Why care about rotating frames? Well, the first and foremost reason is that we live on a large rotating frame. The rotation of the Earth about its axis may not seem to be of direct importance in our day-to-day lives, but the Coriolis force due to the planet’s rotation has its effects. It shapes the patterns of the weather and plays an intimate role in long-range warfare. And who hasn’t been at least a little enchanted by the motion of Foucault’s pendulum? So, a basic understanding of rotating frames may be motivated by these reasons alone.

That said, I am actually thinking of more mundane and more focused applications of rotating frames. Control of the orbital motion of spacecraft is best understood and implemented in frames that are attached to the motion itself. For example, suppose we want to change the energy of a trajectory. The best way to perform this adjustment is by firing thrusters so that they are aligned with the instantaneous direction of the velocity. Other mechanical applications include understanding the effects of high accelerations to the occupants of cars, planes, roller coasters, etc.

So, how do we describe motion in a rotating frame? There are many ways to do this, but I’ll confine myself to just two different methods. The first is the traditional method found in many physics textbooks, which I will call the ‘classical’ method. It involves the $$\vec \omega \times$$ terms. In my opinion, this method should be used only with great care, if at all. It is often confusing and ambiguous in most people’s hands. The second method, which I will call ‘semi-classical’, is much more reliable. It can be used by novice and expert alike and it doesn’t require the same amount of care. It does require a bit more in the way of computations, but these are easy to implement. In addition, the very structure of this method opens the doors for the more sophisticated computations of differential geometry, differential forms, and geometric algebras.

The classical picture starts with an object located at a position $$\vec r$$ from some origin, with a corresponding velocity $$\vec v$$ with respect to some fixed frame (inertial) spanned by the unit triad $$ \left\{ \hat \imath, \hat \jmath, \hat k \right\} $$. Co-located at the origin of the first frame is a second one spanned by the unit triad $$\left\{ \hat I, \hat J, \hat K \right\}$$ which is rotating arbitrarily as a function of time. Since the rotation, while time-dependent, is an instantaneous orthogonal transformation, the magnitude of the radius vector $$\vec r$$ remains fixed. Therefore, the velocity $$\vec v$$ must be perpendicular to $$\vec r$$ and, using arguments derived from uniform circular motion, the angular velocity, $$\vec \omega$$, defined implicitly as

\[ \vec v = \vec \omega \times \vec r \; ,\]

shows up on the scene.

Combining the velocity in inertial frame with the apparent velocity caused by the rotating frame, we arrive at the ‘old tried-and-true’ relation

\[ \left. \frac{d}{dt} \right)_{rotating} = \left. \frac{d}{dt} \right)_{fixed} – \vec \omega \times \; .\]

This relationship is completely frame-independent, which is both its strength and its weakness. Common mistakes associated with this frame-independence are either to ‘expect’ a particular result based on physical intuition and to fail to obtain it because the outcome is expressed in the wrong frame, or to use the wrong angular velocity, $$\vec \omega$$, which is easy to do for even the simplest of cases. Even ‘experts’ come to seriously wrong conclusions when applying or deriving the classical relationship (e.g., see Chapter 1 of ‘Classical Dynamics of Particle and Systems’, 2nd edition by Marion). In particular, one has to be very careful about being consistent with which frame is being used.

The better treatment, suitable for beginners and experts alike, is what I call the ‘semi-classical’ treatment, in which transformation matrices play a central role. At any instant, the component of the position vector transform is

\[ \vec r_{rotating} = A(t) \vec r_{fixed} \; .\]

