Temp001

Based on Detailed Force Balance

The Steady-State Condition

As has been pointed out in our introductory discussion of the Principal Governing Equations, quite generally we can write the

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

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

where, ${\overrightarrow{\mathrm{\Omega }}}_f$ specifies the time-invariant rotation frequency of the frame and the orientation of the vector about which the frame spins. The condition for detailed force balance in a steady-state configuration is obtained by setting $[\partial \overrightarrow{v}/\partial t{]}_{\mathrm{r}\mathrm{o}\mathrm{t}}=0$. If we furthermore make the substitution, $\mathrm{\nabla }H=\mathrm{\nabla }P/\rho$, where $H$ is enthalpy — an equation of state relation that is appropriate for a barytropic system — we obtain,

$({\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}\cdot \mathrm{\nabla }){\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}$

$=$

$-\mathrm{\nabla }[H+{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}-\frac{1}{2}{\mathrm{\Omega }}_f^2(x^2+y^2)]-2{\overrightarrow{\mathrm{\Omega }}}_f\times {\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}.$

📚 Ou (2006), p. 550, §2, Eq. (4)

Adopted Velocity Flow-Field

As 📚 Ou (2006) has pointed out [text that is taken directly from that publication appears here in an orange-colored font], 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:

${\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}$

$=$

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

📚 Lebovitz & Lifschitz (1996), p. 700, §2, Eq. (1)
📚 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 ${\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}$ 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.

Plugging Ou's expression for ${\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}$ into the expression on the left-hand side of the steady-state Euler equation, we see that for Riemann S-type ellipsoids, $({\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}\cdot \mathrm{\nabla }){\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}$ $=$ $\left[\lambda \right(\frac{a}{b})y\frac{\partial }{\partial x}-\lambda (\frac{b}{a}\left)x\frac{\partial }{\partial y}\right]\left[\hat{\mathit{ı}}\lambda \right(\frac{a}{b})y-\hat{\mathit{ȷ}}\lambda (\frac{b}{a}\left)x\right]$   $=$ $-\hat{\mathit{ȷ}}\left[\lambda \right(\frac{a}{b}\left)y\right]\frac{\partial }{\partial x}\left[\lambda \right(\frac{b}{a}\left)x\right]-\hat{\mathit{ı}}\left[\lambda \right(\frac{b}{a}\left)x\right]\frac{\partial }{\partial y}\left[\lambda \right(\frac{a}{b}\left)y\right]$   $=$ $-{\lambda }^2[\hat{\mathit{ı}}x+\hat{\mathit{ȷ}}y]=-\mathrm{\nabla }[\frac{1}{2}{\lambda }^2(x^2+y^2)].$ Alternatively, from a separate discussion of vector identities we realize that, $(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{v}=\frac{1}{2}\mathrm{\nabla }(\overrightarrow{v}\cdot \overrightarrow{v})+\overrightarrow{\zeta }\times \overrightarrow{v},$ where, $\overrightarrow{\zeta }\equiv \mathrm{\nabla }\times \overrightarrow{v}$ is the fluid vorticity. Plugging in Ou's expression for ${\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}$, we find that … $\overrightarrow{\zeta }=\mathrm{\nabla }\times {\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}$ $=$ $-\hat{\mathit{k}}\lambda [\frac{b}{a}+\frac{a}{b}];$ 📚 Ou (2006), p. 551, §2, Eq. (17) $\overrightarrow{\zeta }\times {\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}$ $=$ $-{\lambda }^2\left[\hat{\mathit{ȷ}}\right(1+\frac{a^2}{b^2})y+\hat{\mathit{ı}}(1+\frac{b^2}{a^2}\left)x\right];$     and, $\frac{1}{2}\mathrm{\nabla }({\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}\cdot {\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}})$ $=$ ${\lambda }^2\left[\hat{\mathit{ı}}\right(\frac{b}{a}{)}^{2}x+\hat{\mathit{ȷ}}\left(\frac{a}{b}{)}^{2}y\right].$ Hence, we again appreciate that, for Riemann S-type ellipsoids, $({\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}\cdot \mathrm{\nabla }){\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}$ $=$ $-{\lambda }^2[\hat{\mathit{ı}}x+\hat{\mathit{ȷ}}y]$ $=$ $-\mathrm{\nabla }[\frac{1}{2}{\lambda }^2(x^2+y^2)].$

