Appendix/Ramblings/HybridSchemeImplications

Implications of Hybrid Scheme

Background

Key H_Book Chapters

[Ref01]   Inertial-Frame Euler Equation

[Ref02]   Traditional Description of Rotating Reference Frame

[Ref03]   Hybrid Advection Scheme

[Ref04]   Riemann S-type Ellipsoids

[Ref05]   Korycansky and Papaloizou (1996)

Principal Governing Equations

Quoting from [Ref01] … Among the principal governing equations we have included the inertial-frame,

Lagrangian Representation
of the Euler Equation,

$\frac{d\overrightarrow{v}}{dt}=-\frac{1}{\rho }\mathrm{\nabla }P-\mathrm{\nabla }\mathrm{\Phi }$

[EFE], Chap. 2, §11, p. 20, Eq. (38)
[BLRY07], p. 13, Eq. (1.55)

Shifting into a rotating frame characterized by the angular velocity vector,

${\overrightarrow{\mathrm{\Omega }}}_f\equiv \widehat{\mathbf{𝐤}}{\mathrm{\Omega }}_f,$

and applying the operations that are specified in the first few subsections of [Ref02], we recognize the following relationships …

${\overrightarrow{v}}_{\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}$	$=$	${\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}+{\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x},$
$[\frac{d\overrightarrow{v}}{dt}{]}_{\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}$	$=$	$[\frac{d\overrightarrow{v}}{dt}{]}_{\mathrm{r}\mathrm{o}\mathrm{t}}+2{\overrightarrow{\mathrm{\Omega }}}_f\times {\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}+{\overrightarrow{\mathrm{\Omega }}}_f\times ({\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x})$
	$=$	$[\frac{d\overrightarrow{v}}{dt}{]}_{\mathrm{r}\mathrm{o}\mathrm{t}}+2{\overrightarrow{\mathrm{\Omega }}}_f\times {\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}-\frac{1}{2}\mathrm{\nabla }|{\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x}{|}^2$
	$=$	$[\frac{\partial \overrightarrow{v}}{\partial t}{]}_{\mathrm{r}\mathrm{o}\mathrm{t}}+({\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}\cdot \mathrm{\nabla }){\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}+2{\overrightarrow{\mathrm{\Omega }}}_f\times {\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}-\frac{1}{2}\mathrm{\nabla }|{\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x}{|}^2.$

Making this substitution on the left-hand-side of the above-specified "Lagrangian Representation of the Euler Equation," we obtain what we have referred to also in [Ref02] as the,

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

$[\frac{\partial \overrightarrow{v}}{\partial t}{]}_{\mathrm{r}\mathrm{o}\mathrm{t}}+({\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}\cdot \mathrm{\nabla }){\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}=-\frac{1}{\rho }\mathrm{\nabla }P-\mathrm{\nabla }[\mathrm{\Phi }-\frac{1}{2}|{\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x}{|}^2]-2{\overrightarrow{\mathrm{\Omega }}}_f\times {\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}.$

This form of the Euler equation also appears early in [Ref05], where we set up a discussion of the paper by Korycansky & Papaloizou (1996, ApJS, 105, 181; hereafter KP96). But, for now, let's back up a couple of steps and retain the total time derivative on the left-hand-side. That is, let's select as the foundation expression the,

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

$[\frac{d\overrightarrow{v}}{dt}{]}_{\mathrm{r}\mathrm{o}\mathrm{t}}$	$=$	$-\frac{1}{\rho }\mathrm{\nabla }P-\mathrm{\nabla }\mathrm{\Phi }-2{\overrightarrow{\mathrm{\Omega }}}_f\times {\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}-{\overrightarrow{\mathrm{\Omega }}}_f\times ({\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x}),$
[EFE], Chap. 2, §12, p. 25, Eq. (62)

which also serves as the foundation of most of our [Ref03] discussions.

Exercising the Hybrid Scheme

Focus on Tracking Angular Momentum

Let's begin by using ${\mathbf{𝐮}}^{\prime }$, instead of ${\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}$, to represent the fluid velocity vector as viewed from the rotating frame of reference. Our foundation expression becomes,

$\frac{d{\mathbf{𝐮}}^{\prime }}{dt}$

$=$

$-\frac{1}{\rho }\mathrm{\nabla }P-\mathrm{\nabla }\mathrm{\Phi }-2{\overrightarrow{\mathrm{\Omega }}}_f\times {\mathbf{𝐮}}^{\prime }-{\overrightarrow{\mathrm{\Omega }}}_f\times ({\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x}),$

where we appreciate that we can move from the Lagrangian to an Eulerian representation by employing the operator substitution,

$\frac{d}{dt}$

$\to$

$\frac{\partial }{\partial t}+{\mathbf{𝐮}}^{\prime }\cdot \mathrm{\nabla }$

Next, using [Ref03] as a guide, let's focus on tracking angular momentum. We need to break the vector momentum equation, as well as the velocity vectors, into their $({\hat{\mathbf{e}}}_{\varpi },{\hat{\mathbf{e}}}_{\phi },\hat{\mathbf{k}})$ components.

NOTE: For the time being, we will write the velocity vector in terms of generic components, namely, ${\mathbf{𝐮}}^{\prime }={\hat{\mathbf{e}}}_{\varpi }u{\text{'}}_{\varpi }+{\hat{\mathbf{e}}}_{\phi }u{\text{'}}_{\phi }+\hat{\mathbf{k}}u{\text{'}}_z.$ But, eventually, we want to explicitly insert the rotating-frame velocity that underpins the equilibrium properties of Riemann S-type ellipsoids. In Chap. 7, §47, Eq. 1 (p. 130) of [EFE], this is given in Cartesian coordinates, so we will need to convert his expressions to the equivalent cylindrical-coordinate components.

The time-derivative on the left-hand-side of our foundation expression becomes,

$\frac{d{\mathbf{𝐮}}^{\prime }}{dt}$	$=$	$\frac{d}{dt}[{\hat{\mathbf{e}}}_{\varpi }u{\text{'}}_{\varpi }+{\hat{\mathbf{e}}}_{\phi }u{\text{'}}_{\phi }+\hat{\mathbf{k}}u{\text{'}}_z]$
	$=$	${\hat{\mathbf{e}}}_{\varpi }\frac{du{\text{'}}_{\varpi }}{dt}+{\hat{\mathbf{e}}}_{\phi }\frac{du{\text{'}}_{\phi }}{dt}+\hat{\mathbf{k}}\frac{du{\text{'}}_z}{dt}+(u{\text{'}}_{\varpi })\frac{d}{dt}{\hat{\mathbf{e}}}_{\varpi }+(u{\text{'}}_{\phi })\frac{d}{dt}{\hat{\mathbf{e}}}_{\phi }$
	$=$	${\hat{\mathbf{e}}}_{\varpi }\frac{du{\text{'}}_{\varpi }}{dt}+{\hat{\mathbf{e}}}_{\phi }\frac{du{\text{'}}_{\phi }}{dt}+\hat{\mathbf{k}}\frac{du{\text{'}}_z}{dt}+{\hat{\mathbf{e}}}_{\phi }(u{\text{'}}_{\varpi })\frac{u{\text{'}}_{\phi }}{\varpi }-{\hat{\mathbf{e}}}_{\varpi }(u{\text{'}}_{\phi })\frac{u{\text{'}}_{\phi }}{\varpi }.$

We also recognize that, when expressed in cylindrical coordinates,

${\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x}$	$=$	$\widehat{\mathbf{𝐤}}{\mathrm{\Omega }}_f\times ({\hat{\mathbf{e}}}_{\varpi }\varpi +\hat{\mathbf{k}}z)={\hat{\mathbf{e}}}_{\phi }{\mathrm{\Omega }}_f\varpi ,$
${\overrightarrow{\mathrm{\Omega }}}_f\times ({\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x})$	$=$	$\widehat{\mathbf{𝐤}}{\mathrm{\Omega }}_f\times ({\hat{\mathbf{e}}}_{\phi }{\mathrm{\Omega }}_f\varpi )=-{\hat{\mathbf{e}}}_{\varpi }{\mathrm{\Omega }}_f^2\varpi ,$
${\overrightarrow{\mathrm{\Omega }}}_f\times {\mathbf{𝐮}}^{\prime }$	$=$	$\widehat{\mathbf{𝐤}}{\mathrm{\Omega }}_f\times ({\hat{\mathbf{e}}}_{\varpi }u{\text{'}}_{\varpi }+{\hat{\mathbf{e}}}_{\phi }u{\text{'}}_{\phi }+\hat{\mathbf{k}}u{\text{'}}_z)={\hat{\mathbf{e}}}_{\phi }{\mathrm{\Omega }}_{f}u{\text{'}}_{\varpi }-{\hat{\mathbf{e}}}_{\varpi }{\mathrm{\Omega }}_{f}u{\text{'}}_{\phi },$
${\overrightarrow{v}}_{\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}$	$=$	${\mathbf{𝐮}}^{\prime }+{\hat{\mathbf{e}}}_{\phi }{\mathrm{\Omega }}_f\varpi .$

The set of scalar momentum-component equations is obtained by "dotting" each unit vector into the vector equation.

