Rotating Frame of Reference

Newton's 2nd Law

The derivation of the Navier-Stokes equations is based on Newton's second law: It is assumed that the kinematics of a particle is determined by the particle's interaction with its physical environment. Every acceleration (change of velocity) is caused by an external force. If one eventually knows all relevant forces, the acceleration can be calculated and from this also the trajectories (by integrating the acceleration twice).

Non-Inertial Frames of Reference

However, this approach is not valid in non-inertial (accelerated) frames of reference (as for instance the rotating earth). In these reference frames, additional accelerations are observed which are not caused by actual physical forces. Here, Newton's 2_nd law cannot be applied directly to determine the equations of motion!

This issue can be fixed by considering a coordinate transformation between the observer's (accelerated) and any inertial frame of reference (in which Newton's 2nd law applies). Then one can simply apply Newton's 2nd law in the inertial frame and replace the inertial acceleration with other quantities that can be measured directly by the observer.

Coordinate Representation, Change of Basis, Rotation Matrix

Consider two cartesian coordinate system: One is inertial ($\text{in}$) and the other one ($\text{rot}$) rotates with respect to the first one with constant angular velocity: \begin{align} \Omega = \frac{\D{\varphi}}{\D{t}} \label{eq:Omega} \end{align} Without loss of generality, the $z$-axes of both systems can be chosen to be aligned parallel to the axis of rotation. Then the relative orientation between the two coordinate systems is given by the angle $\varphi (t)$ that is spanned by the respective $x$- or $y$-axes.

Depending on the choice of the coordinate system, any vector $\vec{\vartheta}$ can be specified by its coordinates with respect to the corresponding basis. In general, these coordinates differ for the different coordinate systems but are related by a transformation matrix $R$ (and its inverse $R^{-1}$): \begin{align} \left[ \vec{\vartheta} \right]_\text{in} &= R \;\;\;\cdot \left[ \vec{\vartheta} \right]_\text{rot} \label{eq:R} \\ \left[ \vec{\vartheta} \right]_\text{rot}\; &= R^{-1} \cdot \left[ \vec{\vartheta} \right]_\text{in} \label{eq:R^T} \end{align} Here \begin{align} \left[ \vec{\vartheta} \right]_\text{rot} := \begin{pmatrix} x_\text{rot} \\ y_\text{rot} \\ z_\text{rot} \end{pmatrix} \quad \text{and} \quad \left[ \vec{\vartheta} \right]_\text{in} := \begin{pmatrix} x_\text{in} \\ y_\text{in} \\ z_\text{in} \end{pmatrix} \end{align} are the coordinate representations of the position vector $\vec{\vartheta}$ with respect to the rotating basis ($\text{rot}$) and the inertial one ($\text{in}$).

Rotation Matrix

The transformation matrix $R$ depends on the angle of rotation $\varphi (t)$ and contains the coordinates of the rotational unit vectors with respect to the inertial basis: \begin{align} R &= \begin{pmatrix} \left[ \hat{x}_\text{rot} \right]_\text{in} & \left[ \hat{y}_\text{rot} \right]_\text{in} & \left[ \hat{z}_\text{rot} \right]_\text{in} \end{pmatrix} \\[2ex] &= \begin{pmatrix} \cos\varphi & -\sin\varphi & 0 \\ \sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{align} As it is characteristic for a rotation matrix, \begin{align} R^{-1} = R^T \label{eq:inverse-transpose} \end{align} since $R R^{T} = \mathbb{I}$ (as you can easily verify). Thus, instead of calculating the inverse laboriously, one can simply take the transpose.

Useful Relations

At this point it is convenient to calculate two more expressions which will be needed later: \begin{align} \frac{\D{R}}{\D{t}} &\stackrel{\eqref{eq:Omega}}{=} \frac{\D{R}}{\D{\varphi}} \underbrace{\frac{\D{\varphi}}{\D{t}}}_{:= \Omega} = \Omega \cdot \begin{pmatrix} -\sin\varphi & -\cos\varphi & 0 \\ \cos\varphi & -\sin\varphi & 0 \\ 0 & 0 & 0 \end{pmatrix} \label{eq:dRdT} \\[2ex] \frac{\D{R}}{\D{t}} \cdot R^T &\stackrel{\eqref{eq:dRdT}}{=} \Omega \cdot \begin{pmatrix} -\sin\varphi & -\cos\varphi & 0 \\ \cos\varphi & -\sin\varphi & 0 \\ 0 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} \cos\varphi & \sin\varphi & 0 \\ -\sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1 \end{pmatrix} \\ &= \Omega \cdot \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \Omega \end{pmatrix} \times \\ &= \vec{\Omega} \times \label{eq:Omegax} \end{align} The penultimate step may be a bit surprising. It has to be understood in a sense of an operator equation that is completely meaningful only when being applied to a dummy vector. So if you apply both operators to an arbitrary vector, you'll notice that the result is the same.

Accordingly, one finds: \begin{align} \frac{\D{R^T}}{\D{t}} \cdot R = - \vec{\Omega} \times \label{eq:-Omegax} \end{align}

Time Derivatives

When the origins of the coordinates systems coincide, then the position vector $\vec{r}$ is independent of the particular coordinate system, since it connects this common origin with a particle's location in space (which is also independent of the frame of reference).

