Difference between revisions of "User:Tohline/PGE/RotatingFrame"

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<font color="red"><b>
NOTE to Eric Hirschmann &amp; David Neilsen... 
</b></font>
I have move the earlier contents of this page to a new Wiki location called
[[User:Tohline/Apps/RiemannEllipsoids_Compressible|Compressible Riemann Ellipsoids]].
=Rotating Reference Frame=
{{LSU_HBook_header}}
{{LSU_HBook_header}}
 
At times, it can be useful to view the motion of a fluid from a frame of reference that is rotating with a uniform (''i.e.,'' time-independent) angular velocity <math>~\Omega_f</math>In order to transform any one of the [[User:Tohline/PGE#Principal_Governing_Equations|principal governing equations]] from the inertial reference frame to such a rotating reference frame, we must specify the orientation as well as the magnitude of the angular velocity vector about which the frame is spinning, <math>{\vec\Omega}_f</math>; and the <math>~d/dt</math> operator, which denotes Lagrangian time-differentiation in the inertial frame, must everywhere be replaced as follows:
=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) ellipsoidsWe will build upon the recent work of [http://adsabs.harvard.edu/abs/2006ApJ...639..549O 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:
<div align="center">
<div align="center">
<math>
<math>
\nabla\cdot(\rho \vec{v}) = 0
\biggl[\frac{d}{dt} \biggr]_{inertial} \rightarrow \biggl[\frac{d}{dt} \biggr]_{rot} + {\vec{\Omega}}_f \times .
</math>
</math>
 
</div>
Performing this transformation implies, for example, that
<div align="center">
<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}
\vec{v}_{inertial} = \vec{v}_{rot} + {\vec{\Omega}}_f \times \vec{x} ,
</math>
 
<math>
\nabla^2 \Phi = 4\pi G \rho
</math>
</math>
</div>
</div>
 
and,
==Proposed Solution Strategy==
===Preamble:===
Specify the three "polar" boundary locations, <math>a, b,</math> and <math>c</math>; specify the <i>direction</i> 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:
<div align="center">
<div align="center">
<math>
<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}} ;
\biggl[ \frac{d\vec{v}}{dt}\biggr]_{inertial} = \biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} + 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} + {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})
</math>
</math>


<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} .
= \biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} + 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} - \frac{1}{2} \nabla \biggl[ |{\vec{\Omega}}_f \times \vec{x}|^2 \biggr]
</math>
</math>
</div>
</div>
From here, on, spatial operators are assumed to be in terms of the dimensionless coordinates.
(If we were to allow <math>{\vec\Omega}_f</math> to be a function of time, an additional term involving the time-derivative of <math>{\vec\Omega}_f</math> also would appear on the right-hand-side of these last expressions; see, for example, Eq.~1D-42 in [[User:Tohline/Appendix/References|BT87]].) Note as well that the relationship between the fluid [[User:Tohline/PGE/RotatingFrame#WikiVorticity|vorticity]] in the two frames is,
 
===Step #1:===
Guess a 3D density distribution &#8212; such as a uniform-density ellipsoid, or one of the converged models from Ou (2006) &#8212; that conforms to a selected set of <i>positional</i> 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:
<div align="center">
<div align="center">
<math>
<math>
\nabla^2 \Phi^* = 4\pi \rho^* .
[\vec\zeta]_{inertial} = [\vec\zeta]_{rot} + 2{\vec\Omega}_f .
</math>
</math>
</div>
</div>


===Step #2:===
 
Use the continuity equation and the curl of the Euler equation to numerically derive the <i>structure</i> 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:
==Continuity Equation (rotating frame)==
Applying these transformations to the standard, inertial-frame representations of the continuity equation presented [[User:Tohline/PGE/ConservingMass#Continuity_Equation|elsewhere]], we obtain the:
 
