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__FORCETOC__ <!-- __NOTOC__ will force TOC off --> =Properties of Homogeneous Ellipsoids (1)= <!-- {| class="JacobiEllipsoids" style="float:left; margin-right: 20px; border-style: solid; border-width: 3px border-color: black" |- ! style="height: 125px; width: 125px; background-color:white;" |[[H_BookTiledMenu#Equilibrium_Structures_2|<b>The<br />Gravitational<br />Potential]]<br />(A<sub>i</sub> coefficients)</b> |} --> ==Gravitational Potential== ===The Defining Integral Expressions=== As has been shown in a separate discussion titled, [[PGE/PoissonOrigin#Origin_of_the_Poisson_Equation|"Origin of the Poisson Equation,"]] the acceleration due to the gravitational attraction of a distribution of mass {{Math/VAR_Density01}} <math>(\vec{x})</math> can be derived from the gradient of a scalar potential {{Math/VAR_NewtonianPotential01}} <math>(\vec{x})</math> defined as follows: <div align="center"> <math> \Phi(\vec{x}) \equiv - \int \frac{G \rho(\vec{x}')}{|\vec{x}' - \vec{x}|} d^3 x' . </math> </div> As has been explicitly demonstrated in Chapter 3 of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>] and summarized in Table 2-2 (p. 57) of [<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>], for an homogeneous ellipsoid this volume integral can be evaluated analytically in closed form. Specifically, at an internal point or on the surface of an homogeneous ellipsoid with semi-axes <math>(x,y,z) = (a_1,a_2,a_3)</math>, <div align="center"> <math> \Phi(\vec{x}) = -\pi G \rho \biggl[ I_\mathrm{BT} a_1^2 - \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) \biggr], </math><br /> [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 3, Eq. (40)</font><sup>1,2</sup> <br /> [<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>], <font color="#00CC00">Chapter 2, Table 2-2</font> </div> [[File:CommentButton02.png|right|100px|Comment by J. E. Tohline on 15 August 2020: This integral definition of A_i also appears as Eq. (5) of §10.2 (p. 234) of T78, but it contains an error — in the denominator on the right-hand-side, a_1 appears instead of a_i.]]where, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> A_i </math> </td> <td align="center"> <math> \equiv </math> </td> <td align="left"> <math> a_1 a_2 a_3 \int_0^\infty \frac{du}{\Delta (a_i^2 + u )} , </math> </td> </tr> <tr> <td align="right"><math>I_\mathrm{BT}</math> </td> <td align="center"><math>\equiv</math> </td> <td align="left"> <math> \frac{a_2 a_3}{a_1} \int_0^\infty \frac{du}{\Delta} = A_1 + A_2\biggl(\frac{a_2}{a_1}\biggr)^2+ A_3\biggl(\frac{a_3}{a_1}\biggr)^2 , </math> </td> </tr> <tr> <td align="right"> <math> \Delta </math> </td> <td align="center"> <math> \equiv </math> </td> <td align="left"> <math> \biggl[ (a_1^2 + u)(a_2^2 + u)(a_3^2 + u) \biggr]^{1/2} . </math> </td> </tr> </table> <div align="center"> [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 3, Eqs. (18), (15 & 22)</font><sup>1</sup><font color="#00CC00">, & (8)</font>, respectively<br /> [<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>], <font color="#00CC00">Chapter 2, Table 2-2</font> </div> This definite-integral definition of <math>A_i</math> may also be found in: * [<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>]: as Eq. (6) in §114 (p. 153); and as Eq. (5) in §373 (p. 700). * [<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>]: as Eq. (5) in §10.2 (p. 234), but note that there is an error in the denominator of the right-hand-side — <math>a_1</math> appears instead of <math>a_i</math>. ===Evaluation of Coefficients=== As is [[#Derivation_of_Expressions_for_Ai|detailed below]], the integrals defining <math>A_i</math> and <math>I_\mathrm{BT}</math> can be evaluated in terms of the [http://en.wikipedia.org/wiki/Elliptic_integral#Incomplete_elliptic_integral_of_the_first_kind incomplete elliptic integral of the first kind], <div align="center"> <math> F(\theta,k) \equiv \int_0^\theta \frac{d\theta '}{\sqrt{1 - k^2 \sin^2\theta '}} \, , </math> </div> and/or the [http://en.wikipedia.org/wiki/Elliptic_integral#Incomplete_elliptic_integral_of_the_second_kind incomplete elliptic integral of the second kind], <div align="center"> <math> E(\theta,k) \equiv \int_0^\theta {\sqrt{1 - k^2 \sin^2\theta '}}d\theta ' \, , </math> </div> where, for our particular problem, <div align="center"> <math> \theta \equiv \cos^{-1} \biggl(\frac{a_3}{a_1} \biggr) \, , </math><br /> <math> k \equiv \biggl[\frac{a_1^2 - a_2^2}{a_1^2 - a_3^2} \biggr]^{1/2} = \biggl[\frac{1 - (a_2/a_1)^2}{1 - (a_3/a_1)^2} \biggr]^{1 / 2} \, , </math><br /> [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 3, Eq. (32)</font> </div> or the integrals can be evaluated in terms of more elementary functions if either <math>a_2 = a_1</math> ([[#Oblate_Spheroids|oblate spheroids]]) or <math>a_3 = a_2</math> ([[#Prolate_Spheroids|prolate spheroids]]). <span id="triaxial"> </span> ====Triaxial Configurations (a<sub>1</sub> > a<sub>2</sub> > a<sub>3</sub>)==== If the three principal axes of the configuration are unequal in length and related to one another such that <math>a_1 > a_2 > a_3 </math>, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> A_1 </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \frac{2a_2 a_3}{a_1^2} \biggl[ \frac{F(\theta,k) - E(\theta,k)}{k^2 \sin^3\theta} \biggr] \, ; </math> </td> </tr> <tr> <td align="right"> <math> A_2 </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \frac{2a_2 a_3}{a_1^2} \biggl[ \frac{E(\theta,k) - (1-k^2)F(\theta,k) - (a_3/a_2)k^2\sin\theta}{k^2 (1-k^2) \sin^3\theta}\biggr] \, ; </math> </td> </tr> <tr> <td align="right"> <math> A_3 </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \frac{2a_2 a_3}{a_1^2} \biggl[ \frac{(a_2/a_3) \sin\theta - E(\theta,k)}{(1-k^2) \sin^3\theta} \biggr] \, ; </math> </td> </tr> <tr> <td align="right"> <math> I_\mathrm{BT} </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \frac{2a_2 a_3}{a_1^2} \biggl[ \frac{F(\theta,k)}{\sin\theta} \biggr] \, . </math> </td> </tr> </table> <div align="center"> [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 3, Eqs. (33), (34) & (35)</font> </div> Notice that there is no need to specify the actual value of <math>a_1</math> in any of these expressions, as they each can be written in terms of the pair of axis ''ratios'', <math>a_2/a_1</math> and <math>a_3/a_1</math>. As a sanity check, let's see if these three expressions can be related to one another in the manner described by equation (108) in §21 of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], namely, <div align="center"> <math>\sum_{\ell=1}^3 A_\ell = 2 \, .</math> </div> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\frac{a_1^2}{2a_2 a_3} \biggl[A_1 + A_3 + A_2\biggr]</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\frac{F(\theta,k) - E(\theta,k)}{k^2 \sin^3\theta} + \frac{(a_2/a_3) \sin\theta - E(\theta,k)}{(1-k^2) \sin^3\theta} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>+ \frac{E(\theta,k) - (1-k^2)F(\theta,k) - (a_3/a_2)k^2\sin\theta}{k^2 (1-k^2) \sin^3\theta}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\frac{1}{k^2(1-k^2)\sin^3\theta} \biggl\{(1-k^2)F(\theta,k) - (1-k^2)E(\theta,k) + k^2(a_2/a_3) \sin\theta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>- k^2E(\theta,k) + E(\theta,k) - (1-k^2)F(\theta,k) - (a_3/a_2)k^2\sin\theta\biggr\}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\frac{1}{(1-k^2)\sin^2\theta} \biggl[ \frac{a_2}{a_3} - \frac{a_3}{a_2} \biggr]</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\frac{a_1^2}{a_2 a_3} \, .