PGE/RotatingFrame

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

Rotating Reference Frame

Overview

Among the principal governing equations we have included the  
 

Lagrangian Representation
of the inertial-frame Euler Equation,

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

[BT87], p. 671, Appendix Eq. (1E-6)
[BLRY07], p. 13, Eq. (1.55)

Alternatively, a rewrite of the LHS gives what we refer to as the,

Eulerian Representation of the intertial-frame Euler Equation
$\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{v}$	$=$	$-\frac{1}{\rho }\mathrm{\nabla }P-\mathrm{\nabla }\mathrm{\Phi }.$

At times, it can be useful to view the motion of a fluid from a frame of reference that is rotating with an angular velocity

$\overrightarrow{\mathrm{\Omega }}=\hat{ı}{\mathrm{\Omega }}_1+\hat{ȷ}{\mathrm{\Omega }}_2+\hat{k}{\mathrm{\Omega }}_3.$

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Often it suffices to align $\overrightarrow{\mathrm{\Omega }}$ with the z-axis of the chosen coordinate system — in which case, ${\mathrm{\Omega }}_1={\mathrm{\Omega }}_2=0$ — and to set $d\overrightarrow{\mathrm{\Omega }}/dt=0$, in which case the nonzero component of the frame's angular velocity, ${\mathrm{\Omega }}_3$, is independent of time.

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In what follows we show that, when viewed from this rotating reference frame, we have what will be referred to as the

Lagrangian Representation of the rotating-frame Euler Equation
$\frac{d\overrightarrow{u}}{dt}$	$=$	$2\overrightarrow{u}\times \overrightarrow{\mathrm{\Omega }}+\overrightarrow{\mathrm{\Omega }}\times (\overrightarrow{x}\times \overrightarrow{\mathrm{\Omega }})+\overrightarrow{x}\times \frac{d\overrightarrow{\mathrm{\Omega }}}{dt}-[\frac{1}{\rho }\mathrm{\nabla }P+\mathrm{\nabla }\mathrm{\Phi }{]}_{\mathrm{r}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}},$
Template:Rossner67, §II, Eq. (1) [BT87], Appendix 1.D, §3, (p. 664) Eq. (1D-42)

where the difference between the rotating-frame velocity, $\overrightarrow{u}$, and the inertial-frame velocity, $\overrightarrow{v}$, is given by the expression,

$\overrightarrow{v}-\overrightarrow{u}$	$=$	$\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}$
	$=$	$\hat{ı}({\mathrm{\Omega }}_{2}z-{\mathrm{\Omega }}_{3}y)+\hat{ȷ}({\mathrm{\Omega }}_{3}x-{\mathrm{\Omega }}_{1}z)+\hat{k}({\mathrm{\Omega }}_{1}y-{\mathrm{\Omega }}_{2}x)).$

As above, a rewrite of the LHS gives what we will refer to as the,

Eulerian Representation of the rotating-frame Euler Equation
$\frac{\partial \overrightarrow{u}}{\partial t}+(\overrightarrow{u}\cdot \mathrm{\nabla })\overrightarrow{u}$	$=$	$2\overrightarrow{u}\times \overrightarrow{\mathrm{\Omega }}+\overrightarrow{\mathrm{\Omega }}\times (\overrightarrow{x}\times \overrightarrow{\mathrm{\Omega }})+\overrightarrow{x}\times \frac{d\overrightarrow{\mathrm{\Omega }}}{dt}-[\frac{1}{\rho }\mathrm{\nabla }P+\mathrm{\nabla }\mathrm{\Phi }{]}_{\mathrm{r}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}}.$

Along the way, and being guided by Chandrasekhar's presentation in Chapter 4, §25 of [EFE], we appreciate that it can be useful to highlight a hybrid representation of the Euler Equation that involves a mixture of the velocity variables, $\overrightarrow{u}$ along with $\overrightarrow{v}$. For example, beginning with this last expressions, we can write,

$2\overrightarrow{u}\times \overrightarrow{\mathrm{\Omega }}+\overrightarrow{\mathrm{\Omega }}\times (\overrightarrow{x}\times \overrightarrow{\mathrm{\Omega }})+\overrightarrow{x}\times \frac{d\overrightarrow{\mathrm{\Omega }}}{dt}-[\frac{1}{\rho }\mathrm{\nabla }P+\mathrm{\nabla }\mathrm{\Phi }]$	$=$	$\frac{\partial \overrightarrow{u}}{\partial t}+(\overrightarrow{u}\cdot \mathrm{\nabla })\overrightarrow{u}$
	$=$	$\frac{\partial }{\partial t}[\overrightarrow{v}-\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}]+(\overrightarrow{u}\cdot \mathrm{\nabla })[\overrightarrow{v}-\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}]$
	$=$	$[\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{u}\cdot \mathrm{\nabla })\overrightarrow{v}]+\frac{\partial }{\partial t}[-\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}]+(\overrightarrow{u}\cdot \mathrm{\nabla })[-\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}]$
	$=$	$[\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{u}\cdot \mathrm{\nabla })\overrightarrow{v}]-\frac{d}{dt}[\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}]$
	$=$	$[\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{u}\cdot \mathrm{\nabla })\overrightarrow{v}]+\overrightarrow{x}\times \frac{d\overrightarrow{\mathrm{\Omega }}}{\partial t}-\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{v}$
	$=$	$[\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{u}\cdot \mathrm{\nabla })\overrightarrow{v}]+\overrightarrow{x}\times \frac{d\overrightarrow{\mathrm{\Omega }}}{dt}-\overrightarrow{\mathrm{\Omega }}\times [\overrightarrow{u}+\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}]$
	$=$	$[\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{u}\cdot \mathrm{\nabla })\overrightarrow{v}]+\overrightarrow{x}\times \frac{d\overrightarrow{\mathrm{\Omega }}}{dt}+\overrightarrow{u}\times \overrightarrow{\mathrm{\Omega }}+\overrightarrow{\mathrm{\Omega }}\times (\overrightarrow{x}\times \overrightarrow{\mathrm{\Omega }})$
$\Rightarrow [\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{u}\cdot \mathrm{\nabla })\overrightarrow{v}]$	$=$	$\overrightarrow{u}\times \overrightarrow{\mathrm{\Omega }}-[\frac{1}{\rho }\mathrm{\nabla }P+\mathrm{\nabla }\mathrm{\Phi }]$