<div align="center">
<div align="center">
<math>
<font color="#770000">'''Lagrangian Representation'''</font><br />
\nabla\cdot(\rho^* {\vec{v}}^*) = 0 ;
of the Continuity Equation <br />
</math>
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>


<math>
<math>\biggl[ \frac{d\rho}{dt} \biggr]_{rot} + \rho \nabla \cdot {\vec{v}}_{rot} = 0</math> ;
\nabla\times \biggl[({\vec{v}}^*\cdot \nabla){\vec{v}}^* +2 {\vec{\omega}}^* \times {\vec{v}}^* \biggr] = 0 .
</math>
</div>
</div>


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:===
<div align="center">
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> &#8212; hopefully this is analogous to solving for the vorticity parameter <math>\lambda</math> in Ou (2006) &#8212; and the Bernoulli constant (or something equivalent).  The relevant "Poisson"-like equation is:
<font color="#770000">'''Eulerian Representation'''</font><br />
of the Continuity Equation <br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>


<div align="center">
<math>\biggl[ \frac{\partial\rho}{\partial t} \biggr]_{rot} + \nabla \cdot (\rho {\vec{v}}_{rot}) = 0</math> .
<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>
</div>
</div>


==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,
==Euler Equation (rotating frame)==
Applying these transformations to the standard, inertial-frame representations of the Euler equation presented [[User:Tohline/PGE/Euler#Euler_Equation|elsewhere]], we obtain the:
 
