ParabolicDensity/Axisymmetric/Structure

Parabolic Density Distribution

Part I:   Gravitational Potential

Part II:   Spherical Structures

Part III:   Axisymmetric Equilibrium Structures  Old: 1st thru 7th tries  Old: 8th thru 10th tries

Part IV:   Triaxial Equilibrium Structures (Exploration)

Axisymmetric (Oblate) Equilibrium Structures

Tentative Summary

Known Relations

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

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,

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

Drawing from our separate "6^th Try" discussion — and as has been highlighted here for example — for the axisymmetric configurations under consideration, the ${\hat{e}}_z$ and ${\hat{e}}_{\varpi }$ components of the Euler equation become, respectively,

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

Multiplying through by length $(a_{\ell })$ and dividing through by the square of the velocity $(\pi G{\rho }_c{a}_{\ell }^2)$, we have,

${\hat{e}}_z$:	$0$	=	$[\frac{1}{\rho }\frac{\partial P}{\partial z}+\frac{\partial \mathrm{\Phi }}{\partial z}]\frac{a_{\ell }}{(\pi G{\rho }_c{a}_{\ell }^2)}$
		=	$\frac{{\rho }_c}{\rho }\cdot \frac{\partial }{\partial \zeta }\left[\frac{P}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]-\frac{\partial }{\partial \zeta }\left[\frac{\mathrm{\Phi }}{(-\pi G{\rho }_c{a}_{\ell }^2)}\right]$
${\hat{e}}_{\varpi }$:	$\frac{j^2}{{\varpi }^3}\cdot \frac{a_{\ell }}{(\pi G{\rho }_c{a}_{\ell }^2)}$	=	$[\frac{1}{\rho }\frac{\partial P}{\partial \varpi }+\frac{\partial \mathrm{\Phi }}{\partial \varpi }]\frac{a_{\ell }}{(\pi G{\rho }_c{a}_{\ell }^2)}$
	$\Rightarrow \frac{1}{{\chi }^3}\cdot \frac{j^2}{(\pi G{\rho }_c{a}_{\ell }^4)}$	=	$\frac{{\rho }_c}{\rho }\cdot \frac{\partial }{\partial \chi }\left[\frac{P}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]-\frac{\partial }{\partial \chi }\left[\frac{\mathrm{\Phi }}{(-\pi G{\rho }_c{a}_{\ell }^2)}\right]$

8^th Try

Foundation

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

Complete the Square

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

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

$\equiv$

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

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

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

we see that we must have,

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

and we must also have,

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

Hence,

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

The pair of roots of this quadratic expression are,

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

Also, then,

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

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

In summary, then, we can write,

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

$=$

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

and,

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

$=$

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

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

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

both of which are real.

9^th Try

Starting Key Relations

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

Play With Vertical Pressure Gradient

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

Integrate over $\zeta$ gives …

$\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\int \left[\frac{\partial P}{\partial \zeta }\right]d\zeta$	$=$	$[(A_{\ell s}{a}_{\ell }^2{\chi }^2-A_s)-(A_{\ell s}{a}_{\ell }^2{\chi }^4-A_s{\chi }^2)]{\zeta }^2+\frac{1}{2}[A_{ss}{a}_{\ell }^2-A_{ss}{a}_{\ell }^2{\chi }^2-(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2{\chi }^2-A_s)]{\zeta }^4+\frac{1}{3}[-(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2]{\zeta }^6+\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}$
	$=$	$[-A_s{\zeta }^2+\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4+\frac{1}{2}(1-e^2{)}^{-1}{A}_s{\zeta }^4-\frac{1}{3}(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2{\zeta }^6]{\chi }^0+[A_{\ell s}{a}_{\ell }^2{\zeta }^2+A_s{\zeta }^2-\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4-\frac{1}{2}(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2{\zeta }^4)]{\chi }^2+[-A_{\ell s}{a}_{\ell }^2{\zeta }^2]{\chi }^4+\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{.}$