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$-\frac{1}{\rho }\mathrm{\nabla }P-\mathrm{\nabla }\mathrm{\Phi }$	$=$	$\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{v}\cdot \mathrm{\nabla })[\overrightarrow{u}+\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}]$
	$=$	$\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{u}+(\overrightarrow{v}\cdot \mathrm{\nabla })[\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}]$
	$=$	$\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{u}+[v_x\frac{\partial }{\partial x}+v_y\frac{\partial }{\partial y}+v_z\frac{\partial }{\partial z}][\hat{ı}({\mathrm{\Omega }}_{2}z-{\mathrm{\Omega }}_{3}y)+\hat{ȷ}({\mathrm{\Omega }}_{3}x-{\mathrm{\Omega }}_{1}z)+\hat{k}({\mathrm{\Omega }}_{1}y-{\mathrm{\Omega }}_{2}x))]$
	$=$	$\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{u}+v_x[\hat{ȷ}({\mathrm{\Omega }}_3)-\hat{k}({\mathrm{\Omega }}_2)]+v_y[-\hat{ı}({\mathrm{\Omega }}_3)+\hat{k}({\mathrm{\Omega }}_1)]+v_z[\hat{ı}({\mathrm{\Omega }}_2)-\hat{ȷ}({\mathrm{\Omega }}_1)]$
	$=$	$\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{u}+\underset{\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{v}}{\underbrace{\hat{ı}({\mathrm{\Omega }}_2{v}_z-{\mathrm{\Omega }}_3{v}_y)+\hat{ȷ}({\mathrm{\Omega }}_3{v}_x-{\mathrm{\Omega }}_1{v}_z)+\hat{k}({\mathrm{\Omega }}_1{v}_y-{\mathrm{\Omega }}_2{v}_x)}}.$

That is, along the way we derive a,

Hybrid (Eulerian) Representation of the rotating-frame Euler Equation
$\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{v}\cdot \mathrm{\nabla })\overrightarrow{u}$	$=$	$-\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{v}-[\frac{1}{\rho }\mathrm{\nabla }P+\mathrm{\nabla }\mathrm{\Phi }].$

CAUTION! If our interpretation of Chandrasekhar's discussion of "moving frames" is correct — see, Chap. 4, §25 of [EFE] — the RHS of his Eq. (18) should match the LHS of our "hybrid" equation, but it does not: the pair of vector velocities in his advection term are swapped. That is, based on our interpretation, the RHS of his Eq. (18) reads, RHS $=$ $\frac{\partial \overrightarrow{v}}{\partial t}+(\overrightarrow{u}\cdot \mathrm{\nabla })\overrightarrow{v};$ and this expression carries over to the LHS of his Eq. (19). This is either a mistake in his presentation, or our interpretation of his presentation is incorrect.

Coordinate Transformation

Traditional Presentation

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 ${\mathrm{\Omega }}_f$. 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, ${\overrightarrow{\mathrm{\Omega }}}_f$; and the $d/dt$ operator, which denotes Lagrangian time-differentiation in the inertial frame, must everywhere be replaced as follows:

$[\frac{d}{dt}{]}_{inertial}\to [\frac{d}{dt}{]}_{rot}+{\overrightarrow{\mathrm{\Omega }}}_f\times .$

[EFE], Chap. 4, §25, Eq. (11)

Operating on the fluid element's position vector, $\overrightarrow{x}$, we obtain the transformation,

$\frac{d\overrightarrow{x}}{dt}{|}_{\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}$

$\to$

$\frac{d\overrightarrow{x}}{dt}{|}_{\mathrm{r}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}}+\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x},$

that is,

${\overrightarrow{v}}_{\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}$	$\to$	${\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}}+\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}.$
[EFE], Chap. 4, §25, Eq. (15)

Performing this transformation implies, for example, that

${\overrightarrow{v}}_{inertial}={\overrightarrow{v}}_{rot}+{\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x},$

and,

$[\frac{d\overrightarrow{v}}{dt}{]}_{inertial}=[\frac{d\overrightarrow{v}}{dt}{]}_{rot}+2{\overrightarrow{\mathrm{\Omega }}}_f\times {\overrightarrow{v}}_{rot}+{\overrightarrow{\mathrm{\Omega }}}_f\times ({\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x})$

$=[\frac{d\overrightarrow{v}}{dt}{]}_{rot}+2{\overrightarrow{\mathrm{\Omega }}}_f\times {\overrightarrow{v}}_{rot}-\frac{1}{2}\mathrm{\nabla }[|{\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x}{|}^2]$

(If we were to allow ${\overrightarrow{\mathrm{\Omega }}}_f$ to be a function of time, an additional term involving the time-derivative of ${\overrightarrow{\mathrm{\Omega }}}_f$ 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,

$[\overrightarrow{\zeta }{]}_{inertial}=[\overrightarrow{\zeta }{]}_{rot}+2{\overrightarrow{\mathrm{\Omega }}}_f.$

Chandrasekhar's Approach

Here we draw extensively from Chapter 4, §25 of [EFE].