</math> </td> </tr> </table> Q.E.D. <span id="oblate"> </span> ====Oblate Spheroids (a<sub>1</sub> = a<sub>2</sub> > a<sub>3</sub>)==== If the longest axis, <math>a_1</math>, and the intermediate axis, <math>a_2</math>, of the ellipsoid are equal to one another, then an equatorial cross-section of the object presents a circle of radius <math>a_1</math> and the object is referred to as an '''oblate spheroid'''. For homogeneous oblate spheroids, evaluation of the integrals defining <math>A_i</math> and <math>I_\mathrm{BT}</math> gives, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> A_1 </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \frac{1}{e^2} \biggl[ \frac{\sin^{-1}e}{e} - (1-e^2)^{1/2} \biggr] (1-e^2)^{1/2} ~~; </math> </td> </tr> <tr> <td align="right"> <math> A_2 </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> A_1 \, ; </math> </td> </tr> <tr> <td align="right"><math>A_3</math> </td> <td align="center"><math>=</math> </td> <td align="left"> <math> \frac{2}{e^2} \biggl[ (1-e^2)^{-1/2} - \frac{\sin^{-1}e}{e} \biggr] (1-e^2)^{1 / 2} \, ; </math> </td> </tr> <tr> <td align="right"><math>I_\mathrm{BT}</math> </td> <td align="center"><math>=</math> </td> <td align="left"> <math> 2A_1 + A_3 (1-e^2) = 2 (1-e^2)^{1/2} \biggl[ \frac{\sin^{-1}e}{e} \biggr] \, , </math> </td> </tr> </table> <div align="center"> [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 3, Eq. (36)</font><br /> [<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>], <font color="#00CC00">§4.5, Eqs. (48) & (49)</font> </div> where the eccentricity, <div align="center"> <math> e \equiv \biggl[1 - \biggl(\frac{a_3}{a_1}\biggr)^2 \biggr]^{1 / 2} \, . </math> </div> <span id="prolate"> </span> ====Prolate Spheroids (a<sub>1</sub> > a<sub>2</sub> = a<sub>3</sub>)==== If the shortest axis <math>(a_3)</math> and the intermediate axis <math>(a_2)</math> of the ellipsoid are equal to one another — and the symmetry (longest, <math>a_1</math>) axis is aligned with the <math>x</math>-axis — then a cross-section in the <math>y-z</math> plane of the object presents a circle of radius <math>a_3</math> and the object is referred to as a '''prolate spheroid'''. For homogeneous prolate spheroids, evaluation of the integrals defining <math>A_i</math> and <math>I_\mathrm{BT}</math> gives, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> A_1 </math> </td> <td align="center"><math>=</math> </td> <td align="left"> <math> \ln\biggl[ \frac{1+e}{1-e} \biggr] \frac{(1-e^2)}{e^3} - \frac{2(1-e^2)}{e^2} \, ; </math> </td> </tr> <tr> <td align="right"> <math> A_2 </math> </td> <td align="center"><math>=</math> </td> <td align="left"> <math> \frac{1}{e^2} - \ln\biggl[ \frac{1+e}{1-e} \biggr]\frac{(1-e^2)}{2e^3} \, ; </math> </td> </tr> <tr> <td align="right"> <math> A_3 </math> </td> <td align="center"><math>=</math> </td> <td align="left"> <math> A_2 \, ; </math> </td> </tr> <tr> <td align="right"> <math> I_\mathrm{BT} </math> </td> <td align="center"><math>=</math> </td> <td align="left"> <math>~ A_1 + 2(1-e^2)A_2 = \ln\biggl[ \frac{1+e}{1-e} \biggr]\frac{(1-e^2)}{e} \, , </math> </td> </tr> </table> <div align="center"> [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 3, Eq. (38)</font> </div> where, again, the eccentricity, <div align="center"> <math> e \equiv \biggl[1 - \biggl(\frac{a_3}{a_1}\biggr)^2 \biggr]^{1/2} \, . </math> </div> <font color="red">NOTE:</font> If, instead, the longest (and, in this case, symmetry) axis of the prolate mass distribution is aligned with the <math>z</math>-axis — in which case, <math>a_1 = a_2 < a_3</math> — then, <math>e = (1 - a_1^2/a_3^2)^{1 / 2}</math> and the mathematical expressions for the <math>A_i</math> coefficients must be altered; they are essentially "swapped." This modified set of coefficient expressions can be found in a [[Aps/MaclaurinSpheroidFreeFall#Prolate_Spheroids|parallel discussion]] of the potential inside and on the surface of prolate-spheroidal mass distributions, as well as in the second column of Table 2-1 (p. 57) of [<b>[[Appendix/References#BT87|<font color="red">BT87</font>]]</b>]. ==Example Evaluations== Here we adopt the notation mapping, <math>~(a_1, a_2, a_3) ~\leftrightarrow~ (a,b,c)</math>. In general, for a given pair of axis ratios, <math>~(\tfrac{b}{a}, \tfrac{c}{a})</math>, a determination of the coefficients, <math>~A_1</math>, <math>~A_2</math>, and <math>~A_3</math>, requires evaluation of elliptic integrals. For practical applications, we have decided to evaluate these special functions using the fortran functions provided in association with the book, ''Numerical Recipes in Fortran''; in order to obtain the results presented in our Table 2, below, we modified those default (single-precision) routines to generate results with double-precision accuracy. Along the way (see results posted in our Table 1), we pulled cruder evaluations of both elliptic integrals, <math>~F(\theta,k)</math> and <math>~E(\theta,k)</math>, from the printed special-functions table found in a CRC handbook. As we developed/debugged the numerical tool that would allow us to determine the values of these three coefficients for arbitrary choices of the pair of axis ratios, it was important that we compare the results of our calculations to those that have appeared in the published literature. As a primary point of comparison, we chose to use ''The properties of the Jacobi ellipsoids'' as tabulated in §39 (Chapter 6) of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>]. In particular, for twenty-six separate axis-ratio pairs, Chandrasekhar's Table IV lists the values of the square of the angular velocity, <math>~\Omega^2</math>, and the total angular momentum, <math>~L</math>, of an equilibrium Jacobi ellipsoid that is associated with each axis-ratio pair. We should be able to duplicate — or, via double-precision arithmetic, improve — Chandrasekhar's tabulated results using the expressions for "omega2", <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\Omega^2}{\pi G\rho}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2B_{12}</math> </td> </tr> <tr><td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">§39, Eq. (5)</font> </td></tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2\biggl[\frac{A_1 - (b/a)^2A_2}{1-(b/a)^2} \biggr] \, ,</math> </td> </tr> <tr><td align="center" colspan="3">using, [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">§21, Eqs. (105) & (107)</font></td></tr> </table> </div> and, for "angmom", <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{L}{(GM^3)^{1/2}(abc)^{1/6}}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{\sqrt{3}}{10}\biggl[ \frac{a^2 + b^2}{(abc)^{2/3}} \biggr]\biggl(\frac{\Omega^2}{\pi G \rho}\biggr)^{1/2} </math> </td> </tr> <tr><td align="center" colspan="3">[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">§39, Eq. (16)</font></td></tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{\sqrt{3}}{10}\biggl[ \frac{1 + (b/a)^2}{(b/a)^{2/3}(c/a)^{2/3}} \biggr]\biggl(\frac{\Omega^2}{\pi G \rho}\biggr)^{1/2} \, .</math> </td> </tr> </table> </div> Or, in connection with the free-energy discussion found in [http://adsabs.harvard.edu/abs/1995ApJ...446..472C D. M. Christodoulou, D. Kazanas, I. Shlosman, & J. E. Tohline (1995, ApJ, 446, 472)], <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{5L}{M}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~a^2\biggl[ 1 + \biggl(\frac{b}{a}\biggr)^2 \biggr]\biggl[\frac{\Omega^2}{\pi G \rho}\biggr]^{1/2}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{15}{4}\biggl(\frac{b}{a}\biggr)^{-1}\biggl(\frac{c}{a}\biggr)^{-1} \biggr]^{2/3} \biggl[ 1 + \biggl(\frac{b}{a}\biggr)^2 \biggr]\biggl[\frac{\Omega^2}{\pi G \rho}\biggr]^{1/2}</math> </td> </tr> </table> </div> <span id="Table1"> </span> <table align="center" cellpadding="5" border="1"> <tr> <th align="center" colspan="12"><font size="+1">Table 1: Example Evaluations</font></th> </tr> <tr> <th align="center" colspan="2">Given</th> <th align="center" colspan="10">Determined using calculator and (crude) CRC tables of elliptic integrals</th> </tr> <tr> <td align="center" rowspan="2"><math>~\frac{a_2}{a_1}</math></td> <td align="center" rowspan="2"><math>~\frac{a_3}{a_1}</math></td> <td align="center" colspan="2"><math>~\theta</math></td> <td align="center" rowspan="2"><math>~k</math></td> <td align="center" colspan="2"><math>~\sin^{-1}k</math></td> <td align="center" rowspan="2"><math>~F(\theta,k)</math></td> <td align="center" rowspan="2"><math>~E(\theta,k)</math></td> <td align="center" rowspan="2"><math>~A_1</math></td> <td align="center" rowspan="2"><math>~A_2</math></td> <td align="center" rowspan="2"><math>~A_3</math></td> </tr> <tr> <td align="center">radians</td> <td align="center">degrees</td> <td align="center">radians</td> <td align="center">degrees</td> </tr> <tr> <td align="right">1.00</td> <td align="right">0.582724</td> <td align="right">0.94871973</td> <td align="right">54.3576</td> <td align="right">0.00000000</td> <td align="right">0.00000000</td> <td align="right">0.000000</td> <td align="right">0.94871973</td> <td align="right">0.94871973</td> <td align="right">0.51589042</td> <td align="right">0.51589042</td> <td align="right">0.96821916</td> </tr> <tr> <td align="right">0.96</td> <td align="right">0.570801</td> <td align="right">0.96331527</td> <td align="right">55.1939</td> <td align="right">0.34101077</td> <td align="right">0.34799191</td> <td align="right">19.9385</td> <td align="right">0.975</td> <td align="right">0.946</td> <td align="right">+0.4937</td> <td align="right">+0.5319</td> <td align="right">+0.9744</td> </tr> <tr> <td align="right">0.60</td> <td align="right">0.433781</td> <td align="right">1.12211141</td> <td align="right">64.292</td> <td align="right">0.88788426</td> <td align="right">1.09272580</td> <td align="right">62.609</td> <td align="right">1.3375</td> <td align="right">0.9547</td> <td align="right">0.3455</td> <td align="right">0.6741</td> <td align="right">0.9803</td> </tr> </table> <b>With regard to our Table 1 (immediately above):</b> To begin with, we picked three axis-ratio pairs from Table IV of EFE, and considered them to be "given." For each pair, we used a hand-held calculator to calculate the corresponding values of the two arguments of the elliptic integrals, namely, <math>~\theta</math> and <math>~k</math>, as [[#Evaluation_of_Coefficients|defined above]]. By default, each determined value of <math>~\theta</math> is in radians. Because the published CRC special-functions tables quantify both arguments of the special functions in angular ''degrees'', we converted <math>~\theta</math> from radians to degrees (see column 4 of Table 1) and, similarly, we converted <math>~\sin^{-1}k</math> to degrees (see column 7 of Table 1). For the axisymmetric configuration — the first row of numbers in Table 1, for which <math>~a_2/a_1 = 1</math> — the coefficients, <math>~A_1</math>, <math>~A_2</math>, and <math>~A_3</math>, were determined to eight digits of precision using the [[#Oblate_Spheroids|appropriate expressions for oblate spheroids]]. Note that, in this axisymmetric case, <math>~F(\theta,0) = E(\theta,0) = \theta</math>, but these function values are irrelevant with respect to the determination of the <math>~A_\ell</math> coefficients. <div align="center" id="Table2"> <table border="1" cellpadding="5" align="center"> <tr> <th align="center" colspan="1"> <font size="+1">Table 2: Double-Precision Evaluations</font><p></p> Related to Table IV in [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 6, §39 (p. 103)</font> </th> </tr> <tr><td align="left"> <pre> precision b/a c/a F E A1 A2 A3 [2-(A1+A2+A3)]/2 1.00 0.582724 ----- ----- 5.158904180D-01 5.158904180D-01 9.682191640D-01 0.0D+00 0.96 0.570801 9.782631357D-01 9.487496699D-01 5.024584655D-01 5.292952683D-01 9.682462661D-01 4.4D-16 0.92 0.558330 1.009516282D+00 9.489290273D-01 4.884500698D-01 5.432292722D-01 9.683206580D-01 0.0D+00 0.88 0.545263 1.042655826D+00 9.492826127D-01 4.738278227D-01 5.577100115D-01 9.684621658D-01 2.2D-16 0.84 0.531574 1.077849658D+00 9.498068890D-01 4.585648648D-01 5.727687434D-01 9.686663918D-01 2.2D-16 0.80 0.517216 1.115314984D+00 9.505192815D-01 4.426242197D-01 5.884274351D-01 9.689483451D-01 -4.4D-16 0.76 0.502147 1.155290552D+00 9.514282210D-01 4.259717080D-01 6.047127268D-01 9.693155652D-01 2.2D-16 0.72 0.486322 1.198053140D+00 9.525420558D-01 4.085724682D-01 6.216515450D-01 9.697759868D-01 -4.4D-16 0.68 0.469689 1.243931393D+00 9.538724717D-01 3.903895871D-01 6.392680107D-01 9.703424022D-01 2.2D-16 0.64 0.452194 1.293310292D+00 9.554288569D-01 3.713872890D-01 6.575860416D-01 9.710266694D-01 4.4D-16 0.60 0.433781 1.346645618D+00 9.572180643D-01 3.515319835D-01 6.766289416D-01 9.718390749D-01 -3.3D-16 0.56 0.414386 1.404492405D+00 9.592491501D-01 3.307908374D-01 6.964136019D-01 9.727955606D-01 -6.7D-16 0.52 0.393944 1.467522473D+00 9.615263122D-01 3.091371405D-01 7.169543256D-01 9.739085339D-01 4.4D-16 0.48 0.372384 1.536570313D+00 9.640523748D-01 2.865506903D-01 7.382563770D-01 9.751929327D-01 -2.2D-16 0.44 0.349632 1.612684395D+00 9.668252052D-01 2.630231082D-01 7.603153245D-01 9.766615673D-01 8.9D-16 0.40 0.325609 1.697213059D+00 9.698379297D-01 2.385623719D-01 7.831101146D-01 9.783275135D-01 0.0D+00 0.36 0.300232 1.791930117D+00 9.730763540D-01 2.132011181D-01 8.065964525D-01 9.802024294D-01 2.2D-15 0.32 0.273419 1.899227853D+00 9.765135895D-01 1.870102340D-01 8.307027033D-01 9.822870627D-01 -1.3D-15 0.28 0.245083 2.022466812D+00 9.801112910D-01 1.601127311D-01 8.553054155D-01 9.845818534D-01 -2.4D-15 0.24 0.215143 2.166555572D+00 9.838093161D-01 1.327137129D-01 8.802197538D-01 9.870665333D-01 1.4D-14 0.20 0.183524 2.339102805D+00 9.875217566D-01 1.051389104D-01 9.051602520D-01 9.897008376D-01 -1.6D-14 0.16 0.150166 2.552849055D+00 9.911267582D-01 7.790060179D-02 9.296886827D-01 9.924107155D-01 -3.4D-14 0.12 0.115038 2.831664019D+00 9.944537935D-01 5.180880535D-02 9.531203882D-01 9.950708065D-01 1.4D-13 0.08 0.078166 3.229072310D+00 9.972669475D-01 2.817821170D-02 9.743504218D-01 9.974713665D-01 3.9D-13 0.04 0.039688 3.915557866D+00 9.992484565D-01 9.281550546D-03 9.914470033D-01 9.992714461D-01 9.8D-13 </pre> </td></tr> </table> </div> <b>With regard to our Table 2 (immediately above):</b> Next, given each pair of axis ratios, <math>~(\tfrac{b}{a},\tfrac{c}{a})</math> — copied from Table IV of EFE (see columns 1 and 2 of our Table 2) — we used some fortran routines from [http://numerical.recipes/ Numerical Recipes] to calculate <math>~F(\theta,k)</math> and <math>~E(\theta,k)</math> (see columns 3 and 4 of our Table 2); we converted the routines to accommodate double-precision arithmetic. We subsequently evaluated the coefficients, <math>~A_1</math>, <math>~A_2</math>, and <math>~A_3</math>, (columns 5, 6, & 7 of Table 2) using the expressions given above, then demonstrated that, in each case, the three coefficients sum to 2.0 to better than twelve digits accuracy. =Derivation of Expressions for A<sub>i</sub>= Let's carry out the integrals that appear in the definition of the <math>~A_i</math> coefficients, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> ~A_i </math> </td> <td align="center"> <math> ~\equiv </math> </td> <td align="left"> <math> ~a_\ell a_m a_s \int_0^\infty \frac{du}{\Delta (a_i^2 + u )} , </math> </td> </tr> </table> where, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> ~\Delta </math> </td> <td align="center"> <math> ~\equiv </math> </td> <td align="left"> <math> ~\biggl[ (a_\ell^2 + u)(a_m^2 + u)(a_s^2 + u) \biggr]^{1/2} \, . </math> </td> </tr> </table> Here, we are adopting the <math>~(\ell, m, s)</math> subscript notation to identify which semi-axis length is the (largest, medium-length, smallest). ==Evaluating A<sub>ℓ</sub>== First, let's focus on the coefficient associated with the longest axis <math>~(i = \ell)</math>: <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_\ell}{a_\ell a_m a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_0^\infty \biggl[ (a_\ell^2 + u)^3(a_m^2 + u)(a_s^2 + u) \biggr]^{-1 / 2} du </math> </td> </tr> </table> Changing the integration variable to <math>~x \equiv -u</math>, we obtain a definite integral expression that appears as equation (3.133.1) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ I. W. Gradshteyn & I. M. Ryzhik (2007; 7<sup>th</sup> Edition)], ''Table of Integrals, Series, and Products'' — hereafter, GR7<sup>th</sup> — namely, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_\ell}{a_\ell a_m a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_{-\infty}^0 \biggl[ (a_\ell^2 - x)^3(a_m^2 - x)(a_s^2 - x) \biggr]^{-1 / 2} dx </math> </td> <td align="left"> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\frac{2}{(a_\ell^2 - a_m^2) \sqrt{a_\ell^2 - a_s^2}} \biggl[ F(\alpha, p) - E(\alpha, p) \biggr] </math> </td> <td align="left"> … valid for <math>[a_\ell > a_m > a_s \ge 0]</math></td> </tr> <tr> <td align="center" colspan="4">[https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], p. 255, Eq. (3.133.1)</td> </tr> </table> where (see p. 254 of [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>]), <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\sin^2\alpha</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{a_\ell^2 - a_s^2}{a_\ell^2 - 0} = 1 - \frac{a_s^2}{a_\ell^2} \, ,</math> </td> <td align="center" width="20%"> </td> <td align="right"> <math>~p</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\biggl[ \frac{a_\ell^2 - a_m^2}{a_\ell^2 - a_s^2} \biggr]^{1 / 2} \, ,</math> </td> </tr> </table> and where, <math>E(\alpha, p)</math> and <math>F(\alpha, p)</math> are [https://dlmf.nist.gov/19.2#ii Legendre incomplete elliptic integrals of the first and second kind], respectively. (Note that in the notation convention adopted by [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], the order of the argument list, <math>~(\alpha, p)</math>, is flipped relative to the convention that we have adopted [[#Evaluation_of_Coefficients|above]] and elsewhere throughout our online, MediaWiki-based chapters.) Recognizing that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~p^2 \sin^3\alpha</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\frac{a_\ell^2 - a_s^2}{a_\ell^2 }\biggr]^{3 / 2}\biggl[ \frac{a_\ell^2 - a_m^2}{a_\ell^2 - a_s^2} \biggr] = \frac{(a_\ell^2 - a_s^2)^{1 / 2}}{a_\ell^3 } \biggl[ a_\ell^2 - a_m^2 \biggr] \, , </math> </td> </tr> </table> we see that the expression for <math>~A_\ell</math> can be rewritten as, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_\ell}{a_\ell a_m a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\frac{2}{a_\ell^3 ~p^2 \sin^3\alpha} \biggl[ F(\alpha, p) - E(\alpha, p) \biggr] \, . </math> </td> </tr> </table> This matches the expression that we have provided for <math>~A_1</math>, [[#Triaxial_Configurations|above in the context of triaxial configurations]]. ==Evaluating A<sub>m</sub>== Next, let's evaluate the coefficient associated with the axis of intermediate length <math>~(i = m)</math>: <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_m}{a_\ell a_m a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_0^\infty \biggl[ (a_\ell^2 + u)(a_m^2 + u)^3(a_s^2 + u) \biggr]^{-1 / 2} du \, . </math> </td> </tr> </table> This time, by changing the integration variable to <math>~x \equiv -u</math>, we obtain a definite integral expression that appears as equation (3.133.7) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], namely, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_m}{a_\ell a_m a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_{-\infty}^0 \biggl[ (a_\ell^2 - x)(a_m^2 - x)^3(a_s^2 - x) \biggr]^{-1 / 2} dx </math> </td> <td align="left"> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\frac{2\sqrt{a_\ell^2 - a_s^2}}{(a_\ell^2 - a_m^2) (a_m^2 - a_s^2)} E(\alpha, p) - \frac{2}{(a_\ell^2 - a_m^2) \sqrt{a_\ell^2 - a_s^2}} F(\alpha, p) - \frac{2}{a_m^2 - a_s^2} \biggl[ \frac{a_s^2}{a_\ell^2 a_m^2} \biggr]^{1 / 2} </math> </td> <td align="left"> … valid for <math>[a_\ell > a_m > a_s \ge 0]</math></td> </tr> <tr> <td align="center" colspan="4">[https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], p. 256, Eq. (3.133.7)</td> </tr> </table> (Here, the parameters, <math>~\alpha</math> and <math>~p</math>, have the same definitions as in our [[#Evaluating_A.E2.84.93|above evaluation of]] <math>~A_\ell</math>.) This time it is useful to recognize that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~1 -p^2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 1 - \frac{a_\ell^2 - a_m^2}{a_\ell^2 - a_s^2} = \frac{a_m^2 - a_s^2 }{a_\ell^2 - a_s^2} </math> </td> </tr> </table> in which case, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~p^2(1-p^2) \sin^3\alpha</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ ( a_\ell^2 - a_m^2 )( a_m^2 - a_s^2 )}{a_\ell^3 (a_\ell^2 - a_s^2)^{1 / 2}} \, . </math> </td> </tr> </table> So the coefficient, <math>~A_m</math>, may be rewritten as, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~p^2 (1-p^2) \sin^3\alpha\biggl[ \frac{A_m}{a_\ell a_m a_s} \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ ( a_\ell^2 - a_m^2 )( a_m^2 - a_s^2 )}{a_\ell^3 (a_\ell^2 - a_s^2)^{1 / 2}} \biggl\{ \frac{2\sqrt{a_\ell^2 - a_s^2}}{(a_\ell^2 - a_m^2) (a_m^2 - a_s^2)} E(\alpha, p) ~-~ \frac{2}{(a_\ell^2 - a_m^2) \sqrt{a_\ell^2 - a_s^2}} F(\alpha, p) ~-~\frac{2}{a_m^2 - a_s^2} \biggl[ \frac{a_s^2}{a_\ell^2 a_m^2} \biggr]^{1 / 2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ 2}{a_\ell^3 } \biggl\{ E(\alpha, p) \biggr\} -~\frac{ 2( a_m^2 - a_s^2 )}{a_\ell^3 (a_\ell^2 - a_s^2)} \biggl\{ F(\alpha, p) \biggr\} -~\frac{ 2( a_\ell^2 - a_m^2 )}{a_\ell^3 (a_\ell^2 - a_s^2)^{1 / 2}} \biggl[ \frac{a_s}{a_\ell a_m} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ 2}{a_\ell^3 } \biggl\{ E(\alpha, p) -~(1-p^2) F(\alpha, p) -~p^2\sin\alpha \biggl[ \frac{a_s}{a_m} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \frac{A_m}{a_\ell a_m a_s} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ 2}{a_\ell^3 } \biggl[ \frac{ E(\alpha, p) -~(1-p^2) F(\alpha, p) -~(a_s/a_m)p^2\sin\alpha}{p^2 (1-p^2)\sin^3\alpha} \biggr] \, . </math> </td> </tr> </table> This matches the expression that we have provided for <math>~A_2</math>, [[#Triaxial_Configurations|above in the context of triaxial configurations]]. ==Evaluating A<sub>s</sub>== Finally, let's evaluate the coefficient associated with the shortest axis, <math>~(i = s)</math>: <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_s}{a_\ell a_m a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_0^\infty \biggl[ (a_\ell^2 + u)(a_m^2 + u)(a_s^2 + u)^3 \biggr]^{-1 / 2} du \, . </math> </td> </tr> </table> By changing the integration variable to <math>~x \equiv -u</math>, this time we obtain a definite integral expression that appears as equation (3.133.13) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], namely, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_s}{a_\ell a_m a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_{-\infty}^0 \biggl[ (a_\ell^2 - x)(a_m^2 - x)(a_s^2 - x)^3 \biggr]^{-1 / 2} dx </math> </td> <td align="left"> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\frac{2}{(a_s^2 - a_m^2) \sqrt{a_\ell^2 - a_s^2}} E(\alpha, p) + \frac{2}{a_m^2 - a_s^2} \biggl[ \frac{a_m^2}{a_\ell^2 a_s^2} \biggr]^{1 / 2} </math> </td> <td align="left"> … valid for <math>[a_\ell > a_m > a_s > 0]</math></td> </tr> <tr> <td align="center" colspan="4">[https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], p. 256, Eq. (3.133.13)</td> </tr> </table> (And, again, the parameters, <math>~\alpha</math> and <math>~p</math>, have the same definitions as in our [[#Evaluating_A.E2.84.93|above evaluation of]] <math>~A_\ell</math>.) Recognizing that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(1-p^2) \sin^3\alpha</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ (a_\ell^2 - a_s^2)^{1 / 2}( a_m^2 - a_s^2 )}{a_\ell^3 } \, , </math> </td> </tr> </table> the coefficient, <math>~A_s</math>, may be rewritten as, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(1-p^2) \sin^3\alpha\biggl[ \frac{A_s}{a_\ell a_m a_s} \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ (a_\ell^2 - a_s^2)^{1 / 2}( a_m^2 - a_s^2 )}{a_\ell^3 } \biggl\{ ~\frac{2}{(a_s^2 - a_m^2) \sqrt{a_\ell^2 - a_s^2}} E(\alpha, p) + \frac{2}{a_m^2 - a_s^2} \biggl[ \frac{a_m^2}{a_\ell^2 a_s^2} \biggr]^{1 / 2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ 2}{a_\ell^3 } \biggl\{ - E(\alpha, p) \biggr\} ~+~ \frac{ 2(a_\ell^2 - a_s^2)^{1 / 2}}{a_\ell^3 } \biggl[ \frac{a_m}{a_\ell a_s} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ 2}{a_\ell^3 } \biggl\{ \biggl(\frac{a_m}{a_s}\biggr) \sin\alpha - E(\alpha, p) \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \frac{A_s}{a_\ell a_m a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ 2}{a_\ell^3 } \biggl[\frac{ (a_m/a_s) \sin\alpha - E(\alpha, p)}{ (1-p^2) \sin^3\alpha } \biggr] \, . </math> </td> </tr> </table> This matches the expression that we have provided for <math>~A_3</math>, [[#Triaxial_Configurations|above in the context of triaxial configurations]]. ==When a<sub>m</sub> = a<sub>ℓ</sub>== When the length of the intermediate axis is the same as the length of the longest axis — that is, when we are dealing with an oblate spheroid — we can write, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> \Delta </math> </td> <td align="center"> <math> \equiv </math> </td> <td align="left"> <math> \biggl[ (a_\ell^2 + u)(a_m^2 + u)(a_s^2 + u) \biggr]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl[ (a_\ell^2 + u)^2(a_s^2 + u) \biggr]^{1/2} \, . </math> </td> </tr> </table> ===Index Symbols of the 1<sup>st</sup> Order=== Keeping in mind that, generically, the 1<sup>st</sup>-order index symbol is given by the expression, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> ~A_i </math> </td> <td align="center"> <math> ~\equiv </math> </td> <td align="left"> <math> ~a_\ell a_m a_s \int_0^\infty \frac{du}{\Delta (a_i^2 + u )} , </math> </td> </tr> <tr> <td align="center" colspan="3"> [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 3, p. 54, Eq. (103)</font> </td> </tr> </table> the coefficient associated with the longest axis is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_\ell}{a_\ell^2 a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_0^\infty \frac{du}{ (a_\ell^2 + u)^2(a_s^2 + u)^{1 / 2} } \, . </math> </td> </tr> </table> Changing the integration variable to <math>~x \equiv (a_\ell^2 + u)</math>, we obtain an integral expression that appears as equation (2.228.1) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], namely, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_\ell}{a_\ell^2 a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_{a_\ell^2}^\infty \frac{dx}{ x^2(a_s^2 - a_\ell^2 + x)^{1 / 2} } </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> - \biggl[ \frac{\sqrt{a_s^2 - a_\ell^2 + x}}{(a_s^2 - a_\ell^2) x}\biggr]_{a_\ell^2}^\infty - \frac{1}{2(a_s^2 - a_\ell^2)} ~\int_{a_\ell^2}^\infty \frac{dx}{ x(a_s^2 - a_\ell^2 + x)^{1 / 2} } </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>- \frac{a_s}{(a_\ell^2 - a_s^2 ) a_\ell^2} + \frac{1}{2(a_\ell^2 - a_s^2)} ~\int_{a_\ell^2}^\infty \frac{dx}{ x(a_s^2 - a_\ell^2 + x)^{1 / 2} } \, . </math> </td> </tr> </table> The remaining integral in this expression appears as equation (2.224.5) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>]. Its resolution depends on the sign of the constant term in the denominator, <math>~(a_s^2 - a_\ell^2)</math>. Given that this term is negative, the integration gives, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_\ell}{a_\ell^2 a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \frac{a_s}{(a_\ell^2 - a_s^2) a_\ell^2} + \frac{1}{2(a_\ell^2 - a_s^2)} ~\biggl\{ \frac{2}{ (a_\ell^2 - a_s^2)^{1 / 2} } \tan^{-1}\bigg[ \frac{(a_s^2 - a_\ell^2 + x)^{1 / 2} }{ (a_\ell^2 - a_s^2)^{1 / 2}} \biggr] \biggr\}_{a_\ell^2}^\infty </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~A_\ell</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \frac{a_s^2}{(a_\ell^2 - a_s^2) } + ~\frac{a_\ell^2 a_s}{ (a_\ell^2 - a_s^2)^{3 / 2} } \biggl\{\frac{\pi}{2} - \tan^{-1}\bigg[ \frac{a_s }{ (a_\ell^2 - a_s^2)^{1 / 2}} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \frac{(1-e^2)}{e^2} + ~\frac{(1-e^2)^{1 / 2}}{ e^3 } \biggl\{\frac{\pi}{2} - \tan^{-1}\bigg[ \frac{ (1-e^2)^{1 / 2}}{ e} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \frac{(1-e^2)}{e^2} + ~\frac{(1-e^2)^{1 / 2}}{ e^3 } \biggl\{\frac{\pi}{2} - \cos^{-1}e \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \frac{(1-e^2)}{e^2} + ~\frac{(1-e^2)^{1 / 2}}{ e^3 } \biggl\{ \sin^{-1}e \biggr\} \, , </math> </td> </tr> </table> where, <math>~e \equiv (1 - a_s^2/a_\ell^2)^{1 / 2}</math>. Similarly, the coefficient associated with the shortest axis is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_s}{a_\ell^2 a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_0^\infty \frac{du}{ (a_\ell^2 + u)(a_s^2 + u)^{3 / 2} } \, . </math> </td> </tr> </table> This time, after changing the integration variable to <math>~x \equiv (a_\ell^2 + u)</math>, we obtain an integral expression that appears as equation (2.229.1) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], namely, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_s}{a_\ell^2 a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_{a_\ell^2}^\infty \frac{dx}{ x(a_s^2 - a_\ell^2 + x)^{3 / 2} } </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> \biggl[ \frac{2}{(a_s^2 - a_\ell^2) \sqrt{a_s^2 - a_\ell^2 + x}}\biggr]_{a_\ell^2}^\infty + \frac{1}{(a_s^2 - a_\ell^2)} ~\int_{a_\ell^2}^\infty \frac{dx}{ x(a_s^2 - a_\ell^2 + x)^{1 / 2} } \, . </math> </td> </tr> </table> As before, the remaining integral in this expression appears as equation (2.224.5) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>]; and, as before, the sign of the constant term in the denominator, <math>~(a_s^2 - a_\ell^2)</math>, is negative. Hence, the integration gives, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_s}{a_\ell^2 a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> \frac{2}{(a_\ell^2 - a_s^2) a_s} - \frac{1}{(a_\ell^2 - a_s^2)} ~\biggl\{ \frac{2}{ (a_\ell^2 - a_s^2)^{1 / 2} } \tan^{-1}\bigg[ \frac{(a_s^2 - a_\ell^2 + x)^{1 / 2} }{ (a_\ell^2 - a_s^2)^{1 / 2}} \biggr] \biggr\}_{a_\ell^2}^\infty </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ A_s</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> \frac{2a_\ell^2 a_s}{(a_\ell^2 - a_s^2) a_s} - \frac{2a_\ell^2 a_s}{(a_\ell^2 - a_s^2)^{3 / 2}} \biggl\{\frac{\pi}{2} - \tan^{-1}\bigg[ \frac{a_s }{ (a_\ell^2 - a_s^2)^{1 / 2}} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> \frac{2}{e^2} - \frac{2(1-e^2)^{1 / 2}}{e^3} \biggl\{\frac{\pi}{2} - \tan^{-1}\bigg[ \frac{ (1-e^2)^{1 / 2}}{ e} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> \frac{2}{e^2} - \frac{2(1-e^2)^{1 / 2}}{e^3} \biggl\{\sin^{-1}e \biggr\} \, . </math> </td> </tr> </table> Because we are evaluating the case where <math>~A_m = A_\ell</math>, we alternatively should have been able to obtain the expression for <math>~A_s</math> immediately from our derived expression for <math>~A_\ell</math> via the known relation, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~A_\ell + A_m + A_s = 2A_\ell + A_s \, .</math> </td> </tr> </table> This approach gives, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~A_s</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2 - 2A_\ell </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2 + 2\biggl\{ \frac{(1-e^2)}{e^2} - ~\frac{(1-e^2)^{1 / 2}}{ e^3 } \biggl[ \sin^{-1}e \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2}{e^2} - ~\frac{2(1-e^2)^{1 / 2}}{ e^3 } \biggl[ \sin^{-1}e \biggr] \, , </math> </td> </tr> </table> which, indeed, matches our separately derived expression for <math>~A_s</math>. ===Index Symbols of the 2<sup>nd</sup> Order=== Keeping in mind that, generically, the 2<sup>nd</sup>-order index symbol is given by the expression, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> A_{ij} </math> </td> <td align="center"> <math> \equiv </math> </td> <td align="left"> <math> a_\ell a_m a_s \int_0^\infty \frac{du}{\Delta (a_i^2 + u )(a_j^2 + u )} , </math> </td> </tr> <tr> <td align="center" colspan="3"> [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 3, p. 54, Eq. (103)</font> </td> </tr> </table> and, in addition, for oblate-spheroidal configurations, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> \Delta </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl[ (a_\ell^2 + u)^2(a_s^2 + u) \biggr]^{1/2} \, , </math> </td> </tr> </table> we have the following three independent expressions: <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> A_{\ell \ell} = A_{m m} = A_{m \ell} </math> </td> <td align="center"> <math> \equiv </math> </td> <td align="left"> <math> a_\ell^2 a_s \int_0^\infty \frac{du}{(a_\ell^2 + u)^3(a_s^2 + u)^{1 / 2} } \, ; </math> </td> </tr> <tr> <td align="right"> <math> A_{s\ell} = A_{sm} </math> </td> <td align="center"> <math> \equiv </math> </td> <td align="left"> <math> a^2_\ell a_s \int_0^\infty \frac{du}{(a_\ell^2 + u)^2(a_s^2 + u)^{3 / 2} } \, ; </math> </td> </tr> <tr> <td align="right"> <math> A_{ss} </math> </td> <td align="center"> <math> \equiv </math> </td> <td align="left"> <math> a_\ell^2 a_s \int_0^\infty \frac{du}{(a_\ell^2 + u)(a_s^2 + u)^{5 / 2} } \, . </math> </td> </tr> </table> Setting <math>x \equiv (a_\ell^2 + u)</math> in each expression gives: <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> \frac{A_{\ell \ell}}{a_\ell^2 a_s } = \frac{A_{m m}}{a_\ell^2 a_s } = \frac{A_{m \ell}}{a_\ell^2 a_s } </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \int_{a_\ell^2}^\infty \frac{dx}{x^3(a_s^2 - a_\ell^2 + x)^{1 / 2} } </math> </td> <td align="right"> <math>\cdots</math> See equation (2.228.2) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>] </td> </tr> <tr> <td align="right"> <math> \frac{A_{s \ell}}{a_\ell^2 a_s } = \frac{A_{s m}}{a_\ell^2 a_s } </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \int_{a_\ell^2}^\infty \frac{dx}{x^2(a_s^2 - a_\ell^2 + x)^{3 / 2} } </math> </td> <td align="right"> <math>\cdots</math> See equation (2.229.2) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>] </td> </tr> <tr> <td align="right"> <math> \frac{A_{s s}}{a_\ell^2 a_s } </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \int_{a_\ell^2}^\infty \frac{dx}{x(a_s^2 - a_\ell^2 + x)^{5 / 2} } </math> </td> <td align="right"> <math>\cdots</math> See equation (2.227) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>] </td> </tr> </table> Completing these integrals one at a time — while realizing that <math>z \equiv (a + bx)</math>, with <math>b=1</math> and <math>a \equiv (a_s^2 - a_\ell^2) < 0</math> — we have: <font color="red">FIRST INTEGRAL …</font> <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> \frac{A_{\ell \ell}}{a_\ell^2 a_s } = \frac{A_{m m}}{a_\ell^2 a_s } = \frac{A_{m \ell}}{a_\ell^2 a_s } </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \int_{a_\ell^2}^\infty \frac{dx}{x^3 z^{1 / 2} } </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl[ \biggl( - \frac{1}{2ax^2} + \frac{3b}{4a^2 x}\biggr) z^{1 / 2} \biggr]_{a_\ell^2}^\infty + \frac{3b^2}{8a^2}\int_{a_\ell^2}^\infty \frac{dx}{x z^{1 / 2} } </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl[ \biggl( - \frac{1}{2ax^2} + \frac{3b}{4a^2 x}\biggr) [(a_s^2 - a_\ell^2) + x]^{1 / 2} \biggr]_{a_\ell^2}^\infty + \frac{3b^2}{8a^2} \biggl\{ \frac{2}{(-a)^{1 / 2}} \tan^{-1} \biggl[ \frac{ z^{1 / 2} }{ (-a)^{1 / 2} } \biggr] \biggr\}_{a_\ell^2}^\infty </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl[ \biggl( - \frac{3}{4a^2 a_\ell^2} + \frac{1}{2a a_\ell^4} \biggr) [(a_s^2 - a_\ell^2) + a_\ell^2]^{1 / 2} \biggr] + \frac{3}{4(-a)^{5 / 2}} \biggl\{\tan^{-1}\biggl[\infty\biggr] - \tan^{-1} \biggl[ \frac{ a_s }{ (a_\ell^2 - a_s^2)^{1 / 2} } \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl( \frac{2a - 3a_\ell^2 }{4a^2 a_\ell^4} \biggr) a_s + \frac{3}{4(a_\ell^2 - a_s^2)^{5 / 