ParabolicDensity/Axisymmetric/Structure

Parabolic Density Distribution

Part I:   Gravitational Potential

Part II:   Spherical Structures

Part III:   Axisymmetric Equilibrium Structures  Old: 1st thru 7th tries

Part IV:   Triaxial Equilibrium Structures (Exploration)

Axisymmetric (Oblate) Equilibrium Structures

Gravitational Potential

As we have detailed in an accompanying discussion, for an oblate-spheroidal configuration — that is, when $a_s<a_m=a_{\ell }$ — the gravitational potential may be obtained from the expression,

$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\mathbf{𝐱})}{(-\pi G{\rho }_c)}$

$=$

$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}{a}_1^2-(A_1{x}^2+A_2{y}^2+A_3{z}^2)+(A_{12}{x}^2{y}^2+A_{13}{x}^2{z}^2+A_{23}{y}^2{z}^2)+\frac{1}{6}(3A_{11}{x}^4+3A_{22}{y}^4+3A_{33}{z}^4),$

where, in the present context, we can rewrite this expression as,

$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\mathbf{𝐱})}{(-\pi G{\rho }_c)}$	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}{a}_{\ell }^2-[A_{\ell }(x^2+y^2)+A_s{z}^2]+[A_{\ell \ell }{x}^2{y}^2+A_{\ell s}{x}^2{z}^2+A_{\ell s}{y}^2{z}^2]+\frac{1}{6}[3A_{\ell \ell }{x}^4+3A_{\ell \ell }{y}^4+3A_{ss}{z}^4]$
	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}{a}_{\ell }^2-[A_{\ell }{\varpi }^2+A_s{z}^2]+[A_{\ell \ell }{x}^2{y}^2+A_{\ell s}{\varpi }^2{z}^2]+\frac{1}{2}[A_{\ell \ell }(x^4+y^4)+A_{ss}{z}^4]$
	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}{a}_{\ell }^2-[A_{\ell }{\varpi }^2+A_s{z}^2]+\frac{A_{\ell \ell }}{2}[(x^2+y^2{)}^2]+\frac{1}{2}[A_{ss}{z}^4]+[A_{\ell s}{\varpi }^2{z}^2]$
	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}{a}_{\ell }^2-[A_{\ell }{\varpi }^2+A_s{z}^2]+\frac{A_{\ell \ell }}{2}\left[{\varpi }^4\right]+\frac{1}{2}\left[A_{ss}{z}^4\right]+\left[A_{\ell s}{\varpi }^2{z}^2\right]$
$\Rightarrow \frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\mathbf{𝐱})}{(-\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-\left[A_{\ell }\right(\frac{{\varpi }^2}{a_{\ell }^2})+A_s(\frac{z^2}{a_{\ell }^2}\left)\right]+\frac{1}{2}\left[A_{\ell \ell }{a}_{\ell }^2\right(\frac{{\varpi }^4}{a_{\ell }^4})+A_{ss}{a}_{\ell }^2(\frac{z^4}{a_{\ell }^4})+2A_{\ell s}{a}_{\ell }^2(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\left)\right].$

Index Symbol Expressions

The expression for the zeroth-order normalization term $(I_{BT})$, and the relevant pair of 1^st-order index symbol expressions are:

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

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

where the eccentricity,

$e\equiv [1-(\frac{a_s}{a_{\ell }}{)}^2{]}^{1/2}.$

The relevant 2^nd-order index symbol expressions are:

$a_{\ell }^2{A}_{\ell \ell }$	$=$	$\frac{1}{4e^4}\{-(3+2e^2)(1-e^2)+3(1-e^2{)}^{1/2}\left[\frac{{\sin }^{-1}e}{e}\right]\};$
$\frac{3}{2}{a}_{\ell }^2{A}_{ss}$	$=$	$\frac{(4e^2-3)}{e^4(1-e^2)}+\frac{3(1-e^2{)}^{1/2}}{e^4}\left[\frac{{\sin }^{-1}e}{e}\right];$
$a_{\ell }^2{A}_{\ell s}$	$=$	$\frac{1}{e^4}\{(3-e^2)-3(1-e^2{)}^{1/2}\left[\frac{{\sin }^{-1}e}{e}\right]\}.$

We can crosscheck this last expression by drawing on a shortcut expression,

$A_{\ell s}$	$=$	$-\frac{A_{\ell }-A_s}{(a_{\ell }^2-a_s^2)}$
$\Rightarrow a_{\ell }^2{A}_{\ell s}$	$=$	$\frac{1}{e^2}\{A_s-A_{\ell }\}$
	$=$	$\frac{1}{e^2}\left\{\frac{2}{e^2}\right[(1-e^2{)}^{-1/2}-\frac{{\sin }^{-1}e}{e}](1-e^2{)}^{1/2}-\frac{1}{e^2}[\frac{{\sin }^{-1}e}{e}-(1-e^2{)}^{1/2}](1-e^2{)}^{1/2}\}$
	$=$	$\frac{1}{e^4}\left\{\right[2-2(1-e^2{)}^{1/2}\frac{{\sin }^{-1}e}{e}]-[(1-e^2{)}^{1/2}\frac{{\sin }^{-1}e}{e}-(1-e^2)\left]\right\}$
	$=$	$\frac{1}{e^4}\{(3-e^2)-3(1-e^2{)}^{1/2}\frac{{\sin }^{-1}e}{e}\}.$

Meridional Plane Equi-Potential Contours

Here, we follow closely our separate discussion of equipotential surfaces for Maclaurin Spheroids, assuming no rotation.

Configuration Surface

In the meridional $(\varpi ,z)$ plane, the surface of this oblate-spheroidal configuration — identified by the thick, solid-black curve below, in Figure 1 — is defined by the expression,

$\frac{\rho }{{\rho }_c}$	$=$	$1-[\frac{{\varpi }^2}{a_{\ell }^2}+\frac{z^2}{a_s^2}]=0$
$\Rightarrow \frac{{\varpi }^2}{a_{\ell }^2}+\frac{z^2}{a_s^2}$	$=$	$1$
$\Rightarrow z^2$	$=$	$a_s^2[1-\frac{{\varpi }^2}{a_{\ell }^2}]=a_{\ell }^2(1-e^2)[1-\frac{{\varpi }^2}{a_{\ell }^2}]$
$\Rightarrow \frac{z}{a_{\ell }}$	$=$	$\pm (1-e^2{)}^{1/2}[1-\frac{{\varpi }^2}{a_{\ell }^2}{]}^{1/2},$	for $0\le \frac{|\varpi |}{a_{\ell }}\le 1.$

Expression for Gravitational Potential

Throughout the interior of this configuration, each associated ${\mathrm{\Phi }}_{\mathrm{e}\mathrm{f}\mathrm{f}}$ = constant, equipotential surface is defined by the expression,

${\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}\equiv \frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\mathbf{𝐱})}{(\pi G{\rho }_c{a}_{\ell }^2)}+\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}$

$=$

$\left[A_{\ell }\right(\frac{{\varpi }^2}{a_{\ell }^2})+A_s(\frac{z^2}{a_{\ell }^2}\left)\right]-\frac{1}{2}\left[A_{\ell \ell }{a}_{\ell }^2\right(\frac{{\varpi }^4}{a_{\ell }^4})+A_{ss}{a}_{\ell }^2(\frac{z^4}{a_{\ell }^4})+2A_{\ell s}{a}_{\ell }^2(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\left)\right].$

Letting,

$\zeta \equiv \frac{z^2}{a_{\ell }^2}$,

we can rewrite this expression for ${\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}$ as,

${\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}$	$=$	$A_{\ell }\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)+A_s\zeta -\frac{1}{2}{A}_{\ell \ell }{a}_{\ell }^2\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)-\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^2-A_{\ell s}{a}_{\ell }^2\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)\zeta$
	$=$	$-\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^2+[A_s-A_{\ell s}{a}_{\ell }^2(\frac{{\varpi }^2}{a_{\ell }^2}\left)\right]\zeta +A_{\ell }\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)-\frac{1}{2}{A}_{\ell \ell }{a}_{\ell }^2\left(\frac{{\varpi }^4}{a_{\ell }^4}\right).$

