# 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$$(\vec{x})$ can be derived from the gradient of a scalar potential $~\Phi$$(\vec{x})$ defined as follows:

$\Phi(\vec{x}) \equiv - \int \frac{G \rho(\vec{x}')}{|\vec{x}' - \vec{x}|} d^3 x' .$

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 $~(x,y,z) = (a_1,a_2,a_3)$,

$~\Phi(\vec{x}) = -\pi G \rho \biggl[ I_\mathrm{BT} a_1^2 - \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) \biggr],$

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

where,

 $~A_i$ $~\equiv$ $~a_1 a_2 a_3 \int_0^\infty \frac{du}{\Delta (a_i^2 + u )} ,$ $~I_\mathrm{BT}$ $~\equiv$ $~\frac{a_2 a_3}{a_1} \int_0^\infty \frac{du}{\Delta} = A_1 + A_2\biggl(\frac{a_2}{a_1}\biggr)^2+ A_3\biggl(\frac{a_3}{a_1}\biggr)^2 ,$ $~\Delta$ $~\equiv$ $~\biggl[ (a_1^2 + u)(a_2^2 + u)(a_3^2 + u) \biggr]^{1/2} .$

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

### Evaluation of Coefficients

The integrals defining $~A_i$ and $~I_\mathrm{BT}$ can be evaluated in terms of the incomplete elliptic integral of the first kind,

$~F(\theta,k) \equiv \int_0^\theta \frac{d\theta '}{\sqrt{1 - k^2 \sin^2\theta '}} ~~ ,$

$E(\theta,k) \equiv \int_0^\theta {\sqrt{1 - k^2 \sin^2\theta '}}~d\theta ' ~~ ,$

where, for our particular problem,

$~\theta \equiv \cos^{-1} \biggl(\frac{a_3}{a_1} \biggr) ,$

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

[ 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 $~(a_1 > a_2 > a_3)$

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

 $~A_1$ $~=$ $~\frac{2a_2 a_3}{a_1^2} \biggl[ \frac{F(\theta,k) - E(\theta,k)}{k^2 \sin^3\theta} \biggr] ~~;$ $~A_2$ $~=$ $~\frac{2a_2 a_3}{a_1^2} \biggl[ \frac{E(\theta,k) - (1-k^2)F(\theta,k) - (a_3/a_2)k^2\sin\theta}{k^2 (1-k^2) \sin^3\theta}\biggr] ~~;$ $~A_3$ $~=$ $~\frac{2a_2 a_3}{a_1^2} \biggl[ \frac{(a_2/a_3) \sin\theta - E(\theta,k)}{(1-k^2) \sin^3\theta} \biggr] ~~;$ $~I_\mathrm{BT}$ $~=$ $~\frac{2a_2 a_3}{a_1^2} \biggl[ \frac{F(\theta,k)}{\sin\theta} \biggr] ~~.$

[ 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,

$~\sum_{\ell=1}^3 A_\ell = 2 \, .$

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

Q.E.D.

#### Oblate Spheroids $~(a_1 = a_2 > a_3)$

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 $~I_\mathrm{BT}$ gives,

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

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

where the eccentricity,

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

#### Prolate Spheroids $~(a_1 > a_2 = a_3)$

If the shortest axis $~(a_3)$ and the intermediate axis $~(a_2)$ of the ellipsoid are equal to one another, then a cross-section in the $~x-y$ 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 $~I_\mathrm{BT}$ gives,

 $~A_1$ $~=$ $\ln\biggl[ \frac{1+e}{1-e} \biggr] \frac{(1-e^2)}{e^3} - \frac{2(1-e^2)}{e^2} ~~;$ $~A_2$ $~=$ $\frac{1}{e^2} - \ln\biggl[ \frac{1+e}{1-e} \biggr]\frac{(1-e^2)}{2e^3} ~~;$ $~A_3$ $~=$ $A_2 ~~;$ $~I_\mathrm{BT}$ $~=$ $~ A_1 + 2(1-e^2)A_2 = \ln\biggl[ \frac{1+e}{1-e} \biggr]\frac{(1-e^2)}{e} ~~,$

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

where, again, the eccentricity,

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

## 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, $~(\tfrac{b}{a}, \tfrac{c}{a})$, 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 Chandrasekhar's EFE. In particular, for twenty-six separate axis-ratio pairs, Chandrasekhar's Table IV lists the values of the square of the angular velocity, $~\Omega^2$, 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",

 $~\frac{\Omega^2}{\pi G\rho}$ $~=$ $~2B_{12}$ [ EFE, §39, Eq. (5) ] $~=$ $~2\biggl[\frac{A_1 - (b/a)^2A_2}{1-(b/a)^2} \biggr] \, ,$ [using, EFE, §21, Eqs. (105) & (107) ]

and, for "angmom",

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

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

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

Table 1:  Example Evaluations
Given Determined using calculator and (crude) CRC tables of elliptic integrals
$~\frac{a_2}{a_1}$ $~\frac{a_3}{a_1}$ $~\theta$ $~k$ $~\sin^{-1}k$ $~F(\theta,k)$ $~E(\theta,k)$ $~A_1$ $~A_2$ $~A_3$
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, $~(\tfrac{b}{a},\tfrac{c}{a})$ — 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.

