Editing Apps/DysonWongTori (section)

===MacMillan (1930)===
====Derivation of the Potential====
In &sect;102 of a book titled, [https://www.amazon.com/Theory-Potential-W-D-Macmillan/dp/0486604861/ref=sr_1_2?s=books&ie=UTF8&qid=1503444466&sr=1-2&keywords=the+theory+of+the+potential ''The Theory of the Potential'', W. D. MacMillan (1958; originally, 1930)] derives an analytic expression for the gravitational potential of a uniform, infinitesimally thin, circular "hoop" of radius, <math>~a</math>; as shown, immediately below, the hoop is labeled, <math>~H</math>, in his Figures 60 and 61.

<table border="0" cellpadding="8" align="center"><tr><td align="center">
<table border="1" cellpadding="5" align="center">
<tr><td align="center" colspan="2">
'''Figures 60 &amp; 61 extracted without modification from, respectively, p. 195 &amp; 196 of [https://www.amazon.com/Theory-Potential-W-D-Macmillan/dp/0486604861/ref=sr_1_2?s=books&ie=UTF8&qid=1503444466&sr=1-2&keywords=the+theory+of+the+potential MacMillan (1958)]'''<p></p>
''The Theory of the Potential'', New York: McGraw-Hill
</td></tr>
<tr>
<td>
[[File:MacMillanFigure60.png|460px|center|MacMillan (1958, ''The Theory of the Potential'', New York: McGraw-Hill)]]
</td>
<td>
[[File:MacMillanFigure61.png|460px|center|MacMillan (1958, ''The Theory of the Potential'', New York: McGraw-Hill)]]
</tr></table>
</td></tr></table>

In setting up this problem, MacMillan (1958) says (verbatim text is typeset in a dark green font), <font color="darkgreen">Let <math>~P</math> be any point in space not in <math>~H</math>.  From <math>~P</math> drop the perpendicular <math>~PQ = z</math> to the plane of the hoop.  Draw the diameter of the circle <math>~BOA</math> which, extended, passes through <math>~Q</math>. Let <math>~m</math> be any point on the circle, and draw</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Pm = \rho \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~PA = \rho_1 \, ,</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~PB = \rho_2 \, .</math>
  </td>
</tr>
</table>
</div>

<font color="darkgreen">Evidently <math>~\rho_1</math> and <math>~\rho_2</math> are the minimum and maximum values of <math>~\rho</math> as the point <math>~m</math> runs around the circle.</font> <font color="darkgreen">If the angle <math>~mOA</math> is represented by <math>~2\omega</math>, the arc element is <math>~ds = 2ad\omega</math>, and</font> &#8212; after multiplying MacMillan's &sect;102, equation (1) through by (conventionally, the negative of) the gravitational constant, <math>~G</math> &#8212; <font color="darkgreen">the expression for the gravitational potential is</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Phi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- 2Ga\sigma \int_0^\pi \frac{d\omega}{\rho}  = - \biggl( \frac{GM}{\pi} \biggr) \int_0^\pi \frac{d\omega}{\rho} \, ,</math>
  </td>
</tr>
</table>
</div>

where, <math>~\sigma</math> is the (uniform) linear mass density around the hoop, hence, the total mass of the hoop is <math>~M = 2\pi a \sigma</math>.

Referring further to MacMillan's Figure 60 &#8212; digitally reproduced, above &#8212; if the lengths <math>~mQ</math> and <math>~OQ</math> are represented by <math>~h</math> and <math>~\varpi</math>, respectively, then
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\rho_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(\varpi-a)^2 + z^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\rho_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(\varpi+a)^2 + z^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\rho^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^2 + h^2 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z^2 + \varpi^2 + a^2 - 2a\varpi \cos(2\omega) \, .</math>
  </td>
</tr>
</table>
</div>
Following [https://www.amazon.com/Theory-Potential-W-D-Macmillan/dp/0486604861/ref=sr_1_2?s=books&ie=UTF8&qid=1503444466&sr=1-2&keywords=the+theory+of+the+potential MacMillan (1958)] (p. 196), the expression for <math>~\rho^2</math> can further be written,
<div align="center">
<table border="0" cellpadding="5" align="center">

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\rho_1^2 \cos^2\omega + \rho_2^2\sin^2\omega \, .
</math>
  </td>
</tr>
</table>
</div>
<span id="DeupreeReference">Hence, the expression for the potential becomes,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Phi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl( \frac{2GM}{\pi} \biggr) \int_0^{\pi/2} \frac{d\omega}{[\rho_1^2 \cos^2\omega + \rho_2^2\sin^2\omega]^{1 / 2}} \, .</math>
  </td>
</tr>
</table>
</div>

As [https://www.amazon.com/Theory-Potential-W-D-Macmillan/dp/0486604861/ref=sr_1_2?s=books&ie=UTF8&qid=1503444466&sr=1-2&keywords=the+theory+of+the+potential MacMillan (1958)] argues, <font color="darkgreen">&hellip; this expression shows that <math>~\Phi</math> is symmetric in <math>~\rho_1</math> and <math>~\rho_2</math>, for if <math>~\omega</math> is replaced by <math>~(\tfrac{\pi}{2} - \psi)</math> it becomes</font>

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

<tr>
  <td align="right">
<math>~\Phi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl( \frac{2GM}{\pi} \biggr) \int_0^{\pi/2} \frac{d\psi}{[\rho_1^2 \sin^2\psi + \rho_2^2\cos^2\psi]^{1 / 2}} \, ,</math>
  </td>
</tr>
</table>
</div>
<font color="darkgreen">and therefore</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Phi(\rho_1, \rho_2)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\Phi(\rho_2, \rho_1) \, .</math>
  </td>
</tr>
</table>
</div>
<font color="darkgreen">Along the axis of the hoop, <math>~\rho_1 = \rho_2</math>, and if <math>~\rho_a</math> is their common value, it is seen at once that the value of the potential along this axis <math>~\Phi_a</math> is,</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Phi_a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{GM}{\rho_a} \, .</math>
  </td>
</tr>
</table>
</div>

Furthermore, according to [https://www.amazon.com/Theory-Potential-W-D-Macmillan/dp/0486604861/ref=sr_1_2?s=books&ie=UTF8&qid=1503444466&sr=1-2&keywords=the+theory+of+the+potential MacMillan (1958)], <font color="darkgreen">&hellip; the function <math>~\Phi(\rho_1, \rho_2)</math> is homogeneous of degree "- 1" in <math>~\rho_1</math> and <math>~\rho_2</math>.  Therefore, <math>~\rho_1 \Phi</math> is homogeneous of degree zero and depends only upon the ratio <math>~\rho_1/\rho_2</math>.</font>  
With this in mind, let's rewrite the expression for the potential in the form,

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

<tr>
  <td align="right">
<math>~\rho_1 \Phi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-  \frac{2GM}{\pi}  \int_0^{\pi/2} \biggl[\sin^2\psi + \biggl( \frac{\rho_2^2}{\rho_1^2} \biggr)\cos^2\psi \biggr]^{-1 / 2} d\psi </math>
  </td>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-  \frac{2GM}{\pi} \biggl( \frac{\rho_1}{\rho_2} \biggr) \int_0^{\pi/2} \biggl[\biggl( \frac{\rho_1^2}{\rho_2^2} \biggr)\sin^2\psi + \biggl( 1 - \sin^2\psi \biggr)\biggr]^{-1 / 2} d\psi </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-  \frac{2GM}{\pi} \biggl( \frac{\rho_1}{\rho_2} \biggr) \int_0^{\pi/2} \biggl[1 - \biggl(1 - \frac{\rho_1^2}{\rho_2^2} \biggr)\sin^2\psi \biggr]^{-1 / 2} d\psi \, .</math>
  </td>
</tr>
</table>
</div>

In addition to the hoop, <math>~H</math>, Figure 61 in &sect;102 of MacMillan (1958) &#8212; digitally reproduced, above &#8212; displays a curve in the meridional plane of the hoop for which the ratio,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\rho_1}{\rho_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~c \, ,</math>
  </td>
</tr>
</table>
</div>
where <math>~c</math> is a constant.  As [https://www.amazon.com/Theory-Potential-W-D-Macmillan/dp/0486604861/ref=sr_1_2?s=books&ie=UTF8&qid=1503444466&sr=1-2&keywords=the+theory+of+the+potential MacMillan (1958)] argues, the displayed curve is a circle because this equation <font color="darkgreen">is the equation of a circle in bipolar coordinates; and this circle &hellip; divides the line <math>~BCAD</math> harmonically, since by</font> this last equation,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{AC}{BC} = \frac{AD}{BD}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~c \, .</math>
  </td>
</tr>
</table>
</div>
<span id="RingPotential">It is clear, therefore, that at every point along this meridional circle, the potential is given by the expression,</span>

<table border="1" cellpadding="8" align="center" width="65%">
<tr>
  <th align="center">Gravitational Potential of a Thin Ring</th>
</tr>
<tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Phi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-  \frac{2GMc}{\pi \rho_1} \int_0^{\pi/2} \frac{d\psi}{ \sqrt{1 - k^2 \sin^2\psi }} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-  \biggl[ \frac{2GMc}{\pi \rho_1} \biggr] K(k^2)  \, ,</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[https://www.amazon.com/Theory-Potential-W-D-Macmillan/dp/0486604861/ref=sr_1_2?s=books&ie=UTF8&qid=1503444466&sr=1-2&keywords=the+theory+of+the+potential W. D. MacMillan (1958)], &sect;102, Eq. (5);<p></p>
see also [https://archive.org/details/foundationsofpot033485mbp O. D. Kellogg (1929)], &sect;III.4, Exercise (4)
  </td>
</tr>
</table>
where, <math>~K(k^2)</math> is the [http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html complete elliptic integral of the first kind] for the ''parameter'',
<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>~1-c^2 \, .</math>
  </td>
</tr>
</table>
</td></tr></table>

<span id="CylindricalLocation">The ''parameter'', <math>~k^2</math>, always lies between zero and unity.  For later reference, we note that,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ \biggl[ \frac{2GMc}{\pi \rho_1} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2GM}{\pi \rho_2} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2GM}{\pi } \biggr]\frac{1}{\sqrt{(\varpi+a)^2 + z^2}} \, ,</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~k^2 = 1-c^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 - \frac{\rho_1^2}{\rho_2^2} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
 </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{[(\varpi+a)^2 + z^2] - [(\varpi-a)^2 + z^2]}{(\varpi+a)^2 + z^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
 </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4a\varpi }{(\varpi+a)^2 + z^2} \, .
</math>
  </td>
</tr>
</table>
</div>
<span id="TRApproximation">This gives us what we will henceforth refer to as the,</span>
<table border="0" cellpadding="5" align="center">
<tr><td align="center" colspan="1"><font color="#770000">'''Gravitational Potential in the Thin Ring (TR) Approximation'''</font></td></tr>
<tr>
  <td align="center">
{{ Math/EQ TRApproximation }}
  </td>
</tr>
</table>
Notice that in finalizing the [https://en.wikipedia.org/wiki/Elliptic_integral#Notational_variants precise ''notational form''] of this "key equation," we have chosen to rewrite the argument of the complete elliptic integral of the first kind in terms of the ''elliptic modulus'', <math>~k</math>, rather than in terms of the ''parameter'' <math>~k^2</math>.

====Some Geometric Relations====
<table border="0" cellpadding="8" align="right"><tr><td align="center">
<table border="1" cellpadding="5" align="center">
<tr><td align="center" colspan="1">
'''Modification of MacMillan's Figure 61'''
</td></tr>
<tr>
<td>
[[File:ModifiedMacMillan61.png|460px|center|Modification of MacMillan's Figure 61]]
</tr></table>
</td></tr></table>
Throughout his derivation, [https://www.amazon.com/Theory-Potential-W-D-Macmillan/dp/0486604861/ref=sr_1_2?s=books&ie=UTF8&qid=1503444466&sr=1-2&keywords=the+theory+of+the+potential MacMillan (1958)] uses the parameter, <math>~a</math>, to represent the radius of the circular "hoop" &#8212; that is, the distance from the center of the hoop to either point <math>~A</math> or point <math>~B</math> as marked in both Figure 60 and Figure 61.  In the diagram presented here, on the right, we have modified his Figure 61 (modifications are in red) to explicitly identify two additional lengths that will come into play when we reference toroidal coordinates, below: &nbsp; The parameter, <math>~R</math>, identifies the distance from the center of the hoop to the center of the meridional-plane circle; and the parameter, <math>~d</math>, identifies the radius of this meridional-plane circle.  (Note that the distance between point <math>~A</math> and the center of the meridional-plane circle is <math>~R-a</math>.)  Given that the meridional-plane circle has been drawn in such a way that the ratio, <math>~\rho_1/\rho_2 = c</math>, at all points <math>~P</math> along the circle, a useful relationship can be derived between the three parameters, <math>~R, d</math> and <math>~a</math> as follows.  

If <math>~P</math> is moved around the circle to align with point <math>~D</math>, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\rho_1 = \rho_2 c = d + (R-a) \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; and, &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\rho_2 = 2a + (R-a) + d </math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
<math>~\Rightarrow ~~~ 
c = \frac{d + (R-a)}{2a + (R-a) + d} \, .
</math>
  </td>
</tr>
</table>
</div>

Similarly, if <math>~P</math> is moved around the circle to align with point <math>~C</math>, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\rho_1 = \rho_2 c = d - (R-a) \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; and, &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\rho_2 = 2a - [ d - (R-a)]</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
<math>~\Rightarrow ~~~ 
c = \frac{d - (R-a)}{ 2a - [ d - (R-a)]} \, .
</math>
  </td>
</tr>
</table>
</div>
Equating these two expressions for <math>~c</math> then gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d + R-a}{a + R + d}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d - R + a}{ a  -d + R}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
(d + R-a)(a  -d + R )
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(d - R + a)(a + R + d )
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
[R - (a- d) ][R + (a  -d) ]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[(a + d) - R ][(a + d) + R ]
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2a^2 + 2d^2 \, ,
</math>
  </td>
</tr>
</table>
</div>
<span id="TorusGeometry">or, finally,</span>
<table border="1" cellpadding="8" align="center" width="25%">
<tr>
  <th align="center">Geometric Relationship</th>
</tr>
<tr>
  <td align="center">
<math>~a^2 = R^2 - d^2 \, .</math>
  </td>
</tr>
</table>
Similarly, it can be shown that,
<div align="center">
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{R}{d} \biggl[ 1 - \sqrt{1 - \frac{d^2}{R^2}} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Thus, the ''aspect ratio'',
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{R}{d}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1+c^2}{2c} = \frac{1}{2}\biggl[ \frac{\rho_2}{\rho_1} + \frac{\rho_1}{\rho_2} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>