ThreeDimensionalConfigurations/HomogeneousEllipsoids

Properties of Homogeneous Ellipsoids (1)

Gravitational Potential

The Defining Integral Expressions

As has been shown in a separate discussion titled, "Origin of the Poisson Equation," the acceleration due to the gravitational attraction of a distribution of mass $\rho$ $(\overrightarrow{x})$ can be derived from the gradient of a scalar potential $\mathrm{\Phi }$ $(\overrightarrow{x})$ defined as follows:

$\mathrm{\Phi }(\overrightarrow{x})\equiv -\int \frac{G\rho ({\overrightarrow{x}}^{\prime })}{|{\overrightarrow{x}}^{\prime }-\overrightarrow{x}|}{d}^3{x}^{\prime }.$

As has been explicitly demonstrated in Chapter 3 of [EFE] and summarized in Table 2-2 (p. 57) of [BT87], 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 $(x,y,z)=(a_1,a_2,a_3)$,

$\mathrm{\Phi }(\overrightarrow{x})=-\pi G\rho [I_{\mathrm{B}\mathrm{T}}{a}_1^2-(A_1{x}^2+A_2{y}^2+A_3{z}^2\left)\right],$

[EFE], Chapter 3, Eq. (40)^1,2
[BT87], Chapter 2, Table 2-2

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,

$A_i$	$\equiv$	$a_1{a}_2{a}_3\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{\mathrm{\Delta }(a_i^2+u)},$
$I_{\mathrm{B}\mathrm{T}}$	$\equiv$	$\frac{a_2{a}_3}{a_1}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{\mathrm{\Delta }}=A_1+A_2(\frac{a_2}{a_1}{)}^2+A_3(\frac{a_3}{a_1}{)}^2,$
$\mathrm{\Delta }$	$\equiv$	$[(a_1^2+u)(a_2^2+u)(a_3^2+u){]}^{1/2}.$

[EFE], Chapter 3, Eqs. (18), (15 & 22)¹, & (8), respectively
[BT87], Chapter 2, Table 2-2

This definite-integral definition of $A_i$ may also be found in:

• [Lamb32]: as Eq. (6) in §114 (p. 153); and as Eq. (5) in §373 (p. 700).
• [T78]: as Eq. (5) in §10.2 (p. 234), but note that there is an error in the denominator of the right-hand-side — $a_1$ appears instead of $a_i$.

Evaluation of Coefficients

As is detailed below, the integrals defining $A_i$ and $I_{\mathrm{B}\mathrm{T}}$ can be evaluated in terms of the incomplete elliptic integral of the first kind,

$F(\theta ,k)\equiv \underset{0}{\overset{\theta }{\int }}\frac{d{\theta }^{\prime }}{\sqrt{1-k^2{\sin }^2{\theta }^{\prime }}},$

and/or the incomplete elliptic integral of the second kind,

$E(\theta ,k)\equiv \underset{0}{\overset{\theta }{\int }}\sqrt{1-k^2{\sin }^2{\theta }^{\prime }}d{\theta }^{\prime },$

where, for our particular problem,

$\theta \equiv {\cos }^{-1}\left(\frac{a_3}{a_1}\right),$

$k\equiv [\frac{a_1^2-a_2^2}{a_1^2-a_3^2}{]}^{1/2}=[\frac{1-(a_2/a_1{)}^2}{1-(a_3/a_1{)}^2}{]}^{1/2},$

[EFE], Chapter 3, Eq. (32)

or the integrals can be evaluated in terms of more elementary functions if either $a_2=a_1$ (oblate spheroids) or $a_3=a_2$ (prolate spheroids).

 

Triaxial Configurations (a₁ > a₂ > a₃)

If the three principal axes of the configuration are unequal in length and related to one another such that $a_1>a_2>a_3$,

$A_1$	$=$	$\frac{2a_2{a}_3}{a_1^2}\left[\frac{F(\theta ,k)-E(\theta ,k)}{k^2{\sin }^3\theta }\right];$
$A_2$	$=$	$\frac{2a_2{a}_3}{a_1^2}\left[\frac{E(\theta ,k)-(1-k^2)F(\theta ,k)-(a_3/a_2)k^2\sin \theta }{k^2(1-k^2){\sin }^3\theta }\right];$
$A_3$	$=$	$\frac{2a_2{a}_3}{a_1^2}\left[\frac{(a_2/a_3)\sin \theta -E(\theta ,k)}{(1-k^2){\sin }^3\theta }\right];$
$I_{\mathrm{B}\mathrm{T}}$	$=$	$\frac{2a_2{a}_3}{a_1^2}\left[\frac{F(\theta ,k)}{\sin \theta }\right].$

[EFE], Chapter 3, Eqs. (33), (34) & (35)

Notice that there is no need to specify the actual value of $a_1$ in any of these expressions, as they each can be written in terms of the pair of axis ratios, $a_2/a_1$ and $a_3/a_1$. 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 [EFE], namely,

$\sum _{\ell =1}^3{A}_{\ell }=2.$

$\frac{a_1^2}{2a_2{a}_3}[A_1+A_3+A_2]$	$=$	$\frac{F(\theta ,k)-E(\theta ,k)}{k^2{\sin }^3\theta }+\frac{(a_2/a_3)\sin \theta -E(\theta ,k)}{(1-k^2){\sin }^3\theta }$
		$+\frac{E(\theta ,k)-(1-k^2)F(\theta ,k)-(a_3/a_2)k^2\sin \theta }{k^2(1-k^2){\sin }^3\theta }$
	$=$	$\frac{1}{k^2(1-k^2){\sin }^3\theta }\{(1-k^2)F(\theta ,k)-(1-k^2)E(\theta ,k)+k^2(a_2/a_3)\sin \theta$
		$-k^{2}E(\theta ,k)+E(\theta ,k)-(1-k^2)F(\theta ,k)-(a_3/a_2)k^2\sin \theta \}$
	$=$	$\frac{1}{(1-k^2){\sin }^2\theta }[\frac{a_2}{a_3}-\frac{a_3}{a_2}]$
	$=$	$\frac{a_1^2}{a_2{a}_3}.$

Q.E.D.

 

Oblate Spheroids (a₁ = a₂ > a₃)

If the longest axis, $a_1$, and the intermediate axis, $a_2$, of the ellipsoid are equal to one another, then an equatorial cross-section of the object presents a circle of radius $a_1$ and the object is referred to as an oblate spheroid. For homogeneous oblate spheroids, evaluation of the integrals defining $A_i$ and $I_{\mathrm{B}\mathrm{T}}$ gives,

$A_1$	$=$	$\frac{1}{e^2}[\frac{{\sin }^{-1}e}{e}-(1-e^2{)}^{1/2}](1-e^2{)}^{1/2};$
$A_2$	$=$	$A_1;$
$A_3$	$=$	$\frac{2}{e^2}[(1-e^2{)}^{-1/2}-\frac{{\sin }^{-1}e}{e}](1-e^2{)}^{1/2};$
$I_{\mathrm{B}\mathrm{T}}$	$=$	$2A_1+A_3(1-e^2)=2(1-e^2{)}^{1/2}[\frac{{\sin }^{-1}e}{e}],$

[EFE], Chapter 3, Eq. (36)
[T78], §4.5, Eqs. (48) & (49)

where the eccentricity,

$e\equiv [1-(\frac{a_3}{a_1}{)}^2{]}^{1/2}.$

 