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## 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,

$~\Phi(\vec{x}) = -\pi G \rho \biggl[ I_\mathrm{BT} a_1^2 - \biggl(A_1 z^2 + A_3 \varpi^2 \biggr) \biggr].$

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,

 $~\mathcal{A} \equiv \biggl|- \frac{\partial \Phi}{\partial z}\biggr|_{a_1}$ $~=$ $~2\pi G \rho A_1 a_1 \, ,$

where, as above,

 $~A_1$ $~=$ $\ln\biggl[ \frac{1+e}{1-e} \biggr] \frac{(1-e^2)}{e^3} - \frac{2(1-e^2)}{e^2} \, .$

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

 $~\mathcal{A}$ $~=$ $~\int \biggl[ \frac{G }{r^2} \cdot \frac{(a_1-z)}{r} \biggr] dm = \int \biggl[ \frac{(a_1-z)G }{r^3} \biggr] 2\pi \varpi d\varpi dz$ $~=$ $~2\pi G\rho \int^{a_1}_{-a_1} (a_1-z)dz \int_0^{a_3\sqrt{1-z^2/a_1^2}} [\varpi^2+(z-a_1)^2]^{-3/2}\varpi d\varpi \, ,$

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,

 $~\mathcal{A}$ $~=$ $~2\pi G\rho \int^{a_1}_{-a_1} (a_1-z)dz \biggl\{ -[\varpi^2+(z-a_1)^2]^{-1/2} \biggr\}_0^{a_3\sqrt{1-z^2/a_1^2}}$ $~=$ $~2\pi G\rho \int^{a_1}_{-a_1} (a_1-z)dz \biggl\{ \frac{1}{z - a_1} -\biggl[ a_3^2 \biggl(1-\frac{z^2}{a_1^2} \biggr) + a_1^2\biggl(1-\frac{z}{a_1}\biggr)^2 \biggr]^{-1/2} \biggr\}$ $~=$ $~ - 2\pi G\rho a_1 \int^{1}_{-1} d\zeta \biggl\{ \frac{1-\zeta}{1-\zeta } - (1-\zeta)\biggl[ \biggl(\frac{a_3}{a_1}\biggr)^2 \biggl(1-\zeta^2 \biggr) + \biggl(1-\zeta\biggr)^2 \biggr]^{-1/2} \biggr\}$ $~=$ $~ 2\pi G\rho a_1 \int^{1}_{-1} d\zeta \biggl\{ (1-\zeta) [ (2-e^2) - 2\zeta + e^2\zeta^2 ]^{-1/2} -1 \biggr\} \, ,$

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

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

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,

 $~\int_{-1}^1 \mathcal{Z} d\zeta$ $~=$ $~- \zeta\biggr|_{-1}^{1} - \biggl\{ \frac{\sqrt{X}}{c} \biggr\}_{-1}^1 +\biggl[1 + \frac{b}{2c} \biggr]\int_{-1}^1 \frac{d\zeta}{\sqrt{X}}$ $~=$ $~- 2 - \biggl\{ \frac{\sqrt{(2-e^2) -2\zeta + e^2\zeta^2}}{e^2} \biggr\}_{-1}^1 +\biggl[1 - \frac{1}{e^2} \biggr] \biggl\{ \frac{1}{\sqrt{c}} \ln \biggl[2\sqrt{cX} + 2c\zeta + b \biggr] \biggr\}_{-1}^1$ $~=$ $~- 2 - \biggl\{ \frac{\sqrt{(2-e^2) -2 + e^2}}{e^2} \biggr\} + \biggl\{ \frac{\sqrt{(2-e^2) +2 + e^2}}{e^2} \biggr\} + \biggl[1 - \frac{1}{e^2} \biggr] \biggl\{ \frac{1}{e} \ln \biggl[2\sqrt{e^2[(2-e^2) -2\zeta + e^2\zeta^2]} + 2e^2\zeta - 2 \biggr] \biggr\}_{-1}^1$ $~=$ $~- 2 + \frac{2}{e^2} +\biggl[\frac{e^2-1}{e^3} \biggr] \biggl\{ \ln \biggl[2e^2 - 2 \biggr] - \ln \biggl[4e - 2e^2 - 2 \biggr] \biggr\}$ $~=$ $~- 2\biggl[\frac{e^2 - 1}{e^2}\biggr] +\biggl[\frac{e^2-1}{e^3} \biggr] \biggl\{ \ln \biggl[-2(1-e^2) \biggr] - \ln \biggl[-2(1-e)^2\biggr] \biggr\}$ $~=$ $~\biggl[\frac{1-e^2}{e^3} \biggr] \ln \biggl[\frac{1+e}{1-e} \biggr] -2\biggl[\frac{1-e^2 }{e^2}\biggr]$ $~=$ $~A_1 \, .$

Hence, we have,

$~\mathcal{A} = 2\pi G\rho a_1 \biggl[ \int_{-1}^1 \mathcal{Z} d\zeta\biggr]= 2\pi G \rho A_1 a_1 \, ,$

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

1. In EFE this equation is written in terms of a variable $~I$ instead of $~I_\mathrm{BT}$ as defined here. The two variables are related to one another straightforwardly through the expression, $~I = I_\mathrm{BT} a_1^2$.