The corresponding relationship for the velocity is obtained by taking the time derivative of the position equation to get

\[ \vec v_{rotating} = A(t) \vec v_{fixed} + \dot A(t) \vec r_{fixed} \; \]

where $$A(t)$$ is the transformation matrix that gives the instantaneous orientation (attitude) of the rotating frame relative to the fixed one. Of course, it takes some effort to construct that matrix, but it is straightforward to do since

\[ A(t) = \left[ \begin{array}{ccc} \hat I \cdot \hat \imath & \hat I \cdot \hat \jmath & \hat I \cdot \hat k \\ \hat J \cdot \hat \imath & \hat J \cdot \hat \jmath & \hat J \cdot \hat k \\ \hat K \cdot \hat \imath & \hat K \cdot \hat \jmath & \hat K \cdot \hat k \end{array} \right] \]

and

\[ \dot A(t) = \left[ \begin{array}{ccc} \dot {\hat I} \cdot \hat \imath & \dot {\hat I} \cdot \hat \jmath & \dot {\hat I} \cdot \hat k \\ \dot {\hat J} \cdot \hat \imath & \dot {\hat J} \cdot \hat \jmath & \dot {\hat J} \cdot \hat k \\ \dot {\hat K} \cdot \hat \imath & \dot {\hat K} \cdot \hat \jmath & \dot {\hat K} \cdot \hat k \end{array} \right] \;.\]

For most cases where one has enough knowledge to construct $$\vec \omega$$ one has enough knowledge to construct $$A(t)$$ since a model of how the rotating frame unit vectors are moving relative to the fixed frame is needed in both cases.

To illustrate the two methods side by side, let’s consider the example of a bead on a helical trajectory (with the ‘fixed’ subscript dropped to keep the notational clutter down)

\[ \vec r = a \cos \omega t \hat \imath + a \sin \omega t \hat \jmath + b t \hat k \; .\]

The fixed-frame velocity is immediately obtained

\[ \vec v = \frac{d}{dt} \vec r = -a \omega \sin \omega t \hat \imath + a \omega \cos \omega t \hat \jmath + b \hat k \; .\]

Now, suppose we want to view the bead’s motion in a frame co-rotating with the bead. The rotating frame’s $$\hat K$$ axis coincides with the fixed frame $$\hat k$$ and the rotating frame spins about this axis with an angular rate of $$\omega$$ and with a corresponding angular velocity of

\[ \vec \omega = \omega \hat k = \omega \hat K \; .\]

Our physical expectation is that, in the rotating frame, the bead should just move vertically along $$\hat K$$. Let’s see if it does.

Application of the classical formula gives

\[ \vec v_{rotating} = \vec v – \vec \omega \times \vec r = \left[ \begin{array}{c} -a \omega \sin \omega t \\ a \omega \cos \omega t \\ b \end{array} \right] – \left| \begin{array}{ccc} \hat \imath & \hat \jmath & \hat k \\ 0 & 0 & \omega \\ a \cos \omega t & a \sin \omega t & 0 \end{array} \right| \; , \]

which becomes

\[ \vec v_{rotating} = \left[ \begin{array}{c} -a \omega \sin \omega t \\ a \omega \cos \omega t \\ b \end{array} \right] – \left[ \begin{array}{c} -a \omega \sin \omega t \\ a \omega \cos \omega t \\ 0 \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ b \end{array} \right] = b \hat k\; .\]

So, from the point of view of an observer rotating with $$\vec \omega = \omega \hat k$$ the bead moves upward at a constant velocity $$b \hat k $$. But is this correct? Technically it is not correct, because our observer should be referencing his observation to the rotating frame. He ‘gets’ the correct answer only because $$\vec v_{rotating} = b \hat K = b \hat k$$.

As pointed out earlier, the classical formula is expressed in a frame-independent way, but it is easy for that distinction to be lost. In this case, mixing the frames results only in minor confusion that can be straightened out with some thought. For more complicated problems, this mistake can grind the computation to a complete halt.

Now let’s try the semi-classical approach. In this approach, the unit vectors of the rotating frame are expressed in the fixed frame as

\[ \hat I = \cos \omega t \hat \imath + \sin \omega t \hat \jmath \; ,\]

\[ \hat J = – \sin \omega t \hat \imath + \cos \omega t \hat \jmath \; ,\]

and

\[ \hat K = \hat k \; .\]

The corresponding attitude matrix is

\[ A(t) = \left[ \begin{array}{ccc} \cos \omega t & \sin \omega t & 0 \\ – \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array} \right] \]

and its time derivative is

\[ \dot A(t) = \left[ \begin{array}{ccc} – \omega \sin \omega t & \omega \cos \omega t & 0 \\ – \omega \cos \omega t & – \omega \sin \omega t & 0 \\ 0 & 0 & 0 \end{array} \right] \; .\]

Application of $$A(t)$$ to the position vector in the fixed frame gives

\[ A(t) \vec r = \left[ \begin{array}{c} a \\ 0 \\ b t \end{array} \right] = a \hat I + b t \hat K \; ,\]

which is the position of the bead seen by an observer co-moving in the rotating frame. As expected of our physical intuition, the bead is moving upward while maintaining a fixed offset of $$a$$ in the $$\hat I-\hat J$$ plane.

The velocity in the rotating frame obtains immediately by differentiating this term

\[ \frac{d}{dt} \left( A(t) \vec r \right) = \left[ \begin{array}{c} 0 \\ 0 \\ b \end{array} \right] = b \hat K \; ,\]

and that’s all that is needed to find the velocity.

That said, we can get some insight into the $$\vec \omega \times$$ term in the classical equation by taking various combinations of $$A(t)$$, $$\dot A(t)$$, $$\vec r$$, $$\vec v$$, and $$\vec \omega$$.

First let’s compare the application of $$A(t)$$ to the fixed-frame velocity $$\vec v$$,

\[ A \vec v = \left[ \begin{array}{c} 0 \\ a \omega \\ b \end{array} \right] = a \omega \hat J + b \hat K \; , \]

to the application of $$\dot A(t)$$ to the fixed-frame position $$\vec r$$,

\[ \dot A \vec r = \left[ \begin{array}{c} 0 \\ – a \omega \\ 0 \end{array} \right] = -a \omega \hat J \; .\]

We interpret the first term as the linear velocity of the bead moving in uniform circular motion in the fixed frame but expressed in terms of the rotating frame’s basis vectors. It is not what a co-moving observer would observe, but rather a convenient way for the fixed-frame observer to talk about the motion by describing it saying something like ‘the bead is moving upward with speed $$b$$ and is moving in uniform circular motion with tangential or linear speed $$a \omega$$’. The second term compensates for what is seen as uniform circular motion by the fixed-frame observer. The sum of these two, $$b \hat K$$, is what the co-moving observer sees in his frame.

Next let’s look at the application of $$A(t)$$ to $$-\vec \omega \times \vec r$$, which yields

\[ -A(t) \left( \vec \omega \times \vec r \right) = \left[ \begin{array}{c} 0 \\- a \omega \\ 0 \end{array} \right] = a \omega \hat J \; ,\]

which is just the frame correction term discussed above. So we can deduce the following ‘frame-correction’ relationship:

\[ -A(t) \left( \vec \omega \times \vec r \right) = \dot A(t) \vec r \]

or

\[ \vec \omega \times \vec r = – A^{-1}(t) \dot A(t) \vec r \; .\]

Another fruitful comparison starts by asking what happens if $$\vec \omega$$ and $$\vec r$$ are first converted to the rotating frame and then inserted into the cross-product. This computation gives

\[ \left( A(t) \vec \omega \right) \times \left( A(t) \vec r \right) = \left| \begin{array}{ccc} \hat I & \hat J & \hat K \\ 0 & 0 & \omega \\ a & 0 & b t \end{array} \right| = \left[ \begin{array}{c} 0 \\ a \omega \\ 0 \end{array} \right] = a \omega \hat J \]

from which we are lead to the interesting relationship

\[ A(t) \left( \vec \omega \times \vec r \right) = \left( A(t) \vec \omega \times A(t) \vec r \right) .\]

Next time I’ll explore why this relation holds and how it leads into more sophisticated ways of thinking about frame transformations.