Challenges Constructing Ellipsoidal-Like Configurations

First, let's review the three different approaches that we have described for constructing Riemann S-type ellipsoids. Then let's see how these relate to the technique that has been used to construct infinitesimally thin, nonaxisymmetric disks.

Riemann S-type Ellipsoids

Usually, the density, <math>~\rho</math>, and the pair of axis ratios, <math>~b/a</math> and <math>~c/a</math>, are specified. Then, the Poisson equation is solved to obtain <math>~\Phi_\mathrm{grav}</math> in terms of <math>~A_1</math>, <math>~A_2</math>, and <math>~A_3</math>. The aim, then, is to determine the value of the central enthalpy, <math>~H_0</math> — alternatively, the thermal energy density, <math>~\Pi</math> — and the two parameters, <math>~\Omega_f</math> and <math>~\lambda</math>, that determine the magnitude of the velocity flow-field. Keep in mind that, as viewed from a frame of reference that is spinning with the ellipsoid (at angular frequency, <math>~\Omega_f</math>), the adopted (rotating-frame) velocity field is,

<math>~\lambda \biggl[ \boldsymbol{\hat\imath} \biggl( \frac{a}{b}\biggr) y - \boldsymbol{\hat\jmath} \biggl( \frac{b}{a} \biggr) x \biggr] \, .</math>

Hence, the inertial-frame velocity is given by the expression,

<math>~\bold{u} + \bold{\hat{e}}_\varphi \Omega_f \varpi \, .</math>

While we will fundamentally rely on the <math>~(\Omega_f, \lambda)</math> parameter pair to define the velocity flow-field, in discussions of Riemann S-type ellipsoids it is customary to also refer to the following two additional parameters: The (rotating-frame) vorticity,

<math>~\boldsymbol{\zeta} \equiv \boldsymbol{\nabla \times}\bold{u}</math>	<math>~=</math>	<math>~ \boldsymbol{\hat\imath} \biggl[ \frac{\partial u_z}{\partial y} - \frac{\partial u_y}{\partial z} \biggr] + \boldsymbol{\hat\jmath} \biggl[ \frac{\partial u_x}{\partial z} - \frac{\partial u_z}{\partial x} \biggr] + \bold{\hat{k}} \biggl[ \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} \biggr] </math>
	<math>~=</math>	<math>~\bold{\hat{k}} \biggl[ - \lambda \biggl(\frac{b}{a} + \frac{a}{b}\biggr) \biggr] \, ;</math>

and the dimensionless frequency ratio,

<math>~\equiv</math>

<math>~\frac{ \zeta}{\Omega_f} \, .</math>

2^nd-Order TVE Expressions

As we have discussed in detail in an accompanying chapter, the three diagonal elements of the <math>~(3 \times 3)</math> 2^nd-order tensor virial equation are sufficient to determine the equilibrium values of <math>~\Pi</math>, <math>~\Omega_3</math>, and <math>~\zeta_3</math>.

Indices

2^nd-Order TVE Expressions that are Relevant to Riemann S-Type Ellipsoids

<math>~ \biggl[ \frac{3\cdot 5}{2^2\pi a b c\rho} \biggr] \Pi +\biggl\{ \Omega_3^2 + 2 \biggl[ \frac{b^2}{b^2+a^2}\biggr] \Omega_3 \zeta_3 ~-~(2\pi G\rho) A_1 \biggr\} a^2 + \biggl[ \frac{a^2}{a^2 + b^2}\biggr]^2 \zeta_3^2 b^2 </math>

<math>~ \biggl[ \frac{3\cdot 5}{2^2\pi a b c \rho} \biggr]\Pi + \biggl[ \frac{b^2}{b^2+a^2}\biggr]^2 \zeta_3^2 a^2 + \biggl\{ \Omega_3^2 + 2 \biggl[ \frac{a^2}{a^2 + b^2}\biggr] \Omega_3 \zeta_3 ~-~( 2\pi G \rho) A_2 \biggr\}b^2 </math>

<math>~ \biggl[ \frac{3\cdot 5}{2^2\pi abc\rho} \biggr]\Pi - (2\pi G \rho)A_3 c^2 </math>

The <math>~(i, j) = (3, 3)</math> element gives <math>~\Pi</math> directly in terms of known parameters. The <math>~(1, 1)</math> and <math>~(2, 2)</math> elements can then be combined in a couple of different ways to obtain a coupled set of expressions that define <math>~\Omega_3</math> and <math>~f \equiv \zeta_3/\Omega_3</math>.