Now Play With Radial Pressure Gradient

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

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

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

Integrate over $\chi$ gives …

$\left[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}\right]\int \left[\frac{\partial P}{\partial \chi }\right]d\chi$	$=$	$[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })+(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)-\frac{1}{2}{j}_4^2{\zeta }^2(1-e^2{)}^{-1}+\frac{1}{2}{j}_4^2]{\chi }^2$
		$+[\frac{1}{2}{A}_{\ell \ell }{a}_{\ell }^2+\frac{1}{2}(A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2)-\frac{1}{2}(1-e^2{)}^{-1}{A}_{\ell \ell }{a}_{\ell }^2{\zeta }^2-\frac{1}{4}{j}_4^2-\frac{1}{4}{j}_6^2{\zeta }^2(1-e^2{)}^{-1}+\frac{1}{4}{j}_6^2]{\chi }^4-[\frac{1}{6}{j}_6^2+\frac{1}{3}{A}_{\ell \ell }{a}_{\ell }^2]{\chi }^6$

Compare Pair of Integrations

	Integration over $\zeta$	Integration over $\chi$
${\chi }^0$	$-A_s{\zeta }^2+\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4+\frac{1}{2}(1-e^2{)}^{-1}{A}_s{\zeta }^4-\frac{1}{3}(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2{\zeta }^6$	none
${\chi }^2$	$A_{\ell s}{a}_{\ell }^2{\zeta }^2+A_s{\zeta }^2-\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4-\frac{1}{2}(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2{\zeta }^4)$	$(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })+(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)-\frac{1}{2}{j}_4^2{\zeta }^2(1-e^2{)}^{-1}+\frac{1}{2}{j}_4^2$
${\chi }^4$	$-A_{\ell s}{a}_{\ell }^2{\zeta }^2$	$\frac{1}{2}{A}_{\ell \ell }{a}_{\ell }^2+\frac{1}{2}(A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2)-\frac{1}{2}(1-e^2{)}^{-1}{A}_{\ell \ell }{a}_{\ell }^2{\zeta }^2-\frac{1}{4}{j}_4^2-\frac{1}{4}{j}_6^2{\zeta }^2(1-e^2{)}^{-1}+\frac{1}{4}{j}_6^2$
${\chi }^6$	none	$-\frac{1}{6}{j}_6^2-\frac{1}{3}{A}_{\ell \ell }{a}_{\ell }^2$

Try, $j_6^2=[-2A_{\ell \ell }{a}_{\ell }^2]$ and $\frac{1}{2}{j}_4^2=[A_{\ell }+(A_{\ell s}{a}_{\ell }^2){\zeta }^2]$.

	Integration over $\zeta$	Integration over $\chi$
${\chi }^0$	$-A_s{\zeta }^2+\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4+\frac{1}{2}(1-e^2{)}^{-1}{A}_s{\zeta }^4-\frac{1}{3}(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2{\zeta }^6$	none
${\chi }^2$	$A_{\ell s}{a}_{\ell }^2{\zeta }^2+A_s{\zeta }^2-\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4-\frac{1}{2}(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2{\zeta }^4)$	$(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })+(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)-\frac{1}{2}{j}_4^2{\zeta }^2(1-e^2{)}^{-1}+\frac{1}{2}{j}_4^2$ $=$ $(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })+(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)-[A_{\ell }+(A_{\ell s}{a}_{\ell }^2){\zeta }^2]{\zeta }^2(1-e^2{)}^{-1}+[A_{\ell }+(A_{\ell s}{a}_{\ell }^2){\zeta }^2]$ $=$ $2(A_{\ell s}{a}_{\ell }^2){\zeta }^2[1-{\zeta }^2(1-e^2{)}^{-1}]$
${\chi }^4$	$-A_{\ell s}{a}_{\ell }^2{\zeta }^2$	$\frac{1}{2}{A}_{\ell \ell }{a}_{\ell }^2+\frac{1}{2}(A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2)-\frac{1}{2}(1-e^2{)}^{-1}{A}_{\ell \ell }{a}_{\ell }^2{\zeta }^2-\frac{1}{4}{j}_4^2-\frac{1}{4}[-2A_{\ell \ell }{a}_{\ell }^2]{\zeta }^2(1-e^2{)}^{-1}+\frac{1}{4}[-2A_{\ell \ell }{a}_{\ell }^2]$ $=$ $\frac{1}{4}[2(A_{\ell }-A_{\ell s}{a}_{\ell }^2{\zeta }^2)-2[A_{\ell }+(A_{\ell s}{a}_{\ell }^2){\zeta }^2]]=-A_{\ell s}{a}_{\ell }^2{\zeta }^2$
${\chi }^6$	none	$0$

What expression for $j_4^2$ is required in order to ensure that the ${\chi }^2$ term is the same in both columns?