Prolate Spheroids (a₁ > a₂ = a₃)

If the shortest axis $(a_3)$ and the intermediate axis $(a_2)$ of the ellipsoid are equal to one another — and the symmetry (longest, $a_1$) axis is aligned with the $x$-axis — then a cross-section in the $y-z$ plane of the object presents a circle of radius $a_3$ and the object is referred to as a prolate spheroid. For homogeneous prolate spheroids, evaluation of the integrals defining $A_i$ and $I_{\mathrm{B}\mathrm{T}}$ gives,

$A_1$	$=$	$\ln \left[\frac{1+e}{1-e}\right]\frac{(1-e^2)}{e^3}-\frac{2(1-e^2)}{e^2};$
$A_2$	$=$	$\frac{1}{e^2}-\ln \left[\frac{1+e}{1-e}\right]\frac{(1-e^2)}{2e^3};$
$A_3$	$=$	$A_2;$
$I_{\mathrm{B}\mathrm{T}}$	$=$	$A_1+2(1-e^2)A_2=\ln \left[\frac{1+e}{1-e}\right]\frac{(1-e^2)}{e},$

[EFE], Chapter 3, Eq. (38)

where, again, the eccentricity,

$e\equiv [1-(\frac{a_3}{a_1}{)}^2{]}^{1/2}.$

NOTE:  If, instead, the longest (and, in this case, symmetry) axis of the prolate mass distribution is aligned with the $z$-axis — in which case, $a_1=a_2<a_3$ — then, $e=(1-a_1^2/a_3^2{)}^{1/2}$ and the mathematical expressions for the $A_i$ coefficients must be altered; they are essentially "swapped." This modified set of coefficient expressions can be found in a 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 [BT87].

Example Evaluations

Here we adopt the notation mapping, $(a_1,a_2,a_3)\leftrightarrow (a,b,c)$. In general, for a given pair of axis ratios, $(\frac{b}{a},\frac{c}{a})$, a determination of the coefficients, $A_1$, $A_2$, and $A_3$, 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, $F(\theta ,k)$ and $E(\theta ,k)$, 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 [EFE]. In particular, for twenty-six separate axis-ratio pairs, Chandrasekhar's Table IV lists the values of the square of the angular velocity, ${\mathrm{\Omega }}^2$, and the total angular momentum, $L$, 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",

$\frac{{\mathrm{\Omega }}^2}{\pi G\rho }$	$=$	$2B_{12}$
[EFE], §39, Eq. (5)
	$=$	$2\left[\frac{A_1-(b/a{)}^2{A}_2}{1-(b/a{)}^2}\right],$
using, [EFE], §21, Eqs. (105) & (107)

and, for "angmom",

$\frac{L}{(G{M}^3{)}^{1/2}(abc{)}^{1/6}}$	$=$	$\frac{\sqrt{3}}{10}\left[\frac{a^2+b^2}{(abc{)}^{2/3}}\right](\frac{{\mathrm{\Omega }}^2}{\pi G\rho }{)}^{1/2}$
[EFE], §39, Eq. (16)
	$=$	$\frac{\sqrt{3}}{10}\left[\frac{1+(b/a{)}^2}{(b/a{)}^{2/3}(c/a{)}^{2/3}}\right](\frac{{\mathrm{\Omega }}^2}{\pi G\rho }{)}^{1/2}.$

Or, in connection with the free-energy discussion found in D. M. Christodoulou, D. Kazanas, I. Shlosman, & J. E. Tohline (1995, ApJ, 446, 472),

$\frac{5L}{M}$	$=$	$a^2[1+(\frac{b}{a}{)}^2\left]\right[\frac{{\mathrm{\Omega }}^2}{\pi G\rho }{]}^{1/2}$
	$=$	$\left[\frac{15}{4}\right(\frac{b}{a}{)}^{-1}\left(\frac{c}{a}{)}^{-1}{]}^{2/3}\right[1+\left(\frac{b}{a}{)}^2\right][\frac{{\mathrm{\Omega }}^2}{\pi G\rho }{]}^{1/2}$

 

Table 1:  Example Evaluations
Given		Determined using calculator and (crude) CRC tables of elliptic integrals
$\frac{a_2}{a_1}$	$\frac{a_3}{a_1}$	$\theta$		$k$	${\sin }^{-1}k$		$F(\theta ,k)$	$E(\theta ,k)$	$A_1$	$A_2$	$A_3$
		radians	degrees		radians	degrees
1.00	0.582724	0.94871973	54.3576	0.00000000	0.00000000	0.000000	0.94871973	0.94871973	0.51589042	0.51589042	0.96821916
0.96	0.570801	0.96331527	55.1939	0.34101077	0.34799191	19.9385	0.975	0.946	+0.4937	+0.5319	+0.9744
0.60	0.433781	1.12211141	64.292	0.88788426	1.09272580	62.609	1.3375	0.9547	0.3455	0.6741	0.9803

With regard to our Table 1 (immediately above): 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, $\theta$ and $k$, as defined above. By default, each determined value of $\theta$ is in radians. Because the published CRC special-functions tables quantify both arguments of the special functions in angular degrees, we converted $\theta$ from radians to degrees (see column 4 of Table 1) and, similarly, we converted ${\sin }^{-1}k$ to degrees (see column 7 of Table 1). For the axisymmetric configuration — the first row of numbers in Table 1, for which $a_2/a_1=1$ — the coefficients, $A_1$, $A_2$, and $A_3$, were determined to eight digits of precision using the appropriate expressions for oblate spheroids. Note that, in this axisymmetric case, $F(\theta ,0)=E(\theta ,0)=\theta$, but these function values are irrelevant with respect to the determination of the $A_{\ell }$ coefficients.

Table 2:  Double-Precision Evaluations Related to Table IV in [EFE], Chapter 6, §39 (p. 103)

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

With regard to our Table 2 (immediately above): Next, given each pair of axis ratios, $(\frac{b}{a},\frac{c}{a})$ — copied from Table IV of EFE (see columns 1 and 2 of our Table 2) — we used some fortran routines from Numerical Recipes to calculate $F(\theta ,k)$ and $E(\theta ,k)$ (see columns 3 and 4 of our Table 2); we converted the routines to accommodate double-precision arithmetic. We subsequently evaluated the coefficients, $A_1$, $A_2$, and $A_3$, (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_i

Let's carry out the integrals that appear in the definition of the $A_i$ coefficients,

$A_i$

$\equiv$

$a_{\ell }{a}_m{a}_s\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{\mathrm{\Delta }(a_i^2+u)},$

where,

$\mathrm{\Delta }$

$\equiv$

$[(a_{\ell }^2+u)(a_m^2+u)(a_s^2+u){]}^{1/2}.$

Here, we are adopting the $(\ell ,m,s)$ subscript notation to identify which semi-axis length is the (largest, medium-length, smallest).