<math>~\biggl[ \frac{b^2 a^2}{b^2+a^2}\biggr] f \Omega_3^2 </math>	<math>~=</math>	<math>~ \pi G\rho \biggl[ \frac{(A_1 - A_2)a^2b^2}{ b^2 - a^2} - A_3 c^2\biggr] \, ; </math>
[ EFE, Chapter 7, §48, Eq. (34) ]

and,

<math>~ \Omega_3^2 \biggl\{1 + \biggl[ \frac{a^2b^2}{(a^2 + b^2)^2}\biggr] f^2 \biggr\} </math>	<math>~=</math>	<math>~ \frac{2\pi G\rho}{ (a^2-b^2) } \biggl[ A_1 a^2 - A_2 b^2 \biggr] \, . </math>
[ EFE, Chapter 7, §48, Eq. (33) ]

Ou's (2006) Detailed Force Balance

In a separate accompanying chapter, we have described in detail how Ou(2006) used, essentially, the HSCF technique to solve the detailed force-balance equations. Beginning with the,

Eulerian Representation
of the Euler Equation
as viewed from a Rotating Reference Frame

<math>\biggl[\frac{\partial\vec{v}}{\partial t}\biggr]_{rot} + ({\vec{v}}_{rot}\cdot \nabla) {\vec{v}}_{rot}= - \frac{1}{\rho} \nabla P - \nabla \Phi_\mathrm{grav}

- {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} \, ,</math>

it can be shown that, for the velocity fields associated with all Riemann S-type ellipsoids,

<math>~({\vec{v}}_{rot}\cdot \nabla) {\vec{v}}_{rot}</math>	<math>~=</math>	<math>~ -\nabla \biggl[ \frac{1}{2} \lambda^2(x^2 + y^2) \biggr] \, ; </math>
<math>~- {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})</math>	<math>~=</math>	<math>~ +\nabla\biggl[\frac{1}{2} \Omega_f^2 (x^2 + y^2) \biggr] \, ; </math>
<math>~- 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} </math>	<math>~=</math>	<math>~ - \nabla\biggl[ \Omega_f \lambda\biggl( \frac{b}{a} x^2 + \frac{a}{b}y^2 \biggr) \biggr] \, . </math>

Hence, within each steady-state configuration the following Bernoulli's function must be uniform in space:

<math>~ H + \Phi_\mathrm{grav} - \frac{1}{2} \Omega_f^2(x^2 + y^2) - \frac{1}{2} \lambda^2(x^2 + y^2) + \Omega_f \lambda \biggl(\frac{b}{a}x^2 + \frac{a}{b}y^2 \biggr) </math>

Ou(2006), p. 550, §2, Eq. (6)

where <math>~C_B</math> is a constant. So, at the surface of the ellipsoid (where the enthalpy H = 0) on each of its three principal axes, the equilibrium conditions demanded by the expression for detailed force balance become, respectively:

On the x-axis, where (x, y, z) = (a, 0, 0):

<math>~2\biggl[ \frac{C_B}{a^2} + (\pi G\rho)I_\mathrm{BT} \biggr]</math>

<math>~ (2\pi G \rho) A_1 - \Omega_f^2 - \lambda^2 + 2\Omega_f \lambda \biggl(\frac{b}{a} \biggr) </math>

On the y-axis, where (x, y, z) = (0, b, 0):