2}} \biggl\{\frac{\pi}{2} - \tan^{-1} \biggl[ \frac{ a_s }{ (a_\ell^2 - a_s^2)^{1 / 2} } \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl[\frac{2a_s^2 - 5a_\ell^2 }{4(a_\ell^2 - a_s^2)^2 a_\ell^4} \biggr] a_s + \frac{3}{4(a_\ell^2 - a_s^2)^{5 / 2}} \biggl\{\frac{\pi}{2} - \tan^{-1} \biggl[ \frac{ a_s }{ (a_\ell^2 - a_s^2)^{1 / 2} } \biggr] \biggr\} \, . </math> </td> </tr> </table> Given that the eccentricity of an oblate-spheroidal configuration is <math>e \equiv (1 - a_s^2/a_\ell^2)^{1 / 2}</math>, this latest expression can be rewritten as, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> A_{\ell \ell} = A_{m m} = A_{m \ell} </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl[ \frac{2(1-e^2) - 5}{4e^4 } \biggr] \frac{a_s^2}{a_\ell^4} + \frac{3 }{4 e^{5}} \biggl\{\frac{\pi}{2} - \tan^{-1} \biggl[ \frac{ a_s/a_\ell }{ e } \biggr] \biggr\} \frac{a_s}{a_\ell^3} </math> </td> </tr> <tr> <td align="right"> <math> \Rightarrow ~~~ a_\ell^2 A_{\ell \ell} = a_\ell^2 A_{m m} = a_\ell^2 A_{m \ell} </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl[ \frac{-2e^2-3}{4e^4 } \biggr] (1-e^2) + \frac{3 (1 - e^2)^{1 / 2}}{4 e^{5}} \biggl\{ \sin^{-1}e \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \frac{1}{4e^4}\biggl\{ - (3 + 2e^2) (1-e^2) + 3 (1 - e^2)^{1 / 2} \biggl[ \frac{\sin^{-1}e}{e} \biggr] \biggr\} \, . </math> </td> </tr> </table> <font color="red">SECOND INTEGRAL …</font> remembering that <math>z \equiv (a + bx)</math>, with <math>b=1</math> and <math>a \equiv (a_s^2 - a_\ell^2) < 0</math> … <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> \frac{A_{s \ell}}{a_\ell^2 a_s } = \frac{A_{s m}}{a_\ell^2 a_s } </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \int_{a_\ell^2}^\infty \frac{dx}{x^2 z^{3 / 2} } </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl[ \biggl(- \frac{1}{ax} - \frac{3b}{a^2}\biggr)\frac{1}{z^{1 / 2}} \biggr]_{a_\ell^2}^\infty - \frac{3b}{2a^2} \int_{a_\ell^2}^\infty \frac{dx}{x z^{1 / 2} } </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl[ \biggl(- \frac{1}{ax} - \frac{3b}{a^2}\biggr)\frac{1}{z^{1 / 2}} \biggr]_{a_\ell^2}^\infty - \frac{3b}{2a^2} \biggl\{ \frac{2}{(-a)^{1 / 2}} \tan^{-1}\biggl[ \frac{z^{1 / 2}}{(-a)^{1 / 2}} \biggr] \biggr\}_{a_\ell^2}^\infty </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl\{ \biggl[ \frac{1}{(a_\ell^2 - a_s^2)x} - \frac{3}{(a_\ell^2 - a_s^2)^2}\biggr]\frac{1}{[a_s^2 - a_\ell^2 + x]^{1 / 2}} \biggr\}_{a_\ell^2}^\infty - \frac{3}{(a_\ell^2 - a_s^2)^{5 / 2}} \biggl\{ \tan^{-1}\biggl[ \frac{[a_s^2 - a_\ell^2 + x]^{1 / 2}}{(a_\ell^2 - a_s^2)^{1 / 2}} \biggr] \biggr\}_{a_\ell^2}^\infty </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \biggl\{ \biggl[ \frac{3}{(a_\ell^2 - a_s^2)^2} - \frac{1}{(a_\ell^2 - a_s^2)a_\ell^2} \biggr]\frac{1}{a_s} \biggr\} - \frac{3}{(a_\ell^2 - a_s^2)^{5 / 2}} \biggl\{ \frac{\pi}{2} - \tan^{-1}\biggl[ \frac{a_s}{(a_\ell^2 - a_s^2)^{1 / 2}} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \frac{1}{a_s a_\ell^2}\biggl[ \frac{2a_\ell^2 + a_s^2}{(a_\ell^2 - a_s^2)^2 } \biggr] - \frac{3}{(a_\ell^2 - a_s^2)^{5 / 2}} \biggl\{ \frac{\pi}{2} - \tan^{-1}\biggl[ \frac{a_s}{(a_\ell^2 - a_s^2)^{1 / 2}} \biggr] \biggr\} \, . </math> </td> </tr> </table> That is to say, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> A_{s \ell} = A_{s m} </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \frac{2a_\ell^2 + a_s^2}{(a_\ell^2 - a_s^2)^2 } - \frac{3a_\ell^2 a_s}{(a_\ell^2 - a_s^2)^{5 / 2}} \biggl\{ \frac{\pi}{2} - \tan^{-1}\biggl[ \frac{a_s}{(a_\ell^2 - a_s^2)^{1 / 2}} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \frac{2 + (1-e^2)}{a_\ell^2~ e^4 } - \frac{3a_s}{a_\ell^3 e^5} \biggl\{ \frac{\pi}{2} - \tan^{-1}\biggl[ \frac{a_s/a_\ell}{e} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \frac{1}{a_\ell^2 e^4} \biggl\{ (3-e^2) - \frac{3 (1-e^2)^{1 / 2}}{e} \biggl[\sin^{-1}e\biggr] \biggr\} \, . </math> </td> </tr> </table> <table border="1" align="center" cellpadding="8" width="80%"><tr><td align="left"> <div align="center"><font color="red">CROSSCHECK</font></div> Now, according to the relations stated in [[ParabolicDensity/GravPot#Parabolic_Density_Distribution_2|an accompanying discussion]], we should find that, for <math>i = j</math>, <table border="0" align="center" cellpadding="8"> <tr> <td align="right"> <math>2A_{ii} + \sum_{j=1}^3 A_{ij}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\frac{2}{a_i^2} \, .</math> </td> </tr> </table> Let's check. Specifically, for an oblate spheroid when <math> i = \ell</math>, the summation takes the form, <table border="0" align="center" cellpadding="8"> <tr> <td align="right"> LHS </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>2A_{\ell \ell} + (A_{\ell s} + A_{\ell m} + A_{\ell \ell})</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>4A_{\ell \ell} + A_{\ell s} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{1}{a_\ell^2e^4}\biggl\{ \biggl[ -3 -2e^2 \biggr] (1-e^2) + 3 (1 - e^2)^{1 / 2} \biggl[ \frac{\sin^{-1}e}{e} \biggr] \biggr\} + \frac{1}{a_\ell^2 e^4} \biggl\{ (3-e^2) - 3 (1-e^2)^{1 / 2} \biggl[\frac{\sin^{-1}e}{e}\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{1}{a_\ell^2e^4}\biggl\{ ( - 3 - 2e^2 ) + ( 3 + 2e^2 )e^2 + (3-e^2) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{1}{a_\ell^2e^4} \biggl\{ 2e^4 \biggr\} = \frac{2}{a_\ell^2} \, . </math> </td> </tr> </table> Q. E. D. <font color="red">Yeah!</font> </td></tr></table> <font color="red">THIRD INTEGRAL …</font> <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> \frac{A_{s s}}{a_\ell^2 a_s } </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \int_{a_\ell^2}^\infty \frac{dx}{x z^{5 / 2} } </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \sum_{k=0}^{+1}\biggl[\frac{2}{ (2k+1)a^{2-k} z^{k+1 / 2} } \biggr]_{a_\ell^2}^\infty + \frac{1}{a^2}\int_{a_\ell^2}^\infty \frac{dx}{x z^{1 / 2} } </math> </td> </tr> <tr> <td align="center" colspan="3"> … on my whiteboard I completed the evaluation<br />of this integral and obtained … </td> </tr> <tr> <td align="right"> <math>\frac{3}{2} a_\ell^2 A_{ss} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{( 4e^2 - 3 )}{e^4(1-e^2)} + \frac{3 (1-e^2)^{1 / 2}}{e^4} \biggl[\frac{\sin^{-1}e}{e}\biggr] \, . </math> </td> </tr> </table> <table border="1" align="center" cellpadding="8" width="80%"><tr><td align="left"> <div align="center"><font color="red">CROSSCHECK</font></div> Again, according to the relations stated in [[ParabolicDensity/GravPot#Parabolic_Density_Distribution_2|an accompanying discussion]], we should find that, for <math>i = j</math>, <table border="0" align="center" cellpadding="8"> <tr> <td align="right"> <math>2A_{ii} + \sum_{j=1}^3 A_{ij}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\frac{2}{a_i^2} \, .</math> </td> </tr> </table> This time, specifically for an oblate spheroid, when <math> i = s</math>, this gives, <table border="0" align="center" cellpadding="8"> <tr> <td align="right"> <math>3A_{ss} + 2A_{s \ell} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\frac{2}{a_s^2}</math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{3}{2} a_\ell^2 A_{ss} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\frac{1}{(1-e^2)} - a_\ell^2 A_{s \ell} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{1}{(1-e^2)} - \frac{1}{e^4} \biggl\{ (3-e^2) - 3 (1-e^2)^{1 / 2} \biggl[\frac{\sin^{-1}e}{e}\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{1}{(1-e^2)} - \frac{(3-e^2)}{e^4} + \frac{3 (1-e^2)^{1 / 2}}{e^4} \biggl[\frac{\sin^{-1}e}{e}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{e^4 - (1-e^2)(3-e^2)}{e^4(1-e^2)} + \frac{3 (1-e^2)^{1 / 2}}{e^4} \biggl[\frac{\sin^{-1}e}{e}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{( 4e^2 - 3 )}{e^4(1-e^2)} + \frac{3 (1-e^2)^{1 / 2}}{e^4} \biggl[\frac{\sin^{-1}e}{e}\biggr] </math> </td> </tr> </table> Q. E. D. <font color="red">Yeah!