${\hat{\mathbf{e}}}_{\varpi }:$	$\frac{du{\text{'}}_{\varpi }}{dt}-\frac{(u{\text{'}}_{\phi }{)}^2}{\varpi }$	$=$	$-{\hat{\mathbf{e}}}_{\varpi }\cdot \frac{\mathrm{\nabla }P}{\rho }-{\hat{\mathbf{e}}}_{\varpi }\cdot \mathrm{\nabla }\mathrm{\Phi }+2\left[{\mathrm{\Omega }}_{f}u{\text{'}}_{\phi }\right]+{\mathrm{\Omega }}_f^2\varpi$
$\Rightarrow \frac{du{\text{'}}_{\varpi }}{dt}$		$=$	$-{\hat{\mathbf{e}}}_{\varpi }\cdot \frac{\mathrm{\nabla }P}{\rho }-{\hat{\mathbf{e}}}_{\varpi }\cdot \mathrm{\nabla }\mathrm{\Phi }+\frac{1}{\varpi }[(u{\text{'}}_{\phi }{)}^2+2{\mathrm{\Omega }}_{f}u{\text{'}}_{\phi }\varpi +{\mathrm{\Omega }}_f^2{\varpi }^2]$
		$=$	$-{\hat{\mathbf{e}}}_{\varpi }\cdot \frac{\mathrm{\nabla }P}{\rho }-{\hat{\mathbf{e}}}_{\varpi }\cdot \mathrm{\nabla }\mathrm{\Phi }+\frac{1}{\varpi }(u{\text{'}}_{\phi }+{\mathrm{\Omega }}_f\varpi {)}^2;$
${\hat{\mathbf{e}}}_{\phi }:$	$\frac{du{\text{'}}_{\phi }}{dt}+\frac{u{\text{'}}_{\varpi }u{\text{'}}_{\phi }}{\varpi }$	$=$	$-{\hat{\mathbf{e}}}_{\phi }\cdot \frac{\mathrm{\nabla }P}{\rho }-{\hat{\mathbf{e}}}_{\phi }\cdot \mathrm{\nabla }\mathrm{\Phi }-2\left[{\mathrm{\Omega }}_{f}u{\text{'}}_{\varpi }\right]$
(mult. thru by ϖ)   $\Rightarrow \frac{d(\varpi u{\text{'}}_{\phi })}{dt}$		$=$	$-{\hat{\mathbf{e}}}_{\phi }\cdot \frac{\varpi \mathrm{\nabla }P}{\rho }-{\hat{\mathbf{e}}}_{\phi }\cdot \varpi \mathrm{\nabla }\mathrm{\Phi }-2{\mathrm{\Omega }}_f\varpi u{\text{'}}_{\varpi };$
$\hat{\mathbf{k}}:$	$\frac{du{\text{'}}_z}{dt}$	$=$	$-\hat{\mathbf{k}}\cdot \frac{\mathrm{\nabla }P}{\rho }-\hat{\mathbf{k}}\cdot \mathrm{\nabla }\mathrm{\Phi }.$

Now, recalling that ${\mathbf{𝐮}}^{\prime }=(\mathbf{𝐯}-{\hat{\mathbf{e}}}_{\phi }\varpi {\mathrm{\Omega }}_f)$, let's make the substitutions …

$u{\text{'}}_{\varpi }\to v_{\varpi },$

$u{\text{'}}_{\phi }\to (v_{\phi }-\varpi {\mathrm{\Omega }}_f),$      and,

$u{\text{'}}_z\to v_z.$

This mapping gives,

${\hat{\mathbf{e}}}_{\phi }:$	$\frac{d[\varpi v_{\phi }-{\varpi }^2{\mathrm{\Omega }}_f]}{dt}$	$=$	$-{\hat{\mathbf{e}}}_{\phi }\cdot \frac{\varpi \mathrm{\nabla }P}{\rho }-{\hat{\mathbf{e}}}_{\phi }\cdot \varpi \mathrm{\nabla }\mathrm{\Phi }-2{\mathrm{\Omega }}_f\varpi v_{\varpi };$
$\Rightarrow \frac{d(\varpi v_{\phi })}{dt}$		$=$	$-{\hat{\mathbf{e}}}_{\phi }\cdot \frac{\varpi \mathrm{\nabla }P}{\rho }-{\hat{\mathbf{e}}}_{\phi }\cdot \varpi \mathrm{\nabla }\mathrm{\Phi };$
$\Rightarrow \frac{1}{\varpi }\frac{d(\varpi v_{\phi })}{dt}$		$=$	$-{\hat{\mathbf{e}}}_{\phi }\cdot [\frac{\mathrm{\nabla }P}{\rho }+\mathrm{\nabla }\mathrm{\Phi }];$
$\hat{\mathbf{k}}:$	$\frac{d{v}_z}{dt}$	$=$	$-\hat{\mathbf{k}}\cdot [\frac{\mathrm{\nabla }P}{\rho }+\mathrm{\nabla }\mathrm{\Phi }].$
${\hat{\mathbf{e}}}_{\varpi }:$	$\frac{d{v}_{\varpi }}{dt}$	$=$	$-{\hat{\mathbf{e}}}_{\varpi }\cdot [\frac{\mathrm{\nabla }P}{\rho }+\mathrm{\nabla }\mathrm{\Phi }]+\frac{v_{\phi }^2}{\varpi };$

Steady-State Velocity Field for Jacobi Ellipsoids

In steady-state, the (Lagrangian time-derivative) operator on the left-hand-side of all three component equations maps to the following operator:

${\mathbf{𝐮}}^{\prime }\cdot \mathrm{\nabla }$	$=$	$\sum _{i=1}^{3}u{\text{'}}_i\frac{\partial }{\partial x_i},$	(in Cartesian coordinates);
${\mathbf{𝐮}}^{\prime }\cdot \mathrm{\nabla }$	$=$	$u{\text{'}}_{\varpi }\frac{\partial }{\partial \varpi }+\frac{u{\text{'}}_{\phi }}{\varpi }\frac{\partial }{\partial \phi }+u{\text{'}}_z\frac{\partial }{\partial z},$	(in cylindrical coordinates);

We know, as well, that,

$u{\text{'}}_{\varpi }=u{\text{'}}_x\cos \phi +u{\text{'}}_y\sin \phi ,$

and,

$u{\text{'}}_{\phi }=u{\text{'}}_y\cos \phi -u{\text{'}}_x\sin \phi .$

Hence, the cylindrical-coordinate-based operator may be rewritten as,

${\mathbf{𝐮}}^{\prime }\cdot \mathrm{\nabla }$

$=$

$(u{\text{'}}_x\cos \phi +u{\text{'}}_y\sin \phi )\frac{\partial }{\partial \varpi }+(u{\text{'}}_y\cos \phi -u{\text{'}}_x\sin \phi )\frac{1}{\varpi }\frac{\partial }{\partial \phi }+u{\text{'}}_z\frac{\partial }{\partial z}.$

Drawing from [ Ref04 ] … As Ou(2006) has pointed out, the velocity field of a Riemann S-type ellipsoid as viewed from a frame rotating with angular velocity ${\overrightarrow{\mathrm{\Omega }}}_f=\hat{\mathit{k}}{\mathrm{\Omega }}_f$ takes the following form:

${\mathbf{𝐮}}^{\prime }$

$=$

$\lambda \left[\hat{\mathit{ı}}\right(\frac{a}{b})y-\hat{\mathit{ȷ}}(\frac{b}{a}\left)x\right],$

Ou(2006), p. 550, §2, Eq. (3)

where $\lambda$ is a constant that determines the magnitude of the internal motion of the fluid, and the origin of the x-y coordinate system is at the center of the ellipsoid. This velocity field, ${\mathbf{𝐮}}^{\prime }$, is designed so that velocity vectors everywhere are always aligned with elliptical stream lines by demanding that they be tangent to the equi-effective-potential contours, which are concentric ellipses. Hence, for Riemann S-type ellipsoids, we have,

$u{\text{'}}_x=\lambda \left(\frac{a}{b}\right)y=\lambda \left(\frac{a}{b}\right)\varpi \sin \phi ;$

$u{\text{'}}_y=-\lambda \left(\frac{b}{a}\right)x=-\lambda \left(\frac{b}{a}\right)\varpi \cos \phi ;$

$u{\text{'}}_z=0.$

So, for the velocity flow that underpins Riemann S-type ellipsoids, the cylindrical-coordinate-based operator is

${\mathbf{𝐮}}^{\prime }\cdot \mathrm{\nabla }$	$=$	$\left[\lambda \right(\frac{a}{b})\varpi \sin \phi \cos \phi -\lambda (\frac{b}{a})\varpi \cos \phi \sin \phi ]\frac{\partial }{\partial \varpi }+[-\lambda (\frac{b}{a})\varpi \cos \phi \cos \phi -\lambda (\frac{a}{b})\varpi \sin \phi \sin \phi ]\frac{1}{\varpi }\frac{\partial }{\partial \phi }$
	$=$	$\left[\right(\frac{a}{b})-(\frac{b}{a}\left)\right]\lambda \varpi \sin \phi \cos \phi \frac{\partial }{\partial \varpi }-\left[\right(\frac{b}{a}){\cos }^{}\phi +(\frac{a}{b}\left){\sin }^{}\phi \right]\lambda \frac{\partial }{\partial \phi }.$

And, given that,

${\hat{\mathbf{e}}}_{\phi }{\mathrm{\Omega }}_f\varpi$

$=$

${\mathrm{\Omega }}_f\varpi [\hat{\mathit{ȷ}}\cos \phi -\hat{\mathit{ı}}\sin \phi ],$

the inertial-frame velocity components are,

$v_x=\lambda \left(\frac{a}{b}\right)\varpi \sin \phi -{\mathrm{\Omega }}_f\varpi \sin \phi =\left[\lambda \right(\frac{a}{b})-{\mathrm{\Omega }}_f]\varpi \sin \phi ;$

$v_y=-\lambda \left(\frac{b}{a}\right)\varpi \cos \phi +{\mathrm{\Omega }}_f\varpi \cos \phi =[{\mathrm{\Omega }}_f-\lambda (\frac{b}{a}\left)\right]\varpi \cos \phi ;$