<div align="center">
<div align="center">
<math>
<font color="#770000">'''Lagrangian Representation'''</font><br />
\vec{A} \equiv ({\vec{v}}\cdot \nabla){\vec{v}} +2 {\vec{\omega}} \times {\vec{v}} .
of the Euler Equation <br />
</math>
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>
</div>
The components are:
<div align="left">
<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><br />
<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><br />
<math>
~~~~~\hat{k}:~~~~~A_z = 0 .
</math>
</div>
The curl of <math>\vec{A}</math> (needed in Step #2, above) produces a vector with the following three Cartesian components:
<div align="left">
<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><br />
<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><br />
<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>
</div>
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{k}:~~~~~ \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>.


 
<math>\biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} = - \frac{1}{\rho} \nabla P - \nabla \Phi - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})</math> ;
Generically, the continuity equation demands,
<div align="center">
<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>
</div>
</div>




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>
<font color="#770000">'''Eulerian Representation'''</font><br />
\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]
of the Euler Equation <br />
</math><br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>
<math>
= \frac{\partial}{\partial x} \biggl[ C_{z2}(x,y) \biggr] + \frac{\partial}{\partial y} \biggl[C_{z1}(x,y)  \biggr]  .
</math>
</div>


===Riemann Flow-Field===
<math>\biggl[\frac{\partial\vec{v}}{\partial t}\biggr]_{rot} + ({\vec{v}}_{rot}\cdot \nabla) {\vec{v}}_{rot}= - \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}}_{rot} </math> ;
In Riemann S-Type ellipsoids, the adopted planar flow-field as viewed from the rotating reference frame is,
<div align="center">
<math>
\vec{v} = \hat{i} \biggl( \frac{\lambda a y}{b} \biggr) + \hat{j} \biggl( - \frac{\lambda b x}{a} \biggr) .
</math>
</div>
Hence,
<div align="center">
<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><br />
<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>
</div>
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,
<div align="center">
<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>
</div>
However, the right-hand-side of our Poisson-like equation is not zero; rather, it is,
<div align="center">
<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>
</div>
</div>


===Summary===
What can we learn from the Riemann S-Type ellipsoids?  Well, let's assume that the <i>structure</i> 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 <i>planar</i>, 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:


====<font color="darkblue">Step #1 (simplified):</font>====
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:
<div align="center">
<div align="center">
<math>
Euler Equation<br />
\nabla^2 \Phi = 4\pi G \rho .
written <font color="#770000">'''in terms of the Vorticity'''</font> and<br />
</math>
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>
</div>


====<font color="darkblue">Step #2 (simplified):</font>====
<math>\biggl[\frac{\partial\vec{v}}{\partial t}\biggr]_{rot} + ({\vec{\zeta}}_{rot}+2{\vec\Omega}_f) \times {\vec{v}}_{rot}= - \frac{1}{\rho} \nabla P - \nabla \biggl[\Phi + \frac{1}{2}v_{rot}^2 - \frac{1}{2}|{\vec{\Omega}}_f \times \vec{x}|^2 \biggr]</math> .
Use the continuity equation and the curl of the Euler equation to numerically derive the <i>structure</i> 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:
<div align="center">
<math>
\frac{\partial}{\partial x}\biggl[ \rho v_x \biggr] + \frac{\partial}{\partial y}\biggl[ \rho v_y \biggr] = 0 .
</math>
</div>
<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{k}:~~~~~ \frac{\partial}{\partial x} \biggl[ C_{z1}(x,y) \biggr] = \frac{\partial}{\partial y} \biggl[C_{z2}(x,y)  \biggr] ,
</math>
</div>
</div>


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> &#8212; 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.


====<font color="darkblue">Step #3 (simplified):</font>====
==Centrifugal and Coriolis Accelerations==
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:


Following along the lines of the discussion presented in Appendix 1.D, &sect;3 of [[User:Tohline/Appendix/References|BT87]], in a rotating reference frame the Lagrangian representation of the Euler equation may be written in the form,
<div align="center">
<div align="center">
<math>
<math>\biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} = - \frac{1}{\rho} \nabla P - \nabla \Phi + {\vec{a}}_{fict} </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>
</div>
</div>
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 <i>algebraic</i> condition:
where,
<div align="center">
<div align="center">
<math>
<math>
H + \Phi -\frac{1}{2}\omega^2 (x^2 + y^2) + f(x,y) = C_\mathrm{Bernoulli} ,
{\vec{a}}_{fict} \equiv - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) .
</math>
</math>
</div>
</div>
where the function <math>f(x,y)</math> contains only quadratic terms in <math>x</math> and <math>y</math>. Specifically [see Eq. (6) of Ou (2006)],
So, as viewed from a rotating frame of reference, material moves as if it were subject to two ''fictitious accelerations'' which traditionally are referred to as the,
<div align="center">
<div align="center">
<font color="#770000">'''Coriolis Acceleration'''</font>
<math>
<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] .
{\vec{a}}_{Coriolis} \equiv - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} ,
</math>
</math>
</div>
</div>
This makes sense because in the Riemann case the equipotential surfaces were perfect ellipsoids, just like the shape of the centrifugal potential term.  So the remaining "source" terms could adopt the same "shape".  But in our more general case, surfaces of constant <math>\Phi</math> will have more complicated shapes, so it will be necessary for the function <math>f(x,y)</math> (or equivalent) to incorporate higher order terms in <math>x</math> and <math>y</math>.  Is it possible that equilibrium configurations can be found by including additional terms just in <math>x^4</math> and <math>y^4</math>?


 
(see the related [[User:Tohline/PGE/RotatingFrame#WikiCoriolis|Wikipedia discussion]]) and the
==A Related Two-dimensional Treatment==
===Preamble===
In a related work, [http://adsabs.harvard.edu/abs/1996ApJS..105..181K Korycansky &amp; Papaloizou] (1996, ApJS, 105, 181; hereafter KP96) developed a method to find nontrivial, nonaxisymmetric steady-state flows in a <i>two-dimensional</i> 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 <i>compressible</i> fluid may very well be instructive.  What follows is a summary of their approach.
 
===Governing Steady-State Equations===
 
As in our above [http://www.vistrails.org/index.php/User:Tohline/PGE/RotatingFrame#Example_of_Riemann_S-Type_Ellipsoids 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),
<div align="center">
<div align="center">
<math>
<font color="#770000">'''Centrifugal Acceleration'''</font>  
(\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>
<math>
\nabla\cdot(\rho \vec{v}) = \vec{v}\cdot\nabla\rho + \rho\nabla\cdot\vec{v} = 0 .
{\vec{a}}_{Centrifugal} \equiv - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})
= \frac{1}{2} \nabla\biggl[ |{\vec{\Omega}}_f \times \vec{x}|^2  \biggr]
</math>
</math>
</div>
</div>
Note that the KP96 notation is slightly different from ours:
(see the related [[User:Tohline/PGE/RotatingFrame#WikiCentrifugal|Wikipedia discussion]]).
* <math>\Sigma</math> is used in place of <math>\rho</math> to denote a two-dimensional <i>surface</i> 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 <font color="darkgreen">vector identity</font>,
==Nonlinear Velocity Cross-Product==
<div align="center">
<math>
(\vec{v}\cdot \nabla)\vec{v} = \frac{1}{2}\nabla(v^2) - \vec{v}\times(\nabla\times\vec{v}) ,
</math>
</div>
the above steady-state Euler equation can also be written as,
<div align="center">
<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>
</div>
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,
In some contexts &#8212; for example, our discussion of [[User:Tohline/Apps/RiemannEllipsoids_Compressible|Riemann ellipsoids]] or the analysis by [[User:Tohline/Apps/Korycansky_Papaloizou_1996|Korycansky &amp; Papaloizou (1996)]] of nonaxisymmetric disk structures &#8212; it proves useful to isolate and analyze the term in the "vorticity formulation" of the Euler equation that involves a nonlinear cross-product of the rotating-frame velocity vector, namely,
<div align="center">
<div align="center">
<math>
