# Silly question about Galilei Group

I have an silly doubt about Galilei Group.

From Wikipedia:

"The Galilean symmetries can be uniquely written as the composition of a rotation, a translation and a uniform motion of spacetime. Let x represent a point in three-dimensional space, and t a point in one-dimensional time. A general point in spacetime is given by an ordered pair (x, t).

A uniform motion, with velocity v, is given by $({\mathbf {x}},t)\mapsto ({\mathbf {x}}+t{\mathbf {v}},t),$ where $v ∈ ℝ^3$.

A translation is given by $({\mathbf {x}},t)\mapsto ({\mathbf {x}}+{\mathbf {a}},t+s),$ where $a ∈ ℝ^3$ and $s ∈ ℝ$.

A rotation is given by $({\mathbf {x}},t)\mapsto (G{\mathbf {x}},t),$ where $G : ℝ^3 → ℝ^3$ is an orthogonal transformation."

What suppose to mean, physically, the $t+s$ (I mean: $t'=t$?) and $\textbf{a}$ (on $\textbf{x+a}$) on translation? And why we are considering rotations (I mean, these kind of motion are non inertial)?

Actually it is a fair question, and even though it is a conceptual and basic question, it is well stated and precisely written. It makes clear what you are confused about. It's two factors, one is thinking of active transformations instead of passive ones (see below)[and the two are equivalent once understood], and the other one is, thinki g of the transformations, at least for G, as a continuously rotating body.

First, the most obvious, the rotation G is not rotational motion (which would have you in a non-inertial reference frame, as you indicated). It is instead a simple rotation of the axis of reference, where your motion is described in the statically defined rotated frame. Ie, you can draw the reference axis with any direction being the x axis, and just rigidly draw the other two so that it is a rotation from my x, y and z axis.

[hopefully not to confuse you, but what I described is a passive rotation, i.e., the reference frames, and what you described was active instead, but also continui got rotate. You could have just rotated the body and anything else in your system, a single rotation and not a continuous one, and still describe it as inertial]

The second question of why and what is s, also shows the same confusion. Think of it as not going to the past or future, but recalibrate get your clocks (or calendars) so that t=0 in the new clock is t=-s in the old one. That's a passive transformation. You can make it active by considering t=s in the new clock as t=0 in the old clock (coordinate system with that clock; or alternatively do t-s).

So the best way to get in-confused is to think of coordinate transformation for a new coordinate frame (=system).

For a continuous rotation you'd need the elements of G to be functions of t, where instead of a fixed rotation by $\theta$ It would have been something like $\omega t$.

Once you understand the passive coordinate transformation it's not too hard conceptually to consider them active, they are equivalent.

Hope this helps, and I can assure you that others have had the same silly confusion.

The translation just adds a constant vector a to any vector in 3d and a constant time shift $s$ to the time $t$. This amounts to translating the origin in space to -a and the origin of time to $-s$.

The rotations are time-independent and are just fixed rotations taking one set of axes to another, i.e. they are fixed rotations between basis vectors of two inertial frames; basically there is no reason to suppose that the inertial frames have basis vectors that are aligned.

If the rotation was time-dependent, i.e. if the set of basis vectors in the second frame rotated in time w/r to the first frame, then the second frame would not be inertial.