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Hence, demanding (as in Step #2) that <math>\nabla\times\vec{A} = 0</math> means that each of these components independently must be zero. This, in turn, implies: | |||
<div align="left"> | |||
<math> | |||
~~~~~\hat{i}:~~~~~ v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x = C_{z1}(x,y); | |||
</math><br /> | |||
<math> | |||
~~~~~\hat{j}:~~~~~ v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y = C_{z2}(x,y) ; | |||
</math><br /> | |||
<math> | |||
~~~~~\hat{j}:~~~~~ \frac{\partial}{\partial x} \biggl[ C_{z1}(x,y) \biggr] = \frac{\partial}{\partial y} \biggl[C_{z2}(x,y) \biggr] , | |||
</math> | |||
</div> | |||
where the integration "constants" <math>C_{z1}</math> and <math>C_{z2}</math> may be functions of <math>x</math> and/or <math>y</math> but they must be independent of <math>z</math>. | |||
The divergence of <math>\vec{A}</math> (providing the right-hand-side of the Poisson-like equation in Step #3, above) generates: | |||
<div align="center"> | <div align="center"> | ||
<math> | <math> | ||
\nabla\cdot\vec{A} = \frac{\partial}{\partial x} \biggl[ v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y \biggr] + \frac{\partial}{\partial y} \biggl[ v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x \biggr] + \frac{\partial}{\partial z} \biggl[ 0 \biggr] . | \nabla\cdot\vec{A} = \frac{\partial}{\partial x} \biggl[ v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y \biggr] + \frac{\partial}{\partial y} \biggl[ v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x \biggr] + \frac{\partial}{\partial z} \biggl[ 0 \biggr] | ||
</math><br /> | |||
<math> | |||
= \frac{\partial}{\partial x} \biggl[ C_{z2}(x,y) \biggr] + \frac{\partial}{\partial y} \biggl[C_{z1}(x,y) \biggr] . | |||
</math> | </math> | ||
</div> | </div> |
Revision as of 22:39, 12 March 2010
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Compressible Analogs of Riemann S-Type Ellipsoids
Here we attempt to develop a self-consistent-field type, iterative technique that will permit the construction of steady-state structures that are compressible analogs of Riemann S-Type (incompressible) ellipsoids. We will build upon the recent work of Ou (2006).
Standard Steady-State Governing Relations
As viewed from a rotating frame of reference and written in Eulerian form, the steady-state version of the three-dimensional principal governing equations are:
<math> \nabla\cdot(\rho \vec{v}) = 0 </math>
<math> (\vec{v}\cdot \nabla)\vec{v} = -\nabla \biggl[H + \Phi -\frac{1}{2}\omega^2 R^2 \biggr] -2\vec{\omega}\times\vec{v} </math>
<math> \nabla^2 \Phi = 4\pi G \rho </math>
Proposed Solution Strategy
Preamble:
Specify the three "polar" boundary locations, <math>a, b,</math> and <math>c</math>; specify the direction but not the magnitude of the rotating frame's angular velocity, for example, <math>(\vec{\omega}/\omega) = \hat{k}</math>; pin the central density to the value <math>\rho_c = 1</math>. Define the following dimensionless density, velocity vector, angular velocity, enthalpy, gravitational potential, and position vector:
<math> \rho^* \equiv \frac{\rho}{\rho_c} ; ~~~~~{\vec{v}}^* \equiv \frac{\vec{v}}{[a^2G\rho_c]^{1/2}} ; ~~~~~\omega^* \equiv \frac{\omega}{[G\rho_c]^{1/2}} ; </math>
<math> H^* \equiv \frac{H}{[a^2G\rho_c]} ; ~~~~~\Phi^* \equiv \frac{\Phi}{[a^2G\rho_c]} ; ~~~~~{\vec{x}}^* \equiv \frac{\vec{x}}{a} . </math>
From here, on, spatial operators are assumed to be in terms of the dimensionless coordinates.
