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

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

$I_{\mathrm{B}\mathrm{T}}$	$=$	$2A_{\ell }+A_s(1-e^2)=2(1-e^2{)}^{1/2}[\frac{{\sin }^{-1}e}{e}];$	[1.7160030]
$A_{\ell }$	$=$	$\frac{1}{e^2}[\frac{{\sin }^{-1}e}{e}-(1-e^2{)}^{1/2}](1-e^2{)}^{1/2};$	[0.6055597]
$A_s$	$=$	$\frac{2}{e^2}[(1-e^2{)}^{-1/2}-\frac{{\sin }^{-1}e}{e}](1-e^2{)}^{1/2};$	[0.7888807]
$a_{\ell }^2{A}_{\ell \ell }$	$=$	$\frac{1}{4e^4}\{-(3+2e^2)(1-e^2)+3(1-e^2{)}^{1/2}\left[\frac{{\sin }^{-1}e}{e}\right]\}=[\frac{1}{2}-\frac{(A_s-A_{\ell })}{4e^2}];$	[0.3726937]
$a_{\ell }^2{A}_{ss}$	$=$	$\frac{2}{3}\{\frac{(4e^2-3)}{e^4(1-e^2)}+\frac{3(1-e^2{)}^{1/2}}{e^4}[\frac{{\sin }^{-1}e}{e}\left]\right\}=\frac{2}{3}[(1-e^2{)}^{-1}-\frac{(A_s-A_{\ell })}{e^2}];$	[0.7021833]
$a_{\ell }^2{A}_{\ell s}$	$=$	$\frac{1}{e^4}\{(3-e^2)-3(1-e^2{)}^{1/2}\left[\frac{{\sin }^{-1}e}{e}\right]\}=\frac{(A_s-A_{\ell })}{e^2},$	[0.5092250]

where the eccentricity,

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

NOTE:   The posted numerical evaluations (inside square brackets) assume that the configuration's eccentricity is $e=0.6\Rightarrow a_s/a_{\ell }=0.8$.

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

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

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

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

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 …

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

If I am interpreting this correctly, $P_{\mathrm{d}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}}^{*}$ should tell how the normalized pressure varies with $\zeta$, for a fixed choice of $0\le \chi \le 1$. Again, for a fixed choice of $\chi$, we want to specify the value of the "const." — hereafter, $C_{\chi }$ — such that $P_{\mathrm{d}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}}^{*}=0$ at the surface of the configuration; but at the surface, it must also be true that,

at the surface   …

${\zeta }^2$

$=$

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

Hence (numerical evaluations assume χ = 0.6),

$-C_{\chi }$

$=$

$\stackrel{\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f}\mathrm{1}=-0.38756}{\overbrace{[(A_{\ell s}{a}_{\ell }^2{\chi }^2-A_s)-(A_{\ell s}{a}_{\ell }^2{\chi }^4-A_s{\chi }^2)]}}[(1-e^2)(1-{\chi }^2)]+\underset{\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f}\mathrm{2}=0.69779}{\underbrace{\frac{1}{2}[A_{ss}{a}_{\ell }^2-A_{ss}{a}_{\ell }^2{\chi }^2-(1-e^2{)}^{-1}(A_{\ell s}{a}_{\ell }^2{\chi }^2-A_s)]}}[(1-e^2)(1-{\chi }^2){]}^2+\stackrel{\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f}\mathrm{3}=-0.36572}{\overbrace{\frac{1}{3}[-(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2]}}[(1-e^2)(1-{\chi }^2){]}^3=-0.66807.$

Central Pressure At the center of the configuration — where $\zeta =\chi =0$ — we see that, $-C_{\chi }{|}_{\chi =0}$ $=$ $[(-A_s)](1-e^2)+\frac{1}{2}[A_{ss}{a}_{\ell }^2+(1-e^2{)}^{-1}{A}_s](1-e^2{)}^2+\frac{1}{3}[-(1-e^2{)}^{-1}{A}_{ss}{a}_{\ell }^2](1-e^2{)}^3$   $=$ $-A_s(1-e^2)+\frac{1}{2}[A_{ss}{a}_{\ell }^2(1-e^2{)}^2+(1-e^2)A_s]-\frac{1}{3}[(1-e^2{)}^2{A}_{ss}{a}_{\ell }^2]$   $=$ $-\frac{1}{2}[A_s(1-e^2)]+\frac{1}{6}[A_{ss}{a}_{\ell }^2(1-e^2{)}^2]$ Hence, the central pressure is, $P_c^{*}\equiv [P_{\mathrm{d}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}}^{*}{]}_{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{l}}=C_{\chi }{|}_{\chi =0}$ $=$ $\frac{1}{2}[A_s(1-e^2)]-\frac{1}{6}[A_{ss}{a}_{\ell }^2(1-e^2{)}^2].$

Ferrers Vertical Pressure

Note for later use that,

$\frac{\partial C_{\chi }}{\partial \chi }$

$=$

…

Now Play With Radial Pressure Gradient

After multiplying through by $\rho /{\rho }_c$, the last term on the RHS of the ${\hat{e}}_{\varpi }$ component is given by the expression,

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

If we replace the normalized pressure by $P_{\mathrm{d}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}}^{*}$, the first term on the RHS of the ${\hat{e}}_{\varpi }$ component becomes,

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

Hence,

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

=

$\left[\frac{\partial P_{\mathrm{d}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}}^{*}}{\partial \chi }\right]-\frac{\rho }{{\rho }_c}\cdot \frac{\partial }{\partial \chi }\left[\frac{{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{(-\pi G{\rho }_c{a}_{\ell }^2)}\right]$

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!

See Also