I'm reading the article On the Galilean Covariance of Classical Mechanics (pdf link here), in which the authors want to establish the transformation rule for momentum, assuming only that $\vec{F}=d\vec{p}/dt$ and notwithstanding the relation $\vec{p}=m\vec{v}$.

For a quick background, we assume the Galilean transformation defined by $$\vec{x}(t)\to\vec{x}'(t')=R\vec{x}(t)+\vec{u}t+\vec{a}\\t\to t'=t+b$$

where $R$ is the orthogonal matrix characterizing the rotation of the primed frame w.r.t. the unprimed frame, $\vec{u}$ is the velocity of the former w.r.t. the latter, $\vec{a}$ is the translation of the origins of the coordinate systems and $b$ is the time translation of clocks rigidly connected with each frame.

Now for the main part: we consider the fundamental equation of mechanics $\vec{F}=d\vec{p}/dt$.

The acting force $\vec{F}$ always transforms according to the simple rule $$\vec{F}\to\vec{F}'(t')=R\vec{F}$$ since otherwise we would not have equal magnitudes of the forces in all inertial reference frames. It follows (from $\vec{F}=d\vec{p}/dt$) that the transformation rule for momentum is of the form $$\vec{p}(t)\to\vec{p}'(t')=R\vec{p}(t)+\vec{C}(R,\vec{u},\vec{a},b)$$

I understood the parts before and after this paragraph, but have doubts on the above.

  1. Why do we require the force magnitude to be the same in all inertial reference frames? Is it because it's experimentally observed and hence considered a postulate, or some other reason entirely? The reason I suspect that it's based in experimental observation is because there's no reason, mathematically, why the magnitude of the force should be the same in all IRFs. I could just say that IRFs are an equivalence class with the Galilean transformation being the relation between them - this doesn't mathematically imply that the force magnitudes are invariant. But I could be completely wrong.

  2. I didn't follow how the momentum transformation rule follows by integrating both sides - more specifically I don't understand from where $\vec{u}$ and $\vec{a}$ pop up.

Would really appreciate clarifications on both questions!

  • 1
    $\begingroup$ Can you update the link to the paper with a non-PDF link from where people can decide to download the PDF if they want to? It's a 50 pages PDF and I weep for people who make the mistake of clicking on it on a phone. $\endgroup$
    – user87745
    May 5, 2020 at 22:18
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    $\begingroup$ @DvijD.C.: Updated! $\endgroup$ May 5, 2020 at 22:22

1 Answer 1


I agree with your first point. With seeing $\vec{p}$ as a quanitity "with its own rights" it can't be derived mathematically that $\vec{F}$ transforms like that, since you don't have any requirements to $\vec{p}$.

Regarding your second point: Thats just the most general way the constant vector $C$ can look like, since the dependence of $C$ on $\vec{u}$ and $\vec{a}$ won't effect the derivative of $\vec{p}'$ with respect to $t$.


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