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{k}:~~~~~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{k}:~~~~~[\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{k}:~~~~~ \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>.

Generically, the continuity equation demands,

<math> 0 = \frac{\partial}{\partial x}\biggl[ \rho v_x \biggr] + \frac{\partial}{\partial y}\biggl[ \rho v_y \biggr] + \frac{\partial}{\partial z}\biggl[ \rho v_z \biggr] . </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> C_{z1}(x,y) = \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) = \biggl[2\omega\biggl( \frac{\lambda a }{b} \biggr)- \lambda^2 \biggr]y, </math>
<math> C_{z2}(x,y) = \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) = \biggl[2\omega\biggl( \frac{\lambda b }{a} \biggr) - \lambda^2 \biggr]x . </math>

Because <math>C_{z1}</math> is independent of <math>x</math> and <math>C_{z2}</math> is independent of <math>y</math>, we see that <math>[\nabla\times\vec{A}]_z = 0</math>, trivially. With this specified velocity flow-field and the appreciation that Riemann S-type ellipsoids also have uniform density, the continuity equation is also trivially satisfied; specifically,

<math> \frac{\partial}{\partial x}\biggl[ \rho v_x \biggr] + \frac{\partial}{\partial y}\biggl[ \rho v_y \biggr] + \frac{\partial}{\partial z}\biggl[ \rho v_z \biggr] = \rho \biggl[ \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} \biggr] = 0 . </math>

However, the right-hand-side of our Poisson-like equation is not zero; rather, it is,

<math> \nabla\cdot\vec{A} = \biggl[2\omega\biggl( \frac{\lambda b }{a} \biggr) - \lambda^2 \biggr] + \biggl[2\omega\biggl( \frac{\lambda a }{b} \biggr)- \lambda^2 \biggr] = 2 \biggl[\omega\lambda \biggl( \frac{b}{a} + \frac{a}{b} \biggr) - \lambda^2 \biggr]. </math>

Summary

What can we learn from the Riemann S-Type ellipsoids? Well, let's assume that the structure of our equilibrium flow-field will be more complicated than simply "flow along elliptical paths", but let's continue to assume that a solution can be found in which the flow remains planar, that is, <math>v_z=0</math> everywhere. Also, let's continue to align the rotation axis of the frame with the <math>z</math>-axis of the configuration. The three steps in our proposed solution strategy become:

Step #1 (simplified):

Guess a 3D density distribution 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 G \rho . </math>

Step #2 (simplified):

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, assuming <math>\vec{\omega}=\hat{k}\omega</math> and the 3D density distribution is known. Here are the relevant equations:

<math> \frac{\partial}{\partial x}\biggl[ \rho v_x \biggr] + \frac{\partial}{\partial y}\biggl[ \rho v_y \biggr] = 0 . </math>

<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{k}:~~~~~ \frac{\partial}{\partial x} \biggl[ C_{z1}(x,y) \biggr] = \frac{\partial}{\partial y} \biggl[C_{z2}(x,y) \biggr] , </math>

Now, for the adopted flow-field of the Riemann ellipsoids, <math>C_{z1}(x,y)</math> is only a function of <math>y</math> and <math>C_{z2}(x,y)</math> is only a function of <math>x</math> — that is, <math>C_{z1}(x,y) \rightarrow C_{z1}(y)</math> and <math>C_{z2}(x,y) \rightarrow C_{z2}(x)</math>. Hence, the third (<math>\hat{k}</math>) condition is automatically satisfied. I don't know whether or not we will find that our more general velocity fields exhibit the same nice character.

Step #3 (simplified):

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. The relevant "Poisson"-like equation is:

<math> \nabla^2 \biggl[H + \Phi -\frac{1}{2}\omega^2 (x^2 + y^2) \biggr] = - \frac{\partial}{\partial x} \biggl[ C_{z2}(x,y) \biggr] - \frac{\partial}{\partial y} \biggl[C_{z1}(x,y) \biggr] . </math>

In the Riemann ellipsoids, the RHS of this expression is a constant. More importantly, in the Riemann case, it is possible to bring the constants from Step #2 inside the spatial operator on the LHS and establish the following algebraic condition:

<math> H + \Phi -\frac{1}{2}\omega^2 (x^2 + y^2) + f(x,y) = C_\mathrm{Bernoulli} , </math>

where the function <math>f(x,y)</math> contained only quadratic terms in <math>x</math> and <math>y</math>. Specifically,

<math> f(x,y) = x^2 \biggl[\omega\lambda \frac{b}{a}-\frac{\lambda^2}{2} \biggr] + y^2 \biggl[ \omega\lambda \frac{a}{b}-\frac{\lambda^2}{2} \biggr] . </math>

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