<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] .
\vec{A} \equiv ({\vec{\zeta}}_{rot}+2{\vec\Omega}_f) \times {\vec{v}}_{rot} .
</math>
</math>
</div>
</div>


NOTE: To simplify notation, for most of the remainder of this subsection we will drop the subscript "rot" on both the velocity and vorticity vectors.


===No Vertical Motions===
===Align <math>{\vec\Omega}_f</math> with z-axis===
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,
Without loss of generality we can set <math>{\vec\Omega}_f = \hat{k}\Omega_f</math>, that is, we can align the frame rotation axis with the z-axis of a Cartesian coordinate system.    The Cartesian components of <math>{\vec{A}}</math> are then,
<div align="center">
<div align="left">
<math>
<math>
\vec{\zeta} \equiv \nabla\times\vec{v} = \hat{i}\zeta_x + \hat{j}\zeta_y + \hat{k}\zeta_z .
\hat{i}: ~~~~~~ A_x = \zeta_y v_z - (\zeta_z + 2\Omega) v_y  ,
</math>
</math><br />
</div>  
 
[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,
<div align="center">
<math>
<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 .
\hat{j}: ~~~~~~ A_y = (\zeta_z + 2\Omega) v_x - \zeta_x v_z ,
</math>
</math><br />
</div>
 
Continuing to follow the lead KP96, we next <font color="red">take the curl of this Euler equation</font>.  Because the curl of a gradient is always zero, this leads us to the same condition discussed above &#8212; but this time written in terms of the components of the vorticity &#8212; namely,
<div align="center">
<math>
<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}] .
\hat{k}: ~~~~~~ A_z = \zeta_x v_y - \zeta_y v_x ,
</math>
</math>
</div>
</div>
Using another <font color="darkgreen">vector identity</font>, namely,
where it is understood that the three Cartesian components of the vorticity vector are,
<div align="center">
<div align="center">
<math>
<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}),
\zeta_x = \biggl[\frac{\partial v_z}{\partial y} - \frac{\partial v_y}{\partial z} \biggr] ,
  ~~~~~~ \zeta_y = \biggl[ \frac{\partial v_x}{\partial z} - \frac{\partial v_z}{\partial x} \biggr] ,
~~~~~~ \zeta_z = \biggl[ \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} \biggr] .
</math>
</math>
</div>
</div>
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:
In turn, the curl of <math>\vec{A}</math> has the following three Cartesian components:  
<div align="left">
<div align="left">
<math>
<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  ;
\hat{i}: ~~~~~~ [\nabla\times\vec{A}]_x = \frac{\partial}{\partial y}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] - \frac{\partial}{\partial z}\biggl[ (\zeta_z + 2\Omega) v_x - \zeta_x v_z \biggr],
</math><br />
</math><br />
<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 ;
\hat{j}: ~~~~~~ [\nabla\times\vec{A}]_y = \frac{\partial}{\partial z}\biggl[ \zeta_y v_z - (\zeta_z + 2\Omega) v_y \biggr] - \frac{\partial}{\partial x}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] ,
</math><br />
</math><br />
<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 .
\hat{k}: ~~~~~~ [\nabla\times\vec{A}]_z = \frac{\partial}{\partial x}\biggl[ (\zeta_z + 2\Omega) v_x - \zeta_x v_z \biggr] - \frac{\partial}{\partial y}\biggl[ \zeta_y v_z - (\zeta_z + 2\Omega) v_y \biggr] .
</math>
</math>
</div>
</div>
With the understanding that, by definition,
<div align="center">
<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>
</div>
it can be shown that these three constraints are identical to the ones presented in the preamble, above.


===Solution Strategy===
===When <math>v_z = 0</math>===


<font color="darkblue"><b>Constraint #1:</b></font>
If we restrict our discussion to configurations that exhibit only planar flows &#8212; that is, systems in which <math>v_z = 0</math> &#8212; then the Cartesian components of <math>{\vec{A}}</math> and <math>\nabla\times\vec{A}</math> simplify somewhat to give, respectively,
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),
<div align="left">
<div align="center">
<math>
<math>
\nabla\cdot\vec{v} =-\vec{v} \cdot \biggl[ \frac{\nabla(2\omega + \zeta_z)}{(2\omega + \zeta_z)} \biggr]
\hat{i}: ~~~~~~ A_x = - (\zeta_z + 2\Omega) v_y  ,
= -\vec{v} \cdot \nabla[\ln(2\omega + \zeta_z)].
</math><br />
</math>
 
</div>
<font color="darkblue"><b>Constraint #2:</b></font>
But from the continuity equation they also know that,
<div align="center">
<math>
<math>