Potential at the Pole

At the pole, $(\varpi ,z)=(0,a_s)$. Hence,

${\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}{|}_{\mathrm{m}\mathrm{i}\mathrm{d}}$	$=$	$-\frac{1}{2}{A}_{ss}{a}_{\ell }^2(\frac{a_s^2}{a_{\ell }^2}{)}^2+[A_s-A_{\ell s}{a}_{\ell }^2{\xcancel{\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)}}^0\left]\right(\frac{a_s^2}{a_{\ell }^2})+A_{\ell }{\xcancel{\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)}}^0-\frac{1}{2}{A}_{\ell \ell }{a}_{\ell }^2{\xcancel{\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)}}^0$
	$=$	$A_s\left(\frac{a_s^2}{a_{\ell }^2}\right)-\frac{1}{2}{A}_{ss}{a}_{\ell }^2(\frac{a_s^2}{a_{\ell }^2}{)}^2.$

General Determination of Vertical Coordinate (ζ)

Given values of the three parameters, $e$, $\varpi$, and ${\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}$, this last expression can be viewed as a quadratic equation for $\zeta$. Specifically, $0$ $=$ $\alpha {\zeta }^2+\beta \zeta +\gamma ,$ where, $\alpha$ $\equiv$ $\frac{1}{2}{A}_{ss}{a}_{\ell }^2$   $=$ $\frac{1}{3}\{\frac{(4e^2-3)}{e^4(1-e^2)}+\frac{3(1-e^2{)}^{1/2}}{e^4}[\frac{{\sin }^{-1}e}{e}\left]\right\},$ $\beta$ $\equiv$ $A_{\ell s}{a}_{\ell }^2\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)-A_s$   $=$ $\frac{1}{e^4}\{(3-e^2)-3(1-e^2{)}^{1/2}\frac{{\sin }^{-1}e}{e}\left\}\right(\frac{{\varpi }^2}{a_{\ell }^2})-\frac{2}{e^2}[(1-e^2{)}^{-1/2}-\frac{{\sin }^{-1}e}{e}](1-e^2{)}^{1/2},$ $\gamma$ $\equiv$ ${\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}+\frac{1}{2}{A}_{\ell \ell }{a}_{\ell }^2\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)-A_{\ell }\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)$   $=$ ${\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}+\frac{1}{8e^4}\{-(3+2e^2)(1-e^2)+3(1-e^2{)}^{1/2}\left[\frac{{\sin }^{-1}e}{e}\right]\left\}\right(\frac{{\varpi }^4}{a_{\ell }^4})-\frac{1}{e^2}[\frac{{\sin }^{-1}e}{e}-(1-e^2{)}^{1/2}](1-e^2{)}^{1/2}(\frac{{\varpi }^2}{a_{\ell }^2}).$ The solution of this quadratic equation gives, $\zeta$ $=$ $\frac{1}{2\alpha }\{-\beta \pm [{\beta }^2-4\alpha \gamma {]}^{1/2}\}.$ Should we adopt the superior (positive) sign, or is it more physically reasonable to adopt the inferior (negative) sign? As it turns out, $\beta$ is intrinsically negative, so the quantity, $-\beta$, is positive. Furthermore, when $\gamma$ goes to zero, we need $\zeta$ to go to zero as well. This will only happen if we adopt the inferior (negative) sign. Hence, the physically sensible root of this quadratic relation is given by the expression, $\zeta$ $=$ $\frac{1}{2\alpha }\{-\beta -[{\beta }^2-4\alpha \gamma {]}^{1/2}\}.$

Here we present a quantitatively accurate depiction of the shape of the (Ferrers) gravitational potential that arises from oblate-spheroidal configurations having a parabolic density distribution. We closely follow the discussion of equi-gravitational potential contours that arise in (uniform-density) Maclaurin spheroids. In order to facilitate comparison with Maclaurin spheroids, we will focus on a model with …

$\frac{a_s}{a_{\ell }}=0.582724,$	$e=0.81267,$
$A_{\ell }=A_m=0.51589042,$	$A_s=0.96821916,$	$I_{\mathrm{B}\mathrm{T}}=1.360556,$
$a_{\ell }^2{A}_{\ell \ell }=0.3287756,$	$a_{\ell }^2{A}_{ss}=1.5066848,$	$a_{\ell }^2{A}_{\ell s}=0.6848975.$

[NOTE:   Along the Maclaurin spheroid sequence, this is the eccentricity that marks bifurcation to the Jacobi ellipsoid sequence — see the first model listed in Table IV (p. 103) of [EFE] and/or see Tables 1 and 2 of our discussion of the Jacobi ellipsoid sequence. It is unlikely that this same eccentricity has a comparably special physical relevance along the sequence of spheroids having parabolic density distributions.]

The largest value of the gravitational potential that will arise inside (actually, on the surface) of the configuration is at $(\varpi ,z)=(1,0)$. That is, when,

${\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}{|}_{\mathrm{m}\mathrm{a}\mathrm{x}}$

$=$

$A_{\ell }-\frac{1}{2}{A}_{\ell \ell }{a}_{\ell }^2=0.3515026.$

So we will plot various equipotential surfaces having, $0<{\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}<{\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}{|}_{\mathrm{m}\mathrm{a}\mathrm{x}}$, recognizing that they will each cut through the equatorial plane $(z=0)$ at the radial coordinate given by,

${\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}$	$=$	$-\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\xcancel{{\zeta }^2}}^0+[A_s-A_{\ell s}{a}_{\ell }^2(\frac{{\varpi }^2}{a_{\ell }^2}\left)\right]{\xcancel{\zeta }}^0+A_{\ell }\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)-\frac{1}{2}{A}_{\ell \ell }{a}_{\ell }^2\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)$
$\Rightarrow 0$	$=$	$\frac{1}{2}{A}_{\ell \ell }{a}_{\ell }^2{\chi }^2-A_{\ell }\chi +{\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}},$

where,

$\chi \equiv \frac{{\varpi }^2}{a_{\ell }^2}.$

The solution to this quadratic equation gives,

${\chi }_{\mathrm{e}\mathrm{q}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{e}}$	$=$	$\frac{1}{A_{\ell \ell }{a}_{\ell }^2}\{A_{\ell }\pm [A_{\ell }^2-2A_{\ell \ell }{a}_{\ell }^2{\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}{]}^{1/2}\}$
	$=$	$\frac{A_{\ell }}{A_{\ell \ell }{a}_{\ell }^2}\{1-[1-\frac{2A_{\ell \ell }{a}_{\ell }^2{\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}}{A_{\ell }^2}{]}^{1/2}\}.$

Note that, again, the physically relevant root is obtained by adopting the inferior (negative) sign, as has been done in this last expression.

Equipotential Contours that Lie Entirely Within Configuration

For all $0<{\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}\le {\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}{|}_{\mathrm{m}\mathrm{i}\mathrm{d}}$, the equipotential contour will reside entirely within the configuration. In this case, for a given ${\varphi }_{\mathrm{c}\mathrm{h}\mathrm{o}\mathrm{i}\mathrm{c}\mathrm{e}}$, we can plot points along the contour by picking (equally spaced?) values of ${\chi }_{\mathrm{e}\mathrm{q}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{e}}\ge \chi \ge 0$, then solve the above quadratic equation for the corresponding value of $\zeta$.

