ParabolicDensity/Axisymmetric/Structure/Try1thru7

Parabolic Density Distribution (Tries 1 thru 7)

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

Setup

Here we specifically discuss the case of configurations that exhibit concentric ellipsoidal iso-density surfaces of the form,

$\rho$

=

${\rho }_c[1-(\frac{x^2+y^2}{a_{\ell }^2}+\frac{z^2}{a_s^2}\left)\right],$

that is, axisymmetric ($a_m=a_{\ell }$, i.e., oblate) configurations with parabolic density distributions. Much of our presentation, here, is drawn from our separate, detailed description of what we will refer to as Ferrers potential.

This can be rewritten in terms of T1 Coordinates. In particular, defining, $q\equiv a_{\ell }/a_s$ and, ${\xi }_1$ $\equiv$ $[z^2+(\frac{\varpi }{q}{)}^2{]}^{1/2}=a_s\left[\right(\frac{z}{a_s}{)}^2+(\frac{\varpi }{a_{\ell }}{)}^2{]}^{1/2}$ $\Rightarrow \frac{\rho }{{\rho }_c}$ $=$ $[1-(\frac{{\xi }_1}{a_s}{)}^2].$ Because we expect contours of constant enthalpy $(H)$ to coincide with contours of constant density in equilibrium configurations, we should expect to find that, $\frac{H}{H_c}$ $=$ $h({\xi }_1).$ If the "radial" enthalpy profile resembles our derived spherical enthalpy profile, we should expect to find that, $h({\xi }_1)$ $\sim$ $h_0[1-h_2{\xi }_1^2-h_4{\xi }_1^4]$ $\Rightarrow 1-\frac{h({\xi }_1)}{h_0}$ $\sim$ $h_2{\xi }_1^2+h_4{\xi }_1^4$   $=$ $h_2{a}_s^2\left[\right(\frac{z}{a_s}{)}^2+\left(\frac{\varpi }{a_{\ell }}{)}^2\right]+h_4\left\{a_s^2\right[(\frac{z}{a_s}{)}^2+(\frac{\varpi }{a_{\ell }}{)}^2]{\}}^2$   $=$ $h_2{a}_s^2\left[\right(\frac{z}{a_s}{)}^2+\left(\frac{\varpi }{a_{\ell }}{)}^2\right]+h_4{a}_s^4\left[\right(\frac{z}{a_s}{)}^4+2\left(\frac{z}{a_s}{)}^2\right(\frac{\varpi }{a_{\ell }}{)}^2+\left(\frac{\varpi }{a_{\ell }}{)}^4\right]$

Total Mass

The surface of the configuration with eccentricity, $e$, is defined by the expression,

$1$

=

$\frac{{\varpi }^2}{a_{\ell }^2}+\frac{z^2}{a_s^2}={\chi }^2+(1-e^2{)}^{-1}{\zeta }^2.$

When integrating over the volume elements, at each "radial" location, $\chi$,the vertical limit will be given by the expression,

${\zeta }_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}$

=

$(1-e^2{)}^{1/2}(1-{\chi }^2{)}^{1/2}$

Specifically,

$\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}$	=	$a_{\ell }^{3}2\pi \underset{0}{\overset{1}{\int }}\chi d\chi \underset{0}{\overset{{\zeta }_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{\int }}2d\zeta$
	=	$a_{\ell }^{3}4\pi \underset{0}{\overset{1}{\int }}\chi [\zeta {]}_0^{{\zeta }_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}d\chi$
	=	$a_{\ell }^{3}4\pi (1-e^2{)}^{1/2}\underset{0}{\overset{1}{\int }}\chi (1-{\chi }^2{)}^{1/2}d\chi .$

Make the variable substitution, $\chi \to \sin \theta \Rightarrow d\chi =\cos \theta d\theta$ …

$\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}$	=	$a_{\ell }^{3}4\pi (1-e^2{)}^{1/2}\underset{0}{\overset{\pi /2}{\int }}\sin \theta {\cos }^{}\theta d\theta$
	=	$a_{\ell }^{3}4\pi (1-e^2{)}^{1/2}[-\frac{{\cos }^3\theta }{3}{]}_0^{\pi /2}$
	=	$\frac{4\pi a_{\ell }^3}{3}(1-e^2{)}^{1/2}.$

Likewise, the mass is given by the expression,

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}$	=	$\left[{\rho }_c{a}_{\ell }^3\right]2\pi \underset{0}{\overset{1}{\int }}\chi d\chi \underset{0}{\overset{{\zeta }_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{\int }}2\left(\frac{\rho }{{\rho }_c}\right)d\zeta$
	=	$\left[{\rho }_c{a}_{\ell }^3\right]4\pi \underset{0}{\overset{1}{\int }}\chi d\chi \left\{\underset{0}{\overset{{\zeta }_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{\int }}\right[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]d\zeta \}$
	=	$\left[{\rho }_c{a}_{\ell }^3\right]4\pi \underset{0}{\overset{1}{\int }}\chi d\chi \left\{\underset{0}{\overset{{\zeta }_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{\int }}\right[1-{\chi }^2]d\zeta -\underset{0}{\overset{{\zeta }_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{\int }}[{\zeta }^2(1-e^2{)}^{-1}]d\zeta \}$
	=	$\left[{\rho }_c{a}_{\ell }^3\right]4\pi \underset{0}{\overset{1}{\int }}\chi d\chi \left\{\right[(1-{\chi }^2)\zeta {]}_0^{{\zeta }_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}-[\frac{{\zeta }^3}{3}(1-e^2{)}^{-1}{]}_0^{{\zeta }_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}\}$
	=	$\left[{\rho }_c{a}_{\ell }^3\right]4\pi \underset{0}{\overset{1}{\int }}\chi d\chi \{(1-{\chi }^2)[(1-e^2{)}^{1/2}(1-{\chi }^2{)}^{1/2}]-\frac{1}{3(1-e^2)}[(1-e^2{)}^{1/2}(1-{\chi }^2{)}^{1/2}{]}^3\}$
	=	$\left[{\rho }_c{a}_{\ell }^3\right]4\pi \underset{0}{\overset{1}{\int }}\chi d\chi \left\{\right[(1-e^2{)}^{1/2}(1-{\chi }^2{)}^{3/2}]-\frac{1}{3}[(1-e^2{)}^{1/2}(1-{\chi }^2{)}^{3/2}\left]\right\}$
	=	$\left[{\rho }_c{a}_{\ell }^3\right]4\pi \underset{0}{\overset{1}{\int }}\chi d\chi \left\{\frac{2}{3}\right[(1-e^2{)}^{1/2}(1-{\chi }^2{)}^{3/2}\left]\right\}$
	=	$\left[{\rho }_c{a}_{\ell }^3\right]\frac{8\pi }{3}(1-e^2{)}^{1/2}\underset{0}{\overset{1}{\int }}\chi [(1-{\chi }^2{)}^{3/2}]d\chi .$

Again making the variable substitution, $\chi \to \sin \theta \Rightarrow d\chi =\cos \theta d\theta$ …

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}$	=	$\left[{\rho }_c{a}_{\ell }^3\right]\frac{8\pi }{3}(1-e^2{)}^{1/2}\underset{0}{\overset{\pi /2}{\int }}\sin \theta {\cos }^{}\theta d\theta$
	=	$\left[{\rho }_c{a}_{\ell }^3\right]\frac{8\pi }{3}(1-e^2{)}^{1/2}[-\frac{{\cos }^5\theta }{5}{]}_0^{\pi /2}$
	=	$\left[{\rho }_c{a}_{\ell }^3\right]\frac{8\pi }{15}(1-e^2{)}^{1/2}.$

When we set $e=0$, this result matches the expression for the total mass of a spherically symmetric configuration with a parabolic density distribution.

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)

Hydrostatic Balance (Algebraic Condition)

Following our separate discussion of the equilibrium structure of Maclaurin spheroids, and given that our solution of the Poisson equation fixes the expression for ${\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$, the algebraic expression ensuring hydrostatic balance is,

$H(\varpi ,z)$

$=$

$C_B-[{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\varpi ,z)+\mathrm{\Psi }(\varpi ,z)],$

where, $\mathrm{\Psi }$ is the centrifugal potential. NOTE:   Generally when modeling axisymmetric astrophysical systems (see our accompanying discussion of simple rotation profiles) it is assumed that $\mathrm{\Psi }$ does not functionally depend on $z$. Here, our other constraints — for example, demanding that the configuration have a parabolic density distribution — may force us to adopt a $z$-dependent rotation profile.

