# Is moving into a rotating frame a Galilean transformation?

In classical mechanics, we know that laws of physics are invariant in Galilean transformations of the form:

$$x' = x -vt$$

My question is does shifting to rotating frame also count as a Galilean transformation? If it is not so, then are laws of physics invariant when shifting between rotating frames?

• No, since one of these frames is necessarily non-inertial: en.wikipedia.org/wiki/Galilean_transformation Sep 22, 2021 at 11:12
• Well, to be picky, the laws of physics are invariant. But the expression of them when moving between rotating frames is not. The thing about Galilean transformations is that they leave the expressions of the laws invariant. Sep 22, 2021 at 21:19

A full Galilean transformation will consist of a translation and a boost. In 2D (the minimum number of dimensions needed for a rotation, you can write a Galilean transformation as $$\left(\begin{matrix}t'\\x'\\y'\end{matrix}\right) = \left(\begin{matrix}t+\Delta t\\ x+\Delta x-v_xt\\y+\Delta y-v_yt\end{matrix}\right)$$ where the coefficients $$\Delta t$$, $$\Delta x$$, $$\Delta y$$, ​$$v_x$$ and $$v_y$$ are constant: they do not depend on $$x$$, $$y$$ or $$t$$. While a rotating frame will have something like $$\left(\begin{matrix}t'\\x'\\y'\end{matrix}\right) = \left(\begin{matrix}t\\ \cos(\omega t)x-\sin(\omega t)y\\\cos(\omega t)y+\sin(\omega t) x\end{matrix}\right).$$ There is no way to put this transformation in the form of a Galilean transformation.
• The second question, as stated, has an affirmative answer, and one needs to consider it in the context of general relativity. The laws of physics can be formulated in any reference frame (classical dynamics, Maxwell's equations, etc ). They are "invariant" when formulated appropriately. For example the homogeneous Maxwell's equations read $dF=0$, and this holds whatever the reference frame you use. Sep 22, 2021 at 12:28
• Newtons law applies perfectly well in a rotating frame without needing to learn any general relativity. Just use classic rotational kinematics to express the acceleration vector in the rotating frame. If $\vec{r}= x^i \vec{e}_i$ then just use $\dot{\vec{e}}_i= \Omega \cdot \vec{e}_i$ when differentiating. Where $\Omega(t)$ is the usual angular velocity tensor for the rotating basis (with respect to some fixed basis). Jun 3 at 6:05