<math>~2\biggl[ \frac{C_B}{a^2} + (\pi G\rho)I_\mathrm{BT} \biggr]</math>

<math>~ (2\pi G \rho) A_2 \biggl( \frac{b^2}{a^2}\biggr) - \Omega_f^2 \biggl( \frac{b^2}{a^2} \biggr) - \lambda^2\biggl( \frac{b^2}{a^2} \biggr) + 2\Omega_f \lambda \biggl(\frac{b}{a}\biggr) </math>

On the z-axis, where (x, y, z) = (0, 0, c):

<math>~\Rightarrow ~~~ 2 \biggl[ \frac{C_B}{a^2} + (\pi G\rho)I_\mathrm{BT}\biggr]</math>

<math>~=</math>

<math>~ (2\pi G \rho) A_3 \biggl( \frac{c^2}{a^2}\biggr) </math>

This third expression can be used to replace the left-hand-side of the first and second expressions. Then via some additional algebraic manipulation, the first and second expressions can be combined to provide the desired solutions for the parameter pair, <math>~(\Omega_f, \lambda)</math>, namely,

<math>~\frac{\Omega_f^2}{(\pi G \rho)}</math>

<math>~\frac{1}{2} \biggl[M + \sqrt{ M^2 - 4N^2} \biggr] \, ,</math>

and

<math>~\frac{\lambda^2}{(\pi G \rho)}</math>

<math>~\frac{1}{2} \biggl[M - \sqrt{ M^2 - 4N^2} \biggr] \, ,</math>

Ou(2006), p. 551, §2, Eqs. (15) & (16)

where,

<math>~M</math>	<math>~\equiv</math>	<math>~ 2\biggl[ A_1 - A_2 \biggl( \frac{b^2}{a^2}\biggr) \biggr]\biggl[ \frac{a^2}{a^2 - b^2} \biggr] \, ,</math> and,
<math>~N</math>	<math>~\equiv</math>	<math>~ \frac{1}{a b ( a^2 - b^2 )}\biggl[ A_3 ( a^2 - b^2 )c^2 - (A_2 - A_1) a^2 b^2 \biggr] \, . </math>

Hybrid Scheme

In a separate chapter we have detailed the hybrid scheme. For steady-state configurations, the three components of the combined Euler + Continuity equations give,

Hybrid Scheme Summary for Steady-State Configurations

<math>~\boldsymbol{\hat{k}:}</math>	<math>~ \bold\nabla \cdot (\rho v_z \bold{u}) </math>	<math>~=</math>	<math>~\bold{\hat{k}} \cdot (\rho \bold{a}) \, ;</math>
<math>~\bold{\hat{e}_\varpi:}</math>	<math>~ \bold\nabla \cdot (\rho v_\varpi \bold{u}) </math>	<math>~=</math>	<math>~\bold{\hat{e}}_\varpi \cdot (\rho \bold{a}) + \frac{v_\varphi^2}{\varpi} \, ;</math>
<math>~\bold{\hat{e}_\varphi:}</math>	<math>~ \bold\nabla \cdot (\rho \varpi v_\varphi \bold{u}) </math>	<math>~=</math>	<math>~\bold{\hat{e}}_\varphi \cdot (\rho \varpi \bold{a}) \, .</math>