Here, we know that the adopted parabolic density distribution gives rise to a gravitational potential of the form,

$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\mathbf{𝐱})}{(-\pi G{\rho }_c{a}_{\ell }^2)}$

$=$

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

Hence,

$\mathrm{\Psi }(\varpi ,z)$	$=$	$C_B-{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\varpi ,z)-H(\varpi ,z)$
	$=$	$C_B+\pi G{\rho }_c{a}_{\ell }^2\{\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-[A_{\ell }\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)+A_s\left(\frac{z^2}{a_{\ell }^2}\right)]+\frac{1}{2}[A_{\ell \ell }{a}_{\ell }^2\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)+A_{ss}{a}_{\ell }^2\left(\frac{z^4}{a_{\ell }^4}\right)+2A_{\ell s}{a}_{\ell }^2\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)\left]\right\}-H_{c}h({\xi }_1).$

We presume that the enthalpy profile, as well as the density profile, can be rewritten in terms of T1 Coordinates. In particular, defining, $q\equiv a_{\ell }/a_s$ and, ${\xi }_1$ $\equiv$ $[z^2+(\frac{\varpi }{q}{)}^2{]}^{1/2}=a_s\left[\right(\frac{z}{a_s}{)}^2+(\frac{\varpi }{a_{\ell }}{)}^2{]}^{1/2}$ $\Rightarrow \frac{\rho }{{\rho }_c}$ $=$ $[1-(\frac{{\xi }_1}{a_s}{)}^2].$ Because we expect contours of constant enthalpy $(H)$ to coincide with contours of constant density in equilibrium configurations, we should expect to find that, $\frac{H}{H_c}$ $=$ $h({\xi }_1).$ If the "radial" enthalpy profile resembles our derived spherical enthalpy profile, we should expect to find that, $h({\xi }_1)$ $\sim$ $h_0[1-h_2{\xi }_1^2-h_4{\xi }_1^4]$ $\Rightarrow 1-\frac{h({\xi }_1)}{h_0}$ $\sim$ $h_2{\xi }_1^2+h_4{\xi }_1^4$   $=$ $h_2{a}_s^2\left[\right(\frac{z}{a_s}{)}^2+\left(\frac{\varpi }{a_{\ell }}{)}^2\right]+h_4\left\{a_s^2\right[(\frac{z}{a_s}{)}^2+(\frac{\varpi }{a_{\ell }}{)}^2]{\}}^2$   $=$ $h_2{a}_s^2\left[\right(\frac{z}{a_s}{)}^2+\left(\frac{\varpi }{a_{\ell }}{)}^2\right]+h_4{a}_s^4\left[\right(\frac{z}{a_s}{)}^4+2\left(\frac{z}{a_s}{)}^2\right(\frac{\varpi }{a_{\ell }}{)}^2+\left(\frac{\varpi }{a_{\ell }}{)}^4\right]$

Adopting this last expression for the enthalpy, we have,

$\frac{h({\xi }_1)}{h_0}$	$=$	$1-h_2{a}_s^2\left[\right(\frac{z}{a_s}{)}^2+\left(\frac{\varpi }{a_{\ell }}{)}^2\right]-h_4{a}_s^4\left[\right(\frac{z}{a_s}{)}^4+2\left(\frac{z}{a_s}{)}^2\right(\frac{\varpi }{a_{\ell }}{)}^2+\left(\frac{\varpi }{a_{\ell }}{)}^4\right]$
	$=$	$1-h_2{a}_s^2\left[\right(\frac{\varpi }{a_{\ell }}{)}^2+(\frac{z}{a_{\ell }}{)}^2(1-e^2{)}^{-1}]-h_4{a}_s^4[(\frac{\varpi }{a_{\ell }}{)}^4+(\frac{z}{a_{\ell }}{)}^4(1-e^2{)}^{-2}+2(\frac{{\varpi }^2{z}^2}{a_{\ell }^4})(1-e^2{)}^{-1}].$

Hence,

$\mathrm{\Psi }(\varpi ,z)$	$=$	$C_B+\pi G{\rho }_c{a}_{\ell }^2\{\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-[A_{\ell }\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)+A_s\left(\frac{z^2}{a_{\ell }^2}\right)]+\frac{1}{2}[A_{\ell \ell }{a}_{\ell }^2\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)+A_{ss}{a}_{\ell }^2\left(\frac{z^4}{a_{\ell }^4}\right)+2A_{\ell s}{a}_{\ell }^2\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)\left]\right\}$
		$-H_c{h}_0\{1-h_2{a}_s^2[(\frac{\varpi }{a_{\ell }}{)}^2+(\frac{z}{a_{\ell }}{)}^2(1-e^2{)}^{-1}]-h_4{a}_s^4\left[\right(\frac{\varpi }{a_{\ell }}{)}^4+(\frac{z}{a_{\ell }}{)}^4(1-e^2{)}^{-2}+2\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)(1-e^2{)}^{-1}]\}.$

At the pole of the configuration — that is, when $(\varpi ,z)=(0,a_s)$ — this statement of hydrostatic balance becomes,

$\mathrm{\Psi }(\varpi ,z)$	$=$	$C_B+\pi G{\rho }_c{a}_{\ell }^2\{\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-[A_{\ell }{\xcancel{\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)}}^0+A_s\left(\frac{a_s^2}{a_{\ell }^2}\right)]+\frac{1}{2}[A_{\ell \ell }{a}_{\ell }^2{\xcancel{\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)}}^0+A_{ss}{a}_{\ell }^2\left(\frac{a_s^4}{a_{\ell }^4}\right)+2A_{\ell s}{a}_{\ell }^2\left(\frac{{\xcancel{{\varpi }^2}}^0{a}_s^2}{a_{\ell }^4}\right)\left]\right\}$
		$-H_c{h}_0\{1-h_2{a}_s^2[{\xcancel{(\frac{\varpi }{a_{\ell }}{)}^2}}^0+(\frac{a_s}{a_{\ell }}{)}^2(1-e^2{)}^{-1}]-h_4{a}_s^4[{\xcancel{(\frac{\varpi }{a_{\ell }}{)}^4}}^0+(\frac{a_s}{a_{\ell }}{)}^4(1-e^2{)}^{-2}+2\left(\frac{{\xcancel{{\varpi }^2}}^0{a}_s^2}{a_{\ell }^4}\right)(1-e^2{)}^{-1}]\}$
	$=$	$C_B+\pi G{\rho }_c{a}_{\ell }^2[\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-A_s(1-e^2)+\frac{1}{2}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2]-H_c{h}_0[1-h_2{a}_s^2-h_4{a}_s^4].$

For centrally condensed configurations, it is astrophysically reasonable to assume that $\mathrm{\Psi }(\varpi ,z)$ is of the form such that the centrifugal potential goes to zero when $\varpi \to 0$. Adopting that assumption here means that the Bernoulli constant has the value,

$C_B$

$=$

$H_c{h}_0[1-h_2{a}_s^2-h_4{a}_s^4]-\pi G{\rho }_c{a}_{\ell }^2[\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-A_s(1-e^2)+\frac{1}{2}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2].$

Plugging this expression for $C_B$ back into the general statement of hydrostatic balance gives,

$\mathrm{\Psi }(\varpi ,z)$	$=$	$\pi G{\rho }_c{a}_{\ell }^2\{\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-[A_{\ell }\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)+A_s\left(\frac{z^2}{a_{\ell }^2}\right)]+\frac{1}{2}[A_{\ell \ell }{a}_{\ell }^2\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)+A_{ss}{a}_{\ell }^2\left(\frac{z^4}{a_{\ell }^4}\right)+2A_{\ell s}{a}_{\ell }^2\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)\left]\right\}$
		$-\pi G{\rho }_c{a}_{\ell }^2[\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-A_s(1-e^2)+\frac{1}{2}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2]$
		$+H_c{h}_0[1-h_2{a}_s^2-h_4{a}_s^4]$
		$-H_c{h}_0\{1-h_2{a}_s^2[(\frac{\varpi }{a_{\ell }}{)}^2+(\frac{z}{a_{\ell }}{)}^2(1-e^2{)}^{-1}]-h_4{a}_s^4\left[\right(\frac{\varpi }{a_{\ell }}{)}^4+(\frac{z}{a_{\ell }}{)}^4(1-e^2{)}^{-2}+2\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)(1-e^2{)}^{-1}]\}$
	$=$	$\pi G{\rho }_c{a}_{\ell }^2\left\{\right[A_s(1-e^2)-A_{\ell }\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)-A_s\left(\frac{z^2}{a_{\ell }^2}\right)]+\frac{1}{2}[A_{\ell \ell }{a}_{\ell }^2\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)+A_{ss}{a}_{\ell }^2\left(\frac{z^4}{a_{\ell }^4}\right)+2A_{\ell s}{a}_{\ell }^2\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)-A_{ss}{a}_{\ell }^2(1-e^2{)}^2]\}$
		$+H_c{h}_0\left\{h_2{a}_s^2\right[(\frac{\varpi }{a_{\ell }}{)}^2+(\frac{z}{a_{\ell }}{)}^2(1-e^2{)}^{-1}-1]+h_4{a}_s^4\left[\right(\frac{\varpi }{a_{\ell }}{)}^4+(\frac{z}{a_{\ell }}{)}^4(1-e^2{)}^{-2}+2\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)(1-e^2{)}^{-1}-1]\}$
	$=$	$\pi G{\rho }_c{a}_{\ell }^2\left\{A_s\right[-\frac{A_{\ell }}{A_s}\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)-\left(\frac{z^2}{a_{\ell }^2}\right)+(1-e^2)]+\frac{A_{ss}{a}_{\ell }^2}{2}[\frac{A_{\ell \ell }}{A_{ss}}\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)+\left(\frac{z^4}{a_{\ell }^4}\right)+\frac{2A_{\ell s}}{A_{ss}}\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)-(1-e^2{)}^2]\}$
		$+H_c{h}_0\{h_2{a}_s^2(1-e^2{)}^{-1}\left[\right(\frac{\varpi }{a_{\ell }}{)}^2(1-e^2)+(\frac{z}{a_{\ell }}{)}^2-(1-e^2)]+h_4{a}_s^4(1-e^2{)}^{-2}[(\frac{\varpi }{a_{\ell }}{)}^4(1-e^2{)}^2+(\frac{z}{a_{\ell }}{)}^4+2(\frac{{\varpi }^2{z}^2}{a_{\ell }^4})(1-e^2)-(1-e^2{)}^2\left]\right\}$

