Korycansky and Papaloizou (1996)

Korycansky & Papaloizou (1996, ApJS, 105, 181; hereafter KP96) developed a method to find nontrivial, nonaxisymmetric steady-state flows in a two-dimensional setting. Specifically, they constructed infinitesimally thin steady-state disk structures in the presence of a time-independent, nonaxisymmetric perturbing potential. While their problem was only two-dimensional and they did not seek a self-consistent solution of the gravitational Poisson equation, the approach they took to solving the 2D Euler equation in tandem with the continuity equation for a compressible fluid may very well be instructive. What follows is a summary of their approach.

Governing Steady-State Equations

As in our above preamble to discussion of Riemann S-Type Ellipsoids, KP96 set <math>{\vec{\omega}} = \hat{k}\omega</math>. Hence, their steady-state Euler equation and steady-state continuity equation become (see their Eq. 1 or their Eq. 7),

<math> (\vec{v}\cdot \nabla)\vec{v} + 2\omega\hat{k}\times\vec{v} + \nabla \biggl[H + \Phi -\frac{1}{2}\omega^2 R^2 \biggr] = 0 , </math>

<math> \nabla\cdot(\rho \vec{v}) = \vec{v}\cdot\nabla\rho + \rho\nabla\cdot\vec{v} = 0 . </math>

Note that the KP96 notation is slightly different from ours:

<math>\Sigma</math> is used in place of <math>\rho</math> to denote a two-dimensional surface density;
<math>\Omega</math> is used instead of <math>\omega</math> to denote the angular frequency of the rotating reference frame;
<math>W</math> is used instead of <math>H</math> to denote the fluid enthalpy; and
<math>\Phi_g</math> represents the combined, time-independent gravitational and centrifugal potential, that is, <math>\Phi_g = (\Phi - \omega^2 R^2/2)</math>.

Using the vector identity,

<math> (\vec{v}\cdot \nabla)\vec{v} = \frac{1}{2}\nabla(v^2) - \vec{v}\times(\nabla\times\vec{v}) , </math>

the above steady-state Euler equation can also be written as,

<math> 2\omega\hat{k}\times\vec{v} - \vec{v}\times(\nabla\times\vec{v}) + \nabla \biggl[\frac{1}{2}v^2 + H + \Phi -\frac{1}{2}\omega^2 R^2 \biggr] = 0 . </math>

Up to this point, no assumptions have been made regarding the behavior of the vector flow-field; we have only chosen to align the <math>\vec{\omega}</math> with the coordinate unit vector, <math>\hat{k}</math>. In particular, these derived forms for the steady-state Euler and continuity equations can serve to describe a fully 3D problem.

Before proceeding further we should emphasize that, in the context of the Euler equation written in this form (i.e., the form preferred by KP96), the vector <math>\vec{A}</math> defined in the preamble, above, should be written,

<math> \vec{A} = 2\omega\hat{k}\times\vec{v} +(\nabla\times\vec{v})\times\vec{v} + \nabla \biggl[\frac{1}{2}v^2 \biggr] . </math>

No Vertical Motions

Now we restrict the flow by setting <math>v_z = 0</math>, that is, from here on we will assume that all the motion is planar. Also, following the lead of KP96, we define the vorticity of the fluid,

<math> \vec{\zeta} \equiv \nabla\times\vec{v} = \hat{i}\zeta_x + \hat{j}\zeta_y + \hat{k}\zeta_z . </math>

[Note that (unfortunately) KP96 use <math>\omega</math> instead of <math>\zeta</math> to represent the rotating-frame vorticity.] In terms of the components of the vorticity vector, the steady-state Euler equation therefore becomes,

<math> (2\omega + \zeta_z)\hat{k}\times\vec{v} + (\hat{i}\zeta_x + \hat{j}\zeta_y)\times\vec{v} + \nabla \biggl[\frac{1}{2}v^2 + H + \Phi -\frac{1}{2}\omega^2 R^2 \biggr] = 0 . </math>

Continuing to follow the lead KP96, we next take the curl of this Euler equation. Because the curl of a gradient is always zero, this leads us to the same condition discussed above — but this time written in terms of the components of the vorticity — namely,

<math> \nabla\times\vec{A} = 0 = \nabla\times [(2\omega + \zeta_z)\hat{k}\times\vec{v} + (\hat{i}\zeta_x + \hat{j}\zeta_y)\times\vec{v}] . </math>

Using another vector identity, namely,

<math> \nabla\times(\vec{C} \times \vec{B}) = (\vec{B}\cdot\nabla)\vec{C} - (\vec{C}\cdot\nabla)\vec{B} + \vec{C}(\nabla\cdot\vec{B}) - \vec{B}(\nabla\cdot\vec{C}), </math>

and remembering that we are assuming <math>v_z = 0</math>, we see in this case that the vector condition <math>\nabla\times\vec{A}=0</math> leads to the following three independent scalar constraints:

