User:Tohline/Apps/Korycansky Papaloizou 1996
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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
KP96 begin with the standard set of principal governing equations, but choose to work from the set that is expressed in terms of a rotating frame of reference. (Throughout the entire presentation on this page, it is to be understood that all variables are viewed from rotating frame even though the subscript notation rot does not appear in the equations.) They use, for example, the
Eulerian Representation
of the Continuity Equation
as viewed from a Rotating Reference Frame
<math>\frac{\partial\rho}{\partial t} + \nabla \cdot (\rho {\vec{v}}) = 0</math> ,
and the,
Eulerian Representation
of the Euler Equation
as viewed from a Rotating Reference Frame
<math>\frac{\partial\vec{v}}{\partial t} + ({\vec{v}}\cdot \nabla) {\vec{v}} = - \frac{1}{\rho} \nabla P - \nabla \biggl[\Phi - \frac{1}{2}|{\vec{\Omega}}_f \times \vec{x}|^2 \biggr] - 2{\vec{\Omega}}_f \times {\vec{v}} </math> .
Assumption #1: KP96 align the angular velocity vector of the rotating frame of reference with the z-axis of a Cartesian coordinate system. Specifically, they set
<math>{\vec{\Omega}}_f = \hat{k}\Omega</math>.
Assumption #2:Because KP96 are seeking steady-state solutions, they set all Eulerian time-derivatives to zero.
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,
<math> \frac{dF}{d\Psi} = -g(\Psi) , </math>
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.
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