Let's set …

$H_c{h}_0=\pi G{\rho }_c{a}_{\ell }^2;$       $h_2=\frac{A_s(1-e^2)}{a_s^2};$       $h_4=-\frac{A_{ss}{a}_{\ell }^2(1-e^2{)}^2}{2a_s^4}.$

This gives,

$\frac{\mathrm{\Psi }(\varpi ,z)}{\pi G{\rho }_c{a}_{\ell }^2}$	$=$	$\left\{A_s\right[-\frac{A_{\ell }}{A_s}\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)-\left(\frac{z^2}{a_{\ell }^2}\right)+(1-e^2)]+\frac{A_{ss}{a}_{\ell }^2}{2}[\frac{A_{\ell \ell }}{A_{ss}}\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)+\left(\frac{z^4}{a_{\ell }^4}\right)+\frac{2A_{\ell s}}{A_{ss}}\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)-(1-e^2{)}^2]\}$
		$+\left\{A_s\right[(\frac{\varpi }{a_{\ell }}{)}^2(1-e^2)+(\frac{z}{a_{\ell }}{)}^2-(1-e^2)]-\frac{A_{ss}{a}_{\ell }^2}{2}[(\frac{\varpi }{a_{\ell }}{)}^4(1-e^2{)}^2+(\frac{z}{a_{\ell }}{)}^4+2(\frac{{\varpi }^2{z}^2}{a_{\ell }^4})(1-e^2)-(1-e^2{)}^2\left]\right\}$
	$=$	$\left\{A_s\right[-\frac{A_{\ell }}{A_s}\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)]+\frac{A_{ss}{a}_{\ell }^2}{2}[\frac{A_{\ell \ell }}{A_{ss}}\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)+\frac{2A_{\ell s}}{A_{ss}}\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)\left]\right\}+\left\{A_s\right[(\frac{\varpi }{a_{\ell }}{)}^2(1-e^2)]-\frac{A_{ss}{a}_{\ell }^2}{2}\left[\right(\frac{\varpi }{a_{\ell }}{)}^4(1-e^2{)}^2+2(\frac{{\varpi }^2{z}^2}{a_{\ell }^4})(1-e^2)]\}$
	$=$	$\{-A_{\ell }(\frac{{\varpi }^2}{a_{\ell }^2})+[\frac{A_{\ell \ell }{a}_{\ell }^2}{2}\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)+A_{\ell s}{a}_{\ell }^2\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)\left]\right\}+\left\{A_s\right[(\frac{\varpi }{a_{\ell }}{)}^2(1-e^2)]-\frac{A_{ss}{a}_{\ell }^2}{2}\left[\right(\frac{\varpi }{a_{\ell }}{)}^4(1-e^2{)}^2+2(\frac{{\varpi }^2{z}^2}{a_{\ell }^4})(1-e^2)]\}$
	$=$	$[A_s(1-e^2)-A_{\ell }]\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)+\frac{A_{\ell \ell }{a}_{\ell }^2}{2}\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)-\frac{A_{ss}{a}_{\ell }^2}{2}\left[\right(\frac{\varpi }{a_{\ell }}{)}^4(1-e^2{)}^2]+A_{\ell s}{a}_{\ell }^2\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)-A_{ss}{a}_{\ell }^2\left[\right(\frac{{\varpi }^2{z}^2}{a_{\ell }^4})(1-e^2)]$
	$=$	$[A_s(1-e^2)-A_{\ell }]\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)+\frac{1}{2}\{A_{\ell \ell }{a}_{\ell }^2-A_{ss}{a}_{\ell }^2(1-e^2{)}^2\left\}\right(\frac{{\varpi }^4}{a_{\ell }^4})+\{A_{\ell s}{a}_{\ell }^2-A_{ss}{a}_{\ell }^2(1-e^2)\left\}\right(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}).$

2^nd Try

Keep in Mind, from Above $\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\mathbf{𝐱})}{(-\pi G{\rho }_c{a}_{\ell }^2)}$ $=$ $\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-\left[A_{\ell }\right(\frac{{\varpi }^2}{a_{\ell }^2})+A_s(\frac{z^2}{a_{\ell }^2}\left)\right]+\frac{1}{2}\left[A_{\ell \ell }{a}_{\ell }^2\right(\frac{{\varpi }^4}{a_{\ell }^4})+A_{ss}{a}_{\ell }^2(\frac{z^4}{a_{\ell }^4})+2A_{\ell s}{a}_{\ell }^2(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\left)\right].$ $\rho$ = ${\rho }_c[1-(\frac{{\varpi }^2}{a_{\ell }^2}+\frac{z^2}{a_s^2}\left)\right]$ $\Rightarrow \frac{\rho }{{\rho }_c}$ = $1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1},$ ${\xi }_1$ $\equiv$ $[z^2+(\frac{\varpi }{q}{)}^2{]}^{1/2}=a_s\left[\right(\frac{z}{a_s}{)}^2+(\frac{\varpi }{a_{\ell }}{)}^2{]}^{1/2}$ $\Rightarrow \frac{\rho }{{\rho }_c}$ $=$ $[1-(\frac{{\xi }_1}{a_s}{)}^2].$

From our presentation of the Eulerian formulation of the Euler equation in cylindrical coordinates, we see that in steady-state axisymmetric flows, the two relevant equilibrium conditions are,

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

Vertical Component

We will focus, first, on the vertical component. Specifically, since both $\rho$ and ${\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$ are known, the vertical gradient of the (unknown) scalar pressure is

$\frac{\partial P}{\partial z}$

$=$

$-\rho \frac{\partial }{\partial z}\left\{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}\right\}$

Multiply thru by $1/(\pi G{\rho }_c^2{a}_{\ell })$:

$\Rightarrow \frac{1}{(\pi G{\rho }_c^2{a}_{\ell })}\cdot \frac{\partial P}{\partial z}$	$=$	$\frac{\rho }{{\rho }_c}\cdot \frac{\partial }{\partial z}\left\{\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(-\pi G{\rho }_c{a}_{\ell })}\right\}$
$\Rightarrow \frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\cdot \frac{\partial P}{\partial \zeta }$	$=$	$\frac{\rho }{{\rho }_c}\cdot \frac{\partial }{\partial \zeta }\left\{\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(-\pi G{\rho }_c{a}_{\ell }^2)}\right\}$
	$=$	$[1-(\frac{{\varpi }^2}{a_{\ell }^2}+\frac{z^2}{a_s^2}\left)\right]\cdot \frac{\partial }{\partial \zeta }\{\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-[A_{\ell }\left(\frac{{\varpi }^2}{a_{\ell }^2}\right)+A_s\left(\frac{z^2}{a_{\ell }^2}\right)]+\frac{1}{2}[A_{\ell \ell }{a}_{\ell }^2\left(\frac{{\varpi }^4}{a_{\ell }^4}\right)+A_{ss}{a}_{\ell }^2\left(\frac{z^4}{a_{\ell }^4}\right)+2A_{\ell s}{a}_{\ell }^2\left(\frac{{\varpi }^2{z}^2}{a_{\ell }^4}\right)\left]\right\}$
	$=$	$\{1-[{\chi }^2+{\zeta }^2(1-e^2{)}^{-1}]\}\cdot \frac{\partial }{\partial \zeta }\{\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-[A_{\ell }{\chi }^2+A_s{\zeta }^2]+\frac{1}{2}[A_{\ell \ell }{a}_{\ell }^2{\chi }^4+A_{ss}{a}_{\ell }^2{\zeta }^4+2A_{\ell s}{a}_{\ell }^2{\chi }^2{\zeta }^2]\}$
	$=$	$\{1-[{\chi }^2+{\zeta }^2(1-e^2{)}^{-1}]\}\cdot \{-2A_s\zeta +[2A_{ss}{a}_{\ell }^2{\zeta }^3+2A_{\ell s}{a}_{\ell }^2{\chi }^2\zeta ]\}$
	$=$	$2\{1-[{\chi }^2+{\zeta }^2(1-e^2{)}^{-1}]\}\cdot [A_{\ell s}{a}_{\ell }^2{\chi }^2\zeta -A_s\zeta +A_{ss}{a}_{\ell }^2{\zeta }^3]$
	$=$	$2\{1-[{\chi }^2+{\zeta }^2(1-e^2{)}^{-1}]\}\cdot [A_{\ell s}{a}_{\ell }^2{\chi }^2\zeta -A_s\zeta +A_{ss}{a}_{\ell }^2{\zeta }^3]$

where (unlike above) we are using the dimensionless lengths, $\chi \equiv \varpi /a_{\ell }$ and $\zeta \equiv z/a_{\ell }$. Continuing to streamline this function, we have,

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

So, let's see what happens if we assume that the pressure has the form,

$\frac{P_{\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}}}{(2\pi G{\rho }_c^2{a}_{\ell }^2)}$	$=$	$p_0+p_2{\zeta }^2+p_4{\zeta }^4+p_6{\zeta }^6$
$\Rightarrow \frac{1}{(2\pi G{\rho }_c^2{a}_{\ell }^2)}\cdot \frac{\partial P_{\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}}}{\partial \zeta }$	$=$	$2p_2\zeta +4p_4{\zeta }^3+6p_6{\zeta }^5,$

in which case,

$\frac{P_{\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}}}{(2\pi G{\rho }_c^2{a}_{\ell }^2)}$

