Can Galilean transformation be derived from length invariance?

Question

Given length invariance in Euclidean 3D space between two inertial frames:$$ds^2=ds'^2$$ Can Galilean transformation be derived like Lorentz transformation derived from space-time interval invariance?

Answer 1

The answer is negative, because that invariance property is also valid regarding two reference systems such that one is inertial and the other is not inertial and thus the coordinate transformations are not Galileian.

If we also impose $\Delta t=\Delta t'$, we find that the admitted transformations are more general than the Galileian ones: $$t'=t+c$$ $$x'= R(t) x + c(t)$$ where $R(t)$ is an orthogonal matrix depending on time in general and $c(t)$ a tridimensional translation depending on time in general.

Answer 2

I think what we should start from is not invariance of a scalar $ds^2$, but co(ntra)variance of a $3$-vector. Note that in terms of split-complex numbers the $2$-dimensional Lorentz transformation$$dt^\prime=\gamma(dt-\beta dx),\,dx^\prime=\gamma(-\beta dt+dx)$$satisfies$$dt^\prime+jdx^\prime=\gamma(1-j\beta)(dt+jdx).$$The Galilean equivalent using dual numbers is$$dt^\prime=dt,\,dx^\prime=-\beta dt+dx\iff dt^\prime+\epsilon dx^\prime=(1-\epsilon\beta)(dt+\epsilon dx).$$Note that $\gamma^2(1-j\beta)(1+j\beta)=(1-\epsilon\beta)(1+\epsilon\beta)=1$ is the reason we can invert these transformations, i.e. swap the roles of the two coordinate systems. It also establishes the theories' respective invariances of$$(dt+jdx)(dt-jdx)=dt^2-dx^2,\,(dt+\epsilon dx)(dt-\epsilon dx)=dt^2.$$That $dx$ isn't even in that last expression is a hint of what's about to go wrong.

I'm tidying things up a bit by nondimensionalising a velocity (though this needn't commit us to an invariant speed in the latter transformation family) and making space $1$-dimensional, but you can fix both of those. The important issue is the invariance of quadratic quantities needn't be equivalent to a specific transformation of linear ones in the $\epsilon$-based case, because $\epsilon^2=0\ne1=j^2$. The linear transformation implies the quadratic invariance but, as others have noted, the converse is false, ultimately because dual numbers don't allow division due to their zero divisors. This can be seen as a motive for expecting nature to be Lorentzian rather than Galilean.