The steady-state Euler-equation specification therefore becomes,

$-\mathrm{\nabla }[\frac{1}{2}{\lambda }^2(x^2+y^2)]$

$=$

$-\mathrm{\nabla }[H+{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}-\frac{1}{2}{\mathrm{\Omega }}_f^2(x^2+y^2)]-\mathrm{\nabla }\left[{\mathrm{\Omega }}_f\lambda \right(\frac{b}{a}{x}^2+\frac{a}{b}{y}^2\left)\right].$

📚 Ou (2006), p. 550, §2, Eq. (5)

Hence, within the configuration the following Bernoulli's function must be uniform in space:

$H+{\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}-\frac{1}{2}{\mathrm{\Omega }}_f^2(x^2+y^2)-\frac{1}{2}{\lambda }^2(x^2+y^2)+{\mathrm{\Omega }}_f\lambda (\frac{b}{a}{x}^2+\frac{a}{b}{y}^2)$

$=$

$C_B,$

📚 Ou (2006), p. 550, §2, Eq. (6)

where $C_B$ is a constant. It is customary to define an effective potential which is the sum of the gravitational potential and the system's centrifugal potential (as viewed from the rotating frame), namely,

${\mathrm{\Phi }}_{\mathrm{e}\mathrm{f}\mathrm{f}}\equiv {\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}+\mathrm{\Psi }$

$=$

${\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}-\frac{1}{2}{\mathrm{\Omega }}_f^2(x^2+y^2)-\frac{1}{2}{\lambda }^2(x^2+y^2)+{\mathrm{\Omega }}_f\lambda (\frac{b}{a}{x}^2+\frac{a}{b}{y}^2),$

📚 Ou (2006), p. 550, §2, Eq. (7)

in which case the statement of detailed force balance in Riemann S-type ellipsoids can be rewritten in the following deceptively simpler form:

$H+{\mathrm{\Phi }}_{\mathrm{e}\mathrm{f}\mathrm{f}}$

$=$

$C_B.$

📚 Ou (2006), p. 550, §2, Eq. (8)

Evaluation of the Gravitational Potential

Drawing from a separate discussion of the gravitational potential of homogeneous ellipsoids, we see that for Riemann S-type ellipsoids,

${\mathrm{\Phi }}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}(\overrightarrow{x})=-\pi G\rho [I_{\mathrm{B}\mathrm{T}}{a}^2-(A_1{x}^2+A_2{y}^2+A_3{z}^2\left)\right],$

[ EFE, Chapter 3, Eq. (40)^1,2 ]
[ BT87, Chapter 2, Table 2-2 ]

where, the normalization constant,

$I_{\mathrm{B}\mathrm{T}}$

$=A_1+A_2(\frac{b}{a}{)}^2+A_3(\frac{c}{a}{)}^2.$

[ EFE, Chapter 3, Eq. (22)¹]
[ BT87, Chapter 2, Table 2-2 ]

Implied Parameter Values

So, at the surface of the ellipsoid (where the enthalpy H = 0) on each of its three principal axes, the equilibrium conditions demanded by the expression for detailed force balance become, respectively:

1. On the x-axis, where (x, y, z) = (a, 0, 0):
   

   $C_B$	$=$	$-\pi G\rho [I_{\mathrm{B}\mathrm{T}}{a}^2-A_1{a}^2]-\frac{1}{2}{\mathrm{\Omega }}_f^2(a^2)-\frac{1}{2}{\lambda }^2(a^2)+{\mathrm{\Omega }}_f\lambda (\frac{b}{a}\cdot a^2)$
   $\Rightarrow 2[\frac{C_B}{a^2}+(\pi G\rho )I_{\mathrm{B}\mathrm{T}}]$	$=$	$(2\pi G\rho )A_1-{\mathrm{\Omega }}_f^2-{\lambda }^2+2{\mathrm{\Omega }}_f\lambda \left(\frac{b}{a}\right)$