$=$

$p_0+\frac{1}{2}[-A_s+(A_{\ell s}{a}_{\ell }^2+A_s){\chi }^2-A_{\ell s}{a}_{\ell }^2{\chi }^4]{\zeta }^2+\frac{1}{4}\{[A_s(1-e^2{)}^{-1}+A_{ss}{a}_{\ell }^2]-[A_{ss}{a}_{\ell }^2+A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}]{\chi }^2\}{\zeta }^4+\frac{1}{6}[-A_{ss}{a}_{\ell }^2(1-e^2{)}^{-1}]{\zeta }^6.$

REMINDER: From above … $\frac{\rho }{{\rho }_c}$ $=$ $1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}.$ And, in the case of the spherically symmetric equilibrium configuration, the pressure distribution derived by 📚 Prasad (1949) has the form, $\frac{P}{P_c}$ $\sim$ $\left(\frac{\rho }{{\rho }_c}{)}^2\right[1+\left(\frac{\rho }{{\rho }_c}\right)].$ In the context of rotationally flattened configurations, therefore, we might expect the (vertical) pressure distribution to be of the form, $\frac{P}{P_c}$ $\sim$ $[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}\left]\right[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}][2-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]$   $\sim$ $[2-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}\left]\right\{[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]-{\chi }^2[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]-{\zeta }^2(1-e^2{)}^{-1}[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]\}$   $\sim$ $[2-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}\left]\right\{[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]+[-{\chi }^2+{\chi }^4+{\chi }^2{\zeta }^2(1-e^2{)}^{-1}]+[-{\zeta }^2(1-e^2{)}^{-1}+{\chi }^2{\zeta }^2(1-e^2{)}^{-1}+{\zeta }^4(1-e^2{)}^{-2}\left]\right\}$

Radial Component

Start with,

$-\frac{j^2\rho }{{\varpi }^3}+\frac{\partial P}{\partial \varpi }$

$=$

$-\rho \frac{\partial }{\partial \varpi }\left\{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}\right\}$

Multiply thru by $1/(\pi G{\rho }_c^2{a}_{\ell })$:

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

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

Continuing to streamline this function, we have,

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

Determine Specific Angular Momentum Distribution

Now, from our analysis of the vertical component, we determined that,

$\frac{12P_{\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}}}{(2\pi G{\rho }_c^2{a}_{\ell }^2)}$

$=$

$12p_0+6[-A_s+(A_{\ell s}{a}_{\ell }^2+A_s){\chi }^2-A_{\ell s}{a}_{\ell }^2{\chi }^4]{\zeta }^2+3\{[A_s(1-e^2{)}^{-1}+A_{ss}{a}_{\ell }^2]-[A_{ss}{a}_{\ell }^2+A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}]{\chi }^2\}{\zeta }^4+2[-A_{ss}{a}_{\ell }^2(1-e^2{)}^{-1}]{\zeta }^6.$

The radial derivative of this function is,

$\left[\frac{12}{(2\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P_{\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}}}{\partial \chi }$	$=$	$6[2(A_{\ell s}{a}_{\ell }^2+A_s){\zeta }^2\chi -4A_{\ell s}{a}_{\ell }^2{\zeta }^2{\chi }^3]+6\{-[A_{ss}{a}_{\ell }^2+A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}]{\zeta }^4\chi \}$
$\Rightarrow \left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P_{\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}}}{\partial \chi }$	$=$	$[(2A_{\ell s}{a}_{\ell }^2+2A_s){\zeta }^2-A_{ss}{a}_{\ell }^2{\zeta }^4-A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}{\zeta }^4]\chi -4A_{\ell s}{a}_{\ell }^2{\zeta }^2{\chi }^3.$

We hypothesize that,

$-\frac{\rho }{{\rho }_c}\cdot \frac{j^2}{(\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^3}$	$=$	$\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P_{\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}}}{\partial \chi }-\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right][\frac{\partial P}{\partial \chi }{]}_{\mathrm{r}\mathrm{a}\mathrm{d}}$
	$=$	$[(2A_{\ell s}{a}_{\ell }^2+2A_s){\zeta }^2-A_{ss}{a}_{\ell }^2{\zeta }^4-A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}{\zeta }^4]\chi -4A_{\ell s}{a}_{\ell }^2{\zeta }^2{\chi }^3$
		$-2[-A_{\ell }(1-e^2{)}^{-1}+A_{\ell s}{a}_{\ell }^2{\zeta }^2+A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4(1-e^2{)}^{-1}]\chi -2[A_{\ell \ell }{a}_{\ell }^2+A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell \ell }{a}_{\ell }^2{\zeta }^2]{\chi }^3+2A_{\ell \ell }{a}_{\ell }^2(1-e^2{)}^{-1}{\chi }^5$
	$=$	$2\left\{\right[(A_{\ell s}{a}_{\ell }^2+A_s){\zeta }^2+\frac{1}{2}[-A_{ss}{a}_{\ell }^2-A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}]{\zeta }^4]\chi -2A_{\ell s}{a}_{\ell }^2{\zeta }^2{\chi }^3$
		$+[A_{\ell }(1-e^2{)}^{-1}-A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell }{\zeta }^2+A_{\ell s}{a}_{\ell }^2{\zeta }^4(1-e^2{)}^{-1}]\chi +[-A_{\ell \ell }{a}_{\ell }^2-A_{\ell }+A_{\ell s}{a}_{\ell }^2{\zeta }^2+A_{\ell \ell }{a}_{\ell }^2{\zeta }^2]{\chi }^3+A_{\ell \ell }{a}_{\ell }^2(1-e^2{)}^{-1}{\chi }^5\}$
$\Rightarrow -\frac{\rho }{{\rho }_c}\cdot \frac{j^2}{(2\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^3}$	$=$	$[(A_{\ell s}{a}_{\ell }^2+A_s){\zeta }^2+\frac{1}{2}[-A_{ss}{a}_{\ell }^2-A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}]{\zeta }^4+A_{\ell }(1-e^2{)}^{-1}-A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell }{\zeta }^2+A_{\ell s}{a}_{\ell }^2{\zeta }^4(1-e^2{)}^{-1}]\chi$
		$+[-A_{\ell \ell }{a}_{\ell }^2-A_{\ell }+A_{\ell s}{a}_{\ell }^2{\zeta }^2+A_{\ell \ell }{a}_{\ell }^2{\zeta }^2-2A_{\ell s}{a}_{\ell }^2{\zeta }^2]{\chi }^3+A_{\ell \ell }{a}_{\ell }^2(1-e^2{)}^{-1}{\chi }^5$
	$=$	$\{A_{\ell }(1-e^2{)}^{-1}+(A_{\ell s}{a}_{\ell }^2+A_s-A_{\ell s}{a}_{\ell }^2-A_{\ell }){\zeta }^2+\frac{1}{2}[-A_{ss}{a}_{\ell }^2-A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}+2A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}]{\zeta }^4\}\chi$
		$+\{(-A_{\ell \ell }{a}_{\ell }^2-A_{\ell })+(A_{\ell s}{a}_{\ell }^2+A_{\ell \ell }{a}_{\ell }^2-2A_{\ell s}{a}_{\ell }^2){\zeta }^2\}{\chi }^3+A_{\ell \ell }{a}_{\ell }^2(1-e^2{)}^{-1}{\chi }^5$
	$=$	$\{A_{\ell }(1-e^2{)}^{-1}+(A_s-A_{\ell }){\zeta }^2+\frac{1}{2}[-A_{ss}{a}_{\ell }^2+A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}\left]{\zeta }^4\right\}\chi +\{(-A_{\ell \ell }{a}_{\ell }^2-A_{\ell })+(A_{\ell \ell }{a}_{\ell }^2-A_{\ell s}{a}_{\ell }^2){\zeta }^2\}{\chi }^3+A_{\ell \ell }{a}_{\ell }^2(1-e^2{)}^{-1}{\chi }^5$

Now, from our layout of relevant index symbol expressions, let's see if the coefficients of various ζ-dependent terms go to zero.