In our example configuration, this means … (to be finished)

Tentative Summary

Known Relations

Density:	$\frac{\rho (\varpi ,z)}{{\rho }_c}$	$=$	$[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}],$
Gravitational Potential:	$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\varpi ,z)}{(-\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-A_{\ell }{\chi }^2-A_s{\zeta }^2+\frac{1}{2}[(A_{ss}{a}_{\ell }^2){\zeta }^4+2(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+(A_{\ell \ell }{a}_{\ell }^2){\chi }^4].$
Specific Angular Momentum:	$\frac{j^2}{(\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^3}$	$=$	$2A_{\ell }\chi -2A_{\ell \ell }{a}_{\ell }^2{\chi }^3.$
Centrifugal Potential:	$\frac{\mathrm{\Psi }}{(\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{2}[A_{\ell \ell }{a}_{\ell }^2{\chi }^4-2A_{\ell }{\chi }^2].$
Enthalpy:	$\left[\frac{H(\chi ,\zeta )-C_B}{(\pi G{\rho }_c{a}_{\ell }^2)}\right]-\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}$	$=$	$-A_s{\zeta }^2+\frac{{\zeta }^2}{2}[(A_{ss}{a}_{\ell }^2)({\zeta }^2-3{\chi }^2)+2(1-e^2{)}^{-1}{\chi }^2].$
Vertical Pressure Gradient:	$\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P}{\partial \zeta }$	$=$	$\frac{\rho }{{\rho }_c}\cdot [2A_{\ell s}{a}_{\ell }^2{\chi }^2\zeta -2A_s\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3]$
Radial Pressure Gradient:	$\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P}{\partial \chi }$	$=$	$\frac{\rho }{{\rho }_c}\cdot \left\{2A_{\ell s}{a}_{\ell }^2{\zeta }^2\chi \right\}$

where, $\chi \equiv \varpi /a_{\ell }$ and $\zeta \equiv z/a_{\ell }$, and the relevant index symbol expressions are:

$I_{\mathrm{B}\mathrm{T}}$	$=$	$2A_{\ell }+A_s(1-e^2)=2(1-e^2{)}^{1/2}[\frac{{\sin }^{-1}e}{e}];$
$A_{\ell }$	$=$	$\frac{1}{e^2}[\frac{{\sin }^{-1}e}{e}-(1-e^2{)}^{1/2}](1-e^2{)}^{1/2};$
$A_s$	$=$	$\frac{2}{e^2}[(1-e^2{)}^{-1/2}-\frac{{\sin }^{-1}e}{e}](1-e^2{)}^{1/2};$
$a_{\ell }^2{A}_{\ell \ell }$	$=$	$\frac{1}{4e^4}\{-(3+2e^2)(1-e^2)+3(1-e^2{)}^{1/2}\left[\frac{{\sin }^{-1}e}{e}\right]\};$
$\frac{3}{2}{a}_{\ell }^2{A}_{ss}$	$=$	$\frac{(4e^2-3)}{e^4(1-e^2)}+\frac{3(1-e^2{)}^{1/2}}{e^4}\left[\frac{{\sin }^{-1}e}{e}\right];$
$a_{\ell }^2{A}_{\ell s}$	$=$	$\frac{1}{e^4}\{(3-e^2)-3(1-e^2{)}^{1/2}\left[\frac{{\sin }^{-1}e}{e}\right]\},$

where the eccentricity,

$e\equiv [1-(\frac{a_s}{a_{\ell }}{)}^2{]}^{1/2}.$

6^th Try

Euler Equation

From, for example, here we can write the,

Eulerian Representation
of the Euler Equation,

$\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{v}=-\frac{1}{\rho }\mathrm{\nabla }P-\mathrm{\nabla }\mathrm{\Phi }$

In steady-state, we should set $\partial \overrightarrow{v}/\partial t=0$. There are various ways of expressing the nonlinear term on the LHS; from here, for example, we find,

$(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{v}=\frac{1}{2}\mathrm{\nabla }(\overrightarrow{v}\cdot \overrightarrow{v})-\overrightarrow{v}\times (\mathrm{\nabla }\times \overrightarrow{v})=\frac{1}{2}\mathrm{\nabla }(v^2)+\overrightarrow{\zeta }\times \overrightarrow{v},$

where,

$\overrightarrow{\zeta }\equiv \mathrm{\nabla }\times \overrightarrow{v}$

is commonly referred to as the vorticity.

Axisymmetric Configurations

From, for example, here, we appreciate that, quite generally, for axisymmetric systems when written in cylindrical coordinates,

$(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{v}$

=

${\hat{e}}_{\varpi }[v_{\varpi }\frac{\partial v_{\varpi }}{\partial \varpi }+v_z\frac{\partial v_{\varpi }}{\partial z}-\frac{v_{\phi }{v}_{\phi }}{\varpi }]+{\hat{e}}_{\phi }[v_{\varpi }\frac{\partial v_{\phi }}{\partial \varpi }+v_z\frac{\partial v_{\phi }}{\partial z}+\frac{v_{\phi }{v}_{\varpi }}{\varpi }]+{\hat{e}}_z[v_{\varpi }\frac{\partial v_z}{\partial \varpi }+v_z\frac{\partial v_z}{\partial z}].$

We seek steady-state configurations for which $v_{\varpi }=0$ and $v_z=0$, in which case this expression simplifies considerably to,

$(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{v}$	$=$	${\hat{e}}_{\varpi }[-\frac{v_{\phi }{v}_{\phi }}{\varpi }]$
	$=$	${\hat{e}}_{\varpi }[-\frac{j^2}{{\varpi }^3}],$

where, in this last expression we have replaced $v_{\phi }$ with the specific angular momentum, $j\equiv \varpi v_{\phi }=({\varpi }^2\dot{\phi })$, which is a conserved quantity in dynamically evolving systems. NOTE:   Up to this point in our discussion, $j$ can be a function of both coordinates, that is, $j=j(\varpi ,z)$.

As has been highlighted here for example — for the axisymmetric configurations under consideration — the ${\hat{e}}_{\varpi }$ and ${\hat{e}}_z$ components of the Euler equation become, respectively,

${\hat{e}}_{\varpi }$:     $-\frac{j^2}{{\varpi }^3}$ = $-[\frac{1}{\rho }\frac{\partial P}{\partial \varpi }+\frac{\partial \mathrm{\Phi }}{\partial \varpi }]$ ${\hat{e}}_z$:     $0$ = $-[\frac{1}{\rho }\frac{\partial P}{\partial z}+\frac{\partial \mathrm{\Phi }}{\partial z}]$

7^th Try

Introduction

Density:	$\frac{\rho (\chi ,\zeta )}{{\rho }_c}$	$=$	$[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}],$
Gravitational Potential:	$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\chi ,\zeta )}{(-\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-A_{\ell }{\chi }^2-A_s{\zeta }^2+\frac{1}{2}[(A_{ss}{a}_{\ell }^2){\zeta }^4+2(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+(A_{\ell \ell }{a}_{\ell }^2){\chi }^4].$
Specific Angular Momentum:	$\frac{j^2}{(\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^3}$	$=$	$2j_1\chi -2j_3{\chi }^3.$
Centrifugal Potential:	$\frac{\mathrm{\Psi }}{(\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{2}[j_3{\chi }^4-2j_1{\chi }^2].$

From above, we recall the following relations: $4e^4(A_{\ell \ell }{a}_{\ell }^2)$ $=$ $-(3+2e^2)(1-e^2)+\mathrm{{\rm Y}};$ $\frac{3}{2}{e}^4(A_{ss}{a}_{\ell }^2)$ $=$ $\frac{(4e^2-3)}{(1-e^2)}+\mathrm{{\rm Y}};$ $e^4(A_{\ell s}{a}_{\ell }^2)$ $=$ $(3-e^2)-\mathrm{{\rm Y}}.$ where, $\mathrm{{\rm Y}}$ $\equiv$ $3(1-e^2{)}^{1/2}[\frac{{\sin }^{-1}e}{e}].$ Crosscheck … Given that, $\mathrm{{\rm Y}}$ $=$ $(3-e^2)-e^4(A_{\ell s}{a}_{\ell }^2).$ we obtain the pair of relations, $4e^4(A_{\ell \ell }{a}_{\ell }^2)$ $=$ $-(3+2e^2)(1-e^2)+(3-e^2)-e^4(A_{\ell s}{a}_{\ell }^2)$   $=$ $-(3-3e^2+2e^2-2e^4)+(3-e^2)-e^4(A_{\ell s}{a}_{\ell }^2)$   $=$ $2e^4-e^4(A_{\ell s}{a}_{\ell }^2)$ $\Rightarrow (A_{\ell \ell }{a}_{\ell }^2)$ $=$ $\frac{1}{2}-\frac{1}{4}(A_{\ell s}{a}_{\ell }^2);$ $\frac{3}{2}{e}^4(A_{ss}{a}_{\ell }^2)$ $=$ $\frac{(4e^2-3)}{(1-e^2)}+(3-e^2)-e^4(A_{\ell s}{a}_{\ell }^2)$   $=$ $\frac{(4e^2-3)+(3-e^2)(1-e^2)}{(1-e^2)}-e^4(A_{\ell s}{a}_{\ell }^2)$   $=$ $\frac{e^4}{(1-e^2)}-e^4(A_{\ell s}{a}_{\ell }^2)$ $\Rightarrow (A_{ss}{a}_{\ell }^2)$ $=$ $\frac{2}{3}[\frac{1}{(1-e^2)}-(A_{\ell s}{a}_{\ell }^2)].$

RHS Square Brackets (TERM1)

Let's rewrite the term inside square brackets on the RHS of the expression for the gravitational potential.