Evaluating A_ℓ

First, let's focus on the coefficient associated with the longest axis $(i=\ell )$:

$\frac{A_{\ell }}{a_{\ell }{a}_m{a}_s}$

$=$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}[(a_{\ell }^2+u{)}^3(a_m^2+u)(a_s^2+u){]}^{-1/2}du$

Changing the integration variable to $x\equiv -u$, we obtain a definite integral expression that appears as equation (3.133.1) in I. W. Gradshteyn & I. M. Ryzhik (2007; 7^th Edition), Table of Integrals, Series, and Products — hereafter, GR7^th — namely,

$\frac{A_{\ell }}{a_{\ell }{a}_m{a}_s}$	$=$	$\underset{-\mathrm{\infty }}{\overset{0}{\int }}[(a_{\ell }^2-x{)}^3(a_m^2-x)(a_s^2-x){]}^{-1/2}dx$
	$=$	$\frac{2}{(a_{\ell }^2-a_m^2)\sqrt{a_{\ell }^2-a_s^2}}[F(\alpha ,p)-E(\alpha ,p)]$	… valid for $[a_{\ell }>a_m>a_s\ge 0]$
GR7th, p. 255, Eq. (3.133.1)

where (see p. 254 of GR7^th),

${\sin }^2\alpha$

$\equiv$

$\frac{a_{\ell }^2-a_s^2}{a_{\ell }^2-0}=1-\frac{a_s^2}{a_{\ell }^2},$

$p$

$\equiv$

$[\frac{a_{\ell }^2-a_m^2}{a_{\ell }^2-a_s^2}{]}^{1/2},$

and where, $E(\alpha ,p)$ and $F(\alpha ,p)$ are Legendre incomplete elliptic integrals of the first and second kind, respectively. (Note that in the notation convention adopted by GR7^th, the order of the argument list, $(\alpha ,p)$, is flipped relative to the convention that we have adopted above and elsewhere throughout our online, MediaWiki-based chapters.) Recognizing that,

$p^2{\sin }^3\alpha$

$=$

$\left[\frac{a_{\ell }^2-a_s^2}{a_{\ell }^2}{]}^{3/2}\right[\frac{a_{\ell }^2-a_m^2}{a_{\ell }^2-a_s^2}]=\frac{(a_{\ell }^2-a_s^2{)}^{1/2}}{a_{\ell }^3}[a_{\ell }^2-a_m^2],$

we see that the expression for $A_{\ell }$ can be rewritten as,

$\frac{A_{\ell }}{a_{\ell }{a}_m{a}_s}$

$=$

$\frac{2}{a_{\ell }^3{p}^2{\sin }^3\alpha }[F(\alpha ,p)-E(\alpha ,p)].$

This matches the expression that we have provided for $A_1$, above in the context of triaxial configurations.

Evaluating A_m

Next, let's evaluate the coefficient associated with the axis of intermediate length $(i=m)$:

$\frac{A_m}{a_{\ell }{a}_m{a}_s}$

$=$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}[(a_{\ell }^2+u)(a_m^2+u{)}^3(a_s^2+u){]}^{-1/2}du.$

This time, by changing the integration variable to $x\equiv -u$, we obtain a definite integral expression that appears as equation (3.133.7) in GR7^th, namely,

$\frac{A_m}{a_{\ell }{a}_m{a}_s}$	$=$	$\underset{-\mathrm{\infty }}{\overset{0}{\int }}[(a_{\ell }^2-x)(a_m^2-x{)}^3(a_s^2-x){]}^{-1/2}dx$
	$=$	$\frac{2\sqrt{a_{\ell }^2-a_s^2}}{(a_{\ell }^2-a_m^2)(a_m^2-a_s^2)}E(\alpha ,p)-\frac{2}{(a_{\ell }^2-a_m^2)\sqrt{a_{\ell }^2-a_s^2}}F(\alpha ,p)-\frac{2}{a_m^2-a_s^2}[\frac{a_s^2}{a_{\ell }^2{a}_m^2}{]}^{1/2}$	… valid for $[a_{\ell }>a_m>a_s\ge 0]$
GR7th, p. 256, Eq. (3.133.7)

(Here, the parameters, $\alpha$ and $p$, have the same definitions as in our above evaluation of $A_{\ell }$.) This time it is useful to recognize that,

$1-p^2$

$=$

$1-\frac{a_{\ell }^2-a_m^2}{a_{\ell }^2-a_s^2}=\frac{a_m^2-a_s^2}{a_{\ell }^2-a_s^2}$

in which case,

$p^2(1-p^2){\sin }^3\alpha$

$=$

$\frac{(a_{\ell }^2-a_m^2)(a_m^2-a_s^2)}{a_{\ell }^3(a_{\ell }^2-a_s^2{)}^{1/2}}.$

So the coefficient, $A_m$, may be rewritten as,

$p^2(1-p^2){\sin }^3\alpha \left[\frac{A_m}{a_{\ell }{a}_m{a}_s}\right]$	$=$	$\frac{(a_{\ell }^2-a_m^2)(a_m^2-a_s^2)}{a_{\ell }^3(a_{\ell }^2-a_s^2{)}^{1/2}}\{\frac{2\sqrt{a_{\ell }^2-a_s^2}}{(a_{\ell }^2-a_m^2)(a_m^2-a_s^2)}E(\alpha ,p)-\frac{2}{(a_{\ell }^2-a_m^2)\sqrt{a_{\ell }^2-a_s^2}}F(\alpha ,p)-\frac{2}{a_m^2-a_s^2}[\frac{a_s^2}{a_{\ell }^2{a}_m^2}{]}^{1/2}\}$
	$=$	$\frac{2}{a_{\ell }^3}\{E(\alpha ,p)\}-\frac{2(a_m^2-a_s^2)}{a_{\ell }^3(a_{\ell }^2-a_s^2)}\{F(\alpha ,p)\}-\frac{2(a_{\ell }^2-a_m^2)}{a_{\ell }^3(a_{\ell }^2-a_s^2{)}^{1/2}}\left[\frac{a_s}{a_{\ell }{a}_m}\right]$
	$=$	$\frac{2}{a_{\ell }^3}\{E(\alpha ,p)-(1-p^2)F(\alpha ,p)-p^2\sin \alpha [\frac{a_s}{a_m}\left]\right\}$
$\Rightarrow \frac{A_m}{a_{\ell }{a}_m{a}_s}$	$=$	$\frac{2}{a_{\ell }^3}\left[\frac{E(\alpha ,p)-(1-p^2)F(\alpha ,p)-(a_s/a_m)p^2\sin \alpha }{p^2(1-p^2){\sin }^3\alpha }\right].$

This matches the expression that we have provided for $A_2$, above in the context of triaxial configurations.

Evaluating A_s

Finally, let's evaluate the coefficient associated with the shortest axis, $(i=s)$:

$\frac{A_s}{a_{\ell }{a}_m{a}_s}$

$=$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}[(a_{\ell }^2+u)(a_m^2+u)(a_s^2+u{)}^3{]}^{-1/2}du.$

By changing the integration variable to $x\equiv -u$, this time we obtain a definite integral expression that appears as equation (3.133.13) in GR7^th, namely,

$\frac{A_s}{a_{\ell }{a}_m{a}_s}$	$=$	$\underset{-\mathrm{\infty }}{\overset{0}{\int }}[(a_{\ell }^2-x)(a_m^2-x)(a_s^2-x{)}^3{]}^{-1/2}dx$
	$=$	$\frac{2}{(a_s^2-a_m^2)\sqrt{a_{\ell }^2-a_s^2}}E(\alpha ,p)+\frac{2}{a_m^2-a_s^2}[\frac{a_m^2}{a_{\ell }^2{a}_s^2}{]}^{1/2}$	… valid for $[a_{\ell }>a_m>a_s>0]$
GR7th, p. 256, Eq. (3.133.13)