In this context, the vector acceleration that drives the fluid flow is, simply,

<math>~-\nabla(H + \Phi_\mathrm{grav} ) \, .</math>

Then, for the velocity flow-patterns in Riemann S-type ellipsoids, we have,

<math>~\nabla \cdot (\rho v_z \bold{u})</math>	<math>~=</math>	<math>~0</math> (because <math>~v_z = 0</math>);
<math>~\nabla \cdot (\rho v_\varpi \bold{u})</math>	<math>~=</math>	<math>~\frac{\lambda^2}{\varpi^3} \biggl[\frac{a}{b} - \frac{b}{a} \biggr] \biggl\{ y^4 \biggl(\frac{a}{b}\biggr) - x^4 \biggl(\frac{b}{a}\biggr) \biggr\}\rho \, ; </math>
<math>~\nabla \cdot (\rho \varpi v_\varphi \bold{u})</math>	<math>~=</math>	<math>~ 2 \lambda xy \Omega_f \biggl[\frac{a}{b} - \frac{b}{a} \biggr]\rho \, ; </math>
<math>~\varpi v_\varphi</math>	<math>~=</math>	<math>~ - \biggl[ \lambda \biggl(\frac{b}{a}\biggr) - \Omega_f\biggr]x^2 - \biggl[ \lambda \biggl(\frac{a}{b}\biggr) - \Omega_f\biggr]y^2 \, . </math>

Vertical Component: Given that <math>~\bold{\hat{k}}\cdot (\rho \bold{a}) = 0</math>, we deduce that,

Azimuthal Component: Irrespective of the <math>~(x, y, z)</math> location of each fluid element, this component requires,

<math>~ - a b \lambda \Omega_f </math>

<math>~ \pi G \rho \biggl[ \frac{( A_1 - A_2 )a^2b^2}{b^2 - a^2} - c^2 A_3 \biggr] \, . </math>

Radial Component: After inserting the "azimuthal component" relation and marching through a fair amount of algebraic manipulation, we find that Irrespective of the <math>~(x, y, z)</math> location of each fluid element, this component requires,

<math>~ \frac{2\pi G \rho }{(a^2 - b^2) } \biggl[ A_1 a^2 - A_2 b^2 \biggr] </math>

<math>~ \biggl[ \lambda^2 + \Omega_f^2\biggr] \, . </math>

Compressible Structures

Here we draw heavily on the published work of Korycansky & Papaloizou (1996, ApJS, 105, 181; hereafter KP96) that we have reviewed in a separate chapter.

Returning to the above-mentioned,

Eulerian Representation
of the Euler Equation
as viewed from a Rotating Reference Frame

<math>\frac{\partial \bold{u}}{\partial t} + (\bold{u}\cdot \nabla) \bold{u} = - \frac{1}{\rho} \nabla P - \nabla \Phi_\mathrm{grav}

- {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) - 2{\vec{\Omega}}_f \times \bold{u} \, ,</math>

we next note — as we have done in our broader discussion of the Euler equation — that,

<math> (\bold{u} \cdot\nabla)\bold{u} = \frac{1}{2}\nabla(\bold{u} \cdot \bold{u}) - \bold{u} \times(\nabla\times\bold{u}) = \frac{1}{2}\nabla(u^2) + \boldsymbol\zeta \times \bold{u} , </math>

where, as above, <math>\boldsymbol\zeta \equiv \nabla\times \bold{u}</math> is the vorticity. Making this substitution, we obtain what is essentially equation (7) of KP96, that is, the,

Euler Equation
written in terms of the Vorticity and
as viewed from a Rotating Reference Frame

<math>\frac{\partial \bold{u}}{\partial t} + (\boldsymbol\zeta+2{\vec\Omega}_f) \times {\bold{u}}= - \frac{1}{\rho} \nabla P - \nabla \biggl[\Phi + \frac{1}{2}u^2 - \frac{1}{2}|{\vec{\Omega}}_f \times \vec{x}|^2 \biggr]</math> .

Hence, in steady-state, the Euler equation becomes,

<math> \nabla F_B + \vec{A} = 0 , </math>

where, the scalar "Bernoulli" function,

<math> F_B \equiv \frac{1}{2}u^2 + H + \Phi - \frac{1}{2}|\Omega\hat{k} \times \vec{x}|^2 ; </math>

and,

<math> \vec{A} \equiv ({\boldsymbol\zeta}+2{\vec\Omega}_f) \times {\bold{u}} . </math>

For later use …

Curl of steady-state Euler equation:

<math>~0</math>	<math>~=</math>	<math>~\nabla F_B + \bold{A}</math>
<math>~\Rightarrow~~~0</math>	<math>~=</math>	<math>~\nabla \times \biggl[ \nabla F_B + \bold{A} \biggr]</math>
<math>~\Rightarrow~~~0</math>	<math>~=</math>	<math>~\nabla \times \bold{A} \, .</math>

This last step is justified because the curl of any gradient is zero.