Transformation Matrix

Following Chandrasekhar, we let $\overrightarrow{X}$ represent the inertial-frame position vector of a fluid element, in which case $d\overrightarrow{X}/dt$ is the inertial-frame velocity $(\overrightarrow{v})$ of that fluid element, and the acceleration, $d\overrightarrow{v}/dt$, that appears on the LHS of the Lagrangian representation of the (intertial-frame) Euler equation may be rewritten as the second time-derivative of $\overrightarrow{X}$, namely,

Lagrangian Representation of the (inertial-frame) Euler Equation
$\frac{d^2\overrightarrow{X}}{d{t}^2}$	$=$	$-[\frac{1}{\rho }\mathrm{\nabla }P+\mathrm{\nabla }\mathrm{\Phi }{]}_{\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}.$

Chandrasekhar uses the matrix, $\mathbf{𝐓}(t)$, to represent the (time-dependent) linear transformation that relates $\overrightarrow{X}$ to the corresponding moving-frame position vector, $\overrightarrow{x}$. Specifically, he sets,

$\overrightarrow{x}$	$=$	$\mathbf{𝐓}\overrightarrow{X}.$
[EFE], Chap. 4, §25, Eq. (1)

Applying the same transformation to the inertial-frame velocity, $d\overrightarrow{X}/dt$, gives,

$\overrightarrow{U}$	$=$	$\mathbf{𝐓}\frac{d\overrightarrow{X}}{dt},$
[EFE], Chap. 4, §25, Eq. (14a)

which Chandrasekhar refers to as the velocity in the inertial frame that has been "resolved along the instantaneous coordinate axes of the moving frame." And applying this transformation to the inertial-frame acceleration gives the term, $\mathbf{𝐓}[d^2\overrightarrow{X}/d{t}^2]$, which Chandrasekhar describes as representing "… the acceleration in the inertial frame resolved, however, along the instantaneous directions of the coordinate axes of the moving frame." Applying the transformation to both sides of the Lagrangian representation of the Euler equation gives,

$\mathbf{𝐓}\frac{d^2\overrightarrow{X}}{d{t}^2}$	$=$	$-[\frac{1}{\rho }\mathrm{\nabla }P+\mathrm{\nabla }\mathrm{\Phi }{]}_{\mathrm{m}\mathrm{o}\mathrm{v}\mathrm{i}\mathrm{n}\mathrm{g}},$
[EFE], Chap. 4, §25, combination of Eqs. (16) & (17)

where, as Chandrasekhar clarifies, the gradients on the RHS must be "… evaluated in the coordinates of the moving frame."

Rotating-Frame Euler Equation

Foundation

Suppose the 3-component vector, $\overrightarrow{\mathrm{\Omega }}$, represents a general time-dependent rotation of the $(x_1,x_2,x_3)$-frame with respect to the inertial frame. In this context, Chandrasekhar introduces a (3 × 3) matrix, ${\mathrm{\Omega }}^{\mathbf{*}}$, whose nine components can be expressed in terms of the three components of $\overrightarrow{\mathrm{\Omega }}$ via the relations,

$({\mathrm{\Omega }}^{*}{)}_{ij}$	$=$	${ϵ}_{ijk}{\mathrm{\Omega }}_k.$
[EFE], Chap. 4, §25, Eq. (6a)

Alternatively, we may write,

$({\mathrm{\Omega }}^{*}{)}_{ik}$

$=$

${ϵ}_{ikj}{\mathrm{\Omega }}_j=-{ϵ}_{ijk}{\mathrm{\Omega }}_j.$

Both of these expressions make use of the three-element Levi-Civita tensor, ${ϵ}_{ijk}$. Its six nonzero component values are … $ijk$ ${ϵ}_{ijk}$   $ijk$ ${ϵ}_{ijk}$ 123 +1 132 -1 312 321 231 213 Hence, the six nonzero components of the matrix, ${\mathrm{\Omega }}^{\mathbf{*}}$, are, $({\mathrm{\Omega }}^{*}{)}_{12}$ $=$ ${\mathrm{\Omega }}_3;$       $({\mathrm{\Omega }}^{*}{)}_{13}$ $=$ $-{\mathrm{\Omega }}_2;$ $({\mathrm{\Omega }}^{*}{)}_{21}$ $=$ $-{\mathrm{\Omega }}_3;$       $({\mathrm{\Omega }}^{*}{)}_{23}$ $=$ ${\mathrm{\Omega }}_1;$ $({\mathrm{\Omega }}^{*}{)}_{31}$ $=$ ${\mathrm{\Omega }}_2;$       $({\mathrm{\Omega }}^{*}{)}_{32}$ $=$ $-{\mathrm{\Omega }}_1.$ ${\mathrm{\Omega }}^{\mathbf{*}}$ (3 × 3 matrix) $0$ ${\mathrm{\Omega }}_3$ $-{\mathrm{\Omega }}_2$ $-{\mathrm{\Omega }}_3$ $0$ ${\mathrm{\Omega }}_1$ ${\mathrm{\Omega }}_2$ $-{\mathrm{\Omega }}_1$ $0$

For later use, we note as well that for an arbitrary vector — call it, $\overrightarrow{Q}$ — the individual components of the product, ${\mathrm{\Omega }}^{\mathbf{*}}\overrightarrow{Q}$, are given by the expression,

$({\mathrm{\Omega }}^{\mathbf{*}}\overrightarrow{Q}{)}_i$

$=$

$({\mathrm{\Omega }}^{*}{)}_{ij}{Q}_j=({ϵ}_{ijk}{\mathrm{\Omega }}_k)Q_j.$

Compare, for example, Eqs. (17) and (19) in §25 of [EFE].

