\cosh^2\eta_0}\biggr] \frac{ aK(k) }{[ (\varpi + a)^2 + z^2]^{1 / 2} } </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>\approx</math> </td> <td align="left"> <math>- \frac{2}{\pi} \cdot \frac{ aK(k) }{[ (\varpi + a)^2 + z^2]^{1 / 2} } \, , </math> </td> </tr> </table> where, in the next-to-last step, we have inserted, from above, the expression for the product of <math>D_0</math> and <math>C_0</math> in the limit of <math>\eta_0 \rightarrow \infty</math>. This is precisely the expression for the ''Gravitational Potential in the Thin Ring Approximation, <math>\Phi_\mathrm{TR}(\varpi,z)</math>, that we have presented elsewhere.
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