(And, again, the parameters, $\alpha$ and $p$, have the same definitions as in our above evaluation of $A_{\ell }$.) Recognizing that,

$(1-p^2){\sin }^3\alpha$

$=$

$\frac{(a_{\ell }^2-a_s^2{)}^{1/2}(a_m^2-a_s^2)}{a_{\ell }^3},$

the coefficient, $A_s$, may be rewritten as,

$(1-p^2){\sin }^3\alpha \left[\frac{A_s}{a_{\ell }{a}_m{a}_s}\right]$	$=$	$\frac{(a_{\ell }^2-a_s^2{)}^{1/2}(a_m^2-a_s^2)}{a_{\ell }^3}\{\frac{2}{(a_s^2-a_m^2)\sqrt{a_{\ell }^2-a_s^2}}E(\alpha ,p)+\frac{2}{a_m^2-a_s^2}[\frac{a_m^2}{a_{\ell }^2{a}_s^2}{]}^{1/2}\}$
	$=$	$\frac{2}{a_{\ell }^3}\{-E(\alpha ,p)\}+\frac{2(a_{\ell }^2-a_s^2{)}^{1/2}}{a_{\ell }^3}\left[\frac{a_m}{a_{\ell }{a}_s}\right]$
	$=$	$\frac{2}{a_{\ell }^3}\left\{\right(\frac{a_m}{a_s})\sin \alpha -E(\alpha ,p)\}$
$\Rightarrow \frac{A_s}{a_{\ell }{a}_m{a}_s}$	$=$	$\frac{2}{a_{\ell }^3}\left[\frac{(a_m/a_s)\sin \alpha -E(\alpha ,p)}{(1-p^2){\sin }^3\alpha }\right].$

This matches the expression that we have provided for $A_3$, above in the context of triaxial configurations.

When a_m = a_ℓ

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,

$\mathrm{\Delta }$	$\equiv$	$[(a_{\ell }^2+u)(a_m^2+u)(a_s^2+u){]}^{1/2}$
	$=$	$[(a_{\ell }^2+u{)}^2(a_s^2+u){]}^{1/2}.$

Index Symbols of the 1^st Order

Keeping in mind that, generically, the 1^st-order index symbol is given by the expression,

$A_i$	$\equiv$	$a_{\ell }{a}_m{a}_s\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{\mathrm{\Delta }(a_i^2+u)},$
[EFE], Chapter 3, p. 54, Eq. (103)

the coefficient associated with the longest axis is,

$\frac{A_{\ell }}{a_{\ell }^2{a}_s}$

$=$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{(a_{\ell }^2+u{)}^2(a_s^2+u{)}^{1/2}}.$

Changing the integration variable to $x\equiv (a_{\ell }^2+u)$, we obtain an integral expression that appears as equation (2.228.1) in GR7^th, namely,

$\frac{A_{\ell }}{a_{\ell }^2{a}_s}$	$=$	$\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2(a_s^2-a_{\ell }^2+x{)}^{1/2}}$
	$=$	$-[\frac{\sqrt{a_s^2-a_{\ell }^2+x}}{(a_s^2-a_{\ell }^2)x}{]}_{a_{\ell }^2}^{\mathrm{\infty }}-\frac{1}{2(a_s^2-a_{\ell }^2)}\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x(a_s^2-a_{\ell }^2+x{)}^{1/2}}$
	$=$	$-\frac{a_s}{(a_{\ell }^2-a_s^2)a_{\ell }^2}+\frac{1}{2(a_{\ell }^2-a_s^2)}\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x(a_s^2-a_{\ell }^2+x{)}^{1/2}}.$

The remaining integral in this expression appears as equation (2.224.5) in GR7^th. Its resolution depends on the sign of the constant term in the denominator, $(a_s^2-a_{\ell }^2)$. Given that this term is negative, the integration gives,

$\frac{A_{\ell }}{a_{\ell }^2{a}_s}$	$=$	$-\frac{a_s}{(a_{\ell }^2-a_s^2)a_{\ell }^2}+\frac{1}{2(a_{\ell }^2-a_s^2)}\left\{\frac{2}{(a_{\ell }^2-a_s^2{)}^{1/2}}{\tan }^{-1}\right[\frac{(a_s^2-a_{\ell }^2+x{)}^{1/2}}{(a_{\ell }^2-a_s^2{)}^{1/2}}]{\}}_{a_{\ell }^2}^{\mathrm{\infty }}$
$\Rightarrow A_{\ell }$	$=$	$-\frac{a_s^2}{(a_{\ell }^2-a_s^2)}+\frac{a_{\ell }^2{a}_s}{(a_{\ell }^2-a_s^2{)}^{3/2}}\{\frac{\pi }{2}-{\tan }^{-1}[\frac{a_s}{(a_{\ell }^2-a_s^2{)}^{1/2}}\left]\right\}$
	$=$	$-\frac{(1-e^2)}{e^2}+\frac{(1-e^2{)}^{1/2}}{e^3}\{\frac{\pi }{2}-{\tan }^{-1}[\frac{(1-e^2{)}^{1/2}}{e}\left]\right\}$
	$=$	$-\frac{(1-e^2)}{e^2}+\frac{(1-e^2{)}^{1/2}}{e^3}\{\frac{\pi }{2}-{\cos }^{-1}e\}$
	$=$	$-\frac{(1-e^2)}{e^2}+\frac{(1-e^2{)}^{1/2}}{e^3}\left\{{\sin }^{-1}e\right\},$

where, $e\equiv (1-a_s^2/a_{\ell }^2{)}^{1/2}$. Similarly, the coefficient associated with the shortest axis is,

$\frac{A_s}{a_{\ell }^2{a}_s}$

$=$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{(a_{\ell }^2+u)(a_s^2+u{)}^{3/2}}.$

This time, after changing the integration variable to $x\equiv (a_{\ell }^2+u)$, we obtain an integral expression that appears as equation (2.229.1) in GR7^th, namely,

$\frac{A_s}{a_{\ell }^2{a}_s}$	$=$	$\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x(a_s^2-a_{\ell }^2+x{)}^{3/2}}$
	$=$	$[\frac{2}{(a_s^2-a_{\ell }^2)\sqrt{a_s^2-a_{\ell }^2+x}}{]}_{a_{\ell }^2}^{\mathrm{\infty }}+\frac{1}{(a_s^2-a_{\ell }^2)}\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x(a_s^2-a_{\ell }^2+x{)}^{1/2}}.$

As before, the remaining integral in this expression appears as equation (2.224.5) in GR7^th; and, as before, the sign of the constant term in the denominator, $(a_s^2-a_{\ell }^2)$, is negative. Hence, the integration gives,

$\frac{A_s}{a_{\ell }^2{a}_s}$	$=$	$\frac{2}{(a_{\ell }^2-a_s^2)a_s}-\frac{1}{(a_{\ell }^2-a_s^2)}\left\{\frac{2}{(a_{\ell }^2-a_s^2{)}^{1/2}}{\tan }^{-1}\right[\frac{(a_s^2-a_{\ell }^2+x{)}^{1/2}}{(a_{\ell }^2-a_s^2{)}^{1/2}}]{\}}_{a_{\ell }^2}^{\mathrm{\infty }}$
$\Rightarrow A_s$	$=$	$\frac{2a_{\ell }^2{a}_s}{(a_{\ell }^2-a_s^2)a_s}-\frac{2a_{\ell }^2{a}_s}{(a_{\ell }^2-a_s^2{)}^{3/2}}\{\frac{\pi }{2}-{\tan }^{-1}[\frac{a_s}{(a_{\ell }^2-a_s^2{)}^{1/2}}\left]\right\}$
	$=$	$\frac{2}{e^2}-\frac{2(1-e^2{)}^{1/2}}{e^3}\{\frac{\pi }{2}-{\tan }^{-1}[\frac{(1-e^2{)}^{1/2}}{e}\left]\right\}$
	$=$	$\frac{2}{e^2}-\frac{2(1-e^2{)}^{1/2}}{e^3}\left\{{\sin }^{-1}e\right\}.$