$\frac{1}{2}{j}_4^2[1-{\zeta }^2(1-e^2{)}^{-1}]$	$=$	$[A_{\ell s}{a}_{\ell }^2{\zeta }^2+A_s{\zeta }^2-\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4-\frac{1}{2}(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2{\zeta }^4)]-[(A_{\ell s}{a}_{\ell }^2{\zeta }^2-A_{\ell })+(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2-A_{\ell s}{a}_{\ell }^2{\zeta }^4)]$
	$=$	$[A_s{\zeta }^2-\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4-\frac{1}{2}(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2{\zeta }^4)]+[(A_{\ell })-(1-e^2{)}^{-1}(A_{\ell }{\zeta }^2)+(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2{\zeta }^4)]$
	$=$	$[A_s{\zeta }^2-\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4+\frac{1}{2}(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2){\zeta }^4]+A_{\ell }[1-(1-e^2{)}^{-1}{\zeta }^2]$
$\Rightarrow \frac{1}{2}{j}_4^2[1-{\zeta }^2(1-e^2{)}^{-1}]-A_{\ell }[1-{\zeta }^2(1-e^2{)}^{-1}]$	$=$	$\frac{1}{2}(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2){\zeta }^4+[A_s]{\zeta }^2-\frac{1}{2}[A_{ss}{a}_{\ell }^2]{\zeta }^4$

Now, considering the following three relations …

$\frac{3}{2}(A_{ss}{a}_{\ell }^2)$	$=$	$(1-e^2{)}^{-1}-(A_{\ell s}{a}_{\ell }^2);$
$A_s$	$=$	$A_{\ell }+e^2(A_{\ell s}{a}_{\ell }^2);$
$e^2(A_{\ell s}{a}_{\ell }^2)$	$=$	$2-3A_{\ell };$

we can write,

$\frac{1}{2}{j}_4^2[1-{\zeta }^2(1-e^2{)}^{-1}]-A_{\ell }[1-{\zeta }^2(1-e^2{)}^{-1}]$	$=$	$\frac{1}{2}(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2){\zeta }^4+[A_{\ell }+e^2(A_{\ell s}{a}_{\ell }^2)]{\zeta }^2-\frac{1}{3}[(1-e^2{)}^{-1}-(A_{\ell s}{a}_{\ell }^2)]{\zeta }^4$
$\Rightarrow 3j_4^2[1-{\zeta }^2(1-e^2{)}^{-1}]-3A_{\ell }[2-2{\zeta }^2(1-e^2{)}^{-1}]$	$=$	$3(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2){\zeta }^4+6[A_{\ell }+e^2(A_{\ell s}{a}_{\ell }^2)]{\zeta }^2-2[(1-e^2{)}^{-1}-(A_{\ell s}{a}_{\ell }^2)]{\zeta }^4$
	$=$	$(A_{\ell s}{a}_{\ell }^2)\{2{\zeta }^4+3{\zeta }^4(1-e^2{)}^{-1}+6e^2{\zeta }^2\}-2{\zeta }^4(1-e^2{)}^{-1}+6A_{\ell }{\zeta }^2$
	$=$	$-2{\zeta }^4(1-e^2{)}^{-1}+6A_{\ell }{\zeta }^2+[2-3A_{\ell }\left]\right\{2{\zeta }^4+3{\zeta }^4(1-e^2{)}^{-1}+6e^2{\zeta }^2\}\frac{1}{e^2}$
	$=$	$-2{\zeta }^4(1-e^2{)}^{-1}-3A_{\ell }\{2{\zeta }^4+3{\zeta }^4(1-e^2{)}^{-1}+4e^2{\zeta }^2\}\frac{1}{e^2}+\{4{\zeta }^4+6{\zeta }^4(1-e^2{)}^{-1}+12e^2{\zeta }^2\}\frac{1}{e^2}$