Now, if the motion of the moving frame relative to the inertial frame is specified entirely by the vector $\overrightarrow{\mathrm{\Omega }}$, Chandrasekhar proves that any time-dependent vector defined in the inertial frame — call it $\overrightarrow{F}$ — will obey the following operator relation:

$0$	$=$	$[\mathbf{𝐓}\frac{d}{dt}-(\frac{d}{dt}-{\mathrm{\Omega }}^{\mathbf{*}}\left)\mathbf{𝐓}\right]\overrightarrow{F}.$
[EFE], Chap. 4, §25, Eq. (11)

Lagrangian Representation In Terms of Inertial-Frame Velocities

For example, if we set $\overrightarrow{F}=d\overrightarrow{X}/dt$, we find,

$0$	$=$	$[\mathbf{𝐓}\frac{d}{dt}-(\frac{d}{dt}-{\mathrm{\Omega }}^{\mathbf{*}}\left)\mathbf{𝐓}\right]\frac{d\overrightarrow{X}}{dt}$
$\Rightarrow \mathbf{𝐓}\frac{d^2\overrightarrow{X}}{d{t}^2}$	$=$	$(\frac{d}{dt}-{\mathrm{\Omega }}^{\mathbf{*}})\mathbf{𝐓}\frac{d\overrightarrow{X}}{dt}$
	$=$	$\frac{d\overrightarrow{U}}{dt}-{\mathrm{\Omega }}^{\mathbf{*}}\overrightarrow{U}.$
[EFE], Chap. 4, §25, Eqs. (13) & (16)

This allows us to write the,

Lagrangian Representation of the rotating-frame Euler Equation in terms of the (transformed) inertial-frame velocity, $\overrightarrow{U}$
$\frac{d\overrightarrow{U}}{dt}-{\mathrm{\Omega }}^{\mathbf{*}}\overrightarrow{U}$	$=$	$-[\frac{1}{\rho }\mathrm{\nabla }P+\mathrm{\nabla }\mathrm{\Phi }{]}_{\mathrm{r}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}}$
$\Rightarrow \frac{d\overrightarrow{U}}{dt}$	$=$	${\mathrm{\Omega }}^{\mathbf{*}}\overrightarrow{U}-[\frac{1}{\rho }\mathrm{\nabla }P+\mathrm{\nabla }\mathrm{\Phi }{]}_{\mathrm{r}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}}.$
[EFE], Chap. 4, §25, Eq. (17)

Appreciating from above that $({\mathrm{\Omega }}^{\mathbf{*}}\overrightarrow{Q}{)}_i={ϵ}_{ijk}{\mathrm{\Omega }}_k{Q}_j$, in component form this version of the Euler equation reads, $\frac{d{U}_i}{dt}$ $=$ ${ϵ}_{imk}{\mathrm{\Omega }}_k{U}_m-\frac{1}{\rho }\frac{\partial p}{\partial x_i}-\frac{\partial \mathrm{\Phi }}{\partial x_i}.$ That is, Component #1:    $\frac{d{U}_1}{dt}$ $=$ ${\mathrm{\Omega }}_3{U}_2-{\mathrm{\Omega }}_2{U}_3-\frac{1}{\rho }\frac{\partial p}{\partial x_1}-\frac{\partial \mathrm{\Phi }}{\partial x_1}$ Component #2:    $\frac{d{U}_2}{dt}$ $=$ ${\mathrm{\Omega }}_1{U}_3-{\mathrm{\Omega }}_3{U}_1-\frac{1}{\rho }\frac{\partial p}{\partial x_2}-\frac{\partial \mathrm{\Phi }}{\partial x_2}$ Component #3:    $\frac{d{U}_3}{dt}$ $=$ ${\mathrm{\Omega }}_2{U}_1-{\mathrm{\Omega }}_1{U}_2-\frac{1}{\rho }\frac{\partial p}{\partial x_3}-\frac{\partial \mathrm{\Phi }}{\partial x_3}$ Notice as well that the individual components of the cross product of $\overrightarrow{U}$ and $\overrightarrow{\mathrm{\Omega }}$ can be represented by the same summation expression, that is, $(\overrightarrow{U}\times \overrightarrow{\mathrm{\Omega }}{)}_i$ $=$ ${ϵ}_{imk}{\mathrm{\Omega }}_k{U}_m.$ This allows us to rewrite the, Lagrangian Representation of the rotating-frame Euler Equation in terms of the (transformed) inertial-frame velocity, $\overrightarrow{U}$ $\frac{d\overrightarrow{U}}{dt}$ $=$ $-\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{U}-[\frac{1}{\rho }\mathrm{\nabla }P+\mathrm{\nabla }\mathrm{\Phi }{]}_{\mathrm{r}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}}.$ in what is perhaps more recognizable notation.

Lagrangian Representation In Terms of Rotating-Frame Velocities

Alternatively, setting $\overrightarrow{F}=\overrightarrow{X}$ gives,

$0$	$=$	$[\mathbf{𝐓}\frac{d}{dt}-(\frac{d}{dt}-{\mathrm{\Omega }}^{\mathbf{*}}\left)\mathbf{𝐓}\right]\overrightarrow{X}$
$\Rightarrow \mathbf{𝐓}\frac{d\overrightarrow{X}}{dt}$	$=$	$(\frac{d}{dt}-{\mathrm{\Omega }}^{\mathbf{*}})\mathbf{𝐓}\overrightarrow{X}$
$\Rightarrow \overrightarrow{U}$	$=$	$(\frac{d}{dt}-{\mathrm{\Omega }}^{\mathbf{*}})\overrightarrow{x}$
	$=$	$\overrightarrow{u}-{\mathrm{\Omega }}^{\mathbf{*}}\overrightarrow{x},$
[EFE], Chap. 4, §25, Eqs. (12) & (15)

where, adopting Chandrasekhar's notation, the variable,

$\overrightarrow{u}$	$\equiv$	$\frac{d\overrightarrow{x}}{dt},$
[EFE], Chap. 4, §25, Eq. (14b)

denotes the fluid velocity as measured "… with respect to an observer [that is] at rest in the moving frame." This allows us to write the,