</font> </td></tr></table> ==When a<sub>m</sub> = a<sub>s</sub>== When the length of the intermediate axis is the same as the length of the shortest axis — that is, when we are dealing with a prolate spheroid — the coefficient associated with the longest axis is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_\ell}{a_\ell a_s^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_0^\infty \frac{du}{ (a_s^2 + u)(a_\ell^2 + u)^{3 / 2} } \, . </math> </td> </tr> </table> Changing the integration variable to <math>~x \equiv (a_s^2 + u)</math>, we obtain an integral expression that appears as equation (2.229.1) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], namely, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_\ell}{a_\ell a_s^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_{a_s^2}^\infty \frac{dx}{ x^2(a_\ell^2 - a_s^2 + x)^{1 / 2} } </math> </td> </tr> <tr> <td align="right"> <math>~</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \frac{2}{(a_\ell^2 - a_s^2) (a_\ell^2 - a_s^2 + x)^{1 / 2}} \biggr]_{a_s^2}^\infty + \frac{1}{(a_\ell^2 - a_s^2)} \int_{a_s^2}^\infty \frac{dx}{ x (a_\ell^2 - a_s^2 + x)^{1 / 2} } \, . </math> </td> </tr> </table> The remaining integral in this expression appears as equation (2.224.5) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>]. Its resolution depends on the sign of the constant term in the denominator, <math>~(a_\ell^2 - a_s^2)</math>. Given that this term is positive, the integration gives, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_\ell}{a_\ell a_s^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{2}{(a_\ell^2 - a_s^2) a_\ell} + \frac{1}{(a_\ell^2 - a_s^2)} \biggl\{ \frac{1}{\sqrt{(a_\ell^2 - a_s^2)}} \ln \biggl[ \frac{ (a_\ell^2 - a_s^2 + x)^{1 / 2} - \sqrt{(a_\ell^2 - a_s^2)} }{(a_\ell^2 - a_s^2 + x)^{1 / 2} + \sqrt{(a_\ell^2 - a_s^2)} } \biggr] \biggr\}_{a_s^2}^\infty </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ A_\ell</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{2a_\ell a_s^2}{(a_\ell^2 - a_s^2) a_\ell} - \frac{a_\ell a_s^2}{(a_\ell^2 - a_s^2)^{3 / 2}} \biggl\{ \ln \biggl[ \frac{ 1-e }{1+e} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{ (1-e^2)}{e^3} \biggl\{ \ln \biggl[ \frac{1+e}{ 1-e } \biggr] \biggr\} - \frac{2(1-e^2)}{e^2 } \, , </math> </td> </tr> </table> where, as above, <math>~e \equiv (1 - a_s^2/a_\ell^2)^{1 / 2}</math>. Now, given that <math>~A_m = A_s</math>, in this case we appreciate that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~A_\ell + A_m + A_s = A_\ell + 2A_s</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ A_s</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 - \frac{A_\ell}{2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 - \frac{1}{2}\biggl[ \frac{ (1-e^2)}{e^3} \cdot \ln \biggl( \frac{1+e}{ 1-e } \biggr) - \frac{2(1-e^2)}{e^2 } \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{e^2 } - \frac{ (1-e^2)}{2e^3} \cdot \ln \biggl( \frac{1+e}{ 1-e } \biggr) \, . </math> </td> </tr> </table> ==For Spheres (a<sub>ℓ</sub> = a<sub>m</sub> = a<sub>s</sub>)== In the case of a sphere, where <math>a_\ell = a_m = a_s</math>, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math>\Delta</math> </td> <td align="center"> <math> \equiv </math> </td> <td align="left"> <math> \biggl[ (a_\ell^2 + u)(a_m^2 + u)(a_s^2 + u) \biggr]^{1 / 2} = (a_i^2 + u)^{3 / 2} \, , </math> </td> </tr> </table> and the definition of all three <math>A_i</math> coefficients is obtained from the integral, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> A_i </math> </td> <td align="center"> <math> \equiv </math> </td> <td align="left"> <math> a_\ell a_m a_s \int_0^\infty \frac{du}{\Delta (a_i^2 + u )} = a_i^3 \int_0^\infty\frac{du}{(a_i^2 + u)^{5 / 2}} \, . </math> </td> </tr> </table> Analogously, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> A_{ii} </math> </td> <td align="center"> <math> \equiv </math> </td> <td align="left"> <math> a_i^3 \int_0^\infty\frac{du}{(a_i^2 + u)^{7 / 2}} \, . </math> </td> </tr> </table> <table border="1" align="center" width="80%" cellpadding="10"><tr><td align="left"> According to <font color="#00CC00">p. 406, Eq. (139)</font> of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>], <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> \int (a+bu)^{-n / 2} du </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> \frac{2(a+bu)^{(2-n)/2}}{b(2-n)} \, . </math> </td> </tr> </table> </td></tr></table> Hence, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math>\frac{A_i}{a_i^3} \equiv \int_0^\infty (a_i^2 + u)^{-5 / 2} du </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> - \biggl[ \frac{2}{3(a_i^2 + u)^{3 / 2}} \biggr]_0^\infty = \frac{2}{3a_i^3} \, ; </math> </td> </tr> </table> and, <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math>\frac{A_{ii}}{a_i^3} \equiv \int_0^\infty (a_i^2 + u)^{-7 / 2} du </math> </td> <td align="center"> <math> = </math> </td> <td align="left"> <math> - \biggl[ \frac{2}{5(a_i^2 + u)^{5 / 2}} \biggr]_0^\infty = \frac{2}{5a_i^5} \, . </math> </td> </tr> </table> =Derivation of Selected 2<sup>nd</sup>-Order Index Symbols= ==Evaluating A<sub>ℓℓ</sub>== In the case of <math>A_{\ell\ell}</math>, we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_{\ell\ell}}{a_\ell a_m a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_0^\infty \biggl[ (a_\ell^2 + u)^5(a_m^2 + u)(a_s^2 + u) \biggr]^{-1 / 2} du \, . </math> </td> </tr> </table> Changing the integration variable to <math>~x \equiv -u</math>, we obtain a definite integral expression that appears as equation (3.134.1) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ I. W. Gradshteyn & I. M. Ryzhik (2007; 7<sup>th</sup> Edition)], ''Table of Integrals, Series, and Products'' — hereafter, GR7<sup>th</sup> — namely, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_{\ell\ell}}{a_\ell a_m a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\int_{-\infty}^0 \biggl[ (a_\ell^2 - x)^5(a_m^2 - x)(a_s^2 - x) \biggr]^{-1 / 2} dx </math> </td> <td align="left"> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{2}{3(a_\ell - a_m)^2 (a_\ell - a_s)^{3/2}} \biggl[ (3a_\ell - a_m - 2a_s) F(\alpha, p) - 2(2a_\ell - a_m - a_s) E(\alpha, p) \biggr] + \frac{2}{3(a_\ell - a_s)(a_\ell -a_m)}\biggl[ \frac{a_s a_m}{a_\ell^3} \biggr]^{1 / 2} \, , </math> </td> <td align="left"> … valid for <math>[a_\ell > a_m > a_s \ge 0]</math></td> </tr> <tr> <td align="center" colspan="4">[https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], p. 257, Eq. (3.134.1)</td> </tr> </table> where (see p. 254 of [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>]), <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\sin^2\alpha</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{a_\ell^2 - a_s^2}{a_\ell^2 - 0} = 1 - \frac{a_s^2}{a_\ell^2} \, ,</math> </td> <td align="center" width="20%"> </td> <td align="right"> <math>~p</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\biggl[ \frac{a_\ell^2 - a_m^2}{a_\ell^2 - a_s^2} \biggr]^{1 / 2} \, ,</math> </td> </tr> </table> and where, <math>E(\alpha, p)</math> and <math>F(\alpha, p)</math> are [https://dlmf.nist.gov/19.2#ii Legendre incomplete elliptic integrals of the first and second kind], respectively. (Note that in the notation convention adopted by [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], the order of the argument list, <math>~(\alpha, p)</math>, is flipped relative to the convention that we have adopted [[#Evaluation_of_Coefficients|above]] and elsewhere throughout our online, MediaWiki-based chapters.) <!-- Recognizing that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~p^2 \sin^3\alpha</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\frac{a_\ell^2 - a_s^2}{a_\ell^2 }\biggr]^{3 / 2}\biggl[ \frac{a_\ell^2 - a_m^2}{a_\ell^2 - a_s^2} \biggr] = \frac{(a_\ell^2 - a_s^2)^{1 / 2}}{a_\ell^3 } \biggl[ a_\ell^2 - a_m^2 \biggr] \, , </math> </td> </tr> </table> we see that the expression for <math>~A_\ell</math> can be rewritten as, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{A_\ell}{a_\ell a_m a_s}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\frac{2}{a_\ell^3 ~p^2 \sin^3\alpha} \biggl[ F(\alpha, p) - E(\alpha, p) \biggr] \, . </math> </td> </tr> </table> --> {{ SGFworkInProgress }} =Work In Progress= ==Derivation of Expression for Gravitational Potential== In §373 (p. 700) of his book titled, ''Hydrodynamics'', [<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>] states that, "<font color="maroon">The gravitation-potential, at internal points, of a uniform mass enclosed by the surface</font> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1</math> </td> </tr> <tr><td align="center" colspan="3">[<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>], <font color="#00CC00">§373, Eq. (1)</font></td></tr> </table> <font color="maroon">… may be written</font> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\Phi(\vec{x})}{G}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \pi \rho (\alpha_0 x^2 + \beta_0 y^2 + \gamma_0 z^2 - \chi_0) \, , </math> </td> </tr> <tr><td align="center" colspan="3">[<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>], <font color="#00CC00">§373, Eq. (4)</font></td></tr> </table> <font color="maroon">where, as in §114,</font>" <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\alpha_0}{abc} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\int_0^\infty \frac{d\lambda}{(a^2 + \lambda)\Delta} \, ,</math> </td> <td align="center"> </td> <td align="right"> <math>~\frac{\beta_0}{abc} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\int_0^\infty \frac{d\lambda}{(b^2 + \lambda)\Delta} \, ,</math> </td> </tr> <tr> <td align="right"> <math>~\frac{\gamma_0}{abc} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\int_0^\infty \frac{d\lambda}{(c^2 + \lambda)\Delta} \, ,</math> </td> <td align="center"> </td> <td align="right"> <math>~\frac{\chi_0}{abc} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\int_0^\infty \frac{d\lambda}{\Delta} \, ,</math> </td> </tr> <tr><td align="center" colspan="7">[<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>], <font color="#00CC00">§373, Eqs. (5) & (6)</font></td></tr> </table> and, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Delta</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~[(a^2 + \lambda)(b^2 + \lambda)(c^2 + \lambda)]^{1 / 2} \, .