FIRST:

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

SECOND:

$3A_{ss}$	$=$	$\frac{2}{a_s^2}-2A_{s\ell }$
$\Rightarrow \frac{3}{2}{A}_{ss}{a}_{\ell }^2$	$=$	$\frac{a_{\ell }^2}{a_s^2}-A_{s\ell }{a}_{\ell }^2=(1-e^2{)}^{-1}-A_{s\ell }{a}_{\ell }^2$
$\Rightarrow -A_{ss}{a}_{\ell }^2$	$=$	$\frac{2}{3}{A}_{s\ell }{a}_{\ell }^2-\frac{2}{3}(1-e^2{)}^{-1}$
$\Rightarrow [-A_{ss}{a}_{\ell }^2+A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}]$	$=$	$\frac{2}{3}{A}_{s\ell }{a}_{\ell }^2-\frac{2}{3}(1-e^2{)}^{-1}+A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}$
	$=$	$\frac{1}{3(1-e^2)}[2A_{s\ell }{a}_{\ell }^2(1-e^2)-2+3A_{\ell s}{a}_{\ell }^2]$
	$=$	$\frac{1}{3(1-e^2)}[A_{s\ell }{a}_{\ell }^2(5-2e^2)-2];$

THIRD:

$3A_{\ell \ell }$	$=$	$\frac{2}{a_{\ell }^2}-A_{\ell \ell }-A_{s\ell }$
$\Rightarrow 4A_{\ell \ell }{a}_{\ell }^2$	$=$	$2-A_{s\ell }{a}_{\ell }^2$
$\Rightarrow (A_{\ell \ell }{a}_{\ell }^2-A_{\ell s}{a}_{\ell }^2)$	$=$	$\frac{1}{2}-\frac{5}{4}{A}_{s\ell }{a}_{\ell }^2$
	$=$	$\frac{1}{4}[2-5A_{s\ell }{a}_{\ell }^2].$

3^rd Try

From the above, "2^nd Try" discussion of the radial component, we can write the following "EXACT!" relation,

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

Now, our earlier examination of the radial derivative of $P_{\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}}$ suggests that the left-hand-side of this expression should be of the form,

LHS   $\equiv \left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P}{\partial \chi }$

$\sim$

$c_2{\zeta }^2+c_4{\zeta }^4,$

where it is understood that the coefficients, $c_2$ and $c_4$, are both functions of $\chi$. This should be compared with the "EXACT!" expression for the RHS after multiplying through by the expression for the dimensionless density, that is,

RHS	$=$	$[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]\cdot \{[2A_{\ell \ell }{a}_{\ell }^2{\chi }^3-2A_{\ell }\chi +\frac{j^2}{(\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^3}]+2A_{\ell s}{a}_{\ell }^2\chi {\zeta }^2\}$
	$=$	$(1-{\chi }^2)[2A_{\ell \ell }{a}_{\ell }^2{\chi }^3-2A_{\ell }\chi +\frac{j^2}{(\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^3}]+2A_{\ell s}{a}_{\ell }^2\chi (1-{\chi }^2){\zeta }^2$
		$-[2A_{\ell \ell }{a}_{\ell }^2{\chi }^3-2A_{\ell }\chi +\frac{j^2}{(\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^3}](1-e^2{)}^{-1}{\zeta }^2-2A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}\chi {\zeta }^4.$

Because we are not expecting to see a term that is independent of $\zeta$, this suggests that the term inside the large square brackets must be zero. This leads to an expression for the distribution of specific angular momentum of the form,

EXCELLENT !! $0$ $=$ $[2A_{\ell \ell }{a}_{\ell }^2{\chi }^3-2A_{\ell }\chi +\frac{j^2}{(\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^3}]$ $\Rightarrow \frac{j^2}{(\pi G{\rho }_c{a}_{\ell }^4)}$ $=$ $2A_{\ell }{\chi }^4-2A_{\ell \ell }{a}_{\ell }^2{\chi }^6.$ According to our accompanying discussion of Simple rotation profiles, the corresponding centrifugal potential is given by the expression, $\mathrm{\Psi }$ $=$ $-\int \frac{j^2(\varpi )}{{\varpi }^3}d\varpi =-(\pi G{\rho }_c{a}_{\ell }^2)\int \frac{1}{{\chi }^3}[2A_{\ell }{\chi }^4-2A_{\ell \ell }{a}_{\ell }^2{\chi }^6]d\chi$ $\Rightarrow \frac{\mathrm{\Psi }}{(\pi G{\rho }_c{a}_{\ell }^2)}$ $=$ $-\int [2A_{\ell }\chi -2A_{\ell \ell }{a}_{\ell }^2{\chi }^3]d\chi =\frac{1}{2}[A_{\ell \ell }{a}_{\ell }^2{\chi }^4-2A_{\ell }{\chi }^2].$ (Here, we ignore the integration constant because it will be folded in with the Bernoulli constant.)

It also means that the RHS expression simplifies to the form,

RHS

$=$

$2A_{\ell s}{a}_{\ell }^2\chi (1-{\chi }^2){\zeta }^2-2A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}\chi {\zeta }^4.$

This should be compared to our earlier examination of the radial derivative of $P_{\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}}$, namely,

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

4^th Try

In our accompanying discussion of Ferrers Potential, we have derived the expression for the gravitational potential inside (and on the surface of) a triaxial ellipsoid with a parabolic density distribution. Specifically, for

$\rho (\mathbf{𝐱})$

$=$

${\rho }_c[1-(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\left)\right],$

we find,

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

$=$

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

In this same accompanying discussion, we plugged this expression for the gravitational potential into the Poisson equation and demonstrated that it properly generates the expression for the parabolic density distribution. For the axisymmetric configuration being considered here — with the short axis aligned with $c=a_3=a_s$ — these two relations become,

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

where, $\chi \equiv \varpi /a_{\ell }$ and $\zeta \equiv z/a_{\ell }$. (This matches the expression derived above.)

---

Discuss scalar relationship between the enthalpy $(H)$ and the effective potential.

As has been detailed in an accompanying discussion of solution techniques, a configuration will be in dynamic equilibrium if,

$\mathrm{\nabla }[H+{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}+\mathrm{\Psi }]$	$=$	$0$
$\Rightarrow H+{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}+\mathrm{\Psi }$	$=$	constant $=C_B$

Given that, in our particular case, we have analytic expressions for ${\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\chi ,\zeta )$ and for $\mathrm{\Psi }(\chi ,\zeta )$, we deduce that, to within a constant, the enthalpy distribution is given by the expression,

$\left[\frac{H(\chi ,\zeta )-C_B}{(\pi G{\rho }_c{a}_{\ell }^2)}\right]$	$=$	$-\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(\pi G{\rho }_c{a}_{\ell }^2)}-\frac{\mathrm{\Psi }}{(\pi G{\rho }_c{a}_{\ell }^2)}$
	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-A_{\ell }{\chi }^2-A_s{\zeta }^2+\frac{1}{2}[(A_{ss}{a}_{\ell }^2){\zeta }^4+2(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+(A_{\ell \ell }{a}_{\ell }^2){\chi }^4]-\frac{1}{2}[A_{\ell \ell }{a}_{\ell }^2{\chi }^4-2A_{\ell }{\chi }^2]$
	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-A_s{\zeta }^2+\frac{1}{2}[(A_{ss}{a}_{\ell }^2){\zeta }^4+2(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2]$
	$=$	$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-A_s{\zeta }^2+\frac{{\zeta }^2}{2}[(A_{ss}{a}_{\ell }^2){\zeta }^2+2(A_{\ell s}{a}_{\ell }^2){\chi }^2]$

Now, according to our related discussion of index symbols,

$3A_{ss}$	$=$	$\frac{2}{a_s^2}-2A_{\ell s}$
$\Rightarrow 3A_{ss}{a}_{\ell }^2$	$=$	$2(1-e^2{)}^{-1}-2A_{\ell s}{a}_{\ell }^2$
$\Rightarrow 2(A_{\ell s}{a}_{\ell }^2){\chi }^2$	$=$	$2(1-e^2{)}^{-1}{\chi }^2-3(A_{ss}{a}_{\ell }^2){\chi }^2.$

Hence,

$\left[\frac{H(\chi ,\zeta )-C_B}{(\pi G{\rho }_c{a}_{\ell }^2)}\right]-\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}$	$=$	$-A_s{\zeta }^2+\frac{{\zeta }^2}{2}[(A_{ss}{a}_{\ell }^2){\zeta }^2+2(1-e^2{)}^{-1}{\chi }^2-3(A_{ss}{a}_{\ell }^2){\chi }^2]$
	$=$	$-A_s{\zeta }^2+\frac{{\zeta }^2}{2}[(A_{ss}{a}_{\ell }^2)({\zeta }^2-3{\chi }^2)+2(1-e^2{)}^{-1}{\chi }^2].$

Examining the radial derivative … $\frac{1}{(\pi G{\rho }_c{a}_{\ell }^2)}\frac{\partial H}{\partial \chi }$ $=$ $\frac{\partial }{\partial \chi }\{-A_s{\zeta }^2+\frac{{\zeta }^2}{2}[(A_{ss}{a}_{\ell }^2)({\zeta }^2-3{\chi }^2)+2(1-e^2{)}^{-1}{\chi }^2]\}$   $=$ $[-3(A_{ss}{a}_{\ell }^2)+2(1-e^2{)}^{-1}]{\zeta }^2\chi$   $=$ $2(A_{\ell s}{a}_{\ell }^2){\zeta }^2\chi .$ YES !!! This matches the "radial" pressure-gradient, below. Now, examining the vertical derivative … $\frac{1}{(\pi G{\rho }_c{a}_{\ell }^2)}\frac{\partial H}{\partial \zeta }$ $=$ $\frac{\partial }{\partial \zeta }\{-A_s{\zeta }^2+\frac{{\zeta }^2}{2}[(A_{ss}{a}_{\ell }^2)({\zeta }^2-3{\chi }^2)+2(1-e^2{)}^{-1}{\chi }^2]\}$   $=$ $\frac{\partial }{\partial \zeta }\{-A_s{\zeta }^2+\frac{1}{2}[(A_{ss}{a}_{\ell }^2){\zeta }^4+[2(1-e^2{)}^{-1}-3(A_{ss}{a}_{\ell }^2)]{\chi }^2{\zeta }^2]\}$   $=$ $-2A_s\zeta +[2(A_{ss}{a}_{\ell }^2){\zeta }^3+[2(1-e^2{)}^{-1}-3(A_{ss}{a}_{\ell }^2)]{\chi }^2\zeta ]$   $=$ $-2A_s\zeta +[2(A_{ss}{a}_{\ell }^2){\zeta }^3+2(A_{\ell s}{a}_{\ell }^2){\chi }^2\zeta ]$ HURRAY !!! This matches the "vertical" pressure-gradient, below.