Velocities

This is a bit different for velocities and accelerations: They are defined as the change of the coordinates of a position vector in a particular coordinate system. A vector which is at rest in the rotating frame rotates with non-zero velocity in the inertial one. Therefore, we will use another (inner) index for velocities and accelerations indicating the frame of reference in which them are defined. The resulting velocity vector can however be represented as an arrow in space, which in turn can be given either with respect to the inertial or rotating basis (indicated by the outer index). \begin{align} \left[ \vec{v}_\text{rot} \right]_\text{rot} & := \frac{\D{}}{\D{t}} \left[ \vec{r} \right]_\text{rot} \stackrel{\eqref{eq:R^T} \eqref{eq:inverse-transpose}}{=} \frac{\D{}}{\D{t}} \left( R^T \cdot \left[ \vec{r} \right]_\text{in} \right) \\[1ex] &= \frac{\D{R^T}}{\D{t}} \cdot \left[\vec{r}\right]_\text{in} + \underbrace{ R^T \cdot \underbrace{ \frac{\D{}}{\D{t}} \left[ \vec{r} \right]_\text{in} }_{= \left[ \vec{v}_\text{in} \right]_\text{in}} }_{ = \left[ \vec{v}_\text{in} \right]_\text{rot} } \\ &\stackrel{\eqref{eq:R}\eqref{eq:R^T}\eqref{eq:inverse-transpose}}{=} \underbrace{ \frac{\D{R^T}}{\D{t}} R }_{-\vec{\Omega}\times} \cdot \left[\vec{r}\right]_\text{rot} + \left[ \vec{v}_\text{in} \right]_\text{rot} \\ &\stackrel{\eqref{eq:-Omegax}}{=} -\vec{\Omega} \times \left[ \vec{r} \right]_\text{rot} + \left[ \vec{v}_\text{in} \right]_\text{rot} \end{align} Analogously: \begin{align} \left[ \vec{v}_\text{in} \right]_\text{in} & := \frac{\D{}}{\D{t}} \left[ \vec{r} \right]_\text{in} \\ &= \vec{\Omega} \times \left[\vec{r}\right]_\text{in} + \left[ \vec{v}_\text{rot} \right]_\text{in} \label{eq:vRotIn} \end{align}

These expressions can now also be stated in coordinate-independent form:

Thus, the two velocities $\vec{v}_\text{rot}$ and $\vec{v}_\text{in}$ differ by a term $\vec{\Omega} \times \vec{r}$ which accounts for the relative motion of the coordinate systems which respect to each other. It appears because the transformation matrix $R(\varphi (t) )$ is a function of time and therefore the order of the time derivative operator ($\frac{\D{}}{\D{t}}$) and the change of basis operators ($R$,$R^T$) cannot just be reversed. Instead, the product rule has to applied which creates the additional term.

Accelerations

For the accelerations one can proceed in an analogous way as before with the velocities: \begin{align} \left[ \vec{a}_\text{rot} \right]_\text{rot} &:= \frac{ \D{} }{ \D{t} } \left[ \vec{v}_\text{rot} \right]_\text{rot} \stackrel{\eqref{eq:R^T} \eqref{eq:inverse-transpose}}{=} \frac{ \D{} }{ \D{t} } \left( R^T \cdot \left[ \vec{v}_\text{rot} \right]_\text{in} \right) \\[1ex] &\stackrel{\eqref{eq:vRotIn}}{=} \frac{ \D{} }{ \D{t} } \left[ R^T \cdot \left( \left[ \vec{v}_\text{in} \right]_\text{in} + \vec{\Omega} \times \left[ \vec{r} \right]_\text{in} \right) \right] \\[2ex] &= \frac{ \D{R} }{ \D{t} } \left[ \vec{v}_\text{in} \right]_\text{in} + R^T \frac{ \D{} }{ \D{t} } \left[ \vec{v}_\text{in} \right]_\text{in} + \frac{ \D{R^T} }{ \D{t} } \vec{\Omega} \times \left[ \vec{r} \right]_\text{in} + R^T \cdot \vec{\Omega} \times \frac{ \D{} }{ \D{t} } \left[ \vec{r} \right]_\text{in} \\[2ex] &= \underbrace{ \frac{ \D{R^T} }{ \D{t} } R }_{ -\vec{\Omega} \times } \left[ \vec{v}_\text{in} \right]_\text{rot} + \left[ \vec{a}_\text{in} \right]_\text{rot} + \underbrace{ \frac{ \D{R^T} }{ \D{t} } R }_{-\vec{\Omega} \times} \vec{\Omega} \times \left[ \vec{r} \right]_\text{rot} + \vec{\Omega} \times \left[ \vec{v}_\text{in} \right]_\text{rot} \\[2ex] &= -2 \cdot \vec{\Omega} \times \left[ \vec{v}_\text{in} \right]_\text{rot} - \vec{\Omega} \times \left( \vec{\Omega} \times \left[ \vec{r} \right]_\text{rot} \right) + \left[ \vec{a}_\text{in} \right]_\text{rot} \end{align} In coordinate-independent notation this reads: