User:Tohline/Appendix/Ramblings/Dyson1893Part1

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Dyson (1893a) Part I: Some Details

This chapter provides some derivation details relevant to our accompanying discussion of Dyson's analysis of the gravitational potential exterior to an anchor ring.

Whitworth's (1981) Isothermal Free-Energy Surface
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Overview

In his pioneering work, F. W. Dyson (1893a, Philosophical Transactions of the Royal Society of London. A., 184, 43 - 95) and (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. C.-Y. Wong (1974, ApJ, 190, 675 - 694) extended Dyson's work, using numerical techniques to obtain more accurate — but still approximate — equilibrium structures for incompressible tori having solid body rotation. Since then, Y. Eriguchi & D. Sugimoto (1981, Progress of Theoretical Physics, 65, 1870 - 1875) and I. Hachisu, J. E. Tohline & 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

On p. 59, at the end of §6 of Dyson (1893a), we find the following expression for the potential at point "P", anywhere exterior to an anchor ring:

<math>~\frac{\pi V(r,\theta)}{M}</math>

<math>~=</math>

<math>~ \mathfrak{I}(r,\theta,c) ~+~ \frac{1}{2^3}\biggl(\frac{a^2}{c}\biggr) \frac{d}{dc} \biggl[ \mathfrak{I}(r,\theta,c)\biggr] ~-~ \frac{1}{2^6\cdot 3}\biggl(\frac{a^2}{c}\biggr)^2 \frac{d^2}{dc^2} \biggl[ \mathfrak{I}(r,\theta,c)\biggr] ~+~\cdots </math>

 

 

<math>~ ~+~(-1)^{n+1} \frac{2}{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{a^2}{c}\biggr)^n \frac{d^n}{dc^n} \biggl[ \mathfrak{I}(r,\theta,c)\biggr] ~+~ \cdots </math>

where (see beginning of §8 on p. 61),

Anchor Ring Schematic

Caption: Anchor ring schematic, adapted from figure near the top of §2 (on p. 47) of Dyson (1893a)

<math>~\mathfrak{I}(r,\theta,c)</math>

<math>~\equiv</math>

<math>~ \int_0^\pi d\phi \biggl[r^2 - 2cr\sin\theta \cos\phi +c^2\biggr]^{-1 / 2} </math>

 

<math>~=</math>

<math>~ 2\int_0^{\pi/2} d\phi \biggl[ R_1^2 - (R_1^2-R^2)\sin^2\phi \biggr]^{-1 / 2} </math>

 

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ \frac{2K(k)}{R_1} \, , </math>

and, where furthermore,

<math>~K(k)</math>

<math>~=</math>

<math>~ \int_0^{\pi/2} d\phi \biggl[1 - k^2\sin^2\phi \bigg]^{-1 / 2} </math>

      and      

<math>~k</math>

<math>~\equiv</math>

<math>~ \biggl[ \frac{R_1^2-R^2}{R_1^2} \biggr]^{1 / 2} \, . </math>

Taking a queue from our accompanying discussion of toroidal coordinates, if we adopt the variable notation,

<math>~\eta \equiv \ln\biggl(\frac{R_1}{R}\biggr) \, ,</math>

then we can write,

<math>~\cosh\eta = \frac{1}{2}\biggl[e^\eta + e^{-\eta}\biggr]</math>

<math>~=</math>

<math>~\frac{R^2 + R_1^2}{2RR_1} \, ,</math>

which implies that,

<math>~\biggl[ \frac{2}{\coth\eta +1} \biggr]^{1 / 2} = [1 - e^{-2\eta}]^{1 / 2}</math>

<math>~=</math>

<math>~\biggl[ 1 - \biggl(\frac{R}{R_1}\biggr)^2 \biggr]^{1 / 2} = k \, .</math>

Now, if we employ the Descending Landen Transformation for the complete elliptic integral of the first kind, we can make the substitution,

<math>~K(k)</math>

<math>~=</math>

<math>~ (1 + \mu)K(\mu) \, , </math>

      where,      

<math>~\mu</math>

<math>~\equiv</math>

<math>~ \frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}} \, . </math>

But notice that, <math>~\sqrt{1-k^2} = e^{-\eta}</math>, in which case,

<math>~\mu </math>

<math>~=</math>

<math>~ \frac{1-e^{-\eta}}{1+e^{-\eta}} </math>

<math>~=</math>

<math>~ \frac{1-R/R_1}{1+R/R_1} </math>

<math>~=</math>

<math>~ \frac{R_1-R}{R_1+R} \, . </math>

Hence, we can write,

<math>~\mathfrak{I}(r,\theta,c) = \frac{2K(k)}{R_1}</math>

<math>~=</math>

<math>~ \frac{2}{R_1} \biggl[(1+\mu)K(\mu) \biggr] </math>

 

<math>~=</math>

<math>~\frac{2K(\mu)}{R_1} \biggl[1+\frac{R_1-R}{R_1+R} \biggr] </math>

 

<math>~=</math>

<math>~\frac{4K(\mu)}{R_1+R} \, .</math>

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

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 R1 with respect to c; as we have indicated here, the numerator should be 4cR1 instead of 4cR.

<math>~\frac{dR}{dc}</math>

<math>~=</math>

<math>~\frac{4c^2 + R^2 - R_1^2}{4cR} \, ,</math>

<math>~\frac{dR_1}{dc}</math>

<math>~=</math>

<math>~\frac{4c^2 + R_1^2 - R^2}{4cR_1} \, ,</math>       and,

<math>~\frac{d\mu}{dc}</math>

<math>~=</math>

<math>~\frac{\mu}{c} \cos\psi \, ,</math>

where — as shown above in the Anchor ring schematic — <math>~\psi</math> is the angle between <math>~R</math> and <math>~R_1</math> for which (according to the law of cosines),

<math>~\cos\psi</math>

<math>~=</math>

<math>~\frac{R^2 + R_1^2 - 4c^2}{2RR_1} \, .</math>

It will be useful for us to note that,

<math>~\frac{d(\cos\psi)}{dc}</math>

<math>~=</math>

<math>~ \frac{d}{dc}\biggl[ \frac{R^2 + R_1^2 - 4c^2}{2RR_1} \biggr] </math>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<math>~\frac{1}{4cR^2 R_1^2}\biggl\{ \cos\psi \biggl[ R_1^4 + R^4 - 2R_1^2 R^2 - 4c^2(R_1^2 + R^2) \biggr] -~8c^2R R_1 \biggr\} </math>


LaTeX mathematical expressions cut-and-pasted directly from
NIST's Digital Library of Mathematical Functions

According to §19.4 of NIST's Digital Library of Mathematical Functions,

<math>~\frac{\mathrm{d}K\left(k\right)}{\mathrm{d}k}</math>

<math>~=</math>

<math>~ \frac{E\left(k\right)-{k^{\prime}}^{2}K\left(k\right)}{k{k^{\prime}}^{2}}, </math>

<math>~\frac{\mathrm{d}(E\left(k\right)-{k^{\prime}}^{2}K\left(k\right))}{\mathrm{d}k}</math>

<math>~=</math>

<math>~ kK\left(k\right), </math>

<math>~\frac{\mathrm{d}E\left(k\right)}{\mathrm{d}k}</math>

<math>~=</math>

<math>~ \frac{E\left(k\right)-K\left(k\right)}{k}, </math>

<math>~\frac{\mathrm{d}(E\left(k\right)-K\left(k\right))}{\mathrm{d}k}</math>

<math>~=</math>

<math>~ -\frac{kE\left(k\right)}{{k^{\prime}}^{2}}, </math>

<math>~\frac{{\mathrm{d}}^{2}E\left(k\right)}{{\mathrm{d}k}^{2}}</math>

<math>~=</math>

<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>

where,

<math>~k^{\prime} \equiv \sqrt{1 - k^2} \, .</math>

Then, drawing upon known expressions for the derivatives of elliptic integrals, as are now tidily catalogued online in NIST's Digital Library of Mathematical Functions, Dyson showed that,

<math>~\frac{d\mathfrak{I}(r,\theta,c)}{dc} = \frac{d}{dc}\biggl[\frac{4K(\mu)}{R_1+R}\biggr]</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ \biggl[ \frac{(R_1+R)E\left(\mu\right)}{cRR_1}\biggr] \cos\psi ~-~ \frac{K(\mu)}{c (R_1+R)}\biggl[2(1-\cos\psi)+4\cos\psi \biggr] </math>

 

<math>~=</math>

<math>~ \frac{1}{c}\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>

This expression appears at the top of Dyson's p. 62. Differentiating a second time gives,

<math>~\frac{d\mathfrak{I}(r,\theta,c)}{dc} = \frac{d}{dc}\biggl[\frac{4K(\mu)}{R_1+R}\biggr]</math>

<math>~=</math>

<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>


Whitworth's (1981) Isothermal Free-Energy Surface

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