Lagrangian Representation of the rotating-frame Euler Equation in terms of the rotating-frame velocity, $\overrightarrow{u}$
$\frac{d}{dt}[\overrightarrow{u}-{\mathrm{\Omega }}^{\mathbf{*}}\overrightarrow{x}]-{\mathrm{\Omega }}^{\mathbf{*}}[\overrightarrow{u}-{\mathrm{\Omega }}^{\mathbf{*}}\overrightarrow{x}]$	$=$	$-[\frac{1}{\rho }\mathrm{\nabla }P+\mathrm{\nabla }\mathrm{\Phi }{]}_{\mathrm{r}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}}$

Again appreciating from above that $({\mathrm{\Omega }}^{\mathbf{*}}\overrightarrow{Q}{)}_i={ϵ}_{ijk}{\mathrm{\Omega }}_k{Q}_j=-{ϵ}_{ijk}{\mathrm{\Omega }}_j{Q}_k$, in component form this version of the Euler equation reads, $\frac{d}{dt}[u_i+{ϵ}_{ijk}{\mathrm{\Omega }}_j{x}_k]$ $=$ ${ϵ}_{imk}{\mathrm{\Omega }}_k[u_m+{ϵ}_{mjk}{\mathrm{\Omega }}_j{x}_k]-\frac{1}{\rho }\frac{\partial p}{\partial x_i}-\frac{\partial \mathrm{\Phi }}{\partial x_i}$ $\Rightarrow \frac{d{u}_i}{dt}+{ϵ}_{ijk}\left[\right(\frac{d{\mathrm{\Omega }}_j}{dt})x_k+{\mathrm{\Omega }}_j(\frac{d{x}_k}{dt}\left)\right]$ $=$ ${ϵ}_{imk}{u}_m{\mathrm{\Omega }}_k+{ϵ}_{imk}{\mathrm{\Omega }}_k\left[{ϵ}_{mjk}{\mathrm{\Omega }}_j{x}_k\right]-\frac{1}{\rho }\frac{\partial p}{\partial x_i}-\frac{\partial \mathrm{\Phi }}{\partial x_i}.$ Now, if we … Swap the "jk" indices of the various terms on the LHS, which dictates that the leading sign be swapped as well: ${ϵ}_{ijk}\left[\right(\frac{d{\mathrm{\Omega }}_j}{dt})x_k+{\mathrm{\Omega }}_j(\frac{d{x}_k}{dt}\left)\right]$ $\to$ $-{ϵ}_{ijk}\left[x_j\right(\frac{d{\mathrm{\Omega }}_k}{dt})+u_j{\mathrm{\Omega }}_k];$ note also that we have set $d{x}_j/dt\to u_j$; In the first term on the RHS, replace the index, "m", with the index, "j": ${ϵ}_{imk}{u}_m{\mathrm{\Omega }}_k$ $\to$ ${ϵ}_{ijk}{u}_j{\mathrm{\Omega }}_k;$ Inside the square brackets of the second term on the RHS, replace the "jk" indices with "hℓ" in order to avoid confusion, then swap the "hℓ" indices of the two variables, which dictates that the leading sign be swapped as well: $\left[{ϵ}_{mjk}{\mathrm{\Omega }}_j{x}_k\right]$ $\to$ $\left[{ϵ}_{mh\ell }{\mathrm{\Omega }}_h{x}_{\ell }\right]$ $\to$ $[-{ϵ}_{mh\ell }{x}_h{\mathrm{\Omega }}_{\ell }];$ Swap the "mk" indices on the Levi-Civiti tensor that lies just outside the square brackets of the second term on the RHS, which dictates that the leading sign be swapped as well: ${ϵ}_{imk}{\mathrm{\Omega }}_k[-{ϵ}_{mh\ell }{x}_h{\mathrm{\Omega }}_{\ell }]$ $\to$ $-{ϵ}_{ikm}{\mathrm{\Omega }}_k[-{ϵ}_{mh\ell }{x}_h{\mathrm{\Omega }}_{\ell }]$ $\to$ ${ϵ}_{ikm}{\mathrm{\Omega }}_k\left[{ϵ}_{mh\ell }{x}_h{\mathrm{\Omega }}_{\ell }\right];$ the Euler equation becomes, $\frac{d{u}_i}{dt}-{ϵ}_{ijk}\left[x_j\right(\frac{d{\mathrm{\Omega }}_k}{dt})+u_j{\mathrm{\Omega }}_k]$ $=$ ${ϵ}_{ijk}{u}_j{\mathrm{\Omega }}_k+{ϵ}_{ikm}{\mathrm{\Omega }}_k\left[{ϵ}_{mh\ell }{x}_h{\mathrm{\Omega }}_{\ell }\right]-\frac{1}{\rho }\frac{\partial p}{\partial x_i}-\frac{\partial \mathrm{\Phi }}{\partial x_i}$ $\Rightarrow \frac{d{u}_i}{dt}$ $=$ $\underset{[2\overrightarrow{u}\times \overrightarrow{\mathrm{\Omega }}{]}_i}{\underbrace{2{ϵ}_{ijk}{u}_j{\mathrm{\Omega }}_k}}+\underset{[\overrightarrow{\mathrm{\Omega }}\times (\overrightarrow{x}\times \overrightarrow{\mathrm{\Omega }}){]}_i}{\underbrace{{ϵ}_{ikm}{\mathrm{\Omega }}_k\left[{ϵ}_{mh\ell }{x}_h{\mathrm{\Omega }}_{\ell }\right]}}+\underset{[\overrightarrow{x}\times (d\overrightarrow{\mathrm{\Omega }}/dt){]}_i}{\underbrace{{ϵ}_{ijk}\left[x_j\right(\frac{d{\mathrm{\Omega }}_k}{dt}\left)\right]}}-\frac{1}{\rho }\frac{\partial p}{\partial x_i}-\frac{\partial \mathrm{\Phi }}{\partial x_i}.$

