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

Coordinate Transformation

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:

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

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

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. Inside the square brackets of the second term on the RHS, replace the "ij" 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:

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

   $\to$

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

   $\to$

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

3. 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,

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

$=$

${ϵ}_{imk}{u}_m{\mathrm{\Omega }}_k+{ϵ}_{imk}{\mathrm{\Omega }}_k\left[{ϵ}_{mjk}{\mathrm{\Omega }}_j{x}_k\right]-\frac{1}{\rho }\frac{\partial p}{\partial x_i}-\frac{\partial \mathrm{\Phi }}{\partial x_i}.$

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.