KP96 only deal with two-dimensional motion in the equatorial plane and, hence, there is no vertical motion:
Hence, <math>~\bold{u}</math> lies in the equatorial plane; both <math>~\vec\zeta</math> and <math>~\vec\Omega_f</math> only have z-components; and, <math>~\bold{A}</math> is perpendicular to both <math>~\vec\Omega_f</math> and <math>~\bold{u}</math>. Also, given that <math>~\bold{A}</math> necessarily lies in the equatorial plane, its curl will only have a z-component, that is,

<math>~\nabla \times \bold{A} = 0</math>

<math>~\Leftrightarrow</math>

<math>~[\nabla \times \bold{A}]_z = 0 \, .</math>

"Dot" <math>~\bold{u}</math> into the steady-state Euler equation:

<math>~0</math>	<math>~=</math>	<math>~\nabla F_B + \bold{A}</math>
<math>~\Rightarrow~~~0</math>	<math>~=</math>	<math>~\bold{u} \cdot \biggl[ \nabla F_B + \bold{A} \biggr]</math>
<math>~\Rightarrow~~~0</math>	<math>~=</math>	<math>~\bold{u} \cdot \nabla F_B \, .</math>

This last step is justified because <math>~\bold{A}</math> is necessarily always perpendicular to <math>~\bold{u}</math>.

We will leave discussion of the Euler equation, for the moment, and instead look at the continuity equation. As viewed from the rotating frame of reference,

<math>~\frac{\partial (\rho \bold{u})}{\partial t} + \nabla\cdot (\rho\bold{u})</math>

If we are able to write the momentum density (in the rotating frame) in terms of a stream-function, <math>~\Psi</math>, such that,

<math>~ \nabla \times (\boldsymbol{\hat{k}} \Psi) = \boldsymbol{\hat\imath} \biggl[ \frac{\partial \Psi}{\partial y} \biggr] - \boldsymbol{\hat\jmath} \biggl[ \frac{\partial \Psi}{\partial x}\biggr] \, ,</math>

then satisfying the steady-state continuity equation is guaranteed because the divergence of a curl is always zero. Note, as well, that when written in terms of this stream-function, the z-component of the vorticity will be,

<math>~\zeta_z </math>	<math>~=</math>	<math>~ \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} </math>
	<math>~=</math>	<math>~ \frac{\partial }{\partial x}\biggl[- \frac{1}{\rho} \frac{\partial \Psi}{\partial x} \biggr] - \frac{\partial }{\partial y} \biggl[\frac{1}{\rho} \frac{\partial \Psi}{\partial y} \biggr] \, . </math>

Note that the steady-state continuity equation may be rewritten in the form,

<math>~\nabla\cdot \bold{u}</math>

<math>~ - \bold{u} \cdot \nabla [ \ln \rho] \, . </math>

It can also be shown that the condition, <math>~[\nabla \times \bold{A}]_z = 0</math> can be rewritten as,

<math>~\nabla\cdot \bold{u}</math>

<math>~ - \bold{u} \cdot \nabla [ \ln(2\Omega_f + \zeta_z] \, . </math>

By combining these last two expressions, we appreciate that,

<math>~ \bold{u} \cdot \nabla \ln \biggl[ \frac{(2\Omega_f + \zeta_z)}{\rho} \biggr] </math>

This means that, in the steady-state flow whose spatial structure we are seeking, the velocity vector, <math>~\bold{u}</math> (and also the momentum density vector, <math>~\rho \bold{u}</math>) must everywhere be tangent to contours of constant vortensity, <math>~[(2\Omega_f + \zeta_z)/\rho]</math>.