We therefore can rewrite in a more familiar vector formulation, the

Lagrangian Representation of the rotating-frame Euler Equation in terms of the rotating-frame velocity, $\overrightarrow{u}$
$\frac{d\overrightarrow{u}}{dt}$	$=$	$2\overrightarrow{u}\times \overrightarrow{\mathrm{\Omega }}+\overrightarrow{\mathrm{\Omega }}\times (\overrightarrow{x}\times \overrightarrow{\mathrm{\Omega }})+\overrightarrow{x}\times \frac{d\overrightarrow{\mathrm{\Omega }}}{dt}-[\frac{1}{\rho }\mathrm{\nabla }P+\mathrm{\nabla }\mathrm{\Phi }{]}_{\mathrm{r}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}}.$
Template:Rossner67, §II, Eq. (1) [BT87], Appendix 1.D, §3, (p. 664) Eq. (1D-42)

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The RHS of this equation matches the RHS of Eq. (1) from Template:Rossner67full after making the notation switch, $\mathrm{\Phi }\to -\mathfrak{𝔅}$, and after acknowledging that $\mathrm{\nabla }P/\rho \to \mathrm{\nabla }(P/\rho )$ when the mass-density is spatially uniform. The referenced equation from [BT87] contains all of the terms shown here, except there, the effects of pressure are ignored.

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Component Form

EXAMPLE #1: Inertial-frame velocities, $\overrightarrow{U}$, as viewed in the inertial frame.

Adding the Euler equation,

$\rho \frac{d\overrightarrow{U}}{dt}$

$=$

$-\mathrm{\nabla }P-\rho \mathrm{\nabla }\mathrm{\Phi },$

to the product of the inertial-frame velocity and the equation of continuity,

$\overrightarrow{U}\frac{d\rho }{dt}$

$=$

$-\rho \overrightarrow{U}(\mathrm{\nabla }\cdot \overrightarrow{u}),$

gives,

$\frac{d(\rho \overrightarrow{U})}{dt}+\rho \overrightarrow{U}(\mathrm{\nabla }\cdot \overrightarrow{u})$	$=$	$-\mathrm{\nabla }P-\rho \mathrm{\nabla }\mathrm{\Phi }$
$\Rightarrow \frac{\partial (\rho \overrightarrow{U})}{\partial t}+\mathrm{\nabla }\cdot [(\rho \overrightarrow{U})\overrightarrow{u}]$	$=$	$-\mathrm{\nabla }P-\rho \mathrm{\nabla }\mathrm{\Phi }$

In component form, the relation between $\overrightarrow{U}$ and $\overrightarrow{u}$ reads,

$U_i=u_i-({\mathrm{\Omega }}^{*}{)}_{ik}{x}_k=u_i+\stackrel{[\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}{]}_i}{\overbrace{{ϵ}_{ijk}{\mathrm{\Omega }}_j{x}_k}}.$

[EFE], Chap. 4, §25, Eq. (21)

and the rotating-frame Euler equation becomes,

$U_i=u_i-({\mathrm{\Omega }}^{*}{)}_{ik}{x}_k=u_i+\stackrel{[\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}{]}_i}{\overbrace{{ϵ}_{ijk}{\mathrm{\Omega }}_j{x}_k}}.$	$=$	${\displaystyle }$
[EFE], Chap. 4, §25, Eq. (21)

Part B

Drawing from Chapter 4, §25 of [EFE] — where the Cartesian components of the inertial-frame velocity $({\overrightarrow{v}}_{\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}})$ are represented by $U_i$ and the Cartesian components of the rotating-frame velocity $({\overrightarrow{v}}_{\mathrm{r}\mathrm{o}\mathrm{t}})$ are represented by $u_i$ — we begin by restating the Lagrangian representation of the intertial-frame Euler equation:

$\frac{d{U}_i}{dt}{|}_{\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}$

$=$

$-\frac{1}{\rho }\frac{\partial p}{\partial x_i}-\frac{\partial \mathrm{\Phi }}{\partial x_i}.$

The LHS of this (Euler) equation transform as follows:

$\frac{d{U}_i}{dt}{|}_{\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}$

$\to$

$\frac{d{U}_i}{dt}{|}_{\mathrm{r}\mathrm{o}\mathrm{t}}-{ϵ}_{imk}{\mathrm{\Omega }}_k{U}_m,$

where we also recognize that,

$U_i$

$\to$

$u_i+{ϵ}_{ijk}{\mathrm{\Omega }}_j{x}_k.$

Both of these expressions make use of the three-element Levi-Civita tensor, ${ϵ}_{ijk}$. Its six nonzero component values are … $ijk$ ${ϵ}_{ijk}$   $ijk$ ${ϵ}_{ijk}$ 123 +1 132 -1 312 321 231 213 Hence, for example, transforming the x-component $(i=1)$ of $\overrightarrow{U}$ gives, $U_1$ $\to$ $u_1+{ϵ}_{1jk}{\mathrm{\Omega }}_j{x}_k=u_1+{ϵ}_{123}{\mathrm{\Omega }}_2{x}_3+{ϵ}_{132}{\mathrm{\Omega }}_3{x}_2=u_1+{\mathrm{\Omega }}_{2}z-{\mathrm{\Omega }}_{3}y;$ transforming the y-component $(i=2)$ gives, $U_2$ $\to$ $u_2+{ϵ}_{2jk}{\mathrm{\Omega }}_j{x}_k=u_2+{ϵ}_{231}{\mathrm{\Omega }}_3{x}_1+{ϵ}_{213}{\mathrm{\Omega }}_1{x}_3=u_2+{\mathrm{\Omega }}_{3}x-{\mathrm{\Omega }}_{1}z;$ and transforming the z-component $(i=3)$ gives, $U_3$ $\to$ $u_3+{ϵ}_{3jk}{\mathrm{\Omega }}_j{x}_k=u_3+{ϵ}_{312}{\mathrm{\Omega }}_1{x}_2+{ϵ}_{321}{\mathrm{\Omega }}_2{x}_1=u_3+{\mathrm{\Omega }}_{1}y-{\mathrm{\Omega }}_{2}x.$ These are the same three components that arise from the vector expression (from above), ${\overrightarrow{v}}_{inertial}={\overrightarrow{v}}_{rot}+\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x};$ we therefore recognize that, $\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}={ϵ}_{ijk}{\mathrm{\Omega }}_j{x}_k$. We note as well that, $\overrightarrow{\mathrm{\Omega }}\times \overrightarrow{x}=-{ϵ}_{ijk}{\mathrm{\Omega }}_k{x}_j$.