$v_z=0.$

That is,

$v_{\varpi }=v_x\cos \phi +v_y\sin \phi$	$=$	$\left[\lambda \right(\frac{a}{b})-{\mathrm{\Omega }}_f]\varpi \sin \phi \cos \phi +[{\mathrm{\Omega }}_f-\lambda (\frac{b}{a}\left)\right]\varpi \cos \phi \sin \phi$
	$=$	$\left[\right(\frac{a}{b})-(\frac{b}{a}\left)\right]\lambda \varpi \sin \phi \cos \phi ;$
$v_{\phi }=v_y\cos \phi -v_x\sin \phi$	$=$	$[{\mathrm{\Omega }}_f-\lambda (\frac{b}{a}\left)\right]\varpi {\cos }^2\phi -\left[\lambda \right(\frac{a}{b})-{\mathrm{\Omega }}_f]\varpi {\sin }^2\phi$
	$=$	${\mathrm{\Omega }}_f\varpi -\lambda \varpi \left[\right(\frac{b}{a}){\cos }^2\phi +(\frac{a}{b}\left){\sin }^2\phi \right].$

Note, as well, that,

$\frac{v_{\phi }^2}{\varpi }$	$=$	$\frac{1}{\varpi }\{{\mathrm{\Omega }}_f\varpi -\lambda \varpi [\left(\frac{b}{a}\right){\cos }^2\phi +\left(\frac{a}{b}\right){\sin }^2\phi ]{\}}^2$
	$=$	$\varpi \{{\mathrm{\Omega }}_f^2-2\lambda {\mathrm{\Omega }}_f[\left(\frac{b}{a}\right){\cos }^2\phi +\left(\frac{a}{b}\right){\sin }^2\phi ]+{\lambda }^2[\left(\frac{b}{a}\right){\cos }^2\phi +\left(\frac{a}{b}\right){\sin }^2\phi {]}^2\}.$

Finally, then, we find that the left-hand-side of the momentum-component expressions are,

$\hat{\mathbf{k}}:$	$\frac{d{v}_z}{dt}$	$=$	$0;$
${\hat{\mathbf{e}}}_{\varpi }:$	$\frac{d{v}_{\varpi }}{dt}$	$=$	$\left\{\right[\left(\frac{a}{b}\right)-\left(\frac{b}{a}\right)]\lambda \varpi \sin \phi \cos \phi \frac{\partial }{\partial \varpi }-[\left(\frac{b}{a}\right){\cos }^{}\phi +\left(\frac{a}{b}\right){\sin }^{}\phi \left]\lambda \frac{\partial }{\partial \phi }\right\}\left[\right(\frac{a}{b})-(\frac{b}{a}\left)\right]\lambda \varpi \sin \phi \cos \phi$
		$=$	$\lambda \left[\right(\frac{a}{b})-(\frac{b}{a}\left)\right]\left\{\right[\left(\frac{a}{b}\right)-\left(\frac{b}{a}\right)]\lambda \varpi \sin \phi \cos \phi \frac{\partial }{\partial \varpi }[\varpi \sin \phi \cos \phi ]-[\left(\frac{b}{a}\right){\cos }^{}\phi +\left(\frac{a}{b}\right){\sin }^{}\phi \left]\lambda \frac{\partial }{\partial \phi }\right[\varpi \sin \phi \cos \phi \left]\right\}$
		$=$	$\lambda \left[\right(\frac{a}{b})-(\frac{b}{a}\left)\right]\left\{\right[\left(\frac{a}{b}\right)-\left(\frac{b}{a}\right)]\lambda \varpi {\sin }^2\phi {\cos }^2\phi +[\left(\frac{b}{a}\right){\cos }^2\phi +\left(\frac{a}{b}\right){\sin }^2\phi \left]\lambda \varpi \right[{\sin }^2\phi -{\cos }^2\phi \left]\right\}$
		$=$	${\lambda }^2\varpi (\frac{a}{b}-\frac{b}{a})\left[\right(\frac{a}{b}){\sin }^4\phi -(\frac{b}{a}\left){\cos }^4\phi \right];$
${\hat{\mathbf{e}}}_{\phi }:$	$\frac{1}{\varpi }\frac{d(\varpi v_{\phi })}{dt}$	$=$	$\frac{1}{\varpi }\left\{\right[\left(\frac{a}{b}\right)-\left(\frac{b}{a}\right)]\lambda \varpi \sin \phi \cos \phi \frac{\partial }{\partial \varpi }-[\left(\frac{b}{a}\right){\cos }^{}\phi +\left(\frac{a}{b}\right){\sin }^{}\phi \left]\lambda \frac{\partial }{\partial \phi }\right\}\left\{\right[{\mathrm{\Omega }}_f-\lambda \left(\frac{b}{a}\right)]{\varpi }^2{\cos }^{}\phi -[\lambda \left(\frac{a}{b}\right)-{\mathrm{\Omega }}_f\left]{\varpi }^2{\sin }^{}\phi \right\}$
		$=$	$\left[\right(\frac{a}{b})-(\frac{b}{a}\left)\right]2\varpi \lambda \sin \phi \cos \phi \left\{\right[{\mathrm{\Omega }}_f-\lambda \left(\frac{b}{a}\right)]{\cos }^{}\phi -[\lambda \left(\frac{a}{b}\right)-{\mathrm{\Omega }}_f\left]{\sin }^{}\phi \right\}$
			$+\left[\right(\frac{b}{a}){\cos }^2\phi +(\frac{a}{b}\left){\sin }^2\phi \right][\frac{a}{b}-\frac{b}{a}]2\varpi {\lambda }^2\sin \phi \cos \phi$
		$=$	$(\frac{a}{b}-\frac{b}{a})2\varpi \lambda \sin \phi \cos \phi \left\{\right[{\mathrm{\Omega }}_f-\lambda \left(\frac{b}{a}\right)]{\cos }^{}\phi -[\lambda \left(\frac{a}{b}\right)-{\mathrm{\Omega }}_f]{\sin }^{}\phi +2\lambda [\left(\frac{b}{a}\right){\cos }^{}\phi +\left(\frac{a}{b}\right){\sin }^{}\phi \left]\right\}$
		$=$	$(\frac{a}{b}-\frac{b}{a})2\varpi \lambda \sin \phi \cos \phi \left\{\right[{\mathrm{\Omega }}_f+\lambda \left(\frac{b}{a}\right)]{\cos }^{}\phi +[{\mathrm{\Omega }}_f+\lambda \left(\frac{a}{b}\right)\left]{\sin }^{}\phi \right\}.$

Try Again

An example equation of motion is,

$\mathbf{𝐚}=-\frac{1}{\rho }\mathrm{\nabla }P-\mathrm{\nabla }{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}.$

Here we will focus only on the left-hand-side of this equation, examining various ways the vector acceleration may be mathematically expressed. We will consider, in particular, building a model in a curvilinear (cylindrical), rather than a Cartesian, coordinate base; and viewing the model's evolution in a rotating, rather than inertial, frame of reference.

Inertial Frame

As viewed from a cylindrical-coordinate-based $(\varpi ,\phi ,z)$ inertial reference frame, we are interested in specifying the location,

$\mathbf{𝐱}={\hat{\mathbf{e}}}_{\varpi }\varpi +\hat{\mathbf{k}}z,$

[BT87], p. 646, Appendix §1.B.2, Eq. (1B-18)

of a Lagrangian fluid element at time $t=0$ — hereafter denoted by the subscript, $0$ — as well as at later times. Although the position vector, $\mathbf{𝐱}$, does not explicitly display a dependence on the azimuthal coordinate angle, $\phi$, it is important to realize that the orientation in space of the unit vector, ${\hat{\mathbf{e}}}_{\varpi }$, does depend on the value of this coordinate angle.

At any point in time, the instantaneous velocity of this Lagrangian fluid element will correspond precisely with the (total) time-derivative of its instantaneous position vector, that is,

$\mathbf{𝐯}$	$\equiv$	$\frac{d\mathbf{𝐱}}{dt}$	$=$	${\hat{\mathbf{e}}}_{\varpi }\frac{d\varpi }{dt}+\hat{\mathbf{k}}\frac{dz}{dt}+\varpi \frac{d{\hat{\mathbf{e}}}_{\varpi }}{dt}$
			$=$	${\hat{\mathbf{e}}}_{\varpi }\frac{d\varpi }{dt}+\hat{\mathbf{k}}\frac{dz}{dt}+\varpi \left[{\hat{\mathbf{e}}}_{\phi }\frac{d\phi }{dt}\right].$
[BT87], p. 647, Appendix §1.B.2, Eq. (1B-23)

In carrying out this time differentiation, the last term on the right-hand-side accounts for the aforementioned dependence of ${\hat{\mathbf{e}}}_{\varpi }$ on $\phi$. Similarly, the following component breakdown of the Lagrangian fluid element's acceleration takes into account the dependence of ${\hat{\mathbf{e}}}_{\phi }$ on $\phi$:

$\mathbf{𝐚}$	$\equiv$	$\frac{d\mathbf{𝐯}}{dt}$	$=$	${\hat{\mathbf{e}}}_{\varpi }\frac{d^2\varpi }{d{t}^2}+\hat{\mathbf{k}}\frac{d^{2}z}{d{t}^2}+{\hat{\mathbf{e}}}_{\phi }[\frac{d\varpi }{dt}\cdot \frac{d\phi }{dt}+\varpi \frac{d^2\phi }{d{t}^2}]+\varpi \frac{d\phi }{dt}\left[\frac{d{\hat{\mathbf{e}}}_{\phi }}{dt}\right]+\frac{d\varpi }{dt}\left[\frac{d{\hat{\mathbf{e}}}_{\varpi }}{dt}\right]$
			$=$	${\hat{\mathbf{e}}}_{\varpi }\frac{d^2\varpi }{d{t}^2}+\hat{\mathbf{k}}\frac{d^{2}z}{d{t}^2}+{\hat{\mathbf{e}}}_{\phi }[\frac{d\varpi }{dt}\cdot \frac{d\phi }{dt}+\varpi \frac{d^2\phi }{d{t}^2}]+\varpi \frac{d\phi }{dt}[-{\hat{\mathbf{e}}}_{\varpi }\frac{d\phi }{dt}]+\frac{d\varpi }{dt}\left[{\hat{\mathbf{e}}}_{\phi }\frac{d\phi }{dt}\right]$
			$=$	${\hat{\mathbf{e}}}_{\varpi }[\frac{d^2\varpi }{d{t}^2}-\varpi (\frac{d\phi }{dt}{)}^2]+{\hat{\mathbf{e}}}_{\phi }[2(\frac{d\varpi }{dt}\cdot \frac{d\phi }{dt})+\varpi \frac{d^2\phi }{d{t}^2}]+\hat{\mathbf{k}}\frac{d^{2}z}{d{t}^2}.$
[BT87], p. 647, Appendix §1.B.2, Eq. (1B-24)