$[{]}_{\mathrm{R}\mathrm{H}\mathrm{S}}$	$\equiv$	$[(A_{ss}{a}_{\ell }^2){\zeta }^4+2(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+(A_{\ell \ell }{a}_{\ell }^2){\chi }^4]$
	$=$	$e^{-4}\left\{\frac{2}{3}\right[\frac{(4e^2-3)}{(1-e^2)}+\mathrm{{\rm Y}}]{\zeta }^4+2[(3-e^2)-\mathrm{{\rm Y}}]{\chi }^2{\zeta }^2+\frac{1}{4}[-(3+2e^2)(1-e^2)+\mathrm{{\rm Y}}\left]{\chi }^4\right\}$
	$=$	$e^{-4}\left\{\frac{2}{3}\right[\frac{(4e^2-3)}{(1-e^2)}]{\zeta }^4+2[(3-e^2)]{\chi }^2{\zeta }^2+\frac{1}{4}[-(3+2e^2)(1-e^2)]{\chi }^4+\frac{2}{3}[{\zeta }^4-3{\zeta }^2{\chi }^2+\frac{3}{8}{\chi }^4\left]\mathrm{{\rm Y}}\right\}$
	$=$	$-e^{-4}\left\{\frac{2}{3}\right[\frac{(3-4e^2)}{(1-e^2)}]{\zeta }^4-2[(3-e^2)]{\chi }^2{\zeta }^2+\frac{1}{4}[(3+2e^2)(1-e^2)\left]{\chi }^4\right\}$
		$+e^{-4}\left\{\frac{2}{3}\right[({\zeta }^2-{\chi }^2)({\zeta }^2-2{\chi }^2)-\frac{13}{8}{\chi }^4\left]\mathrm{{\rm Y}}\right\}$
	$=$	$-e^{-4}\frac{2}{3(1-e^2)}\left\{\right[(3-4e^2)]{\zeta }^4-3[(3-e^2)](1-e^2){\chi }^2{\zeta }^2+\frac{3}{8}[(3+2e^2)](1-e^2{)}^2{\chi }^4\}$
		$+e^{-4}\left\{\frac{2}{3}\right[({\zeta }^2-{\chi }^2)({\zeta }^2-2{\chi }^2)-\frac{13}{8}{\chi }^4\left]\mathrm{{\rm Y}}\right\}$
	$=$	$-\frac{2e^{-4}}{(1-e^2)}\{{\zeta }^4-3(1-e^2){\chi }^2{\zeta }^2+\frac{3}{8}(1-e^2{)}^2{\chi }^4\}+\frac{8e^{-2}}{3(1-e^2)}\{{\zeta }^4-\frac{3}{4}(1-e^2){\chi }^2{\zeta }^2-\frac{3}{16}(1-e^2{)}^2{\chi }^4\}$
		$+\frac{2e^{-4}}{3}[({\zeta }^2-{\chi }^2)({\zeta }^2-2{\chi }^2)-\frac{13}{8}{\chi }^4]\mathrm{{\rm Y}}$
	$=$	$-\frac{2e^{-4}}{(1-e^2)}\left\{\underset{-0.038855}{\underbrace{[{\zeta }^2-(1-e^2){\chi }^2][{\zeta }^2-2(1-e^2){\chi }^2]-\frac{13}{8}(1-e^2{)}^2{\chi }^4}}\right\}+\frac{8e^{-2}}{3(1-e^2)}\left\{\stackrel{-0.010124}{\overbrace{[{\zeta }^2-(1-e^2){\chi }^2][{\zeta }^2+\frac{1}{4}(1-e^2){\chi }^2]+\frac{1}{16}(1-e^2{)}^2{\chi }^4}}\right\}$
		$+\frac{2e^{-4}}{3}\left[\underset{-0.061608}{\underbrace{({\zeta }^2-{\chi }^2)({\zeta }^2-2{\chi }^2)-\frac{13}{8}{\chi }^4}}\right]\mathrm{{\rm Y}}$
	$=$	$0.212119014$       (example #1, below) .

Check #1:

$({\zeta }^2-{\chi }^2)({\zeta }^2-2{\chi }^2)-\frac{13}{8}{\chi }^4$	$=$	${\zeta }^4-3{\chi }^2{\zeta }^2+2{\chi }^4-\frac{13}{8}{\chi }^4$
	$=$	${\zeta }^4-3{\chi }^2{\zeta }^2+\frac{3}{8}{\chi }^4.$

Check #2:

$({\zeta }^2-{\chi }^2)({\zeta }^2+\frac{1}{4}{\chi }^2)+\frac{1}{16}{\chi }^4$	$=$	${\zeta }^4-\frac{3}{4}{\chi }^2{\zeta }^2-\frac{1}{4}{\chi }^4+\frac{1}{16}{\chi }^4$
	$=$	${\zeta }^4-\frac{3}{4}{\chi }^2{\zeta }^2-\frac{3}{16}{\chi }^4$

RHS Quadratic Terms (TERM2)

The quadratic terms on the RHS can be rewritten as,

$A_{\ell }{\chi }^2+A_s{\zeta }^2$	$=$	$\left\{\frac{1}{e^2}\right[\frac{{\sin }^{-1}e}{e}-(1-e^2{)}^{1/2}](1-e^2{)}^{1/2}\}{\chi }^2+\left\{\frac{2}{e^2}\right[(1-e^2{)}^{-1/2}-\frac{{\sin }^{-1}e}{e}](1-e^2{)}^{1/2}\}{\zeta }^2$
	$=$	$\left\{\frac{1}{e^2}\right[(1-e^2{)}^{1/2}\frac{{\sin }^{-1}e}{e}-(1-e^2)]\}{\chi }^2+\{\frac{2}{e^2}[1-(1-e^2{)}^{1/2}\frac{{\sin }^{-1}e}{e}\left]\right\}{\zeta }^2$
	$=$	$\left\{\frac{1}{3e^2}\right[\mathrm{{\rm Y}}-3(1-e^2)\left]\right\}{\chi }^2+\left\{\frac{2}{3e^2}\right[3-\mathrm{{\rm Y}}\left]\right\}{\zeta }^2$
	$=$	$\frac{(\mathrm{{\rm Y}}-3)}{3e^2}[{\chi }^2-2{\zeta }^2]+{\chi }^2$
	$=$	$\frac{(\mathrm{{\rm Y}}-3)}{3e^2}(\chi +\sqrt{2}\zeta )(\chi -\sqrt{2}\zeta )+{\chi }^2$
$\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{2}$	$=$	$0.401150$ (example #1, below) .

