# Why do I get a non-zero Coriolis acceleration for a reference frame that moves in a straight line without rotating?

When you have a stationary reference frame $$\{O', (\pmb{e}'_1, \pmb{e}'_2, \pmb{e}'_3)\}$$ and a moving reference frame $$\{O, (\pmb{e}_1, \pmb{e}_2, \pmb{e}_3)\}$$, my textbook defines the Coriolis acceleration of a particle as

$$2(\pmb{\omega}\times\pmb{v}_r),$$

where $$\pmb{v}_r$$ is the velocity of the particle as measured from the moving reference frame, whereas $$\pmb{\omega}$$ is the angular velocity of (again) the moving frame, defined as

$$\pmb{\omega}\,:=\,\frac{1}{2}\sum_{j=1}^3\biggl(\pmb{e}_j\times\frac{\text{d}}{\text{d}t}\pmb{e}_j\biggr).$$

So far, so good. However, my textbook also says that the Coriolis acceleration only shows up when the moving frame is rotating... and this is where I have a problem with it. Suppose the moving frame is NOT rotating, but is moving in a simple translational motion along a straight line with respect to the stationary one. Then all three vectors from its basis will move with the same scalar speed in the same direction, and let's say this common velocity vector is $$(1, 2, 0)$$. Finally, let $$\pmb{e}_1 = (1, 0, 0)$$, $$\pmb{e}_2 = (0, 1, 0)$$ and $$\pmb{e}_3 = (0, 0, 1)$$, as usual.

With these values, I get a non-zero vector for the angular velocity of the moving frame, namely

\begin{align} \pmb{\omega} &= \frac{1}{2}\sum_{j=1}^3\biggl(\pmb{e}_j\times\frac{\text{d}}{\text{d}t}\pmb{e}_j\biggr) \\ &= \frac{1}{2}\sum_{j=1}^3\bigl(\pmb{e}_j\times(1, 2, 0)\bigr) \\ &= \frac{1}{2}\bigl((1, 1, 1)\times(1, 2, 0)\bigr) \\ &= \biggl(-1, \frac{1}{2}, \frac{1}{2}\biggr). \\ \end{align}

But then this clearly leads me to a non-zero value for the Coriolis acceleration for pretty much any particle whose velocity, as measured from the moving frame, is not parallel to this vector. But... why? Is my frame actually a rotating one in disguise? Did I mess up my computations somewhere? Have I not understood what these vectors are actually supposed to represent?

• Just because the frame origin is moving does not mean that the basis unit vectors are. They are constant, meaning that $\frac{d}{dt}\pmb{e}_j$ are identically zero. – David Hammen Sep 20 at 10:37
• @DavidHammen Aren't all vectors applied in physics, though? My textbook spends lots of pages coming up with the mathematical framework for theoretical mechanics, including setting up affine spaces and the fact each pair $(P, Q)$ of points in said space has a unique vector assigned to it. So if the origin of the frame is moving, even if the basis unit vectors stay the same length and always have the same direction as before, they are connecting different points of the affine space as time passes and the origin carries them around, so they are different vectors... hence they are "moving", no? – Labba Sep 20 at 10:45
• I’d say that $O$ changes, but the $e_j$ are static. – Ben51 Sep 20 at 10:48
• Relative to the stationary reference frame, the unit vector $e_1$ is always $(1,0,0)$. It is not $(t+1, 2t,0)$ - it has to remain a unit vector. – gandalf61 Sep 20 at 11:12