Let's rewrite the velocity vector as,

$\mathbf{𝐯}$

$=$

${\hat{\mathbf{e}}}_{\varpi }\dot{\varpi }+{\hat{\mathbf{e}}}_{\phi }\varpi \dot{\phi }+\hat{\mathbf{k}}\dot{z},$

and (the second line of) this acceleration expression as,

$\mathbf{𝐚}\equiv \frac{d\mathbf{𝐯}}{dt}={\hat{\mathbf{e}}}_{\varpi }\frac{d\dot{\varpi }}{dt}+{\hat{\mathbf{e}}}_{\phi }\frac{d}{dt}\left[\varpi \dot{\phi }\right]+\hat{\mathbf{k}}\frac{d\dot{z}}{dt}+\underset{\text{curvature terms}}{\underbrace{\dot{\varpi }\left[{\hat{\mathbf{e}}}_{\phi }\frac{d\phi }{dt}\right]-\varpi \dot{\phi }\left[{\hat{\mathbf{e}}}_{\varpi }\frac{d\phi }{dt}\right]}}.$

Now, if $\mathbf{𝐁}$ is a vector quantity that characterizes some property of a fluid element — such as momentum density, velocity, or vorticity — the difference between the Lagrangian and Eulerian time-derivatives of that vector quantity is given by the expression,

$\frac{d\mathbf{𝐁}}{dt}-\frac{\partial \mathbf{𝐁}}{\partial t}$

$=$

$(\mathbf{𝐯}\cdot \mathrm{\nabla })\mathbf{𝐁},$

where the various elements of this right-hand-side mathematical operator can be obtained by replacing $\mathbf{𝐀}$ with $\mathbf{𝐯}$ in the so-called convective operator.

Convective Operator in Cylindrical Coordinates $(\mathbf{𝐀}\cdot \mathrm{\nabla })\mathbf{𝐁}$ $=$ ${\hat{\mathbf{e}}}_{\varpi }[A_{\varpi }\frac{\partial B_{\varpi }}{\partial \varpi }+\frac{A_{\phi }}{\varpi }\frac{\partial B_{\varpi }}{\partial \phi }+A_z\frac{\partial B_{\varpi }}{\partial z}-\frac{A_{\phi }{B}_{\phi }}{\varpi }]$     $+{\hat{\mathbf{e}}}_{\phi }[A_{\varpi }\frac{\partial B_{\phi }}{\partial \varpi }+\frac{A_{\phi }}{\varpi }\frac{\partial B_{\phi }}{\partial \phi }+A_z\frac{\partial B_{\phi }}{\partial z}+\frac{A_{\phi }{B}_{\varpi }}{\varpi }]$     $+{\hat{\mathbf{e}}}_z[A_{\varpi }\frac{\partial B_z}{\partial \varpi }+\frac{A_{\phi }}{\varpi }\frac{\partial B_z}{\partial \phi }+A_z\frac{\partial B_z}{\partial z}].$ [BT87], p. 651, Appendix §1.B.3, Eq. (1B-54) We will adopt the following, more compact notation: $(\mathbf{𝐀}\cdot \mathrm{\nabla })\mathbf{𝐁}$ $=$ ${\hat{\mathbf{e}}}_{\varpi }[{{ℒ}}_A{B}_{\varpi }-\frac{A_{\phi }{B}_{\phi }}{\varpi }]+{\hat{\mathbf{e}}}_{\phi }[{{ℒ}}_A{B}_{\phi }+\frac{A_{\phi }{B}_{\varpi }}{\varpi }]+{\hat{\mathbf{e}}}_z\left[{{ℒ}}_A{B}_z\right],$ where the operator, ${{ℒ}}_A$ $\equiv$ $[A_{\varpi }\frac{\partial }{\partial \varpi }+\frac{A_{\phi }}{\varpi }\frac{\partial }{\partial \phi }+A_z\frac{\partial }{\partial z}].$

In particular, if we are examining the behavior of the fluid velocity $(\mathbf{𝐁}\to \mathbf{𝐯})$, we find that,

$\frac{d\mathbf{𝐯}}{dt}-\frac{\partial \mathbf{𝐯}}{\partial t}$		$=(\mathbf{𝐯}\cdot \mathrm{\nabla })\mathbf{𝐯}$
		$={\hat{\mathbf{e}}}_{\varpi }\left[{{ℒ}}_v\dot{\varpi }\right]+{\hat{\mathbf{e}}}_{\phi }[{{ℒ}}_v(\varpi \dot{\phi })]+{\hat{\mathbf{e}}}_z\left[{{ℒ}}_v\dot{z}\right]+\underset{\text{curvature terms}}{\underbrace{{\hat{\mathbf{e}}}_{\phi }(\dot{\phi }\dot{\varpi })-{\hat{\mathbf{e}}}_{\varpi }(\varpi {\dot{\phi }}^2)}},$

where the operator,

${{ℒ}}_v$

$\equiv$

$[v_{\varpi }\frac{\partial }{\partial \varpi }+\frac{v_{\phi }}{\varpi }\frac{\partial }{\partial \phi }+v_z\frac{\partial }{\partial z}].$

Notice that the pair of "curvature terms" that appear in this expression are identical to the pair of curvature terms that appear in the acceleration expression, above. We conclude, therefore, that for each of the three separate (cylindrical-coordinate-based) components of the vector acceleration, the relationship between the Lagrangian (total) and Eulerian (partial) time derivative is, respectively,

${\hat{\mathbf{e}}}_{\varpi }$:	$\frac{d\dot{\varpi }}{dt}-\varpi {\dot{\phi }}^2$	$=$	$\frac{\partial \dot{\varpi }}{\partial t}+\left[{{ℒ}}_v\dot{\varpi }\right]-\varpi {\dot{\phi }}^2;$
${\hat{\mathbf{e}}}_{\phi }$:	$\frac{d(\varpi \dot{\phi })}{dt}+\dot{\varpi }\dot{\phi }$	$=$	$\frac{\partial (\varpi \dot{\phi })}{\partial t}+[{{ℒ}}_v(\varpi \dot{\phi })]+\dot{\varpi }\dot{\phi };$
$\hat{\mathbf{k}}$:	$\frac{d\dot{z}}{dt}$	$=$	$\frac{\partial \dot{z}}{\partial t}+\left[{{ℒ}}_v\dot{z}\right].$

Rotating Frame

Drawing from an accompanying discussion of rotating reference frames, let's build our model in a cylindrical coordinate system that is spinning about its $\hat{\mathbf{k}}$-axis with a time-independent angular velocity, ${\mathrm{\Omega }}_f=\hat{\mathbf{k}}{\mathrm{\Omega }}_f$. Furthermore, let's use $\mathbf{𝐮}$ — instead of $\mathbf{𝐯}$ — to represent the velocity as viewed in the rotating frame. We know that,

$\mathbf{𝐯}$

$=$

$\mathbf{𝐮}+{\mathrm{\Omega }}_f\times {\mathbf{𝐱}}_{\mathrm{r}\mathrm{o}\mathrm{t}}$

$=$

$\mathbf{𝐮}+{\hat{\mathbf{e}}}_{\phi }\varpi {\mathrm{\Omega }}_f,$

and,

$\mathbf{𝐚}=\frac{d\mathbf{𝐯}}{dt}=\frac{d\mathbf{𝐮}}{dt}-\stackrel{\text{Fictitious accelerations}}{\overbrace{[\underset{\text{Coriolis}}{\underbrace{(-2{\mathrm{\Omega }}_f\times \mathbf{𝐮})}}+\underset{\text{Centrifugal}}{\underbrace{(-{\mathrm{\Omega }}_f\times ({\mathrm{\Omega }}_f\times {\mathbf{𝐱}}_{\mathrm{r}\mathrm{o}\mathrm{t}}))}}]}}.$

[BT87], p. 664, Appendix §1.D.3, Eq. (1D-43)

Note that, in the particular case being considered here, ${\mathbf{𝐚}}_{\mathrm{f}\mathrm{i}\mathrm{c}\mathrm{t}}$ $=$ $-2{\mathrm{\Omega }}_f\times \mathbf{𝐮}-{\mathrm{\Omega }}_f\times ({\mathrm{\Omega }}_f\times {\mathbf{𝐱}}_{\mathrm{r}\mathrm{o}\mathrm{t}})$   $=$ $-2{\mathrm{\Omega }}_f\times [\mathbf{𝐯}-{\mathrm{\Omega }}_f\times {\mathbf{𝐱}}_{\mathrm{r}\mathrm{o}\mathrm{t}}]-{\mathrm{\Omega }}_f\times ({\mathrm{\Omega }}_f\times {\mathbf{𝐱}}_{\mathrm{r}\mathrm{o}\mathrm{t}})$   $=$ $-2{\mathrm{\Omega }}_f\times \mathbf{𝐯}+{\mathrm{\Omega }}_f\times ({\mathrm{\Omega }}_f\times {\mathbf{𝐱}}_{\mathrm{r}\mathrm{o}\mathrm{t}})$   $=$ $-2{\mathrm{\Omega }}_f\times \mathbf{𝐯}+\hat{\mathbf{k}}{\mathrm{\Omega }}_f\times ({\hat{\mathbf{e}}}_{\phi }\varpi {\mathrm{\Omega }}_f)$   $=$ $-2{\mathrm{\Omega }}_f[{\hat{\mathbf{e}}}_{\phi }{v}_{\varpi }-{\hat{\mathbf{e}}}_{\varpi }{v}_{\phi }]-({\hat{\mathbf{e}}}_{\varpi }\varpi {\mathrm{\Omega }}_f^2).$   $=$ $+{\hat{\mathbf{e}}}_{\varpi }[2{\mathrm{\Omega }}_f{v}_{\phi }-\varpi {\mathrm{\Omega }}_f^2]-{\hat{\mathbf{e}}}_{\phi }2{\mathrm{\Omega }}_f{v}_{\varpi }.$