Because we are evaluating the case where $A_m=A_{\ell }$, we alternatively should have been able to obtain the expression for $A_s$ immediately from our derived expression for $A_{\ell }$ via the known relation,

$2$

$=$

$A_{\ell }+A_m+A_s=2A_{\ell }+A_s.$

This approach gives,

$A_s$	$=$	$2-2A_{\ell }$
	$=$	$2+2\{\frac{(1-e^2)}{e^2}-\frac{(1-e^2{)}^{1/2}}{e^3}[{\sin }^{-1}e\left]\right\}$
	$=$	$\frac{2}{e^2}-\frac{2(1-e^2{)}^{1/2}}{e^3}\left[{\sin }^{-1}e\right],$

which, indeed, matches our separately derived expression for $A_s$.

Index Symbols of the 2^nd Order

Keeping in mind that, generically, the 2^nd-order index symbol is given by the expression,

$A_{ij}$	$\equiv$	$a_{\ell }{a}_m{a}_s\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{\mathrm{\Delta }(a_i^2+u)(a_j^2+u)},$
[EFE], Chapter 3, p. 54, Eq. (103)

and, in addition, for oblate-spheroidal configurations,

$\mathrm{\Delta }$

$=$

$[(a_{\ell }^2+u{)}^2(a_s^2+u){]}^{1/2},$

we have the following three independent expressions:

$A_{\ell \ell }=A_{mm}=A_{m\ell }$	$\equiv$	$a_{\ell }^2{a}_s\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{(a_{\ell }^2+u{)}^3(a_s^2+u{)}^{1/2}};$
$A_{s\ell }=A_{sm}$	$\equiv$	$a_{\ell }^2{a}_s\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{(a_{\ell }^2+u{)}^2(a_s^2+u{)}^{3/2}};$
$A_{ss}$	$\equiv$	$a_{\ell }^2{a}_s\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{(a_{\ell }^2+u)(a_s^2+u{)}^{5/2}}.$

Setting $x\equiv (a_{\ell }^2+u)$ in each expression gives:

$\frac{A_{\ell \ell }}{a_{\ell }^2{a}_s}=\frac{A_{mm}}{a_{\ell }^2{a}_s}=\frac{A_{m\ell }}{a_{\ell }^2{a}_s}$	$=$	$\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^3(a_s^2-a_{\ell }^2+x{)}^{1/2}}$	$\cdots$   See equation (2.228.2) in GR7th
$\frac{A_{s\ell }}{a_{\ell }^2{a}_s}=\frac{A_{sm}}{a_{\ell }^2{a}_s}$	$=$	$\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2(a_s^2-a_{\ell }^2+x{)}^{3/2}}$	$\cdots$   See equation (2.229.2) in GR7th
$\frac{A_{ss}}{a_{\ell }^2{a}_s}$	$=$	$\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x(a_s^2-a_{\ell }^2+x{)}^{5/2}}$	$\cdots$   See equation (2.227) in GR7th

Completing these integrals one at a time — while realizing that $z\equiv (a+bx)$, with $b=1$ and $a\equiv (a_s^2-a_{\ell }^2)<0$ — we have:

FIRST INTEGRAL …

$\frac{A_{\ell \ell }}{a_{\ell }^2{a}_s}=\frac{A_{mm}}{a_{\ell }^2{a}_s}=\frac{A_{m\ell }}{a_{\ell }^2{a}_s}$	$=$	$\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^3{z}^{1/2}}$
	$=$	$\left[\right(-\frac{1}{2a{x}^2}+\frac{3b}{4a^{2}x})z^{1/2}{]}_{a_{\ell }^2}^{\mathrm{\infty }}+\frac{3b^2}{8a^2}\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x{z}^{1/2}}$
	$=$	$\left[\right(-\frac{1}{2a{x}^2}+\frac{3b}{4a^{2}x})[(a_s^2-a_{\ell }^2)+x{]}^{1/2}{]}_{a_{\ell }^2}^{\mathrm{\infty }}+\frac{3b^2}{8a^2}\left\{\frac{2}{(-a{)}^{1/2}}{\tan }^{-1}\right[\frac{z^{1/2}}{(-a{)}^{1/2}}]{\}}_{a_{\ell }^2}^{\mathrm{\infty }}$
	$=$	$\left[\right(-\frac{3}{4a^2{a}_{\ell }^2}+\frac{1}{2a{a}_{\ell }^4})[(a_s^2-a_{\ell }^2)+a_{\ell }^2{]}^{1/2}]+\frac{3}{4(-a{)}^{5/2}}\{{\tan }^{-1}\left[\mathrm{\infty }\right]-{\tan }^{-1}\left[\frac{a_s}{(a_{\ell }^2-a_s^2{)}^{1/2}}\right]\}$
	$=$	$\left(\frac{2a-3a_{\ell }^2}{4a^2{a}_{\ell }^4}\right)a_s+\frac{3}{4(a_{\ell }^2-a_s^2{)}^{5/2}}\{\frac{\pi }{2}-{\tan }^{-1}[\frac{a_s}{(a_{\ell }^2-a_s^2{)}^{1/2}}\left]\right\}$
	$=$	$\left[\frac{2a_s^2-5a_{\ell }^2}{4(a_{\ell }^2-a_s^2{)}^2{a}_{\ell }^4}\right]a_s+\frac{3}{4(a_{\ell }^2-a_s^2{)}^{5/2}}\{\frac{\pi }{2}-{\tan }^{-1}[\frac{a_s}{(a_{\ell }^2-a_s^2{)}^{1/2}}\left]\right\}.$

Given that the eccentricity of an oblate-spheroidal configuration is $e\equiv (1-a_s^2/a_{\ell }^2{)}^{1/2}$, this latest expression can be rewritten as,

$A_{\ell \ell }=A_{mm}=A_{m\ell }$	$=$	$\left[\frac{2(1-e^2)-5}{4e^4}\right]\frac{a_s^2}{a_{\ell }^4}+\frac{3}{4e^5}\{\frac{\pi }{2}-{\tan }^{-1}[\frac{a_s/a_{\ell }}{e}\left]\right\}\frac{a_s}{a_{\ell }^3}$
$\Rightarrow a_{\ell }^2{A}_{\ell \ell }=a_{\ell }^2{A}_{mm}=a_{\ell }^2{A}_{m\ell }$	$=$	$\left[\frac{-2e^2-3}{4e^4}\right](1-e^2)+\frac{3(1-e^2{)}^{1/2}}{4e^5}\left\{{\sin }^{-1}e\right\}$
	$=$	$\frac{1}{4e^4}\{-(3+2e^2)(1-e^2)+3(1-e^2{)}^{1/2}\left[\frac{{\sin }^{-1}e}{e}\right]\}.$

