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The definition I have of a Galilean transformation is one that preserves intervals of time and the distance between simultaneous events.

But now I can consider a transformation that takes $(t,\mathbf{x})$ to $(t,\mathbf{x}+\mathbf{v}t^2)$, which certainly preserves times, and also preserves the relative distance between simulatenous events, as the $\mathbf{v}t^2$ cancels out in the subtraction.

I am certain that this is wrong, so is there another constraint on what a Galilean transformation should be (say, an affine transformation)? And if there is such a constraint, why should it exist? I've seen some places simply define Galilean transformations as compositions of bossts, translations, and rotations, but to me that seems a little less motivated.

It's my first time learning these concepts, so I'm a little confused by what the actual first principles are - the "axioms" that I should take for granted.

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  • $\begingroup$ Your transformation does not make sense on dimensional grounds since $vt^2$ is not a distance. Moreover two consecutive transformation do not result in a Galilean transformation because of the non-linear $t^2$ term. $\endgroup$
    – ZeroTheHero
    Commented Feb 27, 2021 at 17:19
  • $\begingroup$ @ZeroTheHero The dimensions idea make sense - but could you elaborate on the latter point? I think I would end up with $(t,\mathbf{x}+\mathbf{v}(t^2+t^4))$, which seems fine in terms of preserving distances of simultaneous events. $\endgroup$
    – Vasting
    Commented Feb 27, 2021 at 17:35
  • $\begingroup$ right... the second part of my comment is in error as I was thinking of something else (basically I was thinking of $t$ as a parameter in the transformation rather than as an element to be transformed). I guess it comes down to dimensional analysis. Alternatively, by definition the transformation of coordinates must be so there is no acceleration, i.e. $d^2x/dt^2 =0$ in both frames since the transformation must connect inertial frames. Note also that your $x$ could be rotated without affecting the result but making the transformation (in particular the composition) more complicated. $\endgroup$
    – ZeroTheHero
    Commented Feb 27, 2021 at 18:14
  • $\begingroup$ @ZeroTheHero That makes sense. So should I be taking as a definition that $d^2x/dt^2=0$ since we are dealing with inertial frames? $\endgroup$
    – Vasting
    Commented Feb 27, 2021 at 20:33
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    $\begingroup$ yes that would be correct. $\endgroup$
    – ZeroTheHero
    Commented Feb 27, 2021 at 20:50

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