ParabolicDensity/Axisymmetric/Structure

Parabolic Density Distribution

Part I:   Gravitational Potential

Part II:   Spherical Structures

Part III:   Axisymmetric Equilibrium Structures

Part IV:   Triaxial Equilibrium Structures (Exploration)

Axisymmetric (Oblate) Equilibrium Structures

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

$\rho$

=

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

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

Gravitational Potential

For an oblate-spheroidal configuration — that is, when $a_s<a_m=a_{\ell }$ — the gravitational potential may be obtained from the expression,

In our accompanying discussion we find that,

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

$=$

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

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

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

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,

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

See Also