2. On the y-axis, where (x, y, z) = (0, b, 0):
   

   $C_B$	$=$	$-\pi G\rho [I_{\mathrm{B}\mathrm{T}}{a}^2-A_2{b}^2]-\frac{1}{2}{\mathrm{\Omega }}_f^2(b^2)-\frac{1}{2}{\lambda }^2(b^2)+{\mathrm{\Omega }}_f\lambda (\frac{a}{b}\cdot b^2)$
   $\Rightarrow 2[\frac{C_B}{a^2}+(\pi G\rho )I_{\mathrm{B}\mathrm{T}}]$	$=$	$(2\pi G\rho )A_2\left(\frac{b^2}{a^2}\right)-{\mathrm{\Omega }}_f^2\left(\frac{b^2}{a^2}\right)-{\lambda }^2\left(\frac{b^2}{a^2}\right)+2{\mathrm{\Omega }}_f\lambda \left(\frac{b}{a}\right)$

3. On the z-axis, where (x, y, z) = (0, 0, c):
   

   $C_B$	$=$	$-\pi G\rho [I_{\mathrm{B}\mathrm{T}}{a}^2-A_3{c}^2]$
   $\Rightarrow 2[\frac{C_B}{a^2}+(\pi G\rho )I_{\mathrm{B}\mathrm{T}}]$	$=$	$(2\pi G\rho )A_3\left(\frac{c^2}{a^2}\right)$

Using the result from "III" to replace the left-hand side of both relation "I" and relation "II", we find that,

$(2\pi G\rho )A_3\left(\frac{c^2}{a^2}\right)$

$=$

$(2\pi G\rho )A_1-{\mathrm{\Omega }}_f^2-{\lambda }^2+2{\mathrm{\Omega }}_f\lambda \left(\frac{b}{a}\right),$

and,

$(2\pi G\rho )A_3\left(\frac{c^2}{a^2}\right)$

$=$

$(2\pi G\rho )A_2\left(\frac{b^2}{a^2}\right)-{\mathrm{\Omega }}_f^2\left(\frac{b^2}{a^2}\right)-{\lambda }^2\left(\frac{b^2}{a^2}\right)+2{\mathrm{\Omega }}_f\lambda \left(\frac{b}{a}\right).$

Multiplying the first of these two expressions by $(b/a{)}^2$ then subtracting it from the second gives,

$(2\pi G\rho )A_3\left(\frac{c^2}{a^2}\right)-(\frac{b}{a}{)}^2(2\pi G\rho )A_3(\frac{c^2}{a^2})$	$=$	$(2\pi G\rho )A_2\left(\frac{b^2}{a^2}\right)+2{\mathrm{\Omega }}_f\lambda \left(\frac{b}{a}\right)-\left(\frac{b}{a}{)}^2\right[(2\pi G\rho )A_1+2{\mathrm{\Omega }}_f\lambda \left(\frac{b}{a}\right)]$
$\Rightarrow \frac{(\pi G\rho )c^2}{ab}[A_3{a}^2-A_3{b}^2]$	$=$	$(\pi G\rho )(A_2-A_1)ab+{\mathrm{\Omega }}_f\lambda a^2-ab\left[{\mathrm{\Omega }}_f\lambda \right(\frac{b}{a}\left)\right]$
	$=$	$(\pi G\rho )(A_2-A_1)ab+{\mathrm{\Omega }}_f\lambda (a^2-b^2)$
$\Rightarrow \frac{{\mathrm{\Omega }}_f\lambda }{\pi G\rho }$	$=$	$\frac{1}{ab(a^2-b^2)}[A_3(a^2-b^2)c^2-(A_2-A_1)a^2{b}^2].$