SECOND INTEGRAL … remembering that $z\equiv (a+bx)$, with $b=1$ and $a\equiv (a_s^2-a_{\ell }^2)<0$ …

$\frac{A_{s\ell }}{a_{\ell }^2{a}_s}=\frac{A_{sm}}{a_{\ell }^2{a}_s}$	$=$	$\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2{z}^{3/2}}$
	$=$	$\left[\right(-\frac{1}{ax}-\frac{3b}{a^2})\frac{1}{z^{1/2}}{]}_{a_{\ell }^2}^{\mathrm{\infty }}-\frac{3b}{2a^2}\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x{z}^{1/2}}$
	$=$	$\left[\right(-\frac{1}{ax}-\frac{3b}{a^2})\frac{1}{z^{1/2}}{]}_{a_{\ell }^2}^{\mathrm{\infty }}-\frac{3b}{2a^2}\{\frac{2}{(-a{)}^{1/2}}{\tan }^{-1}\left[\frac{z^{1/2}}{(-a{)}^{1/2}}\right]{\}}_{a_{\ell }^2}^{\mathrm{\infty }}$
	$=$	$\left\{\right[\frac{1}{(a_{\ell }^2-a_s^2)x}-\frac{3}{(a_{\ell }^2-a_s^2{)}^2}]\frac{1}{[a_s^2-a_{\ell }^2+x{]}^{1/2}}{\}}_{a_{\ell }^2}^{\mathrm{\infty }}-\frac{3}{(a_{\ell }^2-a_s^2{)}^{5/2}}\{{\tan }^{-1}\left[\frac{[a_s^2-a_{\ell }^2+x{]}^{1/2}}{(a_{\ell }^2-a_s^2{)}^{1/2}}\right]{\}}_{a_{\ell }^2}^{\mathrm{\infty }}$
	$=$	$\left\{\right[\frac{3}{(a_{\ell }^2-a_s^2{)}^2}-\frac{1}{(a_{\ell }^2-a_s^2)a_{\ell }^2}\left]\frac{1}{a_s}\right\}-\frac{3}{(a_{\ell }^2-a_s^2{)}^{5/2}}\{\frac{\pi }{2}-{\tan }^{-1}[\frac{a_s}{(a_{\ell }^2-a_s^2{)}^{1/2}}\left]\right\}$
	$=$	$\frac{1}{a_s{a}_{\ell }^2}\left[\frac{2a_{\ell }^2+a_s^2}{(a_{\ell }^2-a_s^2{)}^2}\right]-\frac{3}{(a_{\ell }^2-a_s^2{)}^{5/2}}\{\frac{\pi }{2}-{\tan }^{-1}[\frac{a_s}{(a_{\ell }^2-a_s^2{)}^{1/2}}\left]\right\}.$

That is to say,

$A_{s\ell }=A_{sm}$	$=$	$\frac{2a_{\ell }^2+a_s^2}{(a_{\ell }^2-a_s^2{)}^2}-\frac{3a_{\ell }^2{a}_s}{(a_{\ell }^2-a_s^2{)}^{5/2}}\{\frac{\pi }{2}-{\tan }^{-1}[\frac{a_s}{(a_{\ell }^2-a_s^2{)}^{1/2}}\left]\right\}$
	$=$	$\frac{2+(1-e^2)}{a_{\ell }^2{e}^4}-\frac{3a_s}{a_{\ell }^3{e}^5}\{\frac{\pi }{2}-{\tan }^{-1}[\frac{a_s/a_{\ell }}{e}\left]\right\}$
	$=$	$\frac{1}{a_{\ell }^2{e}^4}\{(3-e^2)-\frac{3(1-e^2{)}^{1/2}}{e}[{\sin }^{-1}e\left]\right\}.$

CROSSCHECK Now, according to the relations stated in an accompanying discussion, we should find that, for $i=j$, $2A_{ii}+\sum _{j=1}^3{A}_{ij}$ $=$ $\frac{2}{a_i^2}.$ Let's check. Specifically, for an oblate spheroid when $i=\ell$, the summation takes the form, LHS $=$ $2A_{\ell \ell }+(A_{\ell s}+A_{\ell m}+A_{\ell \ell })$   $=$ $4A_{\ell \ell }+A_{\ell s}$   $=$ $\frac{1}{a_{\ell }^2{e}^4}\left\{\right[-3-2e^2](1-e^2)+3(1-e^2{)}^{1/2}\left[\frac{{\sin }^{-1}e}{e}\right]\}+\frac{1}{a_{\ell }^2{e}^4}\{(3-e^2)-3(1-e^2{)}^{1/2}[\frac{{\sin }^{-1}e}{e}\left]\right\}$   $=$ $\frac{1}{a_{\ell }^2{e}^4}\{(-3-2e^2)+(3+2e^2)e^2+(3-e^2)\}$   $=$ $\frac{1}{a_{\ell }^2{e}^4}\left\{2e^4\right\}=\frac{2}{a_{\ell }^2}.$ Q. E. D.       Yeah!

THIRD INTEGRAL …

$\frac{A_{ss}}{a_{\ell }^2{a}_s}$	$=$	$\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x{z}^{5/2}}$
	$=$	$\sum _{k=0}^{+1}[\frac{2}{(2k+1)a^{2-k}{z}^{k+1/2}}{]}_{a_{\ell }^2}^{\mathrm{\infty }}+\frac{1}{a^2}\underset{a_{\ell }^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x{z}^{1/2}}$
… on my whiteboard I completed the evaluation of this integral and obtained …
$\frac{3}{2}{a}_{\ell }^2{A}_{ss}$	$=$	$\frac{(4e^2-3)}{e^4(1-e^2)}+\frac{3(1-e^2{)}^{1/2}}{e^4}\left[\frac{{\sin }^{-1}e}{e}\right].$

CROSSCHECK Again, according to the relations stated in an accompanying discussion, we should find that, for $i=j$, $2A_{ii}+\sum _{j=1}^3{A}_{ij}$ $=$ $\frac{2}{a_i^2}.$ This time, specifically for an oblate spheroid, when $i=s$, this gives, $3A_{ss}+2A_{s\ell }$ $=$ $\frac{2}{a_s^2}$ $\Rightarrow \frac{3}{2}{a}_{\ell }^2{A}_{ss}$ $=$ $\frac{1}{(1-e^2)}-a_{\ell }^2{A}_{s\ell }$   $=$ $\frac{1}{(1-e^2)}-\frac{1}{e^4}\{(3-e^2)-3(1-e^2{)}^{1/2}\left[\frac{{\sin }^{-1}e}{e}\right]\}$   $=$ $\frac{1}{(1-e^2)}-\frac{(3-e^2)}{e^4}+\frac{3(1-e^2{)}^{1/2}}{e^4}\left[\frac{{\sin }^{-1}e}{e}\right]$   $=$ $\frac{e^4-(1-e^2)(3-e^2)}{e^4(1-e^2)}+\frac{3(1-e^2{)}^{1/2}}{e^4}\left[\frac{{\sin }^{-1}e}{e}\right]$   $=$ $\frac{(4e^2-3)}{e^4(1-e^2)}+\frac{3(1-e^2{)}^{1/2}}{e^4}\left[\frac{{\sin }^{-1}e}{e}\right]$ Q. E. D.       Yeah!