We may therefore also write,

$\mathbf{𝐚}+{\mathbf{𝐚}}_{\mathrm{f}\mathrm{i}\mathrm{c}\mathrm{t}}=\frac{d\mathbf{𝐮}}{dt}$		$=\frac{\partial \mathbf{𝐮}}{\partial t}+(\mathbf{𝐮}\cdot \mathrm{\nabla })\mathbf{𝐮}$
		$=\frac{\partial \mathbf{𝐮}}{\partial t}+{\hat{\mathbf{e}}}_{\varpi }\left[{{ℒ}}_u{u}_{\varpi }\right]+{\hat{\mathbf{e}}}_{\phi }\left[{{ℒ}}_u{u}_{\phi }\right]+{\hat{\mathbf{e}}}_z\left[{{ℒ}}_u{u}_z\right]+\underset{\text{curvature terms}}{\underbrace{{\hat{\mathbf{e}}}_{\phi }\left(\frac{u_{\varpi }{u}_{\phi }}{\varpi }\right)-{\hat{\mathbf{e}}}_{\varpi }\left(\frac{u_{\phi }^2}{\varpi }\right)}},$

where the operator,

${{ℒ}}_u$

$\equiv$

$[u_{\varpi }\frac{\partial }{\partial \varpi }+\frac{u_{\phi }}{\varpi }\frac{\partial }{\partial \phi }+u_z\frac{\partial }{\partial z}].$

In numerical simulations that are carried out on a cylindrical grid and in a rotating reference frame, it is customary to group the "curvature terms" with the fictitious acceleration to obtain,

$\frac{\partial \mathbf{𝐮}}{\partial t}+{\hat{\mathbf{e}}}_{\varpi }\left[{{ℒ}}_u{u}_{\varpi }\right]+{\hat{\mathbf{e}}}_{\phi }\left[{{ℒ}}_u{u}_{\phi }\right]+{\hat{\mathbf{e}}}_z\left[{{ℒ}}_u{u}_z\right]$	$=$	$\mathbf{𝐚}+{\mathbf{𝐚}}_{\mathrm{f}\mathrm{i}\mathrm{c}\mathrm{t}}-{\hat{\mathbf{e}}}_{\phi }\left(\frac{u_{\varpi }{u}_{\phi }}{\varpi }\right)+{\hat{\mathbf{e}}}_{\varpi }\left(\frac{u_{\phi }^2}{\varpi }\right)$
	$=$	$\mathbf{𝐚}+{\hat{\mathbf{e}}}_{\varpi }[2{\mathrm{\Omega }}_f{v}_{\phi }-\varpi {\mathrm{\Omega }}_f^2+(\frac{u_{\phi }^2}{\varpi }\left)\right]-{\hat{\mathbf{e}}}_{\phi }[2{\mathrm{\Omega }}_f{v}_{\varpi }+(\frac{u_{\varpi }{u}_{\phi }}{\varpi }\left)\right]$
	$=$	$\mathbf{𝐚}+{\hat{\mathbf{e}}}_{\varpi }[2{\mathrm{\Omega }}_f(u_{\phi }+\varpi {\mathrm{\Omega }}_f)-\varpi {\mathrm{\Omega }}_f^2+(\frac{u_{\phi }^2}{\varpi }\left)\right]-{\hat{\mathbf{e}}}_{\phi }[2\varpi {\mathrm{\Omega }}_f+u_{\phi }]\frac{u_{\varpi }}{\varpi }$
	$=$	$\mathbf{𝐚}+{\hat{\mathbf{e}}}_{\varpi }[\varpi {\mathrm{\Omega }}_f+u_{\phi }{]}^2\frac{1}{\varpi }-{\hat{\mathbf{e}}}_{\phi }[2\varpi {\mathrm{\Omega }}_f+u_{\phi }]\frac{u_{\varpi }}{\varpi },$

treating the ensemble as an additional "source" of acceleration.

Example from the Literature (see an accompanying related discussion) We begin with the version of the Euler equation that has just been derived, namely, $\frac{\partial \mathbf{𝐮}}{\partial t}+{\hat{\mathbf{e}}}_{\varpi }\left[{{ℒ}}_u{u}_{\varpi }\right]+{\hat{\mathbf{e}}}_{\phi }\left[{{ℒ}}_u{u}_{\phi }\right]+{\hat{\mathbf{e}}}_z\left[{{ℒ}}_u{u}_z\right]$ $=$ $\mathbf{𝐚}+{\hat{\mathbf{e}}}_{\varpi }[\varpi {\mathrm{\Omega }}_f+u_{\phi }{]}^2\frac{1}{\varpi }-{\hat{\mathbf{e}}}_{\phi }[2\varpi {\mathrm{\Omega }}_f+u_{\phi }]\frac{u_{\varpi }}{\varpi }.$ In examining and rearranging terms in each of the three components of this Euler equation, we will recognize that, ${{ℒ}}_u{u}_i$ $=$ $(\mathbf{𝐮}\cdot \mathrm{\nabla })u_i;$ and that, after multiplying the standard Lagrangian representation of the continuity equation through by $u_i$, we have, $0$ $=$ $u_i[\frac{d\rho }{dt}+\rho \mathrm{\nabla }\cdot \mathbf{𝐯}]=u_i[\frac{d\rho }{dt}+\rho \mathrm{\nabla }\cdot \mathbf{𝐮}+\rho {\xcancel{\mathrm{\nabla }\cdot ({\hat{\mathbf{e}}}_{\phi }\varpi {\mathrm{\Omega }}_f)}}^0]$   $=$ $u_i[\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho \mathbf{𝐮})]$   $=$ $\frac{\partial \rho u_i}{\partial t}-\rho \frac{\partial u_i}{\partial t}+\mathrm{\nabla }\cdot (\rho u_i\mathbf{𝐮})-\rho (\mathbf{𝐮}\cdot \mathrm{\nabla })u_i$ $\Rightarrow \rho \frac{\partial u_i}{\partial t}+\rho \left[{{ℒ}}_u{u}_i\right]$ $=$ $\frac{\partial \rho u_i}{\partial t}+\mathrm{\nabla }\cdot (\rho u_i\mathbf{𝐮}).$ Vertical Component:  Multiplying the $\hat{\mathbf{k}}$ component of this Euler equation through by $\rho$, gives, $\rho \frac{\partial u_z}{\partial t}+\rho \left[{{ℒ}}_u{u}_z\right]$ $=$ $\rho a_z$ $\Rightarrow \frac{\partial \rho u_z}{\partial t}+\mathrm{\nabla }\cdot (\rho u_z\mathbf{𝐮})$ $=$ $\rho a_z.$ Norman & Wilson (1978), ApJ, 224, pp. 497 - 511, §III.b, Eq. (5) New & Tohline (1997), ApJ, 490, pp. 311 - 237, §2, Eq. (3) Radial Component:  Multiplying the ${\hat{\mathbf{e}}}_{\varpi }$ component of this Euler equation through by $\rho$, gives, $\rho \frac{\partial u_{\varpi }}{\partial t}+\rho \left[{{ℒ}}_u{u}_{\varpi }\right]$ $=$ $\rho a_{\varpi }+[u_{\phi }+{\mathrm{\Omega }}_f\varpi {]}^2\frac{\rho }{\varpi }$ $\Rightarrow \frac{\partial \rho u_{\varpi }}{\partial t}+\mathrm{\nabla }\cdot (\rho u_{\varpi }\mathbf{𝐮})$ $=$ $\rho a_{\varpi }+[\frac{(\rho \varpi u_{\phi }{)}^2}{\rho {\varpi }^3}+\rho {\mathrm{\Omega }}_f^2\varpi +\frac{2{\mathrm{\Omega }}_f(\rho \varpi u_{\phi })}{\varpi }].$ Norman & Wilson (1978), ApJ, 224, pp. 497 - 511, §III.b, Eq. (6) New & Tohline (1997), ApJ, 490, pp. 311 - 237, §2.2, Eq. (11) Azimuthal Component:  Multiplying the ${\hat{\mathbf{e}}}_{\phi }$ component of this Euler equation through by $\rho$, gives, $\rho \frac{\partial u_{\phi }}{\partial t}+\rho \left[{{ℒ}}_u{u}_{\phi }\right]$ $=$ $\rho a_{\phi }-[2\varpi {\mathrm{\Omega }}_f+u_{\phi }]\frac{\rho u_{\varpi }}{\varpi }$ $\Rightarrow \frac{\partial \rho u_{\phi }}{\partial t}+\mathrm{\nabla }\cdot (\rho u_{\phi }\mathbf{𝐮})$ $=$ $\rho a_{\phi }-[2\varpi {\mathrm{\Omega }}_f+u_{\phi }]\frac{\rho u_{\varpi }}{\varpi }.$ Then, multiplying through by $\varpi$, gives, $\varpi \frac{\partial \rho u_{\phi }}{\partial t}+\varpi \mathrm{\nabla }\cdot (\rho u_{\phi }\mathbf{𝐮})$ $=$ $\rho \varpi a_{\phi }-[2\varpi {\mathrm{\Omega }}_f+u_{\phi }]\rho u_{\varpi }$ $\Rightarrow \frac{\partial \rho \varpi u_{\phi }}{\partial t}+\mathrm{\nabla }\cdot (\rho \varpi u_{\phi }\mathbf{𝐮})-\rho u_{\phi }{\xcancel{\left[{{ℒ}}_u\varpi \right]}}^{u_{\varpi }}$ $=$ $\rho \varpi a_{\phi }-2{\mathrm{\Omega }}_f\varpi \rho u_{\varpi }-\rho u_{\phi }{u}_{\varpi }$ $\Rightarrow \frac{\partial \rho \varpi u_{\phi }}{\partial t}+\mathrm{\nabla }\cdot (\rho \varpi u_{\phi }\mathbf{𝐮})$ $=$ $\rho \varpi a_{\phi }-2{\mathrm{\Omega }}_f\varpi \rho u_{\varpi }.$ Norman & Wilson (1978), ApJ, 224, pp. 497 - 511, §III.b, Eq. (7) New & Tohline (1997), ApJ, 490, pp. 311 - 237, §2.2, Eq. (12)