$\Rightarrow 3j_4^2[1-{\zeta }^2(1-e^2{)}^{-1}]$

$=$

$-2{\zeta }^4(1-e^2{)}^{-1}+\frac{3A_{\ell }(1-e^2{)}^{-1}}{e^2}\{[2e^2(1-e^2)-2e^2{\zeta }^2]-[2{\zeta }^4(1-e^2)+3{\zeta }^4+4e^2(1-e^2){\zeta }^2]\}+\{4{\zeta }^4+6{\zeta }^4(1-e^2{)}^{-1}+12e^2{\zeta }^2\}\frac{1}{e^2}$

10^th Try

Repeating Key Relations

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

From the above (9^th Try) examination of the vertical pressure gradient, we determined that a reasonably good approximation for the normalized pressure throughout the configuration is given by the expression,

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

$=$

$[-A_s{\zeta }^2+\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4+\frac{1}{2}(1-e^2{)}^{-1}{A}_s{\zeta }^4-\frac{1}{3}(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2{\zeta }^6]{\chi }^0+[A_{\ell s}{a}_{\ell }^2{\zeta }^2+A_s{\zeta }^2-\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4-\frac{1}{2}(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2{\zeta }^4)]{\chi }^2+[-A_{\ell s}{a}_{\ell }^2{\zeta }^2]{\chi }^4+\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{.}$

If we set $\chi =0$ — that is, if we look along the vertical axis — this approximation should be particularly good, resulting in the expression,

$P_z\equiv \left\{\right[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}]\int [\frac{\partial P}{\partial \zeta }]d\zeta {\}}_{\chi =0}$

$=$

$P_c^{*}-A_s{\zeta }^2+\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4+\frac{1}{2}(1-e^2{)}^{-1}{A}_s{\zeta }^4-\frac{1}{3}(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2{\zeta }^6.$

Note that in the limit that $z\to a_s$ — that is, at the pole along the vertical (symmetry) axis where the $P_z$ should drop to zero — we should set $\zeta \to (1-e^2{)}^{1/2}$. This allows us to determine the central pressure. $P_c^{*}$ $=$ $A_s(1-e^2)-\frac{1}{2}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2-\frac{1}{2}(1-e^2{)}^{-1}{A}_s(1-e^2{)}^2+\frac{1}{3}(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^3$   $=$ $A_s(1-e^2)-\frac{1}{2}{A}_s(1-e^2)+\frac{1}{3}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2-\frac{1}{2}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2$   $=$ $\frac{1}{2}{A}_s(1-e^2)-\frac{1}{6}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2.$

This means that, along the vertical axis, the pressure gradient is,

$P_z\equiv \left\{\right[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}]\int [\frac{\partial P}{\partial \zeta }]d\zeta {\}}_{\chi =0}$

$=$

$P_c^{*}-A_s{\zeta }^2+\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4+\frac{1}{2}(1-e^2{)}^{-1}{A}_s{\zeta }^4-\frac{1}{3}(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2{\zeta }^6.$

$\frac{\partial P_z}{\partial \zeta }$

$=$

$-2A_s\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3+2(1-e^2{)}^{-1}{A}_s{\zeta }^3-2(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2{\zeta }^5.$

This should match the more general "vertical pressure gradient" expression when we set, $\chi =0$, that is,

$\left\{\right[\frac{1}{(\pi G{\rho }_c^2{a}_{\ell }^2)}]\frac{\partial P}{\partial \zeta }{\}}_{\chi =0}$	$=$	$[1-{\xcancel{{\chi }^2}}^0-{\zeta }^2(1-e^2{)}^{-1}]\cdot [2A_{\ell s}{a}_{\ell }^2\zeta {\xcancel{{\chi }^2}}^0-2A_s\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3]$
	$=$	$[-2A_s\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3]+{\zeta }^2(1-e^2{)}^{-1}[2A_s\zeta -2A_{ss}{a}_{\ell }^2{\zeta }^3]$

Yes! The expressions match!