When a_m = a_s

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,

$\frac{A_{\ell }}{a_{\ell }{a}_s^2}$

$=$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{(a_s^2+u)(a_{\ell }^2+u{)}^{3/2}}.$

Changing the integration variable to $x\equiv (a_s^2+u)$, we obtain an integral expression that appears as equation (2.229.1) in GR7^th, namely,

$\frac{A_{\ell }}{a_{\ell }{a}_s^2}$	$=$	$\underset{a_s^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2(a_{\ell }^2-a_s^2+x{)}^{1/2}}$
$$	$=$	$[\frac{2}{(a_{\ell }^2-a_s^2)(a_{\ell }^2-a_s^2+x{)}^{1/2}}{]}_{a_s^2}^{\mathrm{\infty }}+\frac{1}{(a_{\ell }^2-a_s^2)}\underset{a_s^2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x(a_{\ell }^2-a_s^2+x{)}^{1/2}}.$

The remaining integral in this expression appears as equation (2.224.5) in GR7^th. Its resolution depends on the sign of the constant term in the denominator, $(a_{\ell }^2-a_s^2)$. Given that this term is positive, the integration gives,

$\frac{A_{\ell }}{a_{\ell }{a}_s^2}$	$=$	$-\frac{2}{(a_{\ell }^2-a_s^2)a_{\ell }}+\frac{1}{(a_{\ell }^2-a_s^2)}\{\frac{1}{\sqrt{(a_{\ell }^2-a_s^2)}}\ln [\frac{(a_{\ell }^2-a_s^2+x{)}^{1/2}-\sqrt{(a_{\ell }^2-a_s^2)}}{(a_{\ell }^2-a_s^2+x{)}^{1/2}+\sqrt{(a_{\ell }^2-a_s^2)}}]{\}}_{a_s^2}^{\mathrm{\infty }}$
$\Rightarrow A_{\ell }$	$=$	$-\frac{2a_{\ell }{a}_s^2}{(a_{\ell }^2-a_s^2)a_{\ell }}-\frac{a_{\ell }{a}_s^2}{(a_{\ell }^2-a_s^2{)}^{3/2}}\{\ln [\frac{1-e}{1+e}\left]\right\}$
	$=$	$\frac{(1-e^2)}{e^3}\{\ln [\frac{1+e}{1-e}\left]\right\}-\frac{2(1-e^2)}{e^2},$

where, as above, $e\equiv (1-a_s^2/a_{\ell }^2{)}^{1/2}$. Now, given that $A_m=A_s$, in this case we appreciate that,

$2$	$=$	$A_{\ell }+A_m+A_s=A_{\ell }+2A_s$
$\Rightarrow A_s$	$=$	$1-\frac{A_{\ell }}{2}$
	$=$	$1-\frac{1}{2}[\frac{(1-e^2)}{e^3}\cdot \ln (\frac{1+e}{1-e})-\frac{2(1-e^2)}{e^2}]$
	$=$	$\frac{1}{e^2}-\frac{(1-e^2)}{2e^3}\cdot \ln \left(\frac{1+e}{1-e}\right).$

For Spheres (a_ℓ = a_m = a_s)

In the case of a sphere, where $a_{\ell }=a_m=a_s$,

$\mathrm{\Delta }$

$\equiv$

$[(a_{\ell }^2+u)(a_m^2+u)(a_s^2+u){]}^{1/2}=(a_i^2+u{)}^{3/2},$

and the definition of all three $A_i$ coefficients is obtained from the integral,

$A_i$

$\equiv$

$a_{\ell }{a}_m{a}_s\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{\mathrm{\Delta }(a_i^2+u)}=a_i^3\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{(a_i^2+u{)}^{5/2}}.$

Analogously,

$A_{ii}$

$\equiv$

$a_i^3\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{du}{(a_i^2+u{)}^{7/2}}.$

According to p. 406, Eq. (139) of [CRC], $\int (a+bu{)}^{-n/2}du$ $=$ $\frac{2(a+bu{)}^{(2-n)/2}}{b(2-n)}.$

Hence,

$\frac{A_i}{a_i^3}\equiv \underset{0}{\overset{\mathrm{\infty }}{\int }}(a_i^2+u{)}^{-5/2}du$

$=$

$-[\frac{2}{3(a_i^2+u{)}^{3/2}}{]}_0^{\mathrm{\infty }}=\frac{2}{3a_i^3};$

and,

$\frac{A_{ii}}{a_i^3}\equiv \underset{0}{\overset{\mathrm{\infty }}{\int }}(a_i^2+u{)}^{-7/2}du$

$=$

$-[\frac{2}{5(a_i^2+u{)}^{5/2}}{]}_0^{\mathrm{\infty }}=\frac{2}{5a_i^5}.$

Derivation of Selected 2^nd-Order Index Symbols

Evaluating A_ℓℓ

In the case of $A_{\ell \ell }$, we have,

$\frac{A_{\ell \ell }}{a_{\ell }{a}_m{a}_s}$

$=$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}[(a_{\ell }^2+u{)}^5(a_m^2+u)(a_s^2+u){]}^{-1/2}du.$

Changing the integration variable to $x\equiv -u$, we obtain a definite integral expression that appears as equation (3.134.1) in I. W. Gradshteyn & I. M. Ryzhik (2007; 7^th Edition), Table of Integrals, Series, and Products — hereafter, GR7^th — namely,

$\frac{A_{\ell \ell }}{a_{\ell }{a}_m{a}_s}$	$=$	$\underset{-\mathrm{\infty }}{\overset{0}{\int }}[(a_{\ell }^2-x{)}^5(a_m^2-x)(a_s^2-x){]}^{-1/2}dx$
	$=$	$\frac{2}{3(a_{\ell }-a_m{)}^2(a_{\ell }-a_s{)}^{3/2}}[(3a_{\ell }-a_m-2a_s)F(\alpha ,p)-2(2a_{\ell }-a_m-a_s)E(\alpha ,p)]+\frac{2}{3(a_{\ell }-a_s)(a_{\ell }-a_m)}[\frac{a_s{a}_m}{a_{\ell }^3}{]}^{1/2},$	… valid for $[a_{\ell }>a_m>a_s\ge 0]$
GR7th, p. 257, Eq. (3.134.1)

where (see p. 254 of GR7^th),

${\sin }^2\alpha$

$\equiv$

$\frac{a_{\ell }^2-a_s^2}{a_{\ell }^2-0}=1-\frac{a_s^2}{a_{\ell }^2},$

$p$

$\equiv$

$[\frac{a_{\ell }^2-a_m^2}{a_{\ell }^2-a_s^2}{]}^{1/2},$

and where, $E(\alpha ,p)$ and $F(\alpha ,p)$ are Legendre incomplete elliptic integrals of the first and second kind, respectively. (Note that in the notation convention adopted by GR7^th, the order of the argument list, $(\alpha ,p)$, is flipped relative to the convention that we have adopted above and elsewhere throughout our online, MediaWiki-based chapters.)

