special relativity - Galilean spacetime interval?

Does it make sense to refer to a single Galilean Invariant spacetime interval?

$$ds^2=dt^2+dr^2$$

Or is the proper approach to describe separate invariant interval for space (3D Euclidean distance) and time?

This may be a trivial distinction but I suspect the answer to the opening question is no for if one is rigorous and considers Galilean transformation one of three possible versions of the general Lorentz transformation where $k=0$ ($c=-\dfrac{1}{k^2}=\infty$). My understanding is that the real counterpart to non-Euclidean Minkowski space ($k=-\dfrac{1}{c^2}<0$) in this construct is not classical Galilean spacetime but a 4D Euclidean space ($k>0$) which is not consistent with physical reality.

Any insights, corrections would be greatly appreciated.

Answer

The Galilean spacetime is a tuple $(\mathbb{R}^4,t_{ab},h^{ab},\nabla)$ where $t_{ab}$ (temporal metric) and $h^{ab}$ (spatial metric) are tensor fields and $\nabla$ is the coordinate derivative operator specifying the geodesic trajectories.

A single metric does not work, because the speed of light is infinite. If you consider:

$$\text{d}\tau^2=\text{d}t^2\pm\left(\frac{\text{d}\mathbf{r}}{c}\right)^2$$

the spatial part on the right vanishes for $c\rightarrow\infty$. Therefore time and space shoulld be treated separately with the temporal metric:

$$t_{ab}=(\text{d}_a t)(\text{d}_b t)$$

and the spatial metric:

$$h^{ab}=\left(\dfrac{\partial}{\partial x}\right)^a\left(\dfrac{\partial}{\partial x}\right)^b+ \left(\dfrac{\partial}{\partial y}\right)^a\left(\dfrac{\partial}{\partial y}\right)^b+ \left(\dfrac{\partial}{\partial z}\right)^a\left(\dfrac{\partial}{\partial z}\right)^b$$

that translate to

$$t'=t$$ $$\text{d}\mathbf{r}'^2=\text{d}\mathbf{r}^2$$

While the space of Galilean 4-coordinates is not a Euclidean space, the space of Galilean velocities is a Euclidean space. Differentiating the Galilean transformation (for simplicity in two dimensions):

$$t'=t$$ $$x'=x-vt$$

we obtain $\text{d}t'=\text{d}t$ and therefore

$$\dfrac{\text{d}x'}{\text{d}t'}=\dfrac{\text{d}x}{\text{d}t}-v$$

If $v_R=\dfrac{\text{d}x}{\text{d}t}$ is the velocity of a body as observed from the frame $R$ and $v_{R'}=\dfrac{\text{d}x'}{\text{d}t'}$ is the velocity of the body as observed from the frame $R'$, then the result reveals the Euclidean symmetry

$$v_R=v_{R'}+v_{R'R}$$

Galilean Transformation