Shift to ξ₁ Coordinate

In an accompanying chapter, we defined the coordinate,

$(\frac{{\xi }_1}{a_s}{)}^2$

$\equiv$

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

Given that we want the pressure to be constant on ${\xi }_1$ surfaces, it seems plausible that ${\zeta }^2$ should be replaced by $(1-e^2)({\xi }_1/a_s{)}^2=[(1-e^2){\chi }^2+{\zeta }^2]$ in the expression for $P_z$. That is, we might expect the expression for the pressure at any point in the meridional plane to be,

$P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{1}}$	$=$	$P_c^{*}-A_s[(1-e^2){\chi }^2+{\zeta }^2{]}^1+\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s][(1-e^2){\chi }^2+{\zeta }^2{]}^2-\frac{1}{3}(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2[(1-e^2){\chi }^2+{\zeta }^2{]}^3$
	$=$	$P_c^{*}-A_s[(1-e^2){\chi }^2+{\zeta }^2{]}^1+\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s][(1-e^2{)}^2{\chi }^4+2(1-e^2){\chi }^2{\zeta }^2+{\zeta }^4]-\frac{1}{3}(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2[(1-e^2){\chi }^2+{\zeta }^2][(1-e^2{)}^2{\chi }^4+2(1-e^2){\chi }^2{\zeta }^2+{\zeta }^4]$
	$=$	$P_c^{*}-A_s[(1-e^2){\chi }^2+{\zeta }^2]+\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s\left]\right[(1-e^2{)}^2{\chi }^4+2(1-e^2){\chi }^2{\zeta }^2+{\zeta }^4]$
		$-\frac{1}{3}{A}_{ss}{a}_{\ell }^2[(1-e^2{)}^2{\chi }^6+2(1-e^2){\chi }^4{\zeta }^2+{\chi }^2{\zeta }^4]-\frac{1}{3}{A}_{ss}{a}_{\ell }^2[(1-e^2){\chi }^4{\zeta }^2+2{\chi }^2{\zeta }^4+(1-e^2{)}^{-1}{\zeta }^6]$
	$=$	${\chi }^0\{P_c^{*}-A_s{\zeta }^2+\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s]{\zeta }^4-\frac{1}{3}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^{-1}{\zeta }^6\}+{\chi }^2\{-A_s(1-e^2)+\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s]2(1-e^2){\zeta }^2-\frac{1}{3}{A}_{ss}{a}_{\ell }^2{\zeta }^4-\frac{2}{3}{A}_{ss}{a}_{\ell }^2{\zeta }^4\}$
		$+{\chi }^4\left\{\frac{1}{2}\right[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s](1-e^2{)}^2-\frac{2}{3}{A}_{ss}{a}_{\ell }^2(1-e^2){\zeta }^2-\frac{1}{3}{A}_{ss}{a}_{\ell }^2(1-e^2){\zeta }^2\}+{\chi }^6\{-\frac{1}{3}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2\}$
	$=$	${\chi }^0\{P_c^{*}-A_s{\zeta }^2+\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s]{\zeta }^4-\frac{1}{3}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^{-1}{\zeta }^6\}+{\chi }^2\{-A_s(1-e^2)+\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s]2(1-e^2){\zeta }^2-A_{ss}{a}_{\ell }^2{\zeta }^4\}$
		$+{\chi }^4\left\{\frac{1}{2}\right[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s](1-e^2{)}^2-A_{ss}{a}_{\ell }^2(1-e^2){\zeta }^2\}+{\chi }^6\{-\frac{1}{3}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2\}$

	Integration over $\zeta$	Pressure Guess
${\chi }^0$	$-A_s{\zeta }^2+\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4+\frac{1}{2}(1-e^2{)}^{-1}{A}_s{\zeta }^4-\frac{1}{3}(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2{\zeta }^6$	$P_c^{*}-A_s{\zeta }^2+\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s]{\zeta }^4-\frac{1}{3}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^{-1}{\zeta }^6$
${\chi }^2$	$A_{\ell s}{a}_{\ell }^2{\zeta }^2+A_s{\zeta }^2-\frac{1}{2}{A}_{ss}{a}_{\ell }^2{\zeta }^4-\frac{1}{2}(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2{\zeta }^4)$	$-A_s(1-e^2)+\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s]2(1-e^2){\zeta }^2-A_{ss}{a}_{\ell }^2{\zeta }^4$
${\chi }^4$	$-A_{\ell s}{a}_{\ell }^2{\zeta }^2$	$\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s](1-e^2{)}^2-A_{ss}{a}_{\ell }^2(1-e^2){\zeta }^2$
${\chi }^6$	none	$-\frac{1}{3}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2$