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Work In Progress

Derivation of Expression for Gravitational Potential

In §373 (p. 700) of his book titled, Hydrodynamics, [Lamb32] states that, "The gravitation-potential, at internal points, of a uniform mass enclosed by the surface

$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}$	$=$	$1$
[Lamb32], §373, Eq. (1)

… may be written

$\frac{\mathrm{\Phi }(\overrightarrow{x})}{G}$	$=$	$\pi \rho ({\alpha }_0{x}^2+{\beta }_0{y}^2+{\gamma }_0{z}^2-{\chi }_0),$
[Lamb32], §373, Eq. (4)

where, as in §114,"

$\frac{{\alpha }_0}{abc}$	$=$	$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{d\lambda }{(a^2+\lambda )\mathrm{\Delta }},$		$\frac{{\beta }_0}{abc}$	$=$	$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{d\lambda }{(b^2+\lambda )\mathrm{\Delta }},$
$\frac{{\gamma }_0}{abc}$	$=$	$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{d\lambda }{(c^2+\lambda )\mathrm{\Delta }},$		$\frac{{\chi }_0}{abc}$	$=$	$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{d\lambda }{\mathrm{\Delta }},$
[Lamb32], §373, Eqs. (5) & (6)

and,

$\mathrm{\Delta }$	$=$	$[(a^2+\lambda )(b^2+\lambda )(c^2+\lambda ){]}^{1/2}.$
[Lamb32], §373, Eq. (3)

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 [EFE] and [T78] and presented above. Here, we are interested in demonstrating how [Lamb32] 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 $x$-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 $z$-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,

$\frac{{\varpi }^2}{a_3^2}+\frac{z^2}{a_1^2}=1\Rightarrow \varpi =a_3\sqrt{1-z^2/a_1^2},$

and the gravitational potential will be given by the expression,

$\mathrm{\Phi }(\overrightarrow{x})=-\pi G\rho [I_{\mathrm{B}\mathrm{T}}{a}_1^2-(A_1{z}^2+A_3{\varpi }^2\left)\right].$

The magnitude of the gravitational acceleration at the pole $(\varpi ,z)=(0,a_1)$ of this prolate spheroid can be obtained from the gravitational potential via the expression,

${𝒜}\equiv |-\frac{\partial \mathrm{\Phi }}{\partial z}{|}_{a_1}$

$=$

$2\pi G\rho A_1{a}_1,$

where, as above,

$A_1$

$=$

$\ln \left[\frac{1+e}{1-e}\right]\frac{(1-e^2)}{e^3}-\frac{2(1-e^2)}{e^2}.$

We should also be able to derive this expression for ${𝒜}$ by integrating the $z$-component of the differential acceleration over the mass distribution, that is,

${𝒜}$	$=$	$\int [\frac{G}{r^2}\cdot \frac{(a_1-z)}{r}]dm=\int \left[\frac{(a_1-z)G}{r^3}\right]2\pi \varpi d\varpi dz$
	$=$	$2\pi G\rho \underset{-a_1}{\overset{a_1}{\int }}(a_1-z)dz\underset{0}{\overset{a_3\sqrt{1-z^2/a_1^2}}{\int }}[{\varpi }^2+(z-a_1{)}^2{]}^{-3/2}\varpi d\varpi ,$

where the distance, $r$, has been measured from the pole, that is,

$r^2={\varpi }^2+(z-a_1{)}^2.$

Performing the integral over $\varpi$ gives,

${𝒜}$	$=$	$2\pi G\rho \underset{-a_1}{\overset{a_1}{\int }}(a_1-z)dz\{-[{\varpi }^2+(z-a_1{)}^2{]}^{-1/2}{\}}_0^{a_3\sqrt{1-z^2/a_1^2}}$
	$=$	$2\pi G\rho \underset{-a_1}{\overset{a_1}{\int }}(a_1-z)dz\{\frac{1}{z-a_1}-[a_3^2(1-\frac{z^2}{a_1^2})+a_1^2(1-\frac{z}{a_1}{)}^2{]}^{-1/2}\}$
	$=$	$-2\pi G\rho a_1\underset{-1}{\overset{1}{\int }}d\zeta \{\frac{1-\zeta }{1-\zeta }-(1-\zeta )[\left(\frac{a_3}{a_1}{)}^2\right(1-{\zeta }^2)+(1-\zeta {)}^2{]}^{-1/2}\}$
	$=$	$2\pi G\rho a_1\underset{-1}{\overset{1}{\int }}d\zeta \{(1-\zeta )[(2-e^2)-2\zeta +e^2{\zeta }^2{]}^{-1/2}-1\},$

where, $\zeta \equiv z/a_1$. For later reference, we will identify the expression inside the curly braces as the function, ${𝒵}$; specifically,

${𝒵}$	$\equiv$	$(1-\zeta )[(2-e^2)-2\zeta +e^2{\zeta }^2{]}^{-1/2}-1$
	$=$	$-1-\frac{\zeta }{\sqrt{X}}+\frac{1}{\sqrt{X}},$

where, in an effort to line up with notation found in integral tables, in this last expression we have used the notation, $X\equiv a+b\zeta +c{\zeta }^2$ and, in our case,

$a\equiv (2-e^2),$       $b\equiv -2,$       and       $c\equiv e^2.$

We find that,

$\underset{-1}{\overset{1}{\int }}{𝒵}d\zeta$	$=$	$-\zeta {|}_{-1}^1-\{\frac{\sqrt{X}}{c}{\}}_{-1}^1+[1+\frac{b}{2c}]\underset{-1}{\overset{1}{\int }}\frac{d\zeta }{\sqrt{X}}$
	$=$	$-2-\{\frac{\sqrt{(2-e^2)-2\zeta +e^2{\zeta }^2}}{e^2}{\}}_{-1}^1+[1-\frac{1}{e^2}\left]\right\{\frac{1}{\sqrt{c}}\ln [2\sqrt{cX}+2c\zeta +b]{\}}_{-1}^1$
	$=$	$-2-\left\{\frac{\sqrt{(2-e^2)-2+e^2}}{e^2}\right\}+\left\{\frac{\sqrt{(2-e^2)+2+e^2}}{e^2}\right\}+[1-\frac{1}{e^2}]\{\frac{1}{e}\ln [2\sqrt{e^2[(2-e^2)-2\zeta +e^2{\zeta }^2]}+2e^2\zeta -2]{\}}_{-1}^1$
	$=$	$-2+\frac{2}{e^2}+\left[\frac{e^2-1}{e^3}\right]\{\ln [2e^2-2]-\ln [4e-2e^2-2\left]\right\}$
	$=$	$-2\left[\frac{e^2-1}{e^2}\right]+\left[\frac{e^2-1}{e^3}\right]\{\ln [-2(1-e^2)]-\ln [-2(1-e{)}^2]\}$
	$=$	$\left[\frac{1-e^2}{e^3}\right]\ln \left[\frac{1+e}{1-e}\right]-2\left[\frac{1-e^2}{e^2}\right]$
	$=$	$A_1.$

Hence, we have,

${𝒜}=2\pi G\rho a_1\left[\underset{-1}{\overset{1}{\int }}{𝒵}d\zeta \right]=2\pi G\rho A_1{a}_1,$

which exactly matches the result obtained, above, by taking the derivative of the potential.

See Also

• Properties of Maclaurin Spheroids
• Excerpts from Maclaurin's (1742) A Treatise of Fluxions
• Inhomogeneous Ellipsoids Leading to Ferrers Potentials

Footnotes

1. In [EFE] this equation is written in terms of a variable $I$ instead of $I_{\mathrm{B}\mathrm{T}}$ as defined here. The two variables are related to one another straightforwardly through the expression, $I=I_{\mathrm{B}\mathrm{T}}{a}_1^2$.
2. Throughout [EFE], Chandrasekhar adopts a sign convention for the scalar gravitational potential that is opposite to the sign convention being used here.