The second Newton's law can be written as (in SI units) $$ \frac{d}{dt}\vec p = \vec F. $$ Newton was considered Galilean transformations and the existence of a "absolute" time. Now suppose that this formulation must be independent (or ignore) Galilean transformations (and any other type of transformations, including Lorentz transformation).

Then my question is: the parameter $t$ in the above expression is the time measured by the clock of the laboratory?

I imagine that the some question can be applied to the expression for the velocity $$ \frac{d}{dt}\vec r = \vec v. $$

I also imagine that the answer is "no", since for a onto function $t\mapsto\tau = s(t)$ the time derivative change as $\frac{d}{dt} = \left(\frac{d}{d\tau}s^{-1}(\tau)\right)\frac{d}{d\tau}$ and therefore two observers can discord about the magnitude of the force/velocity. The parameter is therefore the proper time define such that $dt$ is proportional to line element of the trajectory since this quantity is invariant under reparametrization of the trajectory in any riemannian manifold.


1 Answer 1


In Newton's era, "time" was defined by the rotation of the Earth and the motions of the planets. As physics advanced, that was refined to things like Greenwich Mean Time and Ephemeris Time. These have matured into global coordinate time scales like GPS time. For practical Newtonian purposes (most physics and engineering), GPS time is fine, "laboratory time" for the whole world. If it's not fine, you need post-Newtonian mechanics, not simply a better definition of time.

  • $\begingroup$ My question is in some sense a "theoretical question": A physicist formulating laws of mechanics at Newton's era that agrees with Newton's laws but disagrees with Galilean transformation (and have no other better theory). Or, a very rigorous physicist at Newton's era that have the dream of formulating Newton's laws independent of any other [physical] theory. $\endgroup$ Jan 20, 2023 at 18:52
  • $\begingroup$ @IsmaelDamião You should read Greg Egan's novel "Incandescence". $\endgroup$
    – John Doty
    Jan 20, 2023 at 18:55
  • $\begingroup$ I'm going to read it! $\endgroup$ Jan 20, 2023 at 19:01

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