Revision as of 19:20, 17 September 2020

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, and on the doctoral dissertation of Saied W. Andalib (1998).

Part I

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>.

Part II

Consider a steady-state configuration that is the compressible analog of a Riemann S-type ellipsoid; even better, give the configuration a "peanut" shape rather than the shape of an ellipsoid. As viewed from a frame that is spinning with the configuration's overall angular velocity, <math>~\vec\Omega_f = \boldsymbol{\hat{k}} \Omega_f</math>, generally we expect the configuration's internal (and surface) flow to be represented by a set of nested stream-lines and at every <math>~(x, y)</math> location the fluid's velocity (and its momentum-density vector) will be tangent to the stream-line that runs through that point. It is customary to represent the stream-function by a scalar quantity, <math>~\Psi(x, y)</math>, appreciating that each stream-line will be defined by a curve, <math>~\Psi = \mathrm{constant}</math>; also, the local spatial gradient of <math>~\Psi(x,y)</math> will be perpendicular to the local stream-line, hence it will be perpendicular to the local velocity vector as well. If we specifically introduce the stream-function via the relation,

<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>

it will display all of the just-described attributes and we are also guaranteed that the steady-state continuity equation will be satisfied everywhere, because the divergence of a curl is always zero.

We also have demonstrated that the vector, <math>~\bold{A}</math>, has the right properties if,

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

This means that, at every location in the plane of the fluid flow, the gradient of the vortensity also must be perpendicular to the velocity vector. This constraint can be immediately satisfied if we simply demand that the vortensity be a function of the stream-function, <math>~\Psi</math>, that is, we need,

<math>~\frac{(2\Omega_f + \zeta_z)}{\rho}</math>

In other words, the scalar vortensity is constant along each stream-line. And, once we have determined the mathematical expression for this function, we will know that,

<math>~[ \boldsymbol{\hat{k}} g(\Psi) ] \times \rho\bold{u} \, ;</math>

Furthermore, we should be able to determine the mathematical expression for <math>~F_B(x,y)</math> because,

As a check, we should find that,

<math>~\nabla F_B + \bold{A}</math>

Part III

Here we closely follow Chapter 4 of Saied W. Andalib (1998).

From §4.1 (p. 80): "Euler's equation, the equation of continuity, the Poisson equation and the equation of state … govern the dynamics and evolution of these equilibrium configurations."

Equation of Continuity

In steady state, <math>~\partial (\rho\bold{u})/\partial t = 0</math>. Hence the rotating-frame-based continuity equation becomes,

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

If we insist that the momentum-density vector be expressible in terms of the curl of a vector — for example,

<math>~\rho\bold{u}</math>	<math>~=</math>	<math>~\nabla \times \boldsymbol{\mathfrak{J}} \, ,</math>
Saied W. Andalib (1998), §4.1, p. 80, Eq. (4.1)

then satisfying this steady-state continuity equation is guaranteed because the divergence of a curl is always zero. "The task of satisfying the steady-state equation of continuity then shifts to identifying an appropriate expression for the vector potential, <math>~\boldsymbol{\mathfrak{J}} \, .</math>" In the most general case, in terms of this vector potential the three Cartesian components of the momentum-density vector are,

<math>~\rho u_x</math>	<math>~=</math>	<math>~ \frac{\partial \mathfrak{J}_z}{\partial y} - \frac{\partial \mathfrak{J}_y}{\partial z} \, ; </math>
<math>~\rho u_y</math>	<math>~=</math>	<math>~ \frac{\partial \mathfrak{J}_x}{\partial z} - \frac{\partial \mathfrak{J}_z}{\partial x} \, ; </math>
<math>~\rho u_z</math>	<math>~=</math>	<math>~ \frac{\partial \mathfrak{J}_y}{\partial x} - \frac{\partial \mathfrak{J}_x}{\partial y} \, . </math>

Here, we will follow Andalib's lead and only look for fluid flows with no vertical motions. That is to say, we will set <math>~\rho u_z = 0</math>, in which case this last expression establishes the constraint,

<math>~\frac{\partial \mathfrak{J}_y}{\partial x}</math>

<math>~ \frac{\partial \mathfrak{J}_x}{\partial y} \, . </math>

"A general solution to this equation can be found only if there exists a scalar function <math>~\Gamma(x, y, z)</math> such that …"

<math>~\mathfrak{J}_y = \frac{\partial \Gamma}{\partial y}</math>

and,

<math>~\mathfrak{J}_x = \frac{\partial \Gamma}{\partial x} \, ;</math>

note that this adopted functional behavior works because the constraint becomes,

<math>~\frac{\partial^2 \Gamma}{\partial x \partial y}</math>

<math>~\frac{\partial^2 \Gamma}{\partial y \partial x} \, .</math>

Hence, the expressions for the x- and y-components of the momentum-density vector may be rewritten, respectively, as,

<math>~\rho u_x</math>	<math>~=</math>	<math>~ \frac{\partial \mathfrak{J}_z}{\partial y} - \frac{\partial}{\partial z} \biggl[ \frac{\partial \Gamma}{\partial y} \biggr] = + \frac{\partial}{\partial y}\biggl[ \mathfrak{J}_z - \frac{\partial\Gamma}{\partial z} \biggr] \, ; </math>
<math>~\rho u_y</math>	<math>~=</math>	<math>~ \frac{\partial }{\partial z}\biggl[ \frac{\partial \Gamma}{\partial x} \biggr] - \frac{\partial \mathfrak{J}_z}{\partial x} = - \frac{\partial}{\partial x}\biggl[ \mathfrak{J}_z - \frac{\partial\Gamma}{\partial z} \biggr] \, . </math>