where, again,

$\mathrm{{\rm Y}}$

$\equiv$

$3(1-e^2{)}^{1/2}[\frac{{\sin }^{-1}e}{e}]=2.040835.$

Gravitational Potential Rewritten

In summary, then,

$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\chi ,\zeta )}{(-\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-A_{\ell }{\chi }^2-A_s{\zeta }^2+\frac{1}{2}[(A_{ss}{a}_{\ell }^2){\zeta }^4+2(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+(A_{\ell \ell }{a}_{\ell }^2){\chi }^4]$
	$=$	$\frac{1}{3}\mathrm{{\rm Y}}-\frac{(\mathrm{{\rm Y}}-3)}{3e^2}(\chi +\sqrt{2}\zeta )(\chi -\sqrt{2}\zeta )-{\chi }^2$
		$-\frac{e^{-4}}{(1-e^2)}\left\{\right[{\zeta }^2-(1-e^2){\chi }^2\left]\right[{\zeta }^2-2(1-e^2){\chi }^2]-\frac{13}{8}(1-e^2{)}^2{\chi }^4\}+\frac{4e^{-2}}{3(1-e^2)}\{[{\zeta }^2-(1-e^2){\chi }^2][{\zeta }^2+\frac{1}{4}(1-e^2){\chi }^2]+\frac{1}{16}(1-e^2{)}^2{\chi }^4\}$
		$+\frac{e^{-4}}{3}[({\zeta }^2-{\chi }^2)({\zeta }^2-2{\chi }^2)-\frac{13}{8}{\chi }^4]\mathrm{{\rm Y}}$
	$=$	$\frac{1}{3}\mathrm{{\rm Y}}-\frac{(\mathrm{{\rm Y}}-3)}{3e^2}(\chi +\sqrt{2}\zeta )(\chi -\sqrt{2}\zeta )-{\chi }^2+\frac{4}{3e^2(1-e^2)}\left\{\right[{\zeta }^2-(1-e^2){\chi }^2\left]\right[{\zeta }^2+\frac{1}{4}(1-e^2){\chi }^2]+\frac{1}{16}(1-e^2{)}^2{\chi }^4\}$
		$-\frac{1}{e^4(1-e^2)}\left\{\right[{\zeta }^2-(1-e^2){\chi }^2\left]\right[{\zeta }^2-2(1-e^2){\chi }^2]-\frac{13}{8}(1-e^2{)}^2{\chi }^4\}+\frac{1}{3e^4}[({\zeta }^2-{\chi }^2)({\zeta }^2-2{\chi }^2)-\frac{13}{8}{\chi }^4]\mathrm{{\rm Y}}$
	$=$	$\frac{1}{3}\mathrm{{\rm Y}}-\frac{(\mathrm{{\rm Y}}-3)}{3e^2}(\chi +\sqrt{2}\zeta )(\chi -\sqrt{2}\zeta )+\frac{4}{3e^2(1-e^2)}\left\{\right[{\zeta }^2-(1-e^2){\chi }^2\left]\right[{\zeta }^2+\frac{1}{4}(1-e^2){\chi }^2\left]\right\}$
		$-\frac{1}{e^4(1-e^2)}\left\{\right[{\zeta }^2-(1-e^2){\chi }^2\left]\right[{\zeta }^2-2(1-e^2){\chi }^2\left]\right\}+\frac{\mathrm{{\rm Y}}}{3e^4}[({\zeta }^2-{\chi }^2)({\zeta }^2-2{\chi }^2)]$
		$-{\chi }^2+\frac{4}{3e^2(1-e^2)}\{\frac{1}{16}(1-e^2{)}^2{\chi }^4\}+\frac{1}{e^4(1-e^2)}\{\frac{13}{8}(1-e^2{)}^2{\chi }^4\}-\frac{\mathrm{{\rm Y}}}{3e^4}\left\{\frac{13}{8}{\chi }^4\right\}$
	$=$	$\frac{1}{3}\mathrm{{\rm Y}}-\frac{(\mathrm{{\rm Y}}-3)}{3e^2}(\chi +\sqrt{2}\zeta )(\chi -\sqrt{2}\zeta )+\frac{4(1-e^2)}{3e^2}\left\{\right[(1-e^2{)}^{-1}{\zeta }^2-{\chi }^2][(1-e^2{)}^{-1}{\zeta }^2+\frac{1}{4}{\chi }^2\left]\right\}$
		$-\frac{(1-e^2)}{e^4}\left\{\right[(1-e^2{)}^{-1}{\zeta }^2-{\chi }^2][(1-e^2{)}^{-1}{\zeta }^2-2{\chi }^2\left]\right\}+\frac{\mathrm{{\rm Y}}}{3e^4}[({\zeta }^2-{\chi }^2)({\zeta }^2-2{\chi }^2)]$
		$-{\chi }^2+\{\frac{(1-e^2)}{12e^2}+\frac{13(1-e^2)}{8e^4}-\frac{13\mathrm{{\rm Y}}}{24e^4}\}{\chi }^4.$
	$=$	0.767874 (row 1) + 0.5678833 (row 2) - 0.950574 (row 3) = 0.3851876  .

Example Evaluation

Let's evaluate these expressions, borrowing from the quantitative example specified above. Specifically, we choose,

$\frac{a_s}{a_{\ell }}=0.582724,$	$e=0.81267,$
$A_{\ell }=A_m=0.51589042,$	$A_s=0.96821916,$	$I_{\mathrm{B}\mathrm{T}}=\frac{2}{3}\mathrm{{\rm Y}}=1.360556,$
$a_{\ell }^2{A}_{\ell \ell }=0.3287756,$	$a_{\ell }^2{A}_{ss}=1.5066848,$	$a_{\ell }^2{A}_{\ell s}=0.6848975.$

Also, let's set $\rho /{\rho }_c=0.1$ and $\chi ={\chi }_1=0.75\Rightarrow {\chi }_1^2=0.5625$. This means that,

${\zeta }_1^2$	$=$	$(1-e^2)[1-{\chi }^2-\frac{\rho (\chi ,\zeta )}{{\rho }_c}]=[1-(0.81267{)}^2)][1-0.5625-0.1]=0.11460$
$\Rightarrow {\zeta }_1$	$=$	$0.33853.$

So, let's evaluate the gravitational potential …

$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}({\chi }_1,{\zeta }_1)}{(-\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-\left[\stackrel{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{2}}{\overbrace{A_{\ell }{\chi }^2+A_s{\zeta }^2}}\right]+\frac{1}{2}\left[\underset{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{1}}{\underbrace{(A_{ss}{a}_{\ell }^2){\zeta }^4+2(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+(A_{\ell \ell }{a}_{\ell }^2){\chi }^4}}\right]=0.385187372$
$\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{1}$	$=$	$0.019788921+0.088303509+0.104026655=0.212119085$
$\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{2}$	$=$	$0.290188361+0.110961809=0.401150171.$

Replace ζ With Normalized Density

First, let's readjust the last, 3-row expression for the gravitational potential so that ${\zeta }^2$ can be readily replaced with the normalized density.

$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\chi ,\zeta )}{(-\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{3}\mathrm{{\rm Y}}-\frac{(\mathrm{{\rm Y}}-3)}{3e^2}({\chi }^2-2{\zeta }^2)+\frac{4(1-e^2)}{3e^2}\left\{\right[(1-e^2{)}^{-1}{\zeta }^2-{\chi }^2][(1-e^2{)}^{-1}{\zeta }^2+\frac{1}{4}{\chi }^2\left]\right\}$
		$-\frac{(1-e^2)}{e^4}\left\{\right[(1-e^2{)}^{-1}{\zeta }^2-{\chi }^2][(1-e^2{)}^{-1}{\zeta }^2-2{\chi }^2\left]\right\}+\frac{\mathrm{{\rm Y}}}{3e^4}[({\zeta }^2-{\chi }^2)({\zeta }^2-2{\chi }^2)]$
		$-{\chi }^2+\frac{1}{24e^4}\{2e^2(1-e^2)+39(1-e^2)-13\mathrm{{\rm Y}}\}{\chi }^4.$

