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 …

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

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 where $\rho /{\rho }_c=0$, it must also be true that,

at the surface   …

${\zeta }^2$

$=$

$(1-e^2)[1-{\chi }^2-{\xcancel{\frac{\rho }{{\rho }_c}}}^0]=(1-e^2)(1-{\chi }^2).$

Hence (numerical evaluations assume χ = 0.6 as well as e = 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, $P_c^{*}\equiv [P_{\mathrm{d}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}}^{*}{]}_{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{l}}=C_{\chi }{|}_{\chi =0}$ $=$ $\frac{1}{2}[A_s(1-e^2)]-\frac{1}{6}[A_{ss}{a}_{\ell }^2(1-e^2{)}^2].$      [0.2045061]

For an oblate-spheroidal configuration having eccentricity, $e=0.6\Rightarrow a_s/a_{\ell }=0.8$, the figure displayed here, on the right, shows how the normalized gas pressure $(P_{\mathrm{d}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}}^{*}/P_c^{*})$ varies with height above the mid-plane $(\zeta )$ at three different distances from the symmetry axis: (blue) $\chi =0.0$, (orange) $\chi =0.6$, and (gray) $\chi =0.75$. circular marker color chosen $\chi$ resulting … surface $\zeta$ mid-plane pressure blue $0.00$ $0.8000$ $1.00000$ orange $0.60$ $0.6400$ $0.32667$ gray $0.75$ $0.52915$ $0.13085$

Ferrers Vertical Pressure

Inserting the expression for $C_{\lambda }$ into our derived expression for $P_{\mathrm{d}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}}^{*}$ gives,

$P_{\mathrm{d}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}}^{*}$

$=$

$(\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f}\mathrm{1})\cdot [{\zeta }^2-(1-e^2)(1-{\chi }^2)]+(\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f}\mathrm{2})\cdot [{\zeta }^4-(1-e^2{)}^2(1-{\chi }^2{)}^2]+(\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f}\mathrm{3})\cdot [{\zeta }^6-(1-e^2{)}^3(1-{\chi }^2{)}^3].$

---

Note for later use that,

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

$=$

…

Isobaric Surfaces

By design, the mass within our oblate-spheroidal configuration is distributed in such a way that iso-density surfaces are concentric spheroids. As stated earlier, the relevant mathematically prescribed density distribution is,

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

$=$

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

In order to determine the relative stability of each configuration, it will be important to ascertain whether or not isobaric surfaces are also concentric spheroids. (If they are, then we can say that each configuration obeys a barotropic — but not necessarily a polytropic — equation of state; see, for example, the accompanying relevant excerpt drawn from p. 466 of 📚 Lebovitz (1967).) In an effort to make this determination for our $e=0.6$ spheroid, we first examine the iso-density surface for which $\rho /{\rho }_c=0.3$. Via the expression,

${\zeta }^2$

$=$

$(1-e^2)[1-{\chi }^2-\frac{\rho }{{\rho }_c}]=0.64[1-{\chi }^2-0.3],$

we can immediately determine that our three chosen radial cuts $(\chi =0.0,0.6,0.75)$ intersect this iso-density surface at the vertical locations, respectively, $\zeta =0.66933,0.46648,0.29665$; these numerical values have been recorded in the following table. The table also contains coordinates for the points where our three cuts intersect the $(e=0.6)$ iso-density surface for which $\rho /{\rho }_c=0.6$.

diamond marker color	chosen $\rho /{\rho }_c$	chosen $\chi$	resulting …
			$\zeta$	normalized pressure
green	$0.3$	$0.00$	$0.66933$	$0.060466$
		$0.60$	$0.46648$	$0.057433$
		$0.75$	$0.29665$	$0.055727$
purple	$0.6$	$0.00$	$0.50596$	$0.292493$
		$0.60$	$0.16000$	$0.280361$
		$0.75$	n/a	n/a

For each of these five $(\chi ,\zeta )$ coordinate pairs, we have used our above derived expression for $P_{\mathrm{d}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}}^{*}/P_c^{*}$ to calculate the "normalized pressure" at the relevant point inside the configuration. These results appear in the last column of the table; they also have been marked in the accompanying figure: dark green diamonds mark the points relevant to our choice of $\rho /{\rho }_c=0.3$ and purple diamonds mark the points relevant to our choice of $\rho /{\rho }_c=0.6$. Notice that the normalized density is everywhere lower than $0.6$ along the $\chi =0.75$ cut, so the final row in the table has been marked "n/a" (not applicable).

The dark green diamond-shaped markers in the figure — along with the associated tabular data — show that at three separate points along the $\rho /{\rho }_c=0.3$ iso-density surface, the normalized pressure is nearly — but not exactly — the same; its value is approximately $0.057$. Similarly, the purple diamond-shaped markers show that at two separate points along the $\rho /{\rho }_c=0.6$ iso-density surface, the normalized pressure is nearly the same; in this case its value is approximately $0.28$. This seems to indicate that, throughout our configuration, the isobaric surfaces are almost — but not exactly — aligned with iso-density surfaces.

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