---

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

Plug in …

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

Hence, examination of the radial component leads to the following suggested expression for the pressure:

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

While examination of the vertical component leads to the following suggested expression for the pressure:

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

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,

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

Examine Behavior of Enthalpy

${\xi }_1$	$\equiv$	$[z^2+(\frac{\varpi }{q}{)}^2{]}^{1/2}=a_s\left[\right(\frac{\varpi }{a_{\ell }}{)}^2+(\frac{z}{a_s}{)}^2{]}^{1/2}=a_s[{\chi }^2+{\zeta }^2(1-e^2{)}^{-1}{]}^{1/2}$
$\Rightarrow \frac{\rho }{{\rho }_c}$	$=$	$[1-(\frac{{\xi }_1}{a_s}{)}^2].$

Try to Construct Pressure Distribution

Drawing from the expression for the vertical pressure gradient, namely,

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

try the following pressure expression:

$\frac{P}{(\pi G{\rho }_c^2{a}_{\ell }^2)}$	$=$	$f_0+f_2(\frac{{\xi }_1}{a_s}{)}^2+f_4(\frac{{\xi }_1}{a_s}{)}^4+f_6(\frac{{\xi }_1}{a_s}{)}^6$
	$=$	$f_0+f_2[{\chi }^2+{\zeta }^2(1-e^2{)}^{-1}]+f_4[{\chi }^2+{\zeta }^2(1-e^2{)}^{-1}{]}^2+f_6[{\chi }^2+{\zeta }^2(1-e^2{)}^{-1}{]}^3$
	$=$	$f_0+f_2[{\chi }^2+{\zeta }^2(1-e^2{)}^{-1}]+f_4[{\chi }^4+2{\chi }^2{\zeta }^2(1-e^2{)}^{-1}+{\zeta }^4(1-e^2{)}^{-2}]$
		$+f_6[{\chi }^4+2{\chi }^2{\zeta }^2(1-e^2{)}^{-1}+{\zeta }^4(1-e^2{)}^{-2}][{\chi }^2+{\zeta }^2(1-e^2{)}^{-1}]$
	$=$	$f_0+f_2[{\chi }^2+{\zeta }^2(1-e^2{)}^{-1}]+f_4[{\chi }^4+2{\chi }^2{\zeta }^2(1-e^2{)}^{-1}+{\zeta }^4(1-e^2{)}^{-2}]$
		$+f_6[{\chi }^6+3{\chi }^4{\zeta }^2(1-e^2{)}^{-1}+3{\chi }^2{\zeta }^4(1-e^2{)}^{-2}+{\zeta }^6(1-e^2{)}^{-3}].$

The vertical derivative of this expression is,

$\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P}{\partial \zeta }$	$=$	$\frac{\partial }{\partial \zeta }\left\{f_2\right[{\zeta }^2(1-e^2{)}^{-1}]+f_4[2{\chi }^2{\zeta }^2(1-e^2{)}^{-1}+{\zeta }^4(1-e^2{)}^{-2}]+f_6[3{\chi }^4{\zeta }^2(1-e^2{)}^{-1}+3{\chi }^2{\zeta }^4(1-e^2{)}^{-2}+{\zeta }^6(1-e^2{)}^{-3}\left]\right\}$
	$=$	$\left\{f_2\right[2\zeta (1-e^2{)}^{-1}]+f_4[4{\chi }^2\zeta (1-e^2{)}^{-1}+4{\zeta }^3(1-e^2{)}^{-2}]+f_6[6{\chi }^4\zeta (1-e^2{)}^{-1}+12{\chi }^2{\zeta }^3(1-e^2{)}^{-2}+6{\zeta }^5(1-e^2{)}^{-3}\left]\right\}$
	$=$	$\left\{\right[2f_2(1-e^2{)}^{-1}+4f_4{\chi }^2(1-e^2{)}^{-1}+6f_6{\chi }^4(1-e^2{)}^{-1}]\zeta +[4f_4(1-e^2{)}^{-2}+12f_6{\chi }^2(1-e^2{)}^{-2}]{\zeta }^3+[6f_6(1-e^2{)}^{-3}\left]{\zeta }^5\right\}.$

Matching ${\zeta }^5$ terms gives,

$6f_6(1-e^2{)}^{-3}$	$=$	$-2A_{ss}{a}_{\ell }^2(1-e^2{)}^{-1}$
$\Rightarrow f_6$	$=$	$-\frac{1}{3}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2.$

Matching ${\zeta }^3$ terms gives,

$4f_4(1-e^2{)}^{-2}+12f_6{\chi }^2(1-e^2{)}^{-2}$	$=$	$-2A_{ss}{a}_{\ell }^2{\chi }^2+2A_{ss}{a}_{\ell }^2-2A_{\ell s}{a}_{\ell }^2{\chi }^2(1-e^2{)}^{-1}+2A_s(1-e^2{)}^{-1}$
$\Rightarrow 4f_4(1-e^2{)}^{-2}+12[-\frac{1}{3}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2]{\chi }^2(1-e^2{)}^{-2}$	$=$	$[2A_{ss}{a}_{\ell }^2+2A_s(1-e^2{)}^{-1}]-2A_{ss}{a}_{\ell }^2{\chi }^2-2A_{\ell s}{a}_{\ell }^2{\chi }^2(1-e^2{)}^{-1}$
$\Rightarrow 4f_4(1-e^2{)}^{-2}$	$=$	$[2A_{ss}{a}_{\ell }^2+2A_s(1-e^2{)}^{-1}]+[2A_{ss}{a}_{\ell }^2-2A_{\ell s}{a}_{\ell }^2(1-e^2{)}^{-1}]{\chi }^2.$

Matching ${\zeta }^1$ terms gives,

$2f_2(1-e^2{)}^{-1}+4f_4{\chi }^2(1-e^2{)}^{-1}+6f_6{\chi }^4(1-e^2{)}^{-1}$

$=$

$2A_{\ell s}{a}_{\ell }^2{\chi }^2-2A_s-2A_{\ell s}{a}_{\ell }^2{\chi }^4+2A_s{\chi }^2$

5^th Try

We should leave untouched the form of the expression for the centrifugal potential, but let its coefficient values remain unspecified. The enthalpy function will therefore remain flexible, and, in tern, so will the components of the pressure gradient. We should adjust these new coefficients in such a way that the gradient of the pressure is everywhere perpendicular to the surface of a constant-density contour; this means that the P-constant contours will be identical to the density-constant contours.

Modifiable 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:	$\frac{j^2}{(\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^3}$	$=$	$2j_1\chi -2j_3{\chi }^3.$
Centrifugal Potential:	$\frac{\mathrm{\Psi }}{(\pi G{\rho }_c{a}_{\ell }^2)}$	$=$	$\frac{1}{2}[j_3{\chi }^4-2j_1{\chi }^2].$
Enthalpy:	$\left[\frac{H(\chi ,\zeta )-C_B}{(\pi G{\rho }_c{a}_{\ell }^2)}\right]-\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}$	$=$	$-A_{\ell }{\chi }^2-A_s{\zeta }^2+\frac{1}{2}[(A_{ss}{a}_{\ell }^2){\zeta }^4+2(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+(A_{\ell \ell }{a}_{\ell }^2){\chi }^4]-\frac{1}{2}[j_3{\chi }^4-2j_1{\chi }^2]$
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:	$\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\frac{\partial P}{\partial \chi }$	$=$	$\frac{\rho }{{\rho }_c}\cdot \left\{\right[2j_1-2A_{\ell }+2A_{\ell s}{a}_{\ell }^2{\zeta }^2]\chi +[2A_{\ell \ell }{a}_{\ell }^2-2j_3\left]{\chi }^3\right\}$

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

Desired Slopes of Normal Vectors

A vector that is normal to the surface of a constant-density (oblate-spheroidal) contour has the following components:

$\frac{\partial }{\partial \chi }\left[\frac{\rho (\varpi ,z)}{{\rho }_c}\right]$	$=$	$\frac{\partial }{\partial \chi }[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]=-2\chi ;$
$\frac{\partial }{\partial \zeta }\left[\frac{\rho (\varpi ,z)}{{\rho }_c}\right]$	$=$	$\frac{\partial }{\partial \zeta }[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]=-2\zeta (1-e^2{)}^{-1}.$

Hence, the slope, $m$, of this normal vector is,

$m=\left\{\frac{\partial }{\partial \zeta }\right[\frac{\rho (\varpi ,z)}{{\rho }_c}\left]\right\}\left\{\frac{\partial }{\partial \chi }\right[\frac{\rho (\varpi ,z)}{{\rho }_c}]{\}}^{-1}$

$=$

$\frac{-2\zeta (1-e^2{)}^{-1}}{-2\chi }=\frac{\zeta }{\chi (1-e^2)}.$

Now, if the constant-pressure contours are to lie precisely on top of our constant-density contours, the normals have to have the same slopes. This means that,