Now make the substitution,

${\zeta }^2$

$=$

$(1-e^2)[1-{\chi }^2-{\rho }^{*}],$

where,

${\rho }^{*}$

$\equiv$

$\frac{\rho (\chi ,\zeta )}{{\rho }_c}.$

We have,

$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\chi ,\zeta )}{(-\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{3}\mathrm{{\rm Y}}-\frac{(\mathrm{{\rm Y}}-3)}{3e^2}\{{\chi }^2-2(1-e^2)[1-{\chi }^2-{\rho }^{*}\left]\right\}+\frac{4(1-e^2)}{3e^2}\left\{\right[1-{\chi }^2-{\rho }^{*}]-{\chi }^2\}\left\{\right[1-{\chi }^2-{\rho }^{*}]+\frac{1}{4}{\chi }^2\}$
		$-\frac{(1-e^2)}{e^4}\left\{\right[1-{\chi }^2-{\rho }^{*}]-{\chi }^2\}\left\{\right[1-{\chi }^2-{\rho }^{*}]-2{\chi }^2\}+\frac{\mathrm{{\rm Y}}}{3e^4}\{(1-e^2)[1-{\chi }^2-{\rho }^{*}]-{\chi }^2\}\{(1-e^2)[1-{\chi }^2-{\rho }^{*}]-2{\chi }^2\}$
		$-{\chi }^2+\frac{1}{24e^4}\{2e^2(1-e^2)+39(1-e^2)-13\mathrm{{\rm Y}}\}{\chi }^4$
	$=$	$\frac{1}{3}\mathrm{{\rm Y}}-\frac{(\mathrm{{\rm Y}}-3)}{3e^2}\{-2+2e^2+(3-2e^2){\chi }^2+(2-2e^2){\rho }^{*}\}+\frac{4(1-e^2)}{3e^2}\{1-2{\chi }^2-{\rho }^{*}\}\{1-\frac{3}{4}{\chi }^2-{\rho }^{*}\}$
		$-\frac{(1-e^2)}{e^4}\{1-2{\chi }^2-{\rho }^{*}\}\{1-3{\chi }^2-{\rho }^{*}\}+\frac{\mathrm{{\rm Y}}}{3e^4}\{(1-e^2)-(2-e^2){\chi }^2-(1-e^2){\rho }^{*}\}\{(1-e^2)-(3-e^2){\chi }^2-(1-e^2){\rho }^{*}\}$
		$-{\chi }^2+\frac{1}{24e^4}\{39-37e^2-2e^4-13\mathrm{{\rm Y}}\}{\chi }^4$
	$=$	$\frac{1}{3}\mathrm{{\rm Y}}-\frac{(\mathrm{{\rm Y}}-3)}{3e^2}\{-2+3{\chi }^2+2{\rho }^{*}+2e^2[1-{\chi }^2-{\rho }^{*}\left]\right\}+\frac{4(1-e^2)}{3e^2}\{1-2{\chi }^2-{\rho }^{*}\}\{1-\frac{3}{4}{\chi }^2-{\rho }^{*}\}$
		$-\frac{(1-e^2)}{e^4}\{1-2{\chi }^2-{\rho }^{*}\}\{1-3{\chi }^2-{\rho }^{*}\}+\left\{\frac{\mathrm{{\rm Y}}}{3e^4}\right[1-2{\chi }^2-{\rho }^{*}]+\frac{\mathrm{{\rm Y}}}{3e^2}[-1+{\chi }^2+{\rho }^{*}\left]\right\}\{(1-e^2)-(3-e^2){\chi }^2-(1-e^2){\rho }^{*}\}$
		$-{\chi }^2+\frac{1}{24e^4}\{39-37e^2-2e^4-13\mathrm{{\rm Y}}\}{\chi }^4.$
	$=$	0.767874 (row 1) + 0.5678833 (row 2) - 0.950574 (row 3) = 0.3851876  .

Now, let's group together like terms and examine, in particular, whether the coefficient of the cross-product, ${\chi }^2{\rho }^{*})$, goes to zero.

$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\chi ,\zeta )}{(-\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{3}\mathrm{{\rm Y}}-\frac{(\mathrm{{\rm Y}}-3)}{3e^2}\{2e^2-2+(2-2e^2){\rho }^{*}\}$
		$+[1-2{\chi }^2-{\rho }^{*}]\left\{\frac{4(1-e^2)}{3e^2}\right[1-\frac{3}{4}{\chi }^2-{\rho }^{*}]-\frac{(1-e^2)}{e^4}[1-3{\chi }^2-{\rho }^{*}]+[\frac{\mathrm{{\rm Y}}}{3e^4}-\frac{\mathrm{{\rm Y}}}{3e^2}\left]\right[(1-e^2)-(3-e^2){\chi }^2-(1-e^2){\rho }^{*}\left]\right\}$
		$-\left\{\frac{\mathrm{{\rm Y}}}{3e^2}\right\}[(1-e^2)-(3-e^2){\chi }^2-(1-e^2){\rho }^{*}]{\chi }^2$
		$-{\chi }^2+\frac{1}{24e^4}\{39-37e^2-2e^4-13\mathrm{{\rm Y}}\}{\chi }^4-\frac{(\mathrm{{\rm Y}}-3)}{3e^2}\{3{\chi }^2-2e^2{\chi }^2\}$
	$=$	$\frac{1}{3}\mathrm{{\rm Y}}+\frac{(\mathrm{{\rm Y}}-3)}{3e^2}\{2(1-e^2)(1-{\rho }^{*})\}$
		$+[1-2{\chi }^2-{\rho }^{*}]\frac{(1-e^2)}{3e^4}\left\{4e^2\right[1-\frac{3}{4}{\chi }^2-{\rho }^{*}]-3[1-3{\chi }^2-{\rho }^{*}]+\mathrm{{\rm Y}}[(1-e^2)-(3-e^2){\chi }^2-(1-e^2){\rho }^{*}\left]\right\}$
		$+\left\{\frac{\mathrm{{\rm Y}}}{3e^2}\right\}[(1-e^2){\rho }^{*}]{\chi }^2$
		$-{\chi }^2+\frac{1}{24e^4}\{39-37e^2-2e^4-13\mathrm{{\rm Y}}\}{\chi }^4-\frac{(\mathrm{{\rm Y}}-3)}{3e^2}\{3{\chi }^2-2e^2{\chi }^2\}-\left\{\frac{\mathrm{{\rm Y}}}{3e^2}\right\}[(1-e^2)-(3-e^2){\chi }^2]{\chi }^2$
	$=$	$\frac{1}{3}\mathrm{{\rm Y}}+\frac{(\mathrm{{\rm Y}}-3)}{3e^2}\{2(1-e^2)(1-{\rho }^{*})\}$
		$+[(1-{\rho }^{*})]\frac{(1-e^2)}{3e^4}\left\{\right[4e^2-3+\mathrm{{\rm Y}}(1-e^2)](1-{\rho }^{*})\}+[-2{\chi }^2]\frac{(1-e^2)}{3e^4}\left\{\right[4e^2-3+\mathrm{{\rm Y}}(1-e^2)](1-{\rho }^{*})\}$
		$+[(1-{\rho }^{*})]\frac{(1-e^2)}{3e^4}\left\{\right[-3e^2+9-(3-e^2)\mathrm{{\rm Y}}\left]{\chi }^2\right\}+[-2{\chi }^2]\frac{(1-e^2)}{3e^4}\left\{\right[-3e^2+9-(3-e^2)\mathrm{{\rm Y}}\left]{\chi }^2\right\}$
		$+\left[\frac{\mathrm{{\rm Y}}(1-e^2)}{3e^2}\right]{\rho }^{*}{\chi }^2$
		$-{\chi }^2+\frac{1}{24e^4}\{39-37e^2-2e^4-13\mathrm{{\rm Y}}\}{\chi }^4-\frac{(\mathrm{{\rm Y}}-3)}{3e^2}\{3{\chi }^2-2e^2{\chi }^2\}-\left\{\frac{\mathrm{{\rm Y}}}{3e^2}\right\}[(1-e^2)-(3-e^2){\chi }^2]{\chi }^2$

8^th Try

Foundation

Density:	${\rho }^{*}\equiv \frac{\rho (\chi ,\zeta )}{{\rho }_c}$	$=$	$[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}],$
Gravitational Potential:	$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\chi ,\zeta )}{(-\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-A_{\ell }{\chi }^2-A_s{\zeta }^2+\frac{1}{2}[(A_{ss}{a}_{\ell }^2){\zeta }^4+2(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+(A_{\ell \ell }{a}_{\ell }^2){\chi }^4].$