Therefore, as viewed from the rotating frame of reference, the Euler equation becomes,

$\frac{d{U}_i}{dt}{|}_{\mathrm{r}\mathrm{o}\mathrm{t}}-{ϵ}_{imk}{\mathrm{\Omega }}_k{U}_m$	$=$	$-\frac{1}{\rho }\frac{\partial p}{\partial x_i}-\frac{\partial \mathrm{\Phi }}{\partial x_i}$
$\Rightarrow \frac{d}{dt}[u_i+{ϵ}_{ijk}{\mathrm{\Omega }}_j{x}_k]$	$=$	${ϵ}_{imk}{\mathrm{\Omega }}_k[u_m+{ϵ}_{mjk}{\mathrm{\Omega }}_j{x}_k]-\frac{1}{\rho }\frac{\partial p}{\partial x_i}-\frac{\partial \mathrm{\Phi }}{\partial x_i}$
$\Rightarrow \frac{d{u}_i}{dt}+{ϵ}_{ijk}\left[\right(\frac{d{\mathrm{\Omega }}_j}{dt})x_k+{\mathrm{\Omega }}_j(\frac{d{x}_k}{dt}\left)\right]$	$=$	${ϵ}_{imk}{u}_m{\mathrm{\Omega }}_k+{ϵ}_{imk}{\mathrm{\Omega }}_k\left[{ϵ}_{mjk}{\mathrm{\Omega }}_j{x}_k\right]-\frac{1}{\rho }\frac{\partial p}{\partial x_i}-\frac{\partial \mathrm{\Phi }}{\partial x_i}.$

Now, if we …

1. Swap the "jk" indices of the various terms on the LHS, which dictates that the leading sign be swapped as well:

   ${ϵ}_{ijk}\left[\right(\frac{d{\mathrm{\Omega }}_j}{dt})x_k+{\mathrm{\Omega }}_j(\frac{d{x}_k}{dt}\left)\right]$

   $\to$

   $-{ϵ}_{ijk}\left[x_j\right(\frac{d{\mathrm{\Omega }}_k}{dt})+u_j{\mathrm{\Omega }}_k]$

   note also that we have set $d{x}_j/dt\to u_j$;

2. In the first term on the RHS, replace the index, "m", with the index, "j":

   ${ϵ}_{imk}{u}_m{\mathrm{\Omega }}_k$

   $\to$

   ${ϵ}_{ijk}{u}_j{\mathrm{\Omega }}_k;$

3. Inside the square brackets of the second term on the RHS, replace the "jk" indices with "hℓ" in order to avoid confusion, then swap the "hℓ" indices of the two variables, which dictates that the leading sign be swapped as well:

   $\left[{ϵ}_{mjk}{\mathrm{\Omega }}_j{x}_k\right]$

   $\to$

   $\left[{ϵ}_{mh\ell }{\mathrm{\Omega }}_h{x}_{\ell }\right]$

   $\to$

   $[-{ϵ}_{mh\ell }{x}_h{\mathrm{\Omega }}_{\ell }];$

4. Swap the "mk" indices on the Levi-Civiti tensor that lies just outside the square brackets of the second term on the RHS, which dictates that the leading sign be swapped as well:

   ${ϵ}_{imk}{\mathrm{\Omega }}_k[-{ϵ}_{mh\ell }{x}_h{\mathrm{\Omega }}_{\ell }]$

   $\to$

   $-{ϵ}_{ikm}{\mathrm{\Omega }}_k[-{ϵ}_{mh\ell }{x}_h{\mathrm{\Omega }}_{\ell }]$

   $\to$

   ${ϵ}_{ikm}{\mathrm{\Omega }}_k\left[{ϵ}_{mh\ell }{x}_h{\mathrm{\Omega }}_{\ell }\right];$

the Euler equation becomes,

$\frac{d{u}_i}{dt}-{ϵ}_{ijk}\left[x_j\right(\frac{d{\mathrm{\Omega }}_k}{dt})+u_j{\mathrm{\Omega }}_k]$	$=$	${ϵ}_{ijk}{u}_j{\mathrm{\Omega }}_k+{ϵ}_{ikm}{\mathrm{\Omega }}_k\left[{ϵ}_{mh\ell }{x}_h{\mathrm{\Omega }}_{\ell }\right]-\frac{1}{\rho }\frac{\partial p}{\partial x_i}-\frac{\partial \mathrm{\Phi }}{\partial x_i}$
$\Rightarrow \frac{d{u}_i}{dt}$	$=$	$\underset{[2\overrightarrow{u}\times \overrightarrow{\mathrm{\Omega }}{]}_i}{\underbrace{2{ϵ}_{ijk}{u}_j{\mathrm{\Omega }}_k}}+\underset{[\overrightarrow{\mathrm{\Omega }}\times (\overrightarrow{x}\times \overrightarrow{\mathrm{\Omega }}){]}_i}{\underbrace{{ϵ}_{ikm}{\mathrm{\Omega }}_k\left[{ϵ}_{mh\ell }{x}_h{\mathrm{\Omega }}_{\ell }\right]}}+\underset{[\overrightarrow{x}\times (d\overrightarrow{\mathrm{\Omega }}/dt){]}_i}{\underbrace{{ϵ}_{ijk}\left[x_j\right(\frac{d{\mathrm{\Omega }}_k}{dt}\left)\right]}}-\frac{1}{\rho }\frac{\partial p}{\partial x_i}-\frac{\partial \mathrm{\Phi }}{\partial x_i}.$

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

$[\frac{d\rho }{dt}{]}_{rot}+\rho \mathrm{\nabla }\cdot {\overrightarrow{v}}_{rot}=0$ ;

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

$[\frac{\partial \rho }{\partial t}{]}_{rot}+\mathrm{\nabla }\cdot (\rho {\overrightarrow{v}}_{rot})=0$ .