$\frac{\partial P}{\partial \zeta }$	$=$	$\frac{\zeta }{\chi (1-e^2)}\left[\frac{\partial P}{\partial \chi }\right]$
$\Rightarrow \chi (1-e^2)\{2A_{\ell s}{a}_{\ell }^2{\chi }^2\zeta -2A_s\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3\}$	$=$	$\zeta \left\{\right[2j_1-2A_{\ell }+2A_{\ell s}{a}_{\ell }^2{\zeta }^2]\chi +[2A_{\ell \ell }{a}_{\ell }^2-2j_3\left]{\chi }^3\right\}$
$\Rightarrow 2A_{\ell s}{a}_{\ell }^2(1-e^2){\chi }^3\zeta -2A_s(1-e^2)\chi \zeta +2A_{ss}{a}_{\ell }^2(1-e^2)\chi {\zeta }^3$	$=$	$(2j_1-2A_{\ell })\chi \zeta +2A_{\ell s}{a}_{\ell }^2\chi {\zeta }^3+(2A_{\ell \ell }{a}_{\ell }^2-2j_3){\chi }^3\zeta$
$\Rightarrow [2A_{\ell s}{a}_{\ell }^2(1-e^2)-(2A_{\ell \ell }{a}_{\ell }^2-2j_3)]{\chi }^3\zeta +[-2A_s(1-e^2)-(2j_1-2A_{\ell })]\chi \zeta$	$=$	$[2A_{\ell s}{a}_{\ell }^2-2A_{ss}{a}_{\ell }^2(1-e^2)]\chi {\zeta }^3$

Note … $3A_{ss}$ $=$ $\frac{2}{a_s^2}-2A_{\ell s}$   $=$ $\frac{2}{a_{\ell }^2(1-e^2)}-2A_{\ell s}$ $\Rightarrow 3(1-e^2)(A_{ss}{a}_{\ell }^2)$ $=$ $2-2(1-e^2)(A_{\ell s}{a}_{\ell }^2)$ $\Rightarrow \mathrm{R}\mathrm{H}\mathrm{S}$ $=$ $\{2A_{\ell s}{a}_{\ell }^2-2A_{ss}{a}_{\ell }^2(1-e^2)\}\chi {\zeta }^3$   $=$ $\{2A_{\ell s}{a}_{\ell }^2-\frac{2}{3}[2-2(1-e^2)(A_{\ell s}{a}_{\ell }^2)\left]\right\}\chi {\zeta }^3$   $=$ $\left\{\right[2+\frac{4}{3}(1-e^2)](A_{\ell s}{a}_{\ell }^2)-\frac{4}{3}\}\chi {\zeta }^3$   $=$ $\frac{2}{3}[(5-2e^2)(A_{\ell s}{a}_{\ell }^2)-2]\chi {\zeta }^3.$

In order for the ${\chi }^3\zeta$ term on the LHS to be zero, we should set …

$0$	$=$	$[2A_{\ell s}{a}_{\ell }^2(1-e^2)-2A_{\ell \ell }{a}_{\ell }^2+2j_3]$
$\Rightarrow j_3$	$=$	$A_{\ell \ell }{a}_{\ell }^2-A_{\ell s}{a}_{\ell }^2(1-e^2);$

and in order for the $\chi \zeta$ term on the LHS to be zero, we should set …

$0$	$=$	$[-2A_s(1-e^2)-(2j_1-2A_{\ell })]$
$\Rightarrow j_1$	$=$	$A_{\ell }-A_s(1-e^2).$

Desired Slopes of Tangent Vectors

Alternatively, if the constant-pressure contours are to lie precisely on top of our constant-density contours, the tangent vectors have to have slopes given by $-1/m$. This means that,

$\frac{\partial P}{\partial \zeta }$	$=$	$-\frac{1}{m}\left[\frac{\partial P}{\partial \chi }\right]=-\frac{\chi (1-e^2)}{\zeta }\left[\frac{\partial P}{\partial \chi }\right]$
$\Rightarrow \zeta \{2A_{\ell s}{a}_{\ell }^2{\chi }^2\zeta -2A_s\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3\}$	$=$	$-\chi (1-e^2)\left\{\right[2j_1-2A_{\ell }+2A_{\ell s}{a}_{\ell }^2{\zeta }^2]\chi +[2A_{\ell \ell }{a}_{\ell }^2-2j_3\left]{\chi }^3\right\}$
$\Rightarrow \left\{\right[A_{\ell s}{a}_{\ell }^2{\chi }^2-A_s]{\zeta }^2+A_{ss}{a}_{\ell }^2{\zeta }^4\}$	$=$	$-(1-e^2)\left\{\right[j_1-A_{\ell }+A_{\ell s}{a}_{\ell }^2{\zeta }^2]{\chi }^2+[A_{\ell \ell }{a}_{\ell }^2-j_3\left]{\chi }^4\right\}$

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,

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

Strategy

STEP 1:   For the problem being tackled here, we start by recognizing that when considering hydrostatic balance in the ${\hat{e}}_z$ direction, we have analytically known expressions for both $\rho (\varpi ,z)$ and $\partial \mathrm{\Phi }/\partial z$. This means, therefore, that we can construct an analytical expression for the vertical component of the pressure gradient, specifically,

$\frac{\partial P}{\partial z}$	=	$-\rho \cdot \frac{\partial \mathrm{\Phi }}{\partial z}.$
$\Rightarrow \frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\cdot \frac{\partial P}{\partial \zeta }$	$=$	$\frac{\rho }{{\rho }_c}\cdot \frac{\partial }{\partial \zeta }\left\{\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(-\pi G{\rho }_c{a}_{\ell }^2)}\right\}$

STEP 2:   Because we want the meridional-plane, constant-pressure contours to align with the meridional-plane, constant density contours, we can determine the radial component of the pressure gradient by forcing the slope of the tangent vector to match the tangent vector of the density contour.

$\frac{\partial P}{\partial \zeta }$	$=$	$-\frac{1}{m}\left[\frac{\partial P}{\partial \chi }\right]=-\frac{\chi (1-e^2)}{\zeta }\left[\frac{\partial P}{\partial \chi }\right]$
$\Rightarrow \frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\cdot \frac{\partial P}{\partial \chi }$	$=$	$-\frac{\zeta }{\chi (1-e^2)}[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\cdot \frac{\partial P}{\partial \zeta }]$
	$=$	$-\frac{\zeta }{\chi (1-e^2)}\{\frac{\rho }{{\rho }_c}\cdot \frac{\partial }{\partial \zeta }[\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(-\pi G{\rho }_c{a}_{\ell }^2)}\left]\right\}.$

STEP 3:   Via the radial component of the hydrostatic balance expression, we can determine analytically the distribution of specific angular momentum.

$\frac{j^2}{{\varpi }^3}$	=	$[\frac{1}{\rho }\frac{\partial P}{\partial \varpi }+\frac{\partial \mathrm{\Phi }}{\partial \varpi }]$
$\Rightarrow \frac{j^2}{(\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^3}$	$=$	$\left(\frac{\rho }{{\rho }_c}{)}^{-1}\right[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}]\frac{\partial P}{\partial \chi }-\frac{\partial }{\partial \chi }\{\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(-\pi G{\rho }_c{a}_{\ell }^2)}\}$

STEP 4:   From knowledge of both components of $\mathrm{\nabla }P$, see if the expression for the pressure can be ascertained.

Implication

Hence,

$\frac{j^2}{(\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^3}$

$=$

$-\frac{\zeta }{\chi (1-e^2)}\cdot \frac{\partial }{\partial \zeta }\left[\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(-\pi G{\rho }_c{a}_{\ell }^2)}\right]-\frac{\partial }{\partial \chi }\left\{\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(-\pi G{\rho }_c{a}_{\ell }^2)}\right\}.$

Now, given that,

$\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\varpi ,z)}{(-\pi G{\rho }_c{a}_{\ell }^2)}$

$=$

$\frac{1}{2}{I}_{\mathrm{B}\mathrm{T}}-A_{\ell }{\chi }^2-A_s{\zeta }^2+\frac{1}{2}[(A_{ss}{a}_{\ell }^2){\zeta }^4+2(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+(A_{\ell \ell }{a}_{\ell }^2){\chi }^4],$

we see that the pair of partial derivative expressions are:

$\frac{\partial }{\partial \zeta }\left[\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(-\pi G{\rho }_c{a}_{\ell }^2)}\right]$	$=$	$2[(A_{ss}{a}_{\ell }^2){\zeta }^3+(A_{\ell s}{a}_{\ell }^2){\chi }^2\zeta -A_s\zeta ];$
$\frac{\partial }{\partial \chi }\left[\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(-\pi G{\rho }_c{a}_{\ell }^2)}\right]$	$=$	$2[(A_{\ell \ell }{a}_{\ell }^2){\chi }^3+(A_{\ell s}{a}_{\ell }^2)\chi {\zeta }^2-A_{\ell }\chi ].$

As a result we find,

$\frac{j^2(1-e^2)}{(\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^2}$	$=$	$-2\zeta [(A_{ss}{a}_{\ell }^2){\zeta }^3+(A_{\ell s}{a}_{\ell }^2){\chi }^2\zeta -A_s\zeta ]-2\chi (1-e^2)[(A_{\ell \ell }{a}_{\ell }^2){\chi }^3+(A_{\ell s}{a}_{\ell }^2)\chi {\zeta }^2-A_{\ell }\chi ]$
$\Rightarrow \frac{j^2(1-e^2)}{(2\pi G{\rho }_c{a}_{\ell }^4)}\cdot \frac{1}{{\chi }^2}$	$=$	$[-(A_{ss}{a}_{\ell }^2){\zeta }^4-(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+A_s{\zeta }^2]+(1-e^2)[-(A_{\ell \ell }{a}_{\ell }^2){\chi }^4-(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^2+A_{\ell }{\chi }^2]$