\nabla\cdot\vec{v} = -\vec{v}\cdot\biggl[\frac{\nabla\rho}{\rho} \biggr] = -\vec{v} \cdot \nabla[\ln\rho] .
\hat{j}: ~~~~~~ A_y = (\zeta_z + 2\Omega) v_x  ,
</math>
</math><br />
</div>
 
Hence,
<div align="center">
<math>
<math>
\vec{v} \cdot \nabla[\ln(2\omega + \zeta_z)] = \vec{v} \cdot \nabla[\ln\rho] ,
\hat{k}: ~~~~~~ A_z = \zeta_x v_y - \zeta_y v_x ,
</math>
</math>
</div>
</div>
that is,
 
<div align="center">
and,
<div align="left">
<math>
<math>
\vec{v} \cdot \nabla\ln\biggl[ \frac{(2\omega + \zeta_z)}{\rho} \biggr] = 0  .
\hat{i}: ~~~~~~ [\nabla\times\vec{A}]_x = \frac{\partial}{\partial y}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] - \frac{\partial}{\partial z}\biggl[ (\zeta_z + 2\Omega) v_x \biggr],
</math>
</math><br />
</div>
This is essentially KP96's Eq. (3).


<font color="darkblue"><b>Introduce stream function:</b></font>
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 &#8212; 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),
<div align="center">
<math>
<math>
\rho\vec{v} = \nabla\times(\hat{k}\Psi)  .
\hat{j}: ~~~~~~ [\nabla\times\vec{A}]_y = - \frac{\partial}{\partial z}\biggl[(\zeta_z + 2\Omega) v_y \biggr] - \frac{\partial}{\partial x}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] ,
</math>
</math><br />
</div>
 
in which case,
<div align="center">
<math>
<math>
v_x = \frac{1}{\rho} \frac{\partial \Psi}{\partial y} ~~~~~\mathrm{and}~~~~~ 
\hat{k}: ~~~~~~ [\nabla\times\vec{A}]_z = \frac{\partial}{\partial x}\biggl[ (\zeta_z + 2\Omega) v_x  \biggr] + \frac{\partial}{\partial y}\biggl[ (\zeta_z + 2\Omega) v_y \biggr] ,
v_y = - \frac{1}{\rho} \frac{\partial \Psi}{\partial x} .
</math>
</math>
</div>
</div>
This implies as well that the z-component of the fluid vorticity can be expressed in terms
where, in this case, the three Cartesian components of the vorticity vector are,
of the stream function as follows (see Eq. 5 of KP96):
<div align="center">
<div align="center">
<math>
<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].
\zeta_x = - \frac{\partial v_y}{\partial z} ,
~~~~~~ \zeta_y = \frac{\partial v_x}{\partial z} ,
~~~~~~ \zeta_z = \biggl[ \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} \biggr] .
</math>
</math>
</div>
</div>
Line 364: Line 213:




<font color="darkblue"><b>Constraint #3:</b></font>
=Related Discussions=
 
* <span id="WikiVorticity">Wikipedia discussion of [http://en.wikipedia.org/wiki/Vorticity vorticity].</span>
* <span id="WikiCoriolis">Wikipedia discussion of [http://en.wikipedia.org/wiki/Coriolis_effect#Formula Coriolis Effect].</span>
* <span id="WikiCentrifugal">Wikipedia discussion of [http://en.wikipedia.org/wiki/Centrifugal_force#Derivation_using_vectors Centrifugal acceleration].</span>


===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.




{{LSU_HBook_footer}}
{{LSU_HBook_footer}}

Latest revision as of 17:02, 11 July 2015

NOTE to Eric Hirschmann & David Neilsen... I have move the earlier contents of this page to a new Wiki location called Compressible Riemann Ellipsoids.

Rotating Reference Frame

Whitworth's (1981) Isothermal Free-Energy Surface
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At times, it can be useful to view the motion of a fluid from a frame of reference that is rotating with a uniform (i.e., time-independent) angular velocity <math>~\Omega_f</math>. In order to transform any one of the principal governing equations from the inertial reference frame to such a rotating reference frame, we must specify the orientation as well as the magnitude of the angular velocity vector about which the frame is spinning, <math>{\vec\Omega}_f</math>; and the <math>~d/dt</math> operator, which denotes Lagrangian time-differentiation in the inertial frame, must everywhere be replaced as follows:

<math> \biggl[\frac{d}{dt} \biggr]_{inertial} \rightarrow \biggl[\frac{d}{dt} \biggr]_{rot} + {\vec{\Omega}}_f \times . </math>

Performing this transformation implies, for example, that

<math> \vec{v}_{inertial} = \vec{v}_{rot} + {\vec{\Omega}}_f \times \vec{x} , </math>

and,

<math> \biggl[ \frac{d\vec{v}}{dt}\biggr]_{inertial} = \biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} + 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} + {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) </math>

<math> = \biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} + 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} - \frac{1}{2} \nabla \biggl[ |{\vec{\Omega}}_f \times \vec{x}|^2 \biggr] </math>