<math> ~~~~~\hat{i}:~~~~~ [\nabla\times\vec{A}]_x = - \frac{\partial }{\partial z}\biggl[ (2\omega + \zeta)v_x \biggr] + \frac{\partial}{\partial y} \biggl[ \zeta_x v_y - \zeta_y v_x \biggr] = 0 ; </math>
<math> ~~~~~\hat{j}:~~~~~ [\nabla\times\vec{A}]_y = - \frac{\partial }{\partial z} \biggl[ (2\omega + \zeta)v_y \biggr] - \frac{\partial}{\partial x} \biggl[ \zeta_x v_y - \zeta_y v_x \biggr] = 0 ; </math>
<math> ~~~~~\hat{k}:~~~~~ [\nabla\times\vec{A}]_z = (2\omega + \zeta)\nabla\cdot\vec{v} + \biggl[ v_x \frac{\partial}{\partial x} + v_y \frac{\partial}{\partial y} \biggr](2\omega + \zeta) = 0 . </math>

With the understanding that, by definition,

<math> \zeta_x \equiv - \frac{\partial v_y}{\partial z} , ~~~~~ \zeta_y \equiv + \frac{\partial v_x}{\partial z} , ~~~~~ \mathrm{and} ~~~~~ \zeta_z \equiv \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} , </math>

it can be shown that these three constraints are identical to the ones presented in the preamble, above.

Solution Strategy

Constraint #1: For their two-dimensional disk problem, KP96 focused on the constraint provided by the z-component of the curl of the Euler equation, which can be rewritten as (see above derivation, or Eq. 2 of KP96),

<math> \nabla\cdot\vec{v} =-\vec{v} \cdot \biggl[ \frac{\nabla(2\omega + \zeta_z)}{(2\omega + \zeta_z)} \biggr] = -\vec{v} \cdot \nabla[\ln(2\omega + \zeta_z)]. </math>

Constraint #2: But from the continuity equation they also know that,

<math> \nabla\cdot\vec{v} = -\vec{v}\cdot\biggl[\frac{\nabla\rho}{\rho} \biggr] = -\vec{v} \cdot \nabla[\ln\rho] . </math>

Hence,

<math> \vec{v} \cdot \nabla[\ln(2\omega + \zeta_z)] = \vec{v} \cdot \nabla[\ln\rho] , </math>

that is,

<math> \vec{v} \cdot \nabla\ln\biggl[ \frac{(2\omega + \zeta_z)}{\rho} \biggr] = 0 . </math>

This is essentially KP96's Eq. (3).

Introduce stream function: The constraint implied by the continuity equation also suggests that it might be useful to define a stream function in terms of the momentum density — instead of in terms of just the velocity, which is the natural treatment in the context of incompressible fluid flows. KP96 do this. They define the stream function, <math>\Psi</math>, such that (see their Eq. 4),

<math> \rho\vec{v} = \nabla\times(\hat{k}\Psi) . </math>

in which case,

<math> v_x = \frac{1}{\rho} \frac{\partial \Psi}{\partial y} ~~~~~\mathrm{and}~~~~~ v_y = - \frac{1}{\rho} \frac{\partial \Psi}{\partial x} . </math>

This implies as well that the z-component of the fluid vorticity can be expressed in terms of the stream function as follows (see Eq. 5 of KP96):

<math> \zeta_z = - \nabla\cdot \biggl( \frac{\nabla\Psi}{\rho} \biggr) = - \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>

According to KP96, this expression, taken in combination with the conclusion drawn above from the second constraint — that is, Eq. (3) taken in combination with Eq. (4) from KP96 — "tell us that the 'vortensity' <math>(\zeta_z + 2\omega)/\rho</math> is constant along streamlines which are lines of constant <math>\Psi</math>." The vortensity is therefore a function of <math>\Psi</math> alone, so we can write,

<math> \frac{\zeta_z + 2\omega}{\rho} = g(\Psi) . </math>

Constraint #3:

Taking the scalar product of <math>\vec{v}</math> and the following form of the steady-state Euler equation,

<math> 2\omega\hat{k}\times\vec{v} - \vec{v}\times(\nabla\times\vec{v}) + \nabla \biggl[\frac{1}{2}v^2 + H + \Phi -\frac{1}{2}\omega^2 R^2 \biggr] = 0 , </math>

we obtain the constraint,

<math> \vec{v}\cdot\nabla \biggl[\frac{1}{2}v^2 + H + \Phi -\frac{1}{2}\omega^2 R^2 \biggr] = 0 . </math>

When tied with our earlier discussion, this means that the Bernoulli function also must be constant along streamlines. Hence, we can write,

<math> \frac{1}{2}v^2 + H + \Phi -\frac{1}{2}\omega^2 R^2 = F(\Psi) . </math>

KP96 then go on to demonstrate that the relationship between the functions <math>g(\Psi)</math> and <math>F(\Psi)</math> is,

which allows the determination of <math>F</math> up to a constant of integration.

Summary

In summary, KP96 constrain their flow as follows:

They use the z-component of the curl of the Euler equation;
They use the compressible version of the continuity equation;
Instead of taking the divergence of the Euler equation to obtain a Poisson-like equation, they obtain an algebraic constraint on the Bernoulli function (as in our traditional SCF technique) by simply "dotting" <math>\vec{v}</math> into the Euler equation.

User:Tohline/Apps/Korycansky Papaloizou 1996

Contents

Korycansky and Papaloizou (1996)

Governing Steady-State Equations

No Vertical Motions

Solution Strategy

Summary

Navigation menu

Search