We need a function <math>~g(\Psi) </math> such that,

<math>~g(\Psi) </math>	<math>~=</math>	<math>~ \frac{1}{\rho} \biggl\{ \zeta_z + 2\Omega_f \biggr\} </math>
	<math>~=</math>	<math>~ \frac{1}{\rho} \biggl\{ 2\Omega_f - \frac{\partial }{\partial x}\biggl[\frac{1}{\rho} \frac{\partial \Psi}{\partial x} \biggr] - \frac{\partial }{\partial y} \biggl[\frac{1}{\rho} \frac{\partial \Psi}{\partial y} \biggr] \biggr\} \, . </math>

Let's try, <math>~\Psi = \rho^2</math>, and

<math>~\rho</math>	<math>~=</math>	<math>~\rho_c \biggl\{1 - \biggl[ \frac{y^2}{b^2} + \frac{x^2}{a^2}\biggr]\biggr\} </math>
<math>~\Rightarrow~~~\frac{\partial^2 \rho}{\partial x^2} = -\frac{\partial}{\partial x}\biggl\{ \frac{2\rho_c x}{a^2}\biggr\} = - \frac{2\rho_c}{a^2}</math>	and,	<math>~\frac{\partial^2 \rho}{\partial y^2} = - \frac{\partial}{\partial y}\biggl\{ \frac{2\rho_c y}{b^2} \biggr\} = - \frac{2\rho_c}{b^2} \, .</math>

Then,

<math>~g(\Psi) </math>	<math>~=</math>	<math>~ \frac{1}{\rho} \biggl\{ 2\Omega_f - \frac{\partial }{\partial x}\biggl[\frac{1}{\rho} \frac{\partial \rho^2}{\partial x} \biggr] - \frac{\partial }{\partial y} \biggl[\frac{1}{\rho} \frac{\partial \rho^2}{\partial y} \biggr] \biggr\} </math>
	<math>~=</math>	<math>~ \frac{1}{\rho} \biggl\{ 2\Omega_f - 2 \biggl[ \frac{\partial^2 \rho}{\partial x^2} \biggr] - 2 \biggl[\frac{\partial^2 \rho}{\partial y^2} \biggr] \biggr\} </math>
	<math>~=</math>	<math>~ \frac{1}{\rho} \biggl\{ 2\Omega_f + 4\rho_c \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] \biggr\} \, . </math>

Hence,

<math>~ \biggl\{ 2\Omega_f + 4\rho_c \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] \biggr\} \Psi^{-1 / 2} \, . </math>

Next, given that,

we conclude that, to within an additive constant,

<math>~ F_B(\Psi)</math>	<math>~=</math>	<math>~ - \int g(\Psi) d\Psi = - g_0 \int \Psi^{-1 / 2} d\Psi </math>
	<math>~=</math>	<math>~ - 2g_0 \Psi^{1 / 2} </math>
	<math>~=</math>	<math>~ - 2g_0 \rho \, , </math>

where,

<math>~\equiv</math>

<math>~ \biggl\{ 2\Omega_f + 4\rho_c \biggl[ \frac{1}{a^2} + \frac{1}{b^2} \biggr] \biggr\} \, . </math>

Here's what to do:

Given <math>~g(\Psi)</math>, write out the functional forms of <math>~\bold{A}</math> and <math>~F_B(\Psi)</math>. Then see if <math>~\nabla F_B = - \bold{A}</math>.

User:Tohline/ThreeDimensionalConfigurations/Challenges

Contents

Challenges Constructing Ellipsoidal-Like Configurations

Riemann S-type Ellipsoids

2^nd-Order TVE Expressions

Ou's (2006) Detailed Force Balance

Hybrid Scheme

Compressible Structures

See Also

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User:Tohline/ThreeDimensionalConfigurations/Challenges

Challenges Constructing Ellipsoidal-Like Configurations

Riemann S-type Ellipsoids

2nd-Order TVE Expressions

Ou's (2006) Detailed Force Balance

Hybrid Scheme

Compressible Structures

See Also

Navigation menu

Search

2^nd-Order TVE Expressions