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In special relativity there are scalar quantities which are invariant under any Lorentz transformation, called Lorentz scalars. For example, the magnitude of the four-velocity is a Lorentz scalar. If we consider the usual Galilean transformations, however, magnitude of 3-velocity is not an invariant scalar anymore. Why is this? And are there some analogous "Galilean scalars" which are invariant under any Galilean transformation?

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  • $\begingroup$ and accelerations, forces. On the other hand, things like energy and work are generically frame dependent. $\endgroup$ – secavara Apr 28 '18 at 16:38
  • $\begingroup$ @ZeroTheHero More precisely, the length of a displacement vector, but not the magnitude of a "higher-order" vector like velocity. Not sure whether that distinction was implicit in your use of the word "length" instead of "magnitude". $\endgroup$ – tparker Apr 28 '18 at 17:20
  • $\begingroup$ @tparker good point. Yes I was thinking of $\Delta r$. $\endgroup$ – ZeroTheHero Apr 28 '18 at 17:36
  • $\begingroup$ @ZeroTheHero: The length of a vector is not a scalar in Galilean relativity. What's invariant in Galilean relativity is the length of a vector connecting simultaneous events. $\endgroup$ – Ben Crowell Apr 30 '18 at 1:10
  • $\begingroup$ @BenCrowell yes even more precise and correct statement. I better do away with mine as it will induce confusion, $\endgroup$ – ZeroTheHero Apr 30 '18 at 1:12
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A key point in Galilean relativity is that time is a scalar: Everybody can agree on a single value of the time for an event. That, and the ideas of simultaneity that flows from it, seems to be so obvious that people don’t even think about it.

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In special relativity, $p^2=m^2 c^2$ with $m$ the rest mass. In Galilean relativity, we can take the analogous invariant to be $m$. If you prefer a kinematic expression for it, we could choose $m = \frac{p^2}{2T}$, with $T$ the kinetic energy.

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protected by Qmechanic Apr 28 '18 at 17:54

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