Compare Vertical Pressure Gradient Expressions

From our above (9^th try) derivation we know that the vertical pressure gradient is given by the expression,

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

By comparison, the vertical derivative of our "test01" pressure expression gives,

$P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{1}}$	$=$	${\chi }^0\{P_c^{*}-A_s{\zeta }^2+\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s]{\zeta }^4-\frac{1}{3}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^{-1}{\zeta }^6\}+{\chi }^2\{-A_s(1-e^2)+\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s]2(1-e^2){\zeta }^2-A_{ss}{a}_{\ell }^2{\zeta }^4\}$
		$+{\chi }^4\left\{\frac{1}{2}\right[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s](1-e^2{)}^2-A_{ss}{a}_{\ell }^2(1-e^2){\zeta }^2\}+{\chi }^6\{-\frac{1}{3}{A}_{ss}{a}_{\ell }^2(1-e^2{)}^2\}$
$\Rightarrow \frac{\partial P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{1}}}{\partial \zeta }$	$=$	${\chi }^0\{-2A_s\zeta +2[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s]{\zeta }^3-2A_{ss}{a}_{\ell }^2(1-e^2{)}^{-1}{\zeta }^5\}+{\chi }^2\left\{2\right[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s](1-e^2)\zeta -4A_{ss}{a}_{\ell }^2{\zeta }^3\}+{\chi }^4\{-2A_{ss}{a}_{\ell }^2(1-e^2)\zeta \}$
	$=$	${\zeta }^1\{-2A_s+2[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s](1-e^2){\chi }^2-2A_{ss}{a}_{\ell }^2(1-e^2){\chi }^4\}+{\zeta }^3\{2[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s]-4A_{ss}{a}_{\ell }^2{\chi }^2\}+{\zeta }^5\{-2A_{ss}{a}_{\ell }^2(1-e^2{)}^{-1}\}$
	$=$	${\zeta }^1\{2A_s({\chi }^2-1)+2A_{ss}{a}_{\ell }^2(1-e^2){\chi }^2(1-{\chi }^2)\}+{\zeta }^3\{2A_{ss}{a}_{\ell }^2(1-2{\chi }^2)+2(1-e^2{)}^{-1}{A}_s\}+{\zeta }^5\{-2A_{ss}{a}_{\ell }^2(1-e^2{)}^{-1}\}$

Instead, try …

$\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{2}}}{P_c}$	$=$	$p_2(\frac{\rho }{{\rho }_c}{)}^2+p_3(\frac{\rho }{{\rho }_c}{)}^3$
$\Rightarrow \frac{\partial }{\partial \zeta }\left[\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{2}}}{P_c}\right]$	$=$	$2p_2\left(\frac{\rho }{{\rho }_c}\right)\frac{\partial }{\partial \zeta }\left[\frac{\rho }{{\rho }_c}\right]+3p_3\left(\frac{\rho }{{\rho }_c}{)}^2\frac{\partial }{\partial \zeta }\right[\frac{\rho }{{\rho }_c}]$
	$=$	$\left(\frac{\rho }{{\rho }_c}\right)\{2p_2+3p_3(\frac{\rho }{{\rho }_c}\left)\right\}\frac{\partial }{\partial \zeta }\left[\frac{\rho }{{\rho }_c}\right]$
	$=$	$\left(\frac{\rho }{{\rho }_c}\right)\{2p_2+3p_3[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]\left\}\frac{\partial }{\partial \zeta }\right[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]$
	$=$	$\left(\frac{\rho }{{\rho }_c}\right)\{(2p_2+3p_3)-3p_3{\chi }^2-3p_3{\zeta }^2(1-e^2{)}^{-1}\left\}\right[-2\zeta (1-e^2{)}^{-1}]$
	$=$	$\left(\frac{\rho }{{\rho }_c}\right)(1-e^2{)}^{-2}\{6p_3{\chi }^2\zeta (1-e^2)-2(2p_2+3p_3)(1-e^2)\zeta +6p_3{\zeta }^3\}$

Compare the term inside the curly braces with the term, from the beginning of this subsection, inside the square brackets, namely,