Step #1:
Guess a 3D density distribution — such as a uniform-density ellipsoid, or one of the converged models from Ou (2006) — that conforms to a selected set of positional boundary conditions, that is, where the density goes to zero along the three principal axes at <math>x=a</math>, <math>y = b</math>, and <math>z = c</math>. Solve the Poisson equation in order to derive values for <math>\Phi</math> everywhere inside and on the surface of the 3D configuration:
<math> \nabla^2 \Phi^* = 4\pi \rho^* . </math>
Step #2:
Use the continuity equation and the curl of the Euler equation to numerically derive the structure but not the overall magnitude of the velocity flow-field throughout the 3D configuration. Take advantage of the fact that the direction, <math>(\vec{\omega}/\omega)</math>, has been specified; and assume that the 3D density distribution is known. Here are the relevant equations:
<math> \nabla\cdot(\rho^* {\vec{v}}^*) = 0 ; </math>
<math> \nabla\times \biggl[({\vec{v}}^*\cdot \nabla){\vec{v}}^* +2 {\vec{\omega}}^* \times {\vec{v}}^* \biggr] = 0 . </math>
The first of these is a scalar equation; the second is a vector equation and it will presumably provide two useful scalar equations (perhaps constraining the two components of <math>{\vec{v}}^*</math> that are perpendicular to <math>\hat{k}</math> ?). Since the left-hand-side of the second equation is obviously nonlinear in the velocity, we may have to linearize this set of equations and look for small "corrections" <math>\delta\vec{v}</math> to an initial "guess" for the velocity field, such as the flow field in Riemann S-type ellipsoids, which is also the flow-field adopted by Ou (2006).
Step #3:
Take the divergence of the Euler equation and use it to solve for <math>H</math> throughout the configuration, given the structure of the flow-field obtained in Step #2. Boundary conditions at the three "poles" of the configuration may suffice to uniquely determine <math>\omega</math>, the overall normalization factor for the flow-field <math>\vec\zeta</math> — hopefully this is analogous to solving for the vorticity parameter <math>\lambda</math> in Ou (2006) — and the Bernoulli constant (or something equivalent). The relevant "Poisson"-like equation is:
<math> \nabla^2 \biggl[H^* + \Phi^* -\frac{1}{2}(\omega^*)^2 \biggl(\frac{R}{a}\biggr)^2 \biggr] = - \nabla\cdot [({\vec{v}}^*\cdot \nabla){\vec{v}}^* + 2 {\vec{\omega}}^*\times {\vec{v}}^* ] . </math>
Example of Riemann S-Type Ellipsoids
Preamble
First, set <math>{\vec{\omega}} = \hat{k}\omega</math> and <math>v_z = 0</math>, and write out the Cartesian components of the vector,
<math> \vec{A} \equiv ({\vec{v}}\cdot \nabla){\vec{v}} +2 {\vec{\omega}} \times {\vec{v}} . </math>
The components are:
<math>
~~~~~\hat{i}:~~~~~A_x = v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y ;
</math>
<math>
~~~~~\hat{j}:~~~~~A_y = v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x ;
</math>
<math>
~~~~~\hat{j}:~~~~~A_z = 0 .
</math>
The curl of <math>\vec{A}</math> (needed in Step #2, above) produces a vector with the following three Cartesian components:
<math>
~~~~~\hat{i}:~~~~~[\nabla\times\vec{A}]_x = \frac{\partial}{\partial y} \biggl[0 \biggr] - \frac{\partial}{\partial z} \biggl[ v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x \biggr] ;
</math>
<math>
~~~~~\hat{j}:~~~~~[\nabla\times\vec{A}]_y = \frac{\partial}{\partial z} \biggl[ v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y \biggr] - \frac{\partial}{\partial x} \biggl[0 \biggr] ;
</math>
<math>
~~~~~\hat{j}:~~~~~[\nabla\times\vec{A}]_z = \frac{\partial}{\partial x} \biggl[ v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x \biggr] - \frac{\partial}{\partial y} \biggl[ v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y \biggr] .
</math>
Hence, demanding (as in Step #2) that <math>\nabla\times\vec{A} = 0</math> means that each of these components independently must be zero. This, in turn, implies:
<math>
~~~~~\hat{i}:~~~~~ v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x = C_{z1}(x,y);
</math>
<math>
~~~~~\hat{j}:~~~~~ v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y = C_{z2}(x,y) ;
</math>
<math>
~~~~~\hat{j}:~~~~~ \frac{\partial}{\partial x} \biggl[ C_{z1}(x,y) \biggr] = \frac{\partial}{\partial y} \biggl[C_{z2}(x,y) \biggr] ,
</math>
where the integration "constants" <math>C_{z1}</math> and <math>C_{z2}</math> may be functions of <math>x</math> and/or <math>y</math> but they must be independent of <math>z</math>.
The divergence of <math>\vec{A}</math> (providing the right-hand-side of the Poisson-like equation in Step #3, above) generates:
<math>
\nabla\cdot\vec{A} = \frac{\partial}{\partial x} \biggl[ v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} -2\omega v_y \biggr] + \frac{\partial}{\partial y} \biggl[ v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} +2\omega v_x \biggr] + \frac{\partial}{\partial z} \biggl[ 0 \biggr]
</math>
<math>
= \frac{\partial}{\partial x} \biggl[ C_{z2}(x,y) \biggr] + \frac{\partial}{\partial y} \biggl[C_{z1}(x,y) \biggr] .
</math>
Riemann Flow-Field
In Riemann S-Type ellipsoids, the adopted planar flow-field as viewed from the rotating reference frame is,
<math> \vec{v} = \hat{i} \biggl( \frac{\lambda a y}{b} \biggr) + \hat{j} \biggl( - \frac{\lambda b x}{a} \biggr) . </math>
Hence,
<math>
[\nabla\times\vec{A}]_x = \frac{\partial}{\partial z} \biggl[\biggl( \frac{\lambda ay}{b} \biggr) \frac{\partial }{\partial x}\biggl( - \frac{\lambda b x}{a} \biggr) + \biggl( - \frac{\lambda b x}{a} \biggr) \frac{\partial }{\partial y} \biggl( - \frac{\lambda b x}{a} \biggr) +2\omega \biggl( \frac{\lambda a y}{b} \biggr) \biggr]
</math>
<math>
= \frac{\partial}{\partial z} \biggl[\biggl( \frac{\lambda ay}{b} \biggr) \biggl( - \frac{\lambda b }{a} \biggr) +2\omega \biggl( \frac{\lambda a y}{b} \biggr) \biggr];
</math>
<math>
[\nabla\times\vec{A}]_y = \frac{\partial}{\partial z} \biggl[ \biggl( \frac{\lambda a y}{b} \biggr) \frac{\partial }{\partial x}\biggl( \frac{\lambda a y}{b} \biggr) + \biggl( - \frac{\lambda b x}{a} \biggr) \frac{\partial }{\partial y}\biggl( \frac{\lambda a y}{b} \biggr) -2\omega \biggl( - \frac{\lambda b x}{a} \biggr) \biggr]
</math>
<math>
~~~~~~~~~~ = ;
</math>
<math>
[\nabla\times\vec{A}]_z = \frac{\partial}{\partial x} \biggl[ \biggl( \frac{\lambda a y}{b} \biggr) \frac{\partial }{\partial x}\biggl( - \frac{\lambda b x}{a} \biggr) + \biggl( - \frac{\lambda b x}{a} \biggr) \frac{\partial }{\partial y}\biggl( - \frac{\lambda b x}{a} \biggr) +2\omega \biggl( \frac{\lambda a y}{b} \biggr)\biggr]
</math>
<math>
~~~~~~~~~~ - \frac{\partial}{\partial y} \biggl[ \biggl( \frac{\lambda a y}{b} \biggr) \frac{\partial }{\partial x}\biggl( \frac{\lambda a y}{b} \biggr) + \biggl( - \frac{\lambda b x}{a} \biggr) \frac{\partial }{\partial y} \biggl( \frac{\lambda a y}{b} \biggr) -2\omega \biggl( - \frac{\lambda b x}{a} \biggr) \biggr]
</math>
<math>
~~~~~~~~~~ = ;
</math>
and,
<math>
\nabla\cdot\vec{A} = \frac{\partial}{\partial x} \biggl[ \biggl( \frac{\lambda a y}{b} \biggr) \frac{\partial }{\partial x}\biggl( \frac{\lambda a y}{b} \biggr) + \biggl( - \frac{\lambda b x}{a} \biggr) \frac{\partial }{\partial y}\biggl( \frac{\lambda a y}{b} \biggr) -2\omega \biggl( - \frac{\lambda b x}{a} \biggr) \biggr]
</math>
<math>
~~~~~~~~~~ + \frac{\partial}{\partial y} \biggl[ \biggl( \frac{\lambda a y}{b} \biggr) \frac{\partial }{\partial x}\biggl( - \frac{\lambda b x}{a} \biggr) + \biggl( - \frac{\lambda b x}{a} \biggr) \frac{\partial }{\partial y}\biggl( - \frac{\lambda b x}{a} \biggr) +2\omega \biggl( \frac{\lambda a y}{b} \biggr) \biggr] + \frac{\partial}{\partial z} \biggl[ 0 \biggr]
</math>
<math>
~~~~~~~~~~ = .
</math>
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