</math> </td> </tr> <tr><td align="center" colspan="3">[<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>], <font color="#00CC00">§373, Eq. (3)</font></td></tr> </table> Although different variable names have been used, it is easy to see the correspondence between these expressions and the ''defining integral expressions'' that we have drawn from the more recent publications of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>] and [<b>[[Appendix/References#T78|<font color="red">T78</font>]]</b>] and [[#Gravitational_Potential|presented above]]. Here, we are interested in demonstrating how [<b>[[Appendix/References#Lamb32|<font color="red">Lamb32</font>]]</b>] derived his expression for the potential inside (and on the surface of) an homogeneous ellipsoid. ==Acceleration at the Pole== ===Prolate Spheroids=== In our above review, for consistency, we assumed that the longest axis of the ellipsoid was aligned with the <math>~x</math>-axis in all cases — for prolate spheroids as well as for oblate spheroids and for the more generic, triaxial ellipsoids. In this discussion, in order to better align with the operational features of a standard cylindrical coordinate system, we will orient the prolate-spheroidal configuration such that its major axis and, hence, its axis of symmetry aligns with the <math>~z</math>-axis while the center of the spheroid remains at the center of the (cylindrical) coordinate grid. In this case, the surface will be defined by the ellipse, <div align="center"> <math>~\frac{\varpi^2}{a_3^2} + \frac{z^2}{a_1^2} = 1 ~~~~\Rightarrow ~~~~ \varpi = a_3\sqrt{1-z^2/a_1^2} \, ,</math> </div> and the gravitational potential will be given by the expression, <div align="center"> <math> ~\Phi(\vec{x}) = -\pi G \rho \biggl[ I_\mathrm{BT} a_1^2 - \biggl(A_1 z^2 + A_3 \varpi^2 \biggr) \biggr]. </math> </div> <span id="FirstDetermination"> The magnitude of the gravitational acceleration at the pole <math>~(\varpi, z) = (0, a_1)</math> of this prolate spheroid can be obtained from the gravitational potential via the expression,</span> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathcal{A} \equiv \biggl|- \frac{\partial \Phi}{\partial z}\biggr|_{a_1}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2\pi G \rho A_1 a_1 \, ,</math> </td> </tr> </table> </div> where, [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Prolate_Spheroids_.28a1_.3E_a2_.3D_a3.29|as above]], <table align="center" border=0 cellpadding="3"> <tr> <td align="right"> <math> ~A_1 </math> </td> <td align="center"> <math> ~= </math> </td> <td align="left"> <math> \ln\biggl[ \frac{1+e}{1-e} \biggr] \frac{(1-e^2)}{e^3} - \frac{2(1-e^2)}{e^2} \, . </math> </td> </tr> </table> We should also be able to derive this expression for <math>~\mathcal{A}</math> by integrating the <math>~z</math>-component of the differential acceleration over the mass distribution, that is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathcal{A}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\int \biggl[ \frac{G }{r^2} \cdot \frac{(a_1-z)}{r} \biggr] dm = \int \biggl[ \frac{(a_1-z)G }{r^3} \biggr] 2\pi \varpi d\varpi dz</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2\pi G\rho \int^{a_1}_{-a_1} (a_1-z)dz \int_0^{a_3\sqrt{1-z^2/a_1^2}} [\varpi^2+(z-a_1)^2]^{-3/2}\varpi d\varpi \, ,</math> </td> </tr> </table> </div> where the distance, <math>~r</math>, has been measured from the pole, that is, <div align="center"> <math>~r^2 = \varpi^2 + (z-a_1)^2 \, .</math> </div> Performing the integral over <math>~\varpi</math> gives, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathcal{A}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2\pi G\rho \int^{a_1}_{-a_1} (a_1-z)dz \biggl\{ -[\varpi^2+(z-a_1)^2]^{-1/2} \biggr\}_0^{a_3\sqrt{1-z^2/a_1^2}} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2\pi G\rho \int^{a_1}_{-a_1} (a_1-z)dz \biggl\{ \frac{1}{z - a_1} -\biggl[ a_3^2 \biggl(1-\frac{z^2}{a_1^2} \biggr) + a_1^2\biggl(1-\frac{z}{a_1}\biggr)^2 \biggr]^{-1/2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - 2\pi G\rho a_1 \int^{1}_{-1} d\zeta \biggl\{ \frac{1-\zeta}{1-\zeta } - (1-\zeta)\biggl[ \biggl(\frac{a_3}{a_1}\biggr)^2 \biggl(1-\zeta^2 \biggr) + \biggl(1-\zeta\biggr)^2 \biggr]^{-1/2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\pi G\rho a_1 \int^{1}_{-1} d\zeta \biggl\{ (1-\zeta) [ (2-e^2) - 2\zeta + e^2\zeta^2 ]^{-1/2} -1 \biggr\} \, , </math> </td> </tr> </table> </div> where, <math>~\zeta\equiv z/a_1</math>. For later reference, we will identify the expression inside the curly braces as the function, <math>~\mathcal{Z}</math>; specifically, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathcal{Z}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~(1-\zeta) [ (2-e^2) - 2\zeta + e^2\zeta^2 ]^{-1/2} -1</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- 1 - \frac{\zeta}{\sqrt{X}} + \frac{1}{\sqrt{X}} \, ,</math> </td> </tr> </table> </div> where, in an effort to line up with notation found in integral tables, in this last expression we have used the notation, <math>~X \equiv a + b\zeta + c\zeta^2</math> and, in our case, <div align="center"> <math>a \equiv (2-e^2)\, ,</math> <math>b \equiv -2\, ,</math> and <math>c \equiv e^2\, .</math> </div> We find that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\int_{-1}^1 \mathcal{Z} d\zeta</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \zeta\biggr|_{-1}^{1} - \biggl\{ \frac{\sqrt{X}}{c} \biggr\}_{-1}^1 +\biggl[1 + \frac{b}{2c} \biggr]\int_{-1}^1 \frac{d\zeta}{\sqrt{X}} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- 2 - \biggl\{ \frac{\sqrt{(2-e^2) -2\zeta + e^2\zeta^2}}{e^2} \biggr\}_{-1}^1 +\biggl[1 - \frac{1}{e^2} \biggr] \biggl\{ \frac{1}{\sqrt{c}} \ln \biggl[2\sqrt{cX} + 2c\zeta + b \biggr] \biggr\}_{-1}^1 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- 2 - \biggl\{ \frac{\sqrt{(2-e^2) -2 + e^2}}{e^2} \biggr\} + \biggl\{ \frac{\sqrt{(2-e^2) +2 + e^2}}{e^2} \biggr\} + \biggl[1 - \frac{1}{e^2} \biggr] \biggl\{ \frac{1}{e} \ln \biggl[2\sqrt{e^2[(2-e^2) -2\zeta + e^2\zeta^2]} + 2e^2\zeta - 2 \biggr] \biggr\}_{-1}^1 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- 2 + \frac{2}{e^2} +\biggl[\frac{e^2-1}{e^3} \biggr] \biggl\{ \ln \biggl[2e^2 - 2 \biggr] - \ln \biggl[4e - 2e^2 - 2 \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- 2\biggl[\frac{e^2 - 1}{e^2}\biggr] +\biggl[\frac{e^2-1}{e^3} \biggr] \biggl\{ \ln \biggl[-2(1-e^2) \biggr] - \ln \biggl[-2(1-e)^2\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[\frac{1-e^2}{e^3} \biggr] \ln \biggl[\frac{1+e}{1-e} \biggr] -2\biggl[\frac{1-e^2 }{e^2}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~A_1 \, . </math> </td> </tr> </table> </div> Hence, we have, <div align="center"> <math>~\mathcal{A} = 2\pi G\rho a_1 \biggl[ \int_{-1}^1 \mathcal{Z} d\zeta\biggr]= 2\pi G \rho A_1 a_1 \, ,</math> </div> which exactly matches the result [[#FirstDetermination|obtained, above]], by taking the derivative of the potential. =See Also= * [[Apps/MaclaurinSpheroids#Maclaurin_Spheroids_.28axisymmetric_structure.29|Properties of Maclaurin Spheroids]] * [[Apps/MaclaurinSpheroids/GoogleBooks#Excerpts_from_A_Treatise_of_Fluxions|Excerpts from Maclaurin's (1742) ''A Treatise of Fluxions'']] * [[ThreeDimensionalConfigurations/FerrersPotential#Ferrers_.281877.29_Gravitational_Potential_for_Inhomogeneous_Ellipsoids|Inhomogeneous Ellipsoids Leading to Ferrers Potentials]] =Footnotes= <ol type="1"> <li>In [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>] this equation is written in terms of a variable <math>~I</math> instead of <math>~I_\mathrm{BT}</math> as defined here. The two variables are related to one another straightforwardly through the expression, <math>~I = I_\mathrm{BT} a_1^2</math>.</li> <li>Throughout [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], Chandrasekhar adopts a sign convention for the scalar gravitational potential that is opposite to the sign convention being used here.</li> </ol> {{ SGFfooter }}
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