(If we were to allow <math>{\vec\Omega}_f</math> to be a function of time, an additional term involving the time-derivative of <math>{\vec\Omega}_f</math> also would appear on the right-hand-side of these last expressions; see, for example, Eq.~1D-42 in BT87.) Note as well that the relationship between the fluid vorticity in the two frames is,

<math> [\vec\zeta]_{inertial} = [\vec\zeta]_{rot} + 2{\vec\Omega}_f . </math>


Continuity Equation (rotating frame)

Applying these transformations to the standard, inertial-frame representations of the continuity equation presented elsewhere, we obtain the:

Lagrangian Representation
of the Continuity Equation
as viewed from a Rotating Reference Frame

<math>\biggl[ \frac{d\rho}{dt} \biggr]_{rot} + \rho \nabla \cdot {\vec{v}}_{rot} = 0</math> ;


Eulerian Representation
of the Continuity Equation
as viewed from a Rotating Reference Frame

<math>\biggl[ \frac{\partial\rho}{\partial t} \biggr]_{rot} + \nabla \cdot (\rho {\vec{v}}_{rot}) = 0</math> .


Euler Equation (rotating frame)

Applying these transformations to the standard, inertial-frame representations of the Euler equation presented elsewhere, we obtain the:

Lagrangian Representation
of the Euler Equation
as viewed from a Rotating Reference Frame

<math>\biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} = - \frac{1}{\rho} \nabla P - \nabla \Phi - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})</math> ;


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 \biggl[\Phi - \frac{1}{2}|{\vec{\Omega}}_f \times \vec{x}|^2 \biggr] - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} </math> ;


Euler Equation
written in terms of the Vorticity and
as viewed from a Rotating Reference Frame

<math>\biggl[\frac{\partial\vec{v}}{\partial t}\biggr]_{rot} + ({\vec{\zeta}}_{rot}+2{\vec\Omega}_f) \times {\vec{v}}_{rot}= - \frac{1}{\rho} \nabla P - \nabla \biggl[\Phi + \frac{1}{2}v_{rot}^2 - \frac{1}{2}|{\vec{\Omega}}_f \times \vec{x}|^2 \biggr]</math> .


Centrifugal and Coriolis Accelerations

Following along the lines of the discussion presented in Appendix 1.D, §3 of BT87, in a rotating reference frame the Lagrangian representation of the Euler equation may be written in the form,

<math>\biggl[ \frac{d\vec{v}}{dt}\biggr]_{rot} = - \frac{1}{\rho} \nabla P - \nabla \Phi + {\vec{a}}_{fict} </math>,

where,

<math> {\vec{a}}_{fict} \equiv - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) . </math>

So, as viewed from a rotating frame of reference, material moves as if it were subject to two fictitious accelerations which traditionally are referred to as the,

Coriolis Acceleration

<math> {\vec{a}}_{Coriolis} \equiv - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} , </math>

(see the related Wikipedia discussion) and the

Centrifugal Acceleration

<math> {\vec{a}}_{Centrifugal} \equiv - {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) = \frac{1}{2} \nabla\biggl[ |{\vec{\Omega}}_f \times \vec{x}|^2 \biggr] </math>

(see the related Wikipedia discussion).

Nonlinear Velocity Cross-Product

In some contexts — for example, our discussion of Riemann ellipsoids or the analysis by Korycansky & Papaloizou (1996) of nonaxisymmetric disk structures — it proves useful to isolate and analyze the term in the "vorticity formulation" of the Euler equation that involves a nonlinear cross-product of the rotating-frame velocity vector, namely,

<math> \vec{A} \equiv ({\vec{\zeta}}_{rot}+2{\vec\Omega}_f) \times {\vec{v}}_{rot} . </math>

NOTE: To simplify notation, for most of the remainder of this subsection we will drop the subscript "rot" on both the velocity and vorticity vectors.

Align <math>{\vec\Omega}_f</math> with z-axis

Without loss of generality we can set <math>{\vec\Omega}_f = \hat{k}\Omega_f</math>, that is, we can align the frame rotation axis with the z-axis of a Cartesian coordinate system. The Cartesian components of <math>{\vec{A}}</math> are then,

<math> \hat{i}: ~~~~~~ A_x = \zeta_y v_z - (\zeta_z + 2\Omega) v_y , </math>

<math> \hat{j}: ~~~~~~ A_y = (\zeta_z + 2\Omega) v_x - \zeta_x v_z , </math>

<math> \hat{k}: ~~~~~~ A_z = \zeta_x v_y - \zeta_y v_x , </math>

where it is understood that the three Cartesian components of the vorticity vector are,

<math> \zeta_x = \biggl[\frac{\partial v_z}{\partial y} - \frac{\partial v_y}{\partial z} \biggr] ,