$2A_{\ell s}{a}_{\ell }^2{\chi }^2\zeta -2A_s\zeta +2A_{ss}{a}_{\ell }^2{\zeta }^3$	$=$	$\frac{2}{e^4}[(3-e^2)-\mathrm{{\rm Y}}]{\chi }^2\zeta -\left[\frac{4}{e^2}\right(1-\frac{1}{3}\mathrm{{\rm Y}}\left)\right]\zeta +\frac{4}{3e^4}[\frac{4e^2-3}{(1-e^2)}+\mathrm{{\rm Y}}]{\zeta }^3$
	$=$	$\frac{1}{3e^4(1-e^2)}\left\{6\right[(3-e^2)-\mathrm{{\rm Y}}](1-e^2){\chi }^2\zeta -[12e^2(1-\frac{1}{3}\mathrm{{\rm Y}})](1-e^2)\zeta +4[(4e^2-3)+\mathrm{{\rm Y}}\left]{\zeta }^3\right\}.$

Pretty Close!!

Alternatively:   according to the third term, we need to set, $6p_3$ $=$ $4[(4e^2-3)+\mathrm{{\rm Y}}]$ $\Rightarrow \mathrm{{\rm Y}}$ $=$ $\frac{3}{2}{p}_3+(3-4e^2)$ in which case, the first coefficient must be given by the expression, $[(3-e^2)-\mathrm{{\rm Y}}]$ $=$ $(3-e^2)-\frac{3}{2}{p}_3+(4e^2-3)]=[3e^2-\frac{3}{2}{p}_3].$ And, from the second coefficient, we find, $2(2p_2+3p_3)$ $=$ $\left[12e^2\right(1-\frac{1}{3}\mathrm{{\rm Y}}\left)\right]$ $\Rightarrow 2p_2$ $=$ $2e^2(3-\mathrm{{\rm Y}})-3p_3$   $=$ $-3p_3+6e^2-2e^2[\frac{3}{2}{p}_3+(3-4e^2)]$   $=$ $-3p_3+6e^2-[3e^2{p}_3+6e^2-8e^4]$   $=$ $8e^4-3p_3(1+e^2);$ or, $p_2$ $=$ $4e^4-(1+e^2)[(4e^2-3)+\mathrm{{\rm Y}}]$   $=$ $4e^4-(1+e^2)(4e^2-3)-(1+e^2)\mathrm{{\rm Y}}$   $=$ $4e^4-[4e^2-3+4e^4-3e^2]-(1+e^2)\mathrm{{\rm Y}}$   $=$ $3-e^2-(1+e^2)\mathrm{{\rm Y}}$ SUMMARY: $\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{2}}}{P_c}$ $=$ $p_2(\frac{\rho }{{\rho }_c}{)}^2+p_3(\frac{\rho }{{\rho }_c}{)}^3,$ $p_2$ $=$ $3-e^2-(1+e^2)\mathrm{{\rm Y}}=e^4(A_{\ell s}{a}_{\ell }^2)-e^2\mathrm{{\rm Y}},$ $p_3$ $=$ $\frac{2}{3}[(4e^2-3)+\mathrm{{\rm Y}}]=e^4(A_{ss}{a}_{\ell }^2)+\frac{2}{3}{e}^2\mathrm{{\rm Y}}.$

Note:   according to the first term, we need to set, $p_3$ $=$ $[(3-e^2)-\mathrm{{\rm Y}}]$ $\Rightarrow \mathrm{{\rm Y}}$ $=$ $[(3-e^2)-p_3],$ in which case, the third coefficient must be given by the expression, $4[(4e^2-3)+\mathrm{{\rm Y}}]$ $=$ $4[(4e^2-3)+(3-e^2)-p_3]=4[3e^2-p_3].$ And, from the second coefficient, we find, $2(2p_2+3p_3)$ $=$ $\left[12e^2\right(1-\frac{1}{3}\mathrm{{\rm Y}}\left)\right]$ $\Rightarrow 2p_2$ $=$ $2e^2(3-\mathrm{{\rm Y}})-3p_3$   $=$ $2e^2[3-[(3-e^2)-p_3]]-3p_3$   $=$ $2e^2[e^2+p_3]-3p_3$   $=$ $2e^4+(2e^2-3)p_3;$ or, $2p_2$ $=$ $2e^4+(2e^2-3)[(3-e^2)-\mathrm{{\rm Y}}]$   $=$ $2e^4+(2e^2-3)(3-e^2)-(2e^2-3)\mathrm{{\rm Y}}$   $=$ $2e^4+(6e^2-2e^4-9+3e^2)-(2e^2-3)\mathrm{{\rm Y}}$   $=$ $9(e^2-1)-(2e^2-3)\mathrm{{\rm Y}}$

