Galilean spacetime interval?

Question

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 1

I wanted to add an answer because I was still somewhat confused after reading the answers here.

The point of the spacetime interval is that it is preserved under the Lorentz transformation.

So the natural criteria for a Galilean spacetime interval is that it also be a 'measure of distance' in some sense (by this I mean, inner product like, i.e. a bi-linear form), and also be preserved under the Galilean transformation.

But, it is not true that, under the Galilean transformation, the quantity

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

is preserved. Indeed, if we take the Galilean transformation

$$\begin{pmatrix}1 & -v \\ 0 & 1\end{pmatrix}\begin{pmatrix}x \\ t\end{pmatrix} = \begin{pmatrix}x' \\ t'\end{pmatrix}$$

and simply plugin $x = 0$ and $t = 1$ we find $x' = -v$ and $t' = 1$.

But for non $0$ values of $v$,

$$x^2 + t^2 = 1$$ while $$x'^2 + t'^2 = v^2 + 1$$

and clearly $v^2 + 1 \neq 1$ when $v \neq 0$.

But then the question becomes, is there any 'measure of distance' (again, formally, I mean a bi-linear form) which is preserved under the Galilean transformation? And the answer is no.

The reason being that, in order for a matrix to be distance preserving, for any bi-linear form, it must be a combination of rotations and reflections (since linear isometries preserve inner products, and thus must preserve the cosine of the angle between the unit basis vectors), but the Galilean transformation matrix cannot be a rotation if $v \neq 0$ since it fixes $\begin{pmatrix} 1 \\ 0\end{pmatrix}$ for all values of $v$. And it is clearly not a reflection.

So as safesphere's answer says, there must be two different metrics, both are preserved separately under the Galilean transformation, but not 'together'.

This seems contradictory at first by dmckee's argument. I.e. $dt^2$ is preserved under the galilean transformation since $t$ is fixed, but also $dr^2$ is 'preserved', so you would think $dt^2 + dr^2$ is preserved, but the problem with this argument is that $dr^2$ is preserved only in the case when $dt^2$ is 0. For example, in the counterexample we showed above, that the 'length' of $\begin{pmatrix}0 \\ 1\end{pmatrix}$ is not preserved under the transformation, $dt^2$ is 1 in this case, since we are comparing it with $\begin{pmatrix}0 \\ 0\end{pmatrix}$.

In contrast, the 'length' of $\begin{pmatrix}1 \\ 0\end{pmatrix}$, is preserved under the transformation, but in this case $dt^2 = 0$.

If you think about this physically for a while, it starts to make sense. If you are moving forward relative to, say, a measuring stick. To you, the measuring stick will appear to be moving. Imagine two light flashes, one at the first end of the measuring stick, and the other at the opposite end. However, each flash happens when they pass you as you are moving past the measuring stick. In the measuring stick's reference frame, the distance between the two events is the length of the measuring stick since they happen at either end of the stick. But in your reference frame, the distance between the two events is 0, since they both happen at exactly your position, so in general $dr^2$ is not preserved unless the two light flashes are simultaneous.

Answer 2

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

Answer 3

So the thing about Galilean space-time is that the time component is the same for everyone, so writing $$ (\mathrm{d}s)^2 = (\mathrm{d}t)^2 + (\mathrm{d}\mathbf{r})^2 \;, $$ and then $(\mathrm{d}s')^2 = (\mathrm{d}s)^2$ isn't wrong, but it is less informative than writing \begin{align} \mathrm{d}t' &= \mathrm{d}t \\ (\mathrm{d}\mathbf{r}')^2 &= (\mathrm{d}\mathbf{r})^2 \;. \end{align}