~~~~~~ \zeta_y = \biggl[ \frac{\partial v_x}{\partial z} - \frac{\partial v_z}{\partial x} \biggr] ,
~~~~~~ \zeta_z = \biggl[ \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} \biggr] .

</math>

In turn, the curl of <math>\vec{A}</math> has the following three Cartesian components:

<math> \hat{i}: ~~~~~~ [\nabla\times\vec{A}]_x = \frac{\partial}{\partial y}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] - \frac{\partial}{\partial z}\biggl[ (\zeta_z + 2\Omega) v_x - \zeta_x v_z \biggr], </math>

<math> \hat{j}: ~~~~~~ [\nabla\times\vec{A}]_y = \frac{\partial}{\partial z}\biggl[ \zeta_y v_z - (\zeta_z + 2\Omega) v_y \biggr] - \frac{\partial}{\partial x}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] , </math>

<math> \hat{k}: ~~~~~~ [\nabla\times\vec{A}]_z = \frac{\partial}{\partial x}\biggl[ (\zeta_z + 2\Omega) v_x - \zeta_x v_z \biggr] - \frac{\partial}{\partial y}\biggl[ \zeta_y v_z - (\zeta_z + 2\Omega) v_y \biggr] . </math>

When <math>v_z = 0</math>

If we restrict our discussion to configurations that exhibit only planar flows — that is, systems in which <math>v_z = 0</math> — then the Cartesian components of <math>{\vec{A}}</math> and <math>\nabla\times\vec{A}</math> simplify somewhat to give, respectively,

<math> \hat{i}: ~~~~~~ A_x = - (\zeta_z + 2\Omega) v_y , </math>

<math> \hat{j}: ~~~~~~ A_y = (\zeta_z + 2\Omega) v_x , </math>

<math> \hat{k}: ~~~~~~ A_z = \zeta_x v_y - \zeta_y v_x , </math>

and,

<math> \hat{i}: ~~~~~~ [\nabla\times\vec{A}]_x = \frac{\partial}{\partial y}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] - \frac{\partial}{\partial z}\biggl[ (\zeta_z + 2\Omega) v_x \biggr], </math>

<math> \hat{j}: ~~~~~~ [\nabla\times\vec{A}]_y = - \frac{\partial}{\partial z}\biggl[(\zeta_z + 2\Omega) v_y \biggr] - \frac{\partial}{\partial x}\biggl[ \zeta_x v_y - \zeta_y v_x \biggr] , </math>

<math> \hat{k}: ~~~~~~ [\nabla\times\vec{A}]_z = \frac{\partial}{\partial x}\biggl[ (\zeta_z + 2\Omega) v_x \biggr] + \frac{\partial}{\partial y}\biggl[ (\zeta_z + 2\Omega) v_y \biggr] , </math>

where, in this case, the three Cartesian components of the vorticity vector are,

<math> \zeta_x = - \frac{\partial v_y}{\partial z} ,

~~~~~~ \zeta_y = \frac{\partial v_x}{\partial z} ,
~~~~~~ \zeta_z = \biggl[ \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} \biggr] .

</math>


Related Discussions


Whitworth's (1981) Isothermal Free-Energy Surface

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