Complete the Square

Again, let's rewrite the term inside square brackets on the RHS of the expression for the gravitational potential,

$[{]}_{\mathrm{R}\mathrm{H}\mathrm{S}}$

$\equiv$

$[(A_{ss}{a}_{\ell }^2){\zeta }^4+2(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+(A_{\ell \ell }{a}_{\ell }^2){\chi }^4],$

in such a way that we effectively "complete the square." Assuming that the desired expression takes the form,

$[{]}_{\mathrm{R}\mathrm{H}\mathrm{S}}$	$=$	$[(A_{ss}{a}_{\ell }^2{)}^{1/2}{\zeta }^2+B{\chi }^2\left]\right[(A_{ss}{a}_{\ell }^2{)}^{1/2}{\zeta }^2+C{\chi }^2]$
	$=$	$(A_{ss}{a}_{\ell }^2){\zeta }^4+(A_{ss}{a}_{\ell }^2{)}^{1/2}(B+C){\zeta }^2{\chi }^2+BC{\chi }^4,$

we see that we must have,

$(A_{ss}{a}_{\ell }^2{)}^{1/2}(B+C)$	$=$	$2(A_{\ell s}{a}_{\ell }^2)$
$\Rightarrow B$	$=$	$\frac{2(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2{)}^{1/2}}-C;$

and we must also have,

$BC$	$=$	$(A_{\ell \ell }{a}_{\ell }^2)$
$\Rightarrow B$	$=$	$\frac{(A_{\ell \ell }{a}_{\ell }^2)}{C}.$

Hence,

$\frac{(A_{\ell \ell }{a}_{\ell }^2)}{C}$	$=$	$\frac{2(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2{)}^{1/2}}-C$
$\Rightarrow 0$	$=$	$C^2-2\left[\frac{(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2{)}^{1/2}}\right]C+(A_{\ell \ell }{a}_{\ell }^2).$

The pair of roots of this quadratic expression are,

$C_{\pm }$	$=$	$\left[\frac{(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2{)}^{1/2}}\right]\pm \frac{1}{2}\left\{4\right[\frac{(A_{\ell s}{a}_{\ell }^2{)}^2}{(A_{ss}{a}_{\ell }^2)}]-4(A_{\ell \ell }{a}_{\ell }^2){\}}^{1/2}$
	$=$	$\frac{(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2{)}^{1/2}}\{1\pm [1-\frac{(A_{ss}{a}_{\ell }^2)(A_{\ell \ell }{a}_{\ell }^2)}{(A_{\ell s}{a}_{\ell }^2{)}^2}{]}^{1/2}\}$
$\Rightarrow \frac{C_{\pm }}{(A_{ss}{a}_{\ell }^2{)}^{1/2}}$	$=$	$\frac{(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2)}\{1\pm [1-\frac{(A_{ss}{a}_{\ell }^2)(A_{\ell \ell }{a}_{\ell }^2)}{(A_{\ell s}{a}_{\ell }^2{)}^2}{]}^{1/2}\}.$

Also, then,

$\frac{B_{\pm }}{(A_{ss}{a}_{\ell }^2{)}^{1/2}}$	$=$	$\frac{2(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2)}-\frac{C_{\pm }}{(A_{ss}{a}_{\ell }^2{)}^{1/2}}$
	$=$	$\frac{2(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2)}-\frac{(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2)}\{1\pm [1-\frac{(A_{ss}{a}_{\ell }^2)(A_{\ell \ell }{a}_{\ell }^2)}{(A_{\ell s}{a}_{\ell }^2{)}^2}{]}^{1/2}\}$
	$=$	$\frac{(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2)}\{1\mp [1-\frac{(A_{ss}{a}_{\ell }^2)(A_{\ell \ell }{a}_{\ell }^2)}{(A_{\ell s}{a}_{\ell }^2{)}^2}{]}^{1/2}\}.$

NOTE: Given that, $(A_{ss}{a}_{\ell }^2)$ $=$ $\frac{2}{3(1-e^2)}-\frac{2}{3}(A_{\ell s}{a}_{\ell }^2)$       and,       $(A_{\ell \ell }{a}_{\ell }^2)$ $=$ $\frac{1}{2}-\frac{1}{4}(A_{\ell s}{a}_{\ell }^2),$ we can write, Lambda vs Eccentricity $\mathrm{\Lambda }\equiv \frac{(A_{ss}{a}_{\ell }^2)(A_{\ell \ell }{a}_{\ell }^2)}{(A_{\ell s}{a}_{\ell }^2{)}^2}$ $=$ $\frac{1}{(A_{\ell s}{a}_{\ell }^2{)}^2}\left\{\right[\frac{2}{3(1-e^2)}-\frac{2}{3}(A_{\ell s}{a}_{\ell }^2)\left]\right[\frac{1}{2}-\frac{1}{4}(A_{\ell s}{a}_{\ell }^2)\left]\right\}$   $=$ $\frac{1}{6(A_{\ell s}{a}_{\ell }^2{)}^2}\left\{\frac{1}{(1-e^2)}\right[2-(A_{\ell s}{a}_{\ell }^2)]-(A_{\ell s}{a}_{\ell }^2)[2-(A_{\ell s}{a}_{\ell }^2)\left]\right\}$   $=$ $\frac{1}{6(A_{\ell s}{a}_{\ell }^2{)}^2}\left\{\right[\frac{1}{(1-e^2)}-(A_{\ell s}{a}_{\ell }^2)\left]\right[2-(A_{\ell s}{a}_{\ell }^2)\left]\right\}$

In summary, then, we can write,

$\frac{B_{\pm }}{(A_{ss}{a}_{\ell }^2{)}^{1/2}}$

$=$

$\frac{(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2)}[1\mp (1-\mathrm{\Lambda }{)}^{1/2}]$

and,

$\frac{C_{\pm }}{(A_{ss}{a}_{\ell }^2{)}^{1/2}}$

$=$

$\frac{(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2)}[1\pm (1-\mathrm{\Lambda }{)}^{1/2}],$

where, as illustrated by the inset "Lambda vs Eccentricity" plot, for all values of the eccentricity $(0<e\le 1)$, the quantity, $\mathrm{\Lambda }$, is greater than unity. It is clear, then, that both roots of the relevant quadratic equation are complex — i.e., they have imaginary components. But that's okay because the coefficients that appear in the right-hand-side, bracketed quartic expression appear in the combinations,

$(BC{)}_{\pm }$	$=$	$\frac{(A_{\ell s}{a}_{\ell }^2{)}^2}{(A_{ss}{a}_{\ell }^2)}[1-(1-\mathrm{\Lambda }{)}^{1/2}\left]\right[1+(1-\mathrm{\Lambda }{)}^{1/2}]=\frac{(A_{\ell s}{a}_{\ell }^2{)}^2}{(A_{ss}{a}_{\ell }^2)}\left[\mathrm{\Lambda }\right]=(A_{\ell \ell }{a}_{\ell }^2),$
$(B+C{)}_{\pm }$	$=$	$\frac{(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2{)}^{1/2}}[1\mp (1-\mathrm{\Lambda }{)}^{1/2}]+\frac{(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2{)}^{1/2}}[1\pm (1-\mathrm{\Lambda }{)}^{1/2}]=\frac{2(A_{\ell s}{a}_{\ell }^2)}{(A_{ss}{a}_{\ell }^2{)}^{1/2}},$

both of which are real.

