Transformation law of momentum under Galilean transformation

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Shirish
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I'm reading the article https://www.researchgate.net/publication/267938119_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##.
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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
$$\vec{p}(t)\to\vec{p}'(t')=R\vec{p}(t)+\vec{C}(R,\vec{u},\vec{a},b)$$
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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!
 
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Make you really clear what rotation in this context mean!Permanently rotations of a point leads to violation of Newtonian axioms for an observer which means the definition of an inertial frame is violated!So ,be careful with this part!

troglodyte
 
Shirish said:
1. Why do we require the force magnitude to be the same in all inertial reference frames?
Inertial reference frames contain only real interaction forces, which the magnitude of can be directly measured in a frame invariant manner, for example with a spring. All frames must agree how long a spring is.
 
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A.T. said:
Inertial reference frames contain only real interaction forces, which the magnitude of can be directly measured in a frame invariant manner, for example with a spring. All frames must agree how long a spring is.
"All frames must agree on how long a spring is" - pardon a potentially stupid question, but that makes total sense in Newtonian mechanics. Does this fail in case of special relativity? I've heard of length contraction in case of SR => all inertial observers won't agree on spring length => all inertial observers won't agree on invariance of force magnitude. Is this right or am I missing something?
 
Shirish said:
Does this fail in case of special relativity? I've heard of length contraction in case of SR => all inertial observers won't agree on spring length => all inertial observers won't agree on invariance of force magnitude. Is this right or am I missing something?
That's right. In SR different inertial observers might not agree on length measurements. The magnitude of the Newtonian force measured by them will not be the same in general.

However a generalization of Newton's law comes about quite naturally in the SR formalism (i.e., in the differential geommetry formalism). So you still have the observers measuring the same magnitude of a vector which can be called "four force".
 
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A.T. said:
All frames must agree how long a spring is.

They also need to agree about the stiffness of the spring. We know it is frame-independent, but can it be derived from classical mechanics? I would do it the other way around by prooving that m is frame-independent under Galilei transformation. But that also requires Newton III (or conservation of momentum).