Hybrid Scheme

In our hybrid scheme, we will continue to use ${{ℒ}}_u$ — that is, an advection operator that incorporates the rotating-frame velocity, $\mathbf{𝐮}$ — but we will switch all other velocity references to the inertial-frame velocity, $\mathbf{𝐯}$, and its components. This will be done via the above-declared mapping,

$\mathbf{𝐮}$

$\to$

$[\mathbf{𝐯}-{\hat{\mathbf{e}}}_{\phi }\varpi {\mathrm{\Omega }}_f],$

that is, $u_{\varpi }\to v_{\varpi }$, $u_z\to v_z$, and $u_{\phi }\to (v_{\phi }-\varpi {\mathrm{\Omega }}_f)$. The Euler equation becomes,

$\frac{\partial [\mathbf{𝐯}-{\hat{\mathbf{e}}}_{\phi }\varpi {\mathrm{\Omega }}_f]}{\partial t}+{\hat{\mathbf{e}}}_{\varpi }\left[{{ℒ}}_u{v}_{\varpi }\right]+{\hat{\mathbf{e}}}_{\phi }[{{ℒ}}_u(v_{\phi }-\varpi {\mathrm{\Omega }}_f)]+{\hat{\mathbf{e}}}_z\left[{{ℒ}}_u{v}_z\right]$

$=$

$\mathbf{𝐚}+{\hat{\mathbf{e}}}_{\varpi }\left[\frac{v_{\phi }^2}{\varpi }\right]-{\hat{\mathbf{e}}}_{\phi }[\varpi {\mathrm{\Omega }}_f+v_{\phi }]\frac{v_{\varpi }}{\varpi },$

where we recognize that,

${{ℒ}}_u{v}_i$

$=$

$(\mathbf{𝐮}\cdot \mathrm{\nabla })v_i.$

Now, given that,

$\frac{\partial }{\partial t}\left[{\hat{\mathbf{e}}}_{\phi }\varpi {\mathrm{\Omega }}_f\right]$

$=$

$0,$

and,

${{ℒ}}_u(\varpi {\mathrm{\Omega }}_f)$

$=$

$[u_{\varpi }\frac{\partial }{\partial \varpi }+\frac{u_{\phi }}{\varpi }\frac{\partial }{\partial \phi }+u_z\frac{\partial }{\partial z}](\varpi {\mathrm{\Omega }}_f)=u_{\varpi }{\mathrm{\Omega }}_f=v_{\varpi }{\mathrm{\Omega }}_f,$

the Euler equation becomes,

$\frac{\partial \mathbf{𝐯}}{\partial t}+{\hat{\mathbf{e}}}_{\varpi }\left[{{ℒ}}_u{v}_{\varpi }\right]+{\hat{\mathbf{e}}}_{\phi }[{{ℒ}}_u{v}_{\phi }-v_{\varpi }{\mathrm{\Omega }}_f]+{\hat{\mathbf{e}}}_z\left[{{ℒ}}_u{v}_z\right]$	$=$	$\mathbf{𝐚}+{\hat{\mathbf{e}}}_{\varpi }\left[\frac{v_{\phi }^2}{\varpi }\right]-{\hat{\mathbf{e}}}_{\phi }[\varpi {\mathrm{\Omega }}_f+v_{\phi }]\frac{v_{\varpi }}{\varpi }$
$\Rightarrow \frac{\partial \mathbf{𝐯}}{\partial t}+{\hat{\mathbf{e}}}_{\varpi }\left[{{ℒ}}_u{v}_{\varpi }\right]+{\hat{\mathbf{e}}}_{\phi }\left[{{ℒ}}_u{v}_{\phi }\right]+{\hat{\mathbf{e}}}_z\left[{{ℒ}}_u{v}_z\right]$	$=$	$\mathbf{𝐚}+{\hat{\mathbf{e}}}_{\varpi }\left[\frac{v_{\phi }^2}{\varpi }\right]-{\hat{\mathbf{e}}}_{\phi }\left[\frac{v_{\varpi }{v}_{\phi }}{\varpi }\right].$

Also, if we multiply the standard Lagrangian representation of the continuity equation through by $v_i$, we have,

$0$	$=$	$v_i[\frac{d\rho }{dt}+\rho \mathrm{\nabla }\cdot \mathbf{𝐯}]=v_i[\frac{d\rho }{dt}+\rho \mathrm{\nabla }\cdot \mathbf{𝐮}+\rho {\xcancel{\mathrm{\nabla }\cdot ({\hat{\mathbf{e}}}_{\phi }\varpi {\mathrm{\Omega }}_f)}}^0]$
	$=$	$v_i[\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho \mathbf{𝐮})]$
	$=$	$\frac{\partial \rho v_i}{\partial t}-\rho \frac{\partial v_i}{\partial t}+\mathrm{\nabla }\cdot (\rho v_i\mathbf{𝐮})-\rho (\mathbf{𝐮}\cdot \mathrm{\nabla })v_i$
$\Rightarrow \rho \frac{\partial v_i}{\partial t}+\rho \left[{{ℒ}}_u{v}_i\right]$	$=$	$\frac{\partial \rho v_i}{\partial t}+\mathrm{\nabla }\cdot (\rho v_i\mathbf{𝐮}).$

---

Vertical Component:  Multiplying the $\hat{\mathbf{k}}$ component of our modified Euler equation through by $\rho$, then incorporating this version of the continuity equation, gives,

$\rho \frac{\partial v_z}{\partial t}+\rho \left[{{ℒ}}_u{v}_z\right]$	$=$	$\rho a_z$
$\Rightarrow \frac{\partial \rho v_z}{\partial t}+\mathrm{\nabla }\cdot (\rho v_z\mathbf{𝐮})$	$=$	$\rho a_z.$

Radial Component:  Multiplying the ${\hat{\mathbf{e}}}_{\varpi }$ component of our modified Euler equation through by $\rho$, then incorporating the continuity equation, gives,

$\rho \frac{\partial v_{\varpi }}{\partial t}+\rho \left[{{ℒ}}_u{v}_{\varpi }\right]$	$=$	$\rho a_{\varpi }+\frac{v_{\phi }^2}{\varpi }$
$\Rightarrow \frac{\partial \rho v_{\varpi }}{\partial t}+\mathrm{\nabla }\cdot (\rho v_{\varpi }\mathbf{𝐮})$	$=$	$\rho a_{\varpi }+\frac{v_{\phi }^2}{\varpi }.$

Azimuthal Component:  Multiplying the ${\hat{\mathbf{e}}}_{\phi }$ component of our modified Euler equation through by $\rho$, then incorporating the continuity equation, gives,

$\rho \frac{\partial v_{\phi }}{\partial t}+\rho \left[{{ℒ}}_u{v}_{\phi }\right]$	$=$	$\rho a_{\phi }-\frac{\rho v_{\phi }{v}_{\varpi }}{\varpi }$
$\Rightarrow \frac{\partial \rho v_{\phi }}{\partial t}+\mathrm{\nabla }\cdot (\rho v_{\phi }\mathbf{𝐮})$	$=$	$\rho a_{\phi }-\frac{\rho v_{\phi }{v}_{\varpi }}{\varpi }.$

Then, multiplying through by $\varpi$, we have,

$\rho \varpi a_{\phi }-\rho v_{\phi }{v}_{\varpi }$	$=$	$\varpi \frac{\partial \rho v_{\phi }}{\partial t}+\varpi \mathrm{\nabla }\cdot (\rho v_{\phi }\mathbf{𝐮})$
	$=$	$\frac{\partial (\rho \varpi v_{\phi })}{\partial t}-(\rho v_{\phi }){\xcancel{\frac{\partial \varpi }{\partial t}}}^0+\mathrm{\nabla }\cdot (\rho \varpi v_{\phi }\mathbf{𝐮})-\rho v_{\phi }(\mathbf{𝐮}\cdot \mathrm{\nabla })\varpi$
	$=$	$\frac{\partial (\rho \varpi v_{\phi })}{\partial t}+\mathrm{\nabla }\cdot (\rho \varpi v_{\phi }\mathbf{𝐮})-\rho v_{\phi }\left[{\xcancel{{{ℒ}}_u\varpi }}^{v_{\varpi }}\right]$
$\Rightarrow \frac{\partial (\rho \varpi v_{\phi })}{\partial t}+\mathrm{\nabla }\cdot (\rho \varpi v_{\phi }\mathbf{𝐮})$	$=$	$\rho \varpi a_{\phi }.$

Compare

Here we consider which formalism is best suited for modeling a fully three-dimensional, nonaxisymmetric configuration that is spinning about (usually) its shortest axis with a uniform and time-independent frequency and which, when viewed from a frame that is rotating with that frequency, exhibits a nontrivial but nevertheless steady-state internal flow. Examples are Riemann S-type ellipsoids, and binary stars in circular orbits.

It is most desirable to choose a formalism that recognizes the steady-state nature of the flow. In the vast majority of cases being considered here, this rules out using any scheme that is designed around an inertial-frame coordinate base. (As a counterexample, Dedekind ellipsoids can be constructed in the inertial frame because ${\mathrm{\Omega }}_f=0$ for all models along the Dedekind equilibrium sequence.) It is quite reasonable, however, to adopt a rotating, cylindrical-coordinate base as has been described above and as is summarized immediately below.

