ParabolicDensity/Axisymmetric/Structure

Parabolic Density Distribution

Part I:   Gravitational Potential

Part II:   Spherical Structures

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

Part IV:   Triaxial Equilibrium Structures (Exploration)

Axisymmetric (Oblate) Equilibrium Structures

Gravitational Potential

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

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

$=$

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

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

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

Index Symbol Expressions

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

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

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

where the eccentricity,

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

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

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

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

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

Meridional Plane Equi-Potential Contours

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

Configuration Surface

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

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

Expression for Gravitational Potential

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

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

$=$

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

Letting,

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

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

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

Potential at the Pole

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

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

General Determination of Vertical Coordinate (ζ)

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

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

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

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

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

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

$=$

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

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

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

where,

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

The solution to this quadratic equation gives,

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

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

Equipotential Contours that Lie Entirely Within Configuration

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

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

Tentative Summary

Known Relations

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

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

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

where the eccentricity,

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

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,

See Also