9^th Try

Starting Key Relations

Density:	$\frac{\rho (\varpi ,z)}{{\rho }_c}$	$=$	$[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}],$
Gravitational Potential:	$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\varpi ,z)}{(-\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-A_{\ell }{\chi }^2-A_s{\zeta }^2+\frac{1}{2}[(A_{ss}{a}_{\ell }^2){\zeta }^4+2(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+(A_{\ell \ell }{a}_{\ell }^2){\chi }^4].$
Vertical Pressure Gradient:	$\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P}{\partial \zeta }$	$=$	$\frac{\rho }{{\rho }_c}\cdot [2A_{\ell s}{a}_{\ell }^2{\chi }^2\zeta -2A_s\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3]$

Play With Vertical Pressure Gradient

$\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P}{\partial \zeta }$	$=$	$[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}\left]\right[2A_{\ell s}{a}_{\ell }^2{\chi }^2\zeta -2A_s\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3]$
	$=$	$[(2A_{\ell s}{a}_{\ell }^2{\chi }^2-2A_s)\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3]-{\chi }^2[(2A_{\ell s}{a}_{\ell }^2{\chi }^2-2A_s)\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3]-{\zeta }^2(1-e^2{)}^{-1}[(2A_{\ell s}{a}_{\ell }^2{\chi }^2-2A_s)\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3]$
	$=$	$(2A_{\ell s}{a}_{\ell }^2{\chi }^2-2A_s)\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3-(2A_{\ell s}{a}_{\ell }^2{\chi }^4-2A_s{\chi }^2)\zeta -2A_{ss}{a}_{\ell }^2{\chi }^2{\zeta }^3-(1-e^2{)}^{-1}[(2A_{\ell s}{a}_{\ell }^2{\chi }^2-2A_s){\zeta }^3+2A_{ss}{a}_{\ell }^2{\zeta }^5]$
	$=$	$[(2A_{\ell s}{a}_{\ell }^2{\chi }^2-2A_s)-(2A_{\ell s}{a}_{\ell }^2{\chi }^4-2A_s{\chi }^2)]\zeta +[2A_{ss}{a}_{\ell }^2-2A_{ss}{a}_{\ell }^2{\chi }^2-(1-e^2{)}^{-1}(2A_{\ell s}{a}_{\ell }^2{\chi }^2-2A_s)]{\zeta }^3+[-(1-e^2{)}^{-1}2A_{ss}{a}_{\ell }^2]{\zeta }^5.$

Integrate over $\zeta$ gives …

$\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\int \left[\frac{\partial P}{\partial \zeta }\right]d\zeta$

$=$

$[(A_{\ell s}{a}_{\ell }^2{\chi }^2-A_s)-(A_{\ell s}{a}_{\ell }^2{\chi }^4-A_s{\chi }^2)]{\zeta }^2+\frac{1}{2}[A_{ss}{a}_{\ell }^2-A_{ss}{a}_{\ell }^2{\chi }^2-(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2{\chi }^2-A_s)]{\zeta }^4+\frac{1}{3}[-(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2]{\zeta }^6+\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}.$

Now Play With Radial Pressure Gradient

$\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P}{\partial \chi }$	$=$	$\frac{\rho }{{\rho }_c}\cdot \{-2A_{\ell }\chi +\frac{1}{2}[4(A_{\ell s}{a}_{\ell }^2){\zeta }^2\chi +4(A_{\ell \ell }{a}_{\ell }^2){\chi }^3\left]\right\}$
	$=$	$2[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}\left]\right[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })\chi +A_{\ell \ell }{a}_{\ell }^2{\chi }^3]$
$\Rightarrow \frac{1}{2}\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P}{\partial \chi }$	$=$	$[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })\chi +A_{\ell \ell }{a}_{\ell }^2{\chi }^3]-{\chi }^2[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })\chi +A_{\ell \ell }{a}_{\ell }^2{\chi }^3]-{\zeta }^2(1-e^2{)}^{-1}[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })\chi +A_{\ell \ell }{a}_{\ell }^2{\chi }^3]$
	$=$	$(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })\chi +[A_{\ell \ell }{a}_{\ell }^2+(A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2)]{\chi }^3-A_{\ell \ell }{a}_{\ell }^2{\chi }^5+(1-e^2{)}^{-1}[(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)\chi -A_{\ell \ell }{a}_{\ell }^2{\zeta }^2{\chi }^3]$
	$=$	$[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })+(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)]\chi +[A_{\ell \ell }{a}_{\ell }^2+(A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2)-(1-e^2{)}^{-1}{A}_{\ell \ell }{a}_{\ell }^2{\zeta }^2]{\chi }^3-A_{\ell \ell }{a}_{\ell }^2{\chi }^5$

Subtract a term $j^2\sim (j_4^2{\chi }^4+j_6^2{\chi }^6)$ to account for centrifugal acceleration …

$\frac{1}{2}\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P}{\partial \chi }-\frac{j^2}{{\chi }^3}\left[\frac{\rho }{{\rho }_c}\right]$	$=$	$[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })+(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)]\chi +[A_{\ell \ell }{a}_{\ell }^2+(A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2)-(1-e^2{)}^{-1}{A}_{\ell \ell }{a}_{\ell }^2{\zeta }^2]{\chi }^3-A_{\ell \ell }{a}_{\ell }^2{\chi }^5-\frac{j^2}{{\chi }^3}[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]$
	$=$	$[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })+(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)]\chi +[A_{\ell \ell }{a}_{\ell }^2+(A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2)-(1-e^2{)}^{-1}{A}_{\ell \ell }{a}_{\ell }^2{\zeta }^2]{\chi }^3-A_{\ell \ell }{a}_{\ell }^2{\chi }^5$
		$-\frac{(j_4^2{\chi }^4+j_6^2{\chi }^6)}{{\chi }^3}+\frac{(j_4^2{\chi }^4+j_6^2{\chi }^6)}{{\chi }^3}\left[{\chi }^2\right]+\frac{(j_4^2{\chi }^4+j_6^2{\chi }^6)}{{\chi }^3}[{\zeta }^2(1-e^2{)}^{-1}]$
	$=$	$[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })+(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)]\chi +[A_{\ell \ell }{a}_{\ell }^2+(A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2)-(1-e^2{)}^{-1}{A}_{\ell \ell }{a}_{\ell }^2{\zeta }^2]{\chi }^3-A_{\ell \ell }{a}_{\ell }^2{\chi }^5$
		$-(j_4^2\chi +j_6^2{\chi }^3)+(j_4^2\chi +j_6^2{\chi }^3)[{\zeta }^2(1-e^2{)}^{-1}]+(j_4^2{\chi }^3+j_6^2{\chi }^5)$
	$=$	$[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })+(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)]\chi +[A_{\ell \ell }{a}_{\ell }^2+(A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2)-(1-e^2{)}^{-1}{A}_{\ell \ell }{a}_{\ell }^2{\zeta }^2]{\chi }^3-A_{\ell \ell }{a}_{\ell }^2{\chi }^5$
		$+[j_4^2{\zeta }^2(1-e^2{)}^{-1}-j_4^2]\chi +[j_4^2+j_6^2{\zeta }^2(1-e^2{)}^{-1}-j_6^2]{\chi }^3+\left[j_6^2\right]{\chi }^5$
	$=$	$[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })+(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)+j_4^2{\zeta }^2(1-e^2{)}^{-1}-j_4^2]\chi$
		$+[A_{\ell \ell }{a}_{\ell }^2+(A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2)-(1-e^2{)}^{-1}{A}_{\ell \ell }{a}_{\ell }^2{\zeta }^2+j_4^2+j_6^2{\zeta }^2(1-e^2{)}^{-1}-j_6^2]{\chi }^3+[j_6^2-A_{\ell \ell }{a}_{\ell }^2]{\chi }^5$

Integrate over $\chi$ gives …

$\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\int \left[\frac{\partial P}{\partial \chi }\right]d\chi$	$=$	$[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })+(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)+j_4^2{\zeta }^2(1-e^2{)}^{-1}-j_4^2]{\chi }^2$
		$+\frac{1}{2}[A_{\ell \ell }{a}_{\ell }^2+(A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2)-(1-e^2{)}^{-1}{A}_{\ell \ell }{a}_{\ell }^2{\zeta }^2+j_4^2+j_6^2{\zeta }^2(1-e^2{)}^{-1}-j_6^2]{\chi }^4+\frac{1}{3}[j_6^2-A_{\ell \ell }{a}_{\ell }^2]{\chi }^6.$

See Also