Better yet, try …

$\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{3}}}{P_c}$	$=$	$p_2\left(\frac{\rho }{{\rho }_c}{)}^2\right[1-\beta (1-\frac{\rho }{{\rho }_c})]=p_2(\frac{\rho }{{\rho }_c}{)}^2[(1-\beta )+\beta (\frac{\rho }{{\rho }_c}\left)\right]$
$\Rightarrow \frac{\partial }{\partial \zeta }\left[\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{3}}}{P_c}\right]$	$=$	$\cdots$

where, in the case of a spherically symmetric parabolic-density configuration, $\beta =1/2$. Well … this wasn't a bad idea, but as it turns out, this "test03" expression is no different from the "test02" guess. Specifically, the "test03" expression can be rewritten as,

$\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{3}}}{P_c}$

$=$

$p_2(1-\beta )(\frac{\rho }{{\rho }_c}{)}^2+p_2\beta (\frac{\rho }{{\rho }_c}{)}^3,$

which has the same form as the "test02" expression.

Test04

From above, we understand that, analytically,

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

Also from above, we have shown that if,

$\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{2}}}{P_c}$

$=$

$p_2(\frac{\rho }{{\rho }_c}{)}^2+p_3(\frac{\rho }{{\rho }_c}{)}^3$

SUMMARY from test02: $p_2$ $=$ $3-e^2-(1+e^2)\mathrm{{\rm Y}}=e^4(A_{\ell s}{a}_{\ell }^2)-e^2\mathrm{{\rm Y}},$ $p_3$ $=$ $\frac{2}{3}[(4e^2-3)+\mathrm{{\rm Y}}]=e^4(A_{ss}{a}_{\ell }^2)+\frac{2}{3}{e}^2\mathrm{{\rm Y}}.$

$\Rightarrow \frac{\partial }{\partial \zeta }\left[\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{2}}}{P_c}\right]$	$=$	$\left(\frac{\rho }{{\rho }_c}\right)(1-e^2{)}^{-2}\{6p_3{\chi }^2\zeta (1-e^2)-2(2p_2+3p_3)(1-e^2)\zeta +6p_3{\zeta }^3\}$
	$=$	$\left(\frac{\rho }{{\rho }_c}\right)(1-e^2{)}^{-2}\{6[e^4(A_{ss}{a}_{\ell }^2)+\frac{2}{3}{e}^2\mathrm{{\rm Y}}]{\chi }^2\zeta (1-e^2)-2[2e^4(A_{\ell s}{a}_{\ell }^2)+3e^4(A_{ss}{a}_{\ell }^2)](1-e^2)\zeta +6[e^4(A_{ss}{a}_{\ell }^2)+\frac{2}{3}{e}^2\mathrm{{\rm Y}}]{\zeta }^3\}$

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Here (test04), we add a term that is linear in the normalized density, which means,

$\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{4}}}{P_c}$	$=$	$\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{2}}}{P_c}+p_1\left(\frac{\rho }{{\rho }_c}\right)$
$\Rightarrow \frac{\partial }{\partial \zeta }\left[\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{4}}}{P_c}\right]$	$=$	$\frac{\partial }{\partial \zeta }\left[\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{2}}}{P_c}\right]+\frac{\partial }{\partial \zeta }\left[p_1\right(\frac{\rho }{{\rho }_c}\left)\right]=\frac{\partial }{\partial \zeta }\left[\frac{P_{\mathrm{t}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{0}\mathrm{2}}}{P_c}\right]+p_1\frac{\partial }{\partial \zeta }[1-{\chi }^2-{\zeta }^2(1-e^2{)}^{-1}]$

See Also