Traditional Rotating, Cylindrical-Coordinate Summary $\hat{\mathbf{k}}\mathbf{:}$ $\frac{\partial (\rho u_z)}{\partial t}+\mathrm{\nabla }\cdot (\rho u_z\mathbf{𝐮})$ $=$ $\hat{\mathbf{k}}\cdot (\rho \mathbf{𝐚});$ ${\hat{\mathbf{e}}}_{\mathit{\varpi }}\mathbf{:}$ $\frac{\partial (\rho u_{\varpi })}{\partial t}+\mathrm{\nabla }\cdot (\rho u_{\varpi }\mathbf{𝐮})$ $=$ ${\hat{\mathbf{e}}}_{\varpi }\cdot (\rho \mathbf{𝐚})+[\frac{(\rho \varpi u_{\phi }{)}^2}{\rho {\varpi }^3}+\rho {\mathrm{\Omega }}_f^2\varpi +\frac{2{\mathrm{\Omega }}_f(\rho \varpi u_{\phi })}{\varpi }];$ ${\hat{\mathbf{e}}}_{\mathit{\phi }}\mathbf{:}$ $\frac{\partial (\rho \varpi u_{\phi })}{\partial t}+\mathrm{\nabla }\cdot (\rho \varpi u_{\phi }\mathbf{𝐮})$ $=$ ${\hat{\mathbf{e}}}_{\phi }\cdot (\rho \varpi \mathbf{𝐚})-2{\mathrm{\Omega }}_f\varpi \rho u_{\varpi }.$

In this scheme, all of the velocities and associated momentum densities in all three components of the Euler equation are expressed in terms of the rotating-frame velocity vector, $\mathbf{𝐮}$, or its cylindrical-coordinate-based components, $(u_{\varpi },v_{\phi },v_z)$. When the configuration's distorted (nonaxisymmetric) shape is largely supported by rapid rotation, this scheme provides an advantage over other — for example, inertial-frame-based — schemes because the fraction of the fluid's total momentum that is being advected across the grid is often quite small. There is a penalty to be paid, however. Additional "source" terms appear on the right-hand-side of the radial- and azimuthal-component expressions; they are nonlinear in the velocity and introduce cross-talk between the component expressions.

Hybrid Scheme Summary $\hat{\mathbf{k}}\mathbf{:}$ $\frac{\partial (\rho v_z)}{\partial t}+\mathrm{\nabla }\cdot (\rho v_z\mathbf{𝐮})$ $=$ $\hat{\mathbf{k}}\cdot (\rho \mathbf{𝐚});$ ${\hat{\mathbf{e}}}_{\mathit{\varpi }}\mathbf{:}$ $\frac{\partial (\rho v_{\varpi })}{\partial t}+\mathrm{\nabla }\cdot (\rho v_{\varpi }\mathbf{𝐮})$ $=$ ${\hat{\mathbf{e}}}_{\varpi }\cdot (\rho \mathbf{𝐚})+\frac{v_{\phi }^2}{\varpi };$ ${\hat{\mathbf{e}}}_{\mathit{\phi }}\mathbf{:}$ $\frac{\partial (\rho \varpi v_{\phi })}{\partial t}+\mathrm{\nabla }\cdot (\rho \varpi v_{\phi }\mathbf{𝐮})$ $=$ ${\hat{\mathbf{e}}}_{\phi }\cdot (\rho \varpi \mathbf{𝐚}).$

For Riemann S-type ellipsoids,

$u_x$	$=$	$\lambda \left(\frac{a}{b}\right)y,$
$u_y$	$=$	$-\lambda \left(\frac{b}{a}\right)x,$
$v_{\varpi }$	$=$	$\lambda [\frac{a}{b}-\frac{b}{a}]xy(x^2+y^2{)}^{-1/2},$
$\varpi v_{\phi }$	$=$	$-\left[\lambda \right(\frac{b}{a})-{\mathrm{\Omega }}_f]x^2-\left[\lambda \right(\frac{a}{b})-{\mathrm{\Omega }}_f]y^2.$

Hence,

$\mathrm{\nabla }\cdot (v_{\varpi }\mathbf{𝐮})$	$=$	$\frac{\partial }{\partial x}\left\{v_{\varpi }{u}_x\right\}+\frac{\partial }{\partial y}\left\{v_{\varpi }{u}_y\right\}$
	$=$	${\lambda }^2\left(\frac{a}{b}\right)[\frac{a}{b}-\frac{b}{a}]\frac{\partial }{\partial x}\{x{y}^2(x^2+y^2{)}^{-1/2}\}-{\lambda }^2(\frac{b}{a}\left)\right[\frac{a}{b}-\frac{b}{a}\left]\frac{\partial }{\partial y}\right\{x^{2}y(x^2+y^2{)}^{-1/2}\}$
	$=$	$y^2{\lambda }^2\left(\frac{a}{b}\right)[\frac{a}{b}-\frac{b}{a}]\{(x^2+y^2{)}^{-1/2}-x^2(x^2+y^2{)}^{-3/2}\}-x^2{\lambda }^2\left(\frac{b}{a}\right)[\frac{a}{b}-\frac{b}{a}]\{(x^2+y^2{)}^{-1/2}-y^2(x^2+y^2{)}^{-3/2}\}$
	$=$	$\frac{{\lambda }^2}{{\varpi }^3}[\frac{a}{b}-\frac{b}{a}]\left\{y^4\right(\frac{a}{b})-x^4(\frac{b}{a}\left)\right\}$

And,

$\mathrm{\nabla }\cdot (\varpi v_{\phi }\mathbf{𝐮})$	$=$	$\frac{\partial }{\partial x}\{-\lambda (\frac{a}{b}\left)y\right[\lambda \left(\frac{b}{a}\right)-{\mathrm{\Omega }}_f\left]x^2\right\}+\frac{\partial }{\partial y}\left\{\lambda \right(\frac{b}{a}\left)x\right[\lambda \left(\frac{a}{b}\right)-{\mathrm{\Omega }}_f\left]y^2\right\}$
	$=$	$\{-2\lambda (\frac{a}{b}\left)\right[\lambda \left(\frac{b}{a}\right)-{\mathrm{\Omega }}_f\left]xy\right\}+\left\{2\lambda \right(\frac{b}{a}\left)\right[\lambda \left(\frac{a}{b}\right)-{\mathrm{\Omega }}_f\left]xy\right\}$
	$=$	$\{-(\frac{a}{b}\left)\right[\lambda \left(\frac{b}{a}\right)-{\mathrm{\Omega }}_f]+(\frac{b}{a}\left)\right[\lambda \left(\frac{a}{b}\right)-{\mathrm{\Omega }}_f\left]\right\}2\lambda xy$
	$=$	$\{-[\lambda -\left(\frac{a}{b}\right){\mathrm{\Omega }}_f]+[\lambda -\left(\frac{b}{a}\right){\mathrm{\Omega }}_f\left]\right\}2\lambda xy$
	$=$	$2\lambda xy{\mathrm{\Omega }}_f[\frac{a}{b}-\frac{b}{a}].$

To within an additive constant — see, for example, our associated discussion of Maclaurin spheroids — the gravitational potential and the enthalpy are, respectively,

${\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$	$=$	$\pi G\rho [A_1{x}^2+A_2{y}^2+A_3{z}^2]=\pi G\rho [A_1{\varpi }^2{\cos }^2\phi +A_2{\varpi }^2{\sin }^2\phi +A_3{z}^2],$
$H$	$=$	$H_0[1-\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}]=H_0[1-\frac{{\varpi }^2{\cos }^2\phi }{a^2}-\frac{{\varpi }^2{\sin }^2\phi }{b^2}-\frac{z^2}{c^2}].$

Hence,

$\mathbf{𝐚}=-\mathrm{\nabla }(H+{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}})$	$=$	$-\left[{\hat{\mathbf{e}}}_{\varpi }\frac{\partial }{\partial \varpi }\right(H+{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}})+\frac{{\hat{\mathbf{e}}}_{\phi }}{\varpi }\frac{\partial }{\partial \phi }(H+{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}})+\hat{\mathbf{k}}\frac{\partial }{\partial z}(H+{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}\left)\right]$
	$=$	$-{\hat{\mathbf{e}}}_{\varpi }[-2H_0\varpi (\frac{{\cos }^2\phi }{a^2}+\frac{{\sin }^2\phi }{b^2})+2\pi G\rho \varpi (A_1{\cos }^2\phi +A_2{\sin }^2\phi \left)\right]$
		$-\frac{{\hat{\mathbf{e}}}_{\phi }}{\varpi }[-2H_0{\varpi }^2\sin \phi \cos \phi (\frac{1}{b^2}-\frac{1}{a^2})+2\pi G\rho {\varpi }^2\sin \phi \cos \phi (A_2-A_1\left)\right]$
		$-\hat{\mathbf{k}}[-\frac{2H_{0}z}{c^2}+2\pi G\rho A_{3}z].$

---

Vertical Component:   Because $v_z=0$, it must be true that $\hat{\mathbf{k}}\cdot \mathbf{𝐚}=0$. This, in turn means,

$H_0$

$=$

$\pi G\rho c^2{A}_3.$

Azimuthal Component:   In steady-state, the partial time-derivative must be zero, so we require,

$\mathrm{\nabla }\cdot (\varpi v_{\phi }\mathbf{𝐮})$	$=$	$\varpi {\hat{\mathbf{e}}}_{\phi }\cdot \mathbf{𝐚}$
$\Rightarrow 2\lambda xy{\mathrm{\Omega }}_f[\frac{a}{b}-\frac{b}{a}]$	$=$	$[2H_0{\varpi }^2\sin \phi \cos \phi (\frac{1}{b^2}-\frac{1}{a^2})-2\pi G\rho {\varpi }^2\sin \phi \cos \phi (A_2-A_1\left)\right]$
	$=$	$2xy\left[H_0\right(\frac{1}{b^2}-\frac{1}{a^2})-\pi G\rho (A_2-A_1\left)\right]$
	$=$	$2\pi G\rho xy\left[c^2{A}_3\right(\frac{1}{b^2}-\frac{1}{a^2})-(A_2-A_1\left)\right]$
$\Rightarrow \lambda {\mathrm{\Omega }}_f\left[\frac{a^2-b^2}{ab}\right]$	$=$	$\pi G\rho \left[c^2{A}_3\right(\frac{a^2-b^2}{a^2{b}^2})+(A_1-A_2\left)\right]$
	$=$	$\pi G\rho [(A_1-A_2)-c^2{A}_3(\frac{b^2-a^2}{a^2{b}^2}\left)\right]$
$\Rightarrow -ab\lambda {\mathrm{\Omega }}_f$	$=$	$\pi G\rho [\frac{(A_1-A_2)a^2{b}^2}{b^2-a^2}-c^2{A}_3].$

This expression can be used either (a) to give ${\mathrm{\Omega }}_f$ in terms of a set of known quantities and the unknown parameter, $\lambda$; or (b) to give $\lambda$ in terms of a set of known quantities and the unknown parameter, ${\mathrm{\Omega }}_f$.