Next, regarding STEP 4,

$\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\cdot \frac{\partial P}{\partial \zeta }$	$=$	$\frac{\rho }{{\rho }_c}\cdot \frac{\partial }{\partial \zeta }\left\{\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(-\pi G{\rho }_c{a}_{\ell }^2)}\right\}$
	$=$	$2[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}\left]\right[(A_{ss}{a}_{\ell }^2){\zeta }^3+(A_{\ell s}{a}_{\ell }^2){\chi }^2\zeta -A_s\zeta ]$
	$=$	$2[(A_{ss}{a}_{\ell }^2){\zeta }^3+(A_{\ell s}{a}_{\ell }^2){\chi }^2\zeta -A_s\zeta ]-2[(A_{ss}{a}_{\ell }^2){\chi }^2{\zeta }^3+(A_{\ell s}{a}_{\ell }^2){\chi }^4\zeta -A_s{\chi }^2\zeta ]-2[(A_{ss}{a}_{\ell }^2){\zeta }^5(1-e^2{)}^{-1}+(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^3(1-e^2{)}^{-1}-A_s{\zeta }^3(1-e^2{)}^{-1}]$
	$=$	$2\{(A_{ss}{a}_{\ell }^2){\zeta }^3+(A_{\ell s}{a}_{\ell }^2){\chi }^2\zeta -A_s\zeta -(A_{ss}{a}_{\ell }^2){\chi }^2{\zeta }^3-(A_{\ell s}{a}_{\ell }^2){\chi }^4\zeta +A_s{\chi }^2\zeta -(A_{ss}{a}_{\ell }^2){\zeta }^5(1-e^2{)}^{-1}-(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^3(1-e^2{)}^{-1}+A_s{\zeta }^3(1-e^2{)}^{-1}\}$
	$=$	$2\left\{\right[(A_{\ell s}{a}_{\ell }^2){\chi }^2-A_s-(A_{\ell s}{a}_{\ell }^2){\chi }^4+A_s{\chi }^2]\zeta +[(A_{ss}{a}_{\ell }^2)-(A_{ss}{a}_{\ell }^2){\chi }^2-(A_{\ell s}{a}_{\ell }^2){\chi }^2(1-e^2{)}^{-1}+A_s(1-e^2{)}^{-1}]{\zeta }^3+[-(A_{ss}{a}_{\ell }^2)(1-e^2{)}^{-1}]{\zeta }^5\}$
$\Rightarrow \frac{12P}{(2\pi G{\rho }_c^2{a}_{\ell }^2)}$	$\sim$	$6[(A_{\ell s}{a}_{\ell }^2){\chi }^2-A_s-(A_{\ell s}{a}_{\ell }^2){\chi }^4+A_s{\chi }^2]{\zeta }^2+3[(A_{ss}{a}_{\ell }^2)-(A_{ss}{a}_{\ell }^2){\chi }^2-(A_{\ell s}{a}_{\ell }^2){\chi }^2(1-e^2{)}^{-1}+A_s(1-e^2{)}^{-1}]{\zeta }^4+2[-(A_{ss}{a}_{\ell }^2)(1-e^2{)}^{-1}]{\zeta }^6$
	$=$	$6(A_{\ell s}{a}_{\ell }^2){\zeta }^2{\chi }^2-6A_s{\zeta }^2-6(A_{\ell s}{a}_{\ell }^2){\zeta }^2{\chi }^4+6A_s{\zeta }^2{\chi }^2+3(A_{ss}{a}_{\ell }^2){\zeta }^4-3(A_{ss}{a}_{\ell }^2){\zeta }^4{\chi }^2-3(A_{\ell s}{a}_{\ell }^2){\zeta }^4{\chi }^2(1-e^2{)}^{-1}+3A_s{\zeta }^4(1-e^2{)}^{-1}-2(A_{ss}{a}_{\ell }^2){\zeta }^6(1-e^2{)}^{-1}$
	$=$	$+[3(A_{ss}{a}_{\ell }^2){\zeta }^4+3A_s{\zeta }^4(1-e^2{)}^{-1}-6A_s{\zeta }^2-2(A_{ss}{a}_{\ell }^2){\zeta }^6(1-e^2{)}^{-1}]+[6(A_{\ell s}{a}_{\ell }^2){\zeta }^2+6A_s{\zeta }^2-3(A_{ss}{a}_{\ell }^2){\zeta }^4-3(A_{\ell s}{a}_{\ell }^2){\zeta }^4(1-e^2{)}^{-1}]{\chi }^2+[-6(A_{\ell s}{a}_{\ell }^2){\zeta }^2]{\chi }^4;$

and,

$\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\cdot \frac{\partial P}{\partial \chi }$	$=$	$-\frac{\zeta }{\chi (1-e^2)}\{\frac{\rho }{{\rho }_c}\cdot \frac{\partial }{\partial \zeta }\{\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(-\pi G{\rho }_c{a}_{\ell }^2)}\left\}\right\}$
	$=$	$-\frac{\zeta }{\chi (1-e^2)}\left\{2\right[(A_{ss}{a}_{\ell }^2){\zeta }^3+(A_{\ell s}{a}_{\ell }^2){\chi }^2\zeta -A_s\zeta ]-2[(A_{ss}{a}_{\ell }^2){\chi }^2{\zeta }^3+(A_{\ell s}{a}_{\ell }^2){\chi }^4\zeta -A_s{\chi }^2\zeta ]-2[(A_{ss}{a}_{\ell }^2){\zeta }^5(1-e^2{)}^{-1}+(A_{\ell s}{a}_{\ell }^2){\chi }^2{\zeta }^3(1-e^2{)}^{-1}-A_s{\zeta }^3(1-e^2{)}^{-1}]\}$
	$=$	$-\frac{1}{(1-e^2)}\left\{2\right[(A_{ss}{a}_{\ell }^2){\chi }^{-1}{\zeta }^4+(A_{\ell s}{a}_{\ell }^2)\chi {\zeta }^2-A_s{\chi }^{-1}{\zeta }^2]-2[(A_{ss}{a}_{\ell }^2)\chi {\zeta }^4+(A_{\ell s}{a}_{\ell }^2){\chi }^3{\zeta }^2-A_s\chi {\zeta }^2]-2[(A_{ss}{a}_{\ell }^2){\chi }^{-1}{\zeta }^6(1-e^2{)}^{-1}+(A_{\ell s}{a}_{\ell }^2)\chi {\zeta }^4(1-e^2{)}^{-1}-A_s{\chi }^{-1}{\zeta }^4(1-e^2{)}^{-1}]\}$
	$=$	$\frac{2}{(1-e^2)}\{-(A_{ss}{a}_{\ell }^2){\chi }^{-1}{\zeta }^4-(A_{\ell s}{a}_{\ell }^2)\chi {\zeta }^2+A_s{\chi }^{-1}{\zeta }^2+(A_{ss}{a}_{\ell }^2)\chi {\zeta }^4+(A_{\ell s}{a}_{\ell }^2){\chi }^3{\zeta }^2-A_s\chi {\zeta }^2+(A_{ss}{a}_{\ell }^2){\chi }^{-1}{\zeta }^6(1-e^2{)}^{-1}+(A_{\ell s}{a}_{\ell }^2)\chi {\zeta }^4(1-e^2{)}^{-1}-A_s{\chi }^{-1}{\zeta }^4(1-e^2{)}^{-1}\}$
	$=$	$\frac{2}{(1-e^2)}\left\{\right[A_s{\zeta }^2-(A_{ss}{a}_{\ell }^2){\zeta }^4-A_s{\zeta }^4(1-e^2{)}^{-1}+(A_{ss}{a}_{\ell }^2){\zeta }^6(1-e^2{)}^{-1}]{\chi }^{-1}+[-(A_{\ell s}{a}_{\ell }^2){\zeta }^2-A_s{\zeta }^2+(A_{ss}{a}_{\ell }^2){\zeta }^4+(A_{\ell s}{a}_{\ell }^2){\zeta }^4(1-e^2{)}^{-1}]\chi +[(A_{\ell s}{a}_{\ell }^2){\zeta }^2]{\chi }^3\}$
$\Rightarrow \frac{(1-e^2)P}{(2\pi G{\rho }_c^2{a}_{\ell }^2)}$	$\sim$	$[A_s{\zeta }^2-(A_{ss}{a}_{\ell }^2){\zeta }^4-A_s{\zeta }^4(1-e^2{)}^{-1}+(A_{ss}{a}_{\ell }^2){\zeta }^6(1-e^2{)}^{-1}]\ln (\chi )+\frac{1}{2}[-(A_{\ell s}{a}_{\ell }^2){\zeta }^2-A_s{\zeta }^2+(A_{ss}{a}_{\ell }^2){\zeta }^4+(A_{\ell s}{a}_{\ell }^2){\zeta }^4(1-e^2{)}^{-1}]{\chi }^2+\frac{1}{4}[(A_{\ell s}{a}_{\ell }^2){\zeta }^2]{\chi }^4.$

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.

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

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From our examination of spherically symmetric parabolic configurations, we have deduced that the effective enthalpy-density (barotropic) relation is,