Alternatively, just subtracting the first expression from the second gives,

$0$	$=$	$(2\pi G\rho )A_2\left(\frac{b^2}{a^2}\right)-{\mathrm{\Omega }}_f^2\left(\frac{b^2}{a^2}\right)-{\lambda }^2\left(\frac{b^2}{a^2}\right)-[(2\pi G\rho )A_1-{\mathrm{\Omega }}_f^2-{\lambda }^2]$
	$=$	$(2\pi G\rho )\left[A_2\right(\frac{b^2}{a^2})-A_1]+{\mathrm{\Omega }}_f^2[1-\frac{b^2}{a^2}]+{\lambda }^2[1-\frac{b^2}{a^2}]$
$\Rightarrow \frac{{\mathrm{\Omega }}_f^2+{\lambda }^2}{\pi G\rho }$	$=$	$2[A_1-A_2(\frac{b^2}{a^2}\left)\right]\left[\frac{a^2}{a^2-b^2}\right].$

We can eliminate $\lambda$ between these last two expressions as follows: From the first of the two, we have

$\lambda$

$=$

$\frac{1}{{\mathrm{\Omega }}_f}\left\{\frac{\pi G\rho }{ab(a^2-b^2)}\right[A_3(a^2-b^2)c^2-(A_2-A_1)a^2{b}^2\left]\right\}.$

Hence, the second gives,

${\mathrm{\Omega }}_f^2$	$=$	$2(\pi G\rho )[A_1-A_2(\frac{b^2}{a^2}\left)\right]\left[\frac{a^2}{a^2-b^2}\right]-{\lambda }^2$
	$=$	$2(\pi G\rho )[A_1-A_2(\frac{b^2}{a^2}\left)\right]\left[\frac{a^2}{a^2-b^2}\right]-\frac{1}{{\mathrm{\Omega }}_f^2}\left\{\frac{\pi G\rho }{ab(a^2-b^2)}\right[A_3(a^2-b^2)c^2-(A_2-A_1)a^2{b}^2]{\}}^2$
$\Rightarrow 0$	$=$	$\frac{{\mathrm{\Omega }}_f^4}{(\pi G\rho {)}^2}-\frac{2{\mathrm{\Omega }}_f^2}{(\pi G\rho )}[A_1-A_2(\frac{b^2}{a^2}\left)\right]\left[\frac{a^2}{a^2-b^2}\right]+\left\{\frac{1}{ab(a^2-b^2)}\right[A_3(a^2-b^2)c^2-(A_2-A_1)a^2{b}^2]{\}}^2.$

This is a quadratic equation whose solution gives ${\mathrm{\Omega }}_f^2/(\pi G\rho )$ and, in turn, ${\lambda }^2/(\pi G\rho )$. Specifically for Direct configurations, we find that,

$\frac{{\mathrm{\Omega }}_f^2}{(\pi G\rho )}$

$=$

$\frac{1}{2}[M+\sqrt{M^2-4N^2}],$

and

$\frac{{\lambda }^2}{(\pi G\rho )}$

$=$

$\frac{1}{2}[M-\sqrt{M^2-4N^2}],$

📚 Ou (2006), p. 551, §2, Eqs. (15) & (16)

where,

$M$	$\equiv$	$2[A_1-A_2(\frac{b^2}{a^2}\left)\right]\left[\frac{a^2}{a^2-b^2}\right],$     and,
$N$	$\equiv$	$\frac{1}{ab(a^2-b^2)}[A_3(a^2-b^2)c^2-(A_2-A_1)a^2{b}^2].$

TEST (part 3)
$\frac{b}{a}$	$\frac{c}{a}$	$A_1$	$A_2$	$A_3$	$M$	$N$	$\frac{{\mathrm{\Omega }}_f^2}{\pi G\rho }$	$\frac{{\lambda }^2}{\pi G\rho }$	$\frac{{\mathrm{\Omega }}_f}{\sqrt{G\rho }}$	$\frac{\lambda }{\sqrt{G\rho }}$
0.9	0.641	0.521450273	0.595131012	0.883418715	0.414682903	0.054301271	0.407446048	0.007236855	1.131383892	0.150782130

The numerical values listed in the last two columns of this "part 3" test match the values listed above in "part 2" of our test for, respectively, $\omega$ and ${\omega }^{†}$.