If we again follow Andalib's lead and only look for models in which the x-y-plane flow is independent of the vertical coordinate, z, then, <math>~\mathfrak{J}_z</math> and <math>~\partial\Gamma/\partial z</math> must be functions of x and y only. Therefore, <math>~\mathfrak{J}_z</math> is independent of z and <math>~\Gamma</math> is at most linear in z. Now, rather than focusing on the determination of <math>~\Gamma(x,y)</math>, we can just as well define the scalar function,

<math>~\equiv</math>

<math>~ \mathfrak{J}_z - \frac{\partial\Gamma}{\partial z} \, , </math>

in which case "… the components of the momentum density may be written as:"

<math>~ \boldsymbol{\hat\imath} \frac{\partial \Psi}{\partial y} - \boldsymbol{\hat\jmath} \frac{\partial \Psi}{\partial x} \, . </math>

It is straightforward to demonstrate that this expression for the momentum-density vector does satisfy the steady-state continuity equation. "The function <math>~\Psi(x, y)</math> will serve a similar role as the velocity potential for incompressible fluids."

Related Useful Expressions

Given that, by our design, the fluid motion will be confined to the x-y-plane, the fluid vorticity will have only a z-component; that is, <math>~\vec\zeta = \boldsymbol{\hat{k}} \zeta_z</math>. And when it is written in terms of <math>~\Psi(x,y)</math>, this z-component of the vorticity will be obtained from the expression,

<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>

This is useful to know because, in the Euler equation (see immediately below) we will encounter a term that involves the cross product of the vector, <math>~(\vec\zeta + 2\vec\Omega) </math>, with the rotating-frame-based velocity vector. Appreciating as well that the vector, <math>~\vec\Omega = \boldsymbol{\hat{k}} \Omega_f</math>, only has a nonzero z-component, we recognize that this term may be written as,

<math>~(\vec\zeta + 2\vec\Omega) \times \bold{u}</math>	<math>~=</math>	<math>~\boldsymbol{\hat{k}} \biggl[ \frac{ (\zeta_z + 2\Omega_f)}{\rho} \biggr] \times \biggl[ \boldsymbol{\hat\imath} \frac{\partial \Psi}{\partial y} - \boldsymbol{\hat\jmath} \frac{\partial \Psi}{\partial x} \biggr]</math>
	<math>~=</math>	<math>~\biggl[ \frac{ (\zeta_z + 2\Omega_f)}{\rho} \biggr] \biggl[ \boldsymbol{\hat\jmath} \frac{\partial \Psi}{\partial y} + \boldsymbol{\hat\imath} \frac{\partial \Psi}{\partial x} \biggr]</math>
	<math>~=</math>	<math>~\biggl[ \frac{ (\zeta_z + 2\Omega_f)}{\rho} \biggr] \nabla\Psi \, .</math>
Saied W. Andalib (1998), §4.2, p. 83, Eq. (4.13)

Euler Equation

Recognizing that the vorticity, <math>\boldsymbol\zeta \equiv \nabla\times \bold{u}</math>, we begin with 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> .

Next, we rewrite this expression to incorporate the following three realizations:

For a barotropic fluid, the term involving the pressure gradient can be replaced with a term involving the enthalpy via the relation, <math>~\nabla H = \nabla P/\rho</math>.
The expression for the centrifugal potential can be rewritten as, <math>~\tfrac{1}{2}|\vec\Omega_f \times \vec{x}|^2 = \tfrac{1}{2}\Omega_f^2 (x^2 + y^2)</math>.
In steady state, <math>~\partial \bold{u}/\partial t = 0</math>.

This means that,

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

<math>- \nabla \biggl[H + \Phi_\mathrm{grav} + \frac{1}{2}u^2 - \frac{1}{2}\Omega_f^2 (x^2 + y^2) \biggr] \, .</math>

If the term on the left-hand-side of this equation can be expressed in terms of the gradient of a scalar function, then it can be readily grouped with all the other terms on the right-hand-side, which already are in the gradient form.

We have already demonstrated above, the term on the left-hand-side of the Euler equation can be rewritten as,

<math>~(\vec\zeta + 2\vec\Omega) \times \bold{u}</math>

<math>~\biggl[ \frac{ (\zeta_z + 2\Omega_f)}{\rho} \biggr] \nabla\Psi \, .</math>

@@ Line 1,561: / Line 1,561: @@
    </td>
 </tr>
-<tr><td align="center" colspan="3">[https://digitalcommons.lsu.edu/gradschool_disstheses/6650/ Saied W. Andalib (1998)], &sect;4.2, p. 83, Eq. (4.13)
+<tr><td align="center" colspan="3">[https://digitalcommons.lsu.edu/gradschool_disstheses/6650/ Saied W. Andalib (1998)], &sect;4.2, p. 83, Eq. (4.13)</td></tr>
 </table>

Difference between revisions of "User:Tohline/ThreeDimensionalConfigurations/Challenges"

Revision as of 19:20, 17 September 2020

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

Part I

Part II

Part III

Equation of Continuity

Related Useful Expressions

Euler Equation

See Also

Navigation menu

Search

Difference between revisions of "User:Tohline/ThreeDimensionalConfigurations/Challenges"

Revision as of 19:20, 17 September 2020

Challenges Constructing Ellipsoidal-Like Configurations

Riemann S-type Ellipsoids

2nd-Order TVE Expressions

Ou's (2006) Detailed Force Balance

Hybrid Scheme

Compressible Structures

Part I

Part II

Part III

Equation of Continuity

Related Useful Expressions

Euler Equation

See Also

Navigation menu

Search

2^nd-Order TVE Expressions