Recognizing from here that, $f\equiv \zeta /{\mathrm{\Omega }}_f$ and

$\lambda (\frac{a}{b}+\frac{b}{a})$	$=$	$-\zeta =-f{\mathrm{\Omega }}_f$
$\Rightarrow \lambda$	$=$	$-\left(\frac{ab}{a^2+b^2}\right)f{\mathrm{\Omega }}_f,$

this last expression can be rewritten as,

$\left(\frac{a^2{b}^2}{a^2+b^2}\right)f{\mathrm{\Omega }}_f^2$	$=$	$\pi G\rho [\frac{(A_1-A_2)a^2{b}^2}{b^2-a^2}-c^2{A}_3].$
[ EFE, Chapter 7, §48, Eq. (34) ]

Radial Component:   In steady-state, this partial time-derivative also must be zero, so we require,

$\mathrm{\nabla }\cdot (v_{\varpi }\mathbf{𝐮})$	$=$	${\hat{\mathbf{e}}}_{\varpi }\cdot \mathbf{𝐚}+\frac{v_{\phi }^2}{\varpi }$
$\Rightarrow \frac{{\lambda }^2}{{\varpi }^3}[\frac{a}{b}-\frac{b}{a}]\left\{y^4\right(\frac{a}{b})-x^4(\frac{b}{a}\left)\right\}$	$=$	$\frac{2\pi G\rho }{\varpi }\left[c^2{A}_3\right(\frac{x^2}{a^2}+\frac{y^2}{b^2})-(A_1{x}^2+A_2{y}^2\left)\right]$
		$+\frac{1}{{\varpi }^3}\{-[\lambda \left(\frac{b}{a}\right)-{\mathrm{\Omega }}_f]x^2-[\lambda \left(\frac{a}{b}\right)-{\mathrm{\Omega }}_f]y^2{\}}^2$
$\Rightarrow {\lambda }^2\{a^2{y}^4-b^2{x}^4\}$	$=$	$\frac{2\pi G\rho b^2{\varpi }^2}{(a^2-b^2)}[c^2{A}_3{x}^2-A_1{a}^2{x}^2]+\frac{2\pi G\rho a^2{\varpi }^2}{(a^2-b^2)}[c^2{A}_3{y}^2-A_2{b}^2{y}^2]$
		$+\left[\frac{a^2{b}^2}{a^2-b^2}\right]\left\{\right[\lambda \left(\frac{b}{a}\right)-{\mathrm{\Omega }}_f]x^2+[\lambda \left(\frac{a}{b}\right)-{\mathrm{\Omega }}_f]y^2{\}}^2$
$\Rightarrow {\lambda }^2(a^2-b^2)\{a^2{y}^4-b^2{x}^4\}$	$=$	$2\pi G\rho b^2{\varpi }^2{x}^2[c^2{A}_3-A_1{a}^2]+2\pi G\rho a^2{\varpi }^2{y}^2[c^2{A}_3-A_2{b}^2]$
		$+\left\{\right[\lambda b^2-ab{\mathrm{\Omega }}_f]x^2+[\lambda a^2-ab{\mathrm{\Omega }}_f]y^2{\}}^2.$

Also,

$-ab\lambda {\mathrm{\Omega }}_f$	$=$	$\pi G\rho [\frac{(A_1-A_2)a^2{b}^2}{b^2-a^2}-c^2{A}_3]$
$\Rightarrow c^2{A}_3$	$=$	$\frac{(A_1-A_2)a^2{b}^2}{b^2-a^2}+\frac{ab\lambda {\mathrm{\Omega }}_f}{\pi G\rho }.$

Hence,

${\lambda }^2(a^2-b^2)\{a^2{y}^4-b^2{x}^4\}-\left\{\right[\lambda b^2-ab{\mathrm{\Omega }}_f]x^2+[\lambda a^2-ab{\mathrm{\Omega }}_f]y^2{\}}^2$	$=$	$2\pi G\rho b^2{\varpi }^2{x}^2[\frac{(A_1-A_2)a^2{b}^2}{b^2-a^2}+\frac{ab\lambda {\mathrm{\Omega }}_f}{\pi G\rho }-A_1{a}^2]$
		$+2\pi G\rho a^2{\varpi }^2{y}^2[\frac{(A_1-A_2)a^2{b}^2}{b^2-a^2}+\frac{ab\lambda {\mathrm{\Omega }}_f}{\pi G\rho }-A_2{b}^2]$
	$=$	$\frac{2\pi G\rho b^2{\varpi }^2{x}^2}{(b^2-a^2)}[(A_1-A_2)a^2{b}^2-A_1{a}^2(b^2-a^2)]$
		$+\frac{2\pi G\rho a^2{\varpi }^2{y}^2}{(b^2-a^2)}[(A_1-A_2)a^2{b}^2-A_2{b}^2(b^2-a^2)]$
		$+2ab\lambda {\mathrm{\Omega }}_f{\varpi }^2(a^2{y}^2+b^2{x}^2)$
	$=$	$\frac{2\pi G\rho a^2{b}^2{\varpi }^4}{(b^2-a^2)}[A_1{a}^2-A_2{b}^2]+2ab\lambda {\mathrm{\Omega }}_f{\varpi }^2(a^2{y}^2+b^2{x}^2)$

$\Rightarrow \frac{2\pi G\rho }{(a^2-b^2)}[A_1{a}^2-A_2{b}^2]a^2{b}^2{\varpi }^4$	$=$	$2ab\lambda {\mathrm{\Omega }}_f(x^2+y^2)(a^2{y}^2+b^2{x}^2)-{\lambda }^2(a^2-b^2)[a^2{y}^4-b^2{x}^4]$
		$+\left\{\right[\lambda b^2-ab{\mathrm{\Omega }}_f{]}^2{x}^4+2[\lambda b^2-ab{\mathrm{\Omega }}_f][\lambda a^2-ab{\mathrm{\Omega }}_f]y^2{x}^2+[\lambda a^2-ab{\mathrm{\Omega }}_f{]}^2{y}^4\}$
	$=$	$2ab\lambda {\mathrm{\Omega }}_f(a^2{x}^2{y}^2+b^2{x}^4+a^2{y}^4+b^2{x}^2{y}^2)-{\lambda }^2(a^2-b^2)[a^2{y}^4-b^2{x}^4]$
		$+[{\lambda }^2{b}^4-2ab{\mathrm{\Omega }}_f\lambda b^2+a^2{b}^2{\mathrm{\Omega }}_f^2]x^4+2[{\lambda }^2{a}^2{b}^2-\lambda a{b}^3{\mathrm{\Omega }}_f-\lambda a^{3}b{\mathrm{\Omega }}_f+a^2{b}^2{\mathrm{\Omega }}_f^2]y^2{x}^2$
		$+[{\lambda }^2{a}^4-2ab{\mathrm{\Omega }}_f\lambda a^2+a^2{b}^2{\mathrm{\Omega }}_f^2]y^4$
	$=$	$x^2{y}^2\left\{2\right[{\lambda }^2{a}^2{b}^2-\lambda a{b}^3{\mathrm{\Omega }}_f-\lambda a^{3}b{\mathrm{\Omega }}_f+a^2{b}^2{\mathrm{\Omega }}_f^2]+2ab\lambda {\mathrm{\Omega }}_f(a^2+b^2)\}$
		$+[{\lambda }^2{b}^4-2ab{\mathrm{\Omega }}_f\lambda b^2+a^2{b}^2{\mathrm{\Omega }}_f^2+2a{b}^3\lambda {\mathrm{\Omega }}_f+{\lambda }^2(a^2-b^2)b^2]x^4$
		$+[{\lambda }^2{a}^4-2ab{\mathrm{\Omega }}_f\lambda a^2+a^2{b}^2{\mathrm{\Omega }}_f^2+2ab\lambda {\mathrm{\Omega }}_f{a}^2-{\lambda }^2(a^2-b^2)a^2]y^4$
	$=$	$2a^2{b}^2{x}^2{y}^2[{\lambda }^2+{\mathrm{\Omega }}_f^2]+[{\mathrm{\Omega }}_f^2+{\lambda }^2]a^2{b}^2{x}^4+[{\mathrm{\Omega }}_f^2+{\lambda }^2]a^2{b}^2{y}^4$
	$=$	$[{\lambda }^2+{\mathrm{\Omega }}_f^2]a^2{b}^2[x^4+2x^2{y}^2+y^4]$
$\Rightarrow \frac{2\pi G\rho }{(a^2-b^2)}[A_1{a}^2-A_2{b}^2]$	$=$	$[{\lambda }^2+{\mathrm{\Omega }}_f^2].$
[ EFE, Chapter 7, §48, Eq. (33) ]