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

$[\frac{d\overrightarrow{v}}{dt}{]}_{rot}=-\frac{1}{\rho }\mathrm{\nabla }P-\mathrm{\nabla }\mathrm{\Phi }-2{\overrightarrow{\mathrm{\Omega }}}_f\times {\overrightarrow{v}}_{rot}-{\overrightarrow{\mathrm{\Omega }}}_f\times ({\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x})$ ;

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

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

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

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

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,

$[\frac{d\overrightarrow{v}}{dt}{]}_{rot}=-\frac{1}{\rho }\mathrm{\nabla }P-\mathrm{\nabla }\mathrm{\Phi }+{\overrightarrow{a}}_{fict}$,

where,

${\overrightarrow{a}}_{fict}\equiv -2{\overrightarrow{\mathrm{\Omega }}}_f\times {\overrightarrow{v}}_{rot}-{\overrightarrow{\mathrm{\Omega }}}_f\times ({\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x}).$

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

${\overrightarrow{a}}_{Coriolis}\equiv -2{\overrightarrow{\mathrm{\Omega }}}_f\times {\overrightarrow{v}}_{rot},$

(see the related Wikipedia discussion) and the

Centrifugal Acceleration

${\overrightarrow{a}}_{Centrifugal}\equiv -{\overrightarrow{\mathrm{\Omega }}}_f\times ({\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x})=\frac{1}{2}\mathrm{\nabla }[|{\overrightarrow{\mathrm{\Omega }}}_f\times \overrightarrow{x}{|}^2]$

(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,

$\overrightarrow{A}\equiv ({\overrightarrow{\zeta }}_{rot}+2{\overrightarrow{\mathrm{\Omega }}}_f)\times {\overrightarrow{v}}_{rot}.$

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 Ω_f with z-axis

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

$\hat{i}:A_x={\zeta }_y{v}_z-({\zeta }_z+2\mathrm{\Omega })v_y,$

$\hat{j}:A_y=({\zeta }_z+2\mathrm{\Omega })v_x-{\zeta }_x{v}_z,$

$\hat{k}:A_z={\zeta }_x{v}_y-{\zeta }_y{v}_x,$

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

${\zeta }_x=[\frac{\partial v_z}{\partial y}-\frac{\partial v_y}{\partial z}],{\zeta }_y=[\frac{\partial v_x}{\partial z}-\frac{\partial v_z}{\partial x}],{\zeta }_z=[\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}].$

In turn, the curl of $\overrightarrow{A}$ has the following three Cartesian components:

$\hat{i}:[\mathrm{\nabla }\times \overrightarrow{A}{]}_x=\frac{\partial }{\partial y}[{\zeta }_x{v}_y-{\zeta }_y{v}_x]-\frac{\partial }{\partial z}[({\zeta }_z+2\mathrm{\Omega })v_x-{\zeta }_x{v}_z],$

$\hat{j}:[\mathrm{\nabla }\times \overrightarrow{A}{]}_y=\frac{\partial }{\partial z}[{\zeta }_y{v}_z-({\zeta }_z+2\mathrm{\Omega })v_y]-\frac{\partial }{\partial x}[{\zeta }_x{v}_y-{\zeta }_y{v}_x],$

$\hat{k}:[\mathrm{\nabla }\times \overrightarrow{A}{]}_z=\frac{\partial }{\partial x}[({\zeta }_z+2\mathrm{\Omega })v_x-{\zeta }_x{v}_z]-\frac{\partial }{\partial y}[{\zeta }_y{v}_z-({\zeta }_z+2\mathrm{\Omega })v_y].$

When v_z = 0

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

$\hat{i}:A_x=-({\zeta }_z+2\mathrm{\Omega })v_y,$

$\hat{j}:A_y=({\zeta }_z+2\mathrm{\Omega })v_x,$

$\hat{k}:A_z={\zeta }_x{v}_y-{\zeta }_y{v}_x,$

and,

$\hat{i}:[\mathrm{\nabla }\times \overrightarrow{A}{]}_x=\frac{\partial }{\partial y}[{\zeta }_x{v}_y-{\zeta }_y{v}_x]-\frac{\partial }{\partial z}[({\zeta }_z+2\mathrm{\Omega })v_x],$

$\hat{j}:[\mathrm{\nabla }\times \overrightarrow{A}{]}_y=-\frac{\partial }{\partial z}[({\zeta }_z+2\mathrm{\Omega })v_y]-\frac{\partial }{\partial x}[{\zeta }_x{v}_y-{\zeta }_y{v}_x],$

$\hat{k}:[\mathrm{\nabla }\times \overrightarrow{A}{]}_z=\frac{\partial }{\partial x}[({\zeta }_z+2\mathrm{\Omega })v_x]+\frac{\partial }{\partial y}[({\zeta }_z+2\mathrm{\Omega })v_y],$

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

${\zeta }_x=-\frac{\partial v_y}{\partial z},{\zeta }_y=\frac{\partial v_x}{\partial z},{\zeta }_z=[\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}].$

Related Discussions

• Wikipedia discussion of vorticity.
• Wikipedia discussion of Coriolis Effect.
• Wikipedia discussion of Centrifugal acceleration.