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A doubt has arisen for me about the principle of relativity, and being such a fundamental subject I think it only fair to try and clarify it. The following line of reasoning was presented to me in a lecture, but is it really correct? It seems very unstable to me.

Take two frames of reference, $K$ and $K'$, with the latter moving in uniform rectilinear motion with respect to the former. It is possible to express the position of a material point with respect to the system $K \left(\vec{r}\right)$ by knowing the position of the point with respect to the system $K' \left(\vec{r'}\right)$ and knowing the position of the origin of the coordinates of $K'$ with respect to $K$ ($\vec{r_o}$). Thus:

$$\vec{r} = \vec{r'} + \vec{r_o}.$$

Deriving twice with respect to time:

$$\vec{a} = \vec{a'} + \vec{a_o}.$$

But since the system $K'$ moves in a uniform straight line:

$$\vec{a} = \vec{a'}.$$

That is, the acceleration of the material point is the same in the two systems. The equations of motion with respect to the two systems are therefore the same. It would seem that we have thus proved the principle of relativity, but by definition it cannot be proved. I will try to make a few remarks and ask someone to help me.

First objection. In the argument it is implicitly being stated that the properties of the position vectors do not change when passing from a system of fixed to one in uniform rectilinear motion and vice versa. It can be shown that the transformations that are part of Galileo's group are simple displacements of the metric. They therefore preserve the distances between two points and the time intervals. The vectors should therefore not undergo mutation in their length and orientation. However, I have a question to ask with respect to this group. It includes all affine transformations that preserve distances between simultaneous events and time intervals. Is it possible to prove that even in the transition from an inertial to a non-inertial system these characteristics are preserved?

Second objection One is also implicitly assuming that time is equally measured in the two systems, since we derive it without too much trouble. However, this objection too can be dismantled by assuming a priori that the flow of time is the same in all systems.

Third objection I would say the most important one: the statement $\vec{a} = \vec{a'}$ does not mean a priori that the laws of motion are equal in the two systems. Since in fact $f(\vec{r}, \vec{v}) = m \vec{a}$ with the reasoning described above we have shown only that the RHS of the equation is conserved, but nothing is known about the LHS.

Could you help me shed some light on these doubts of mine by explaining to me whether the reasoning presented in the lecture is correct and, if so, where I am wrong; or whether, if not, my objections make sense, or what is the right way to deal with this topic from this presentation?

P.S. I realise that the questions may seem like more than one, but in fact they are all interrelated and are put forward for the purpose of clarification.

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  • $\begingroup$ Newton's laws of motion and Galilean relativity are valid only in inertial frames of reference. Galilean relativity is experimentally verified for objects with speeds much less than the speed of light. $\endgroup$ Commented Dec 28, 2023 at 0:52
  • $\begingroup$ OK. But why is the lecture's explanation correct and my objections are not, or vice versa? If it is true that you answered the question on my first objection, I would like to understand a little more about the general situation I presented. $\endgroup$
    – Bml
    Commented Dec 28, 2023 at 8:42
  • $\begingroup$ The second objection doesn't seem to have any content. For example, the usual assumpiton in Newtonian physics is that time is a universal parameter that holds in all frames of reference, i.e. $\prime t=t$. $\endgroup$ Commented Dec 28, 2023 at 14:35
  • $\begingroup$ In the third objection, your proof is invalid. For example, ${\vec a}^{\prime} =\vec a$ does indeed mean that the laws of motion hold across inertial frames. If you apply Newton's law correctly, you have:$f^{\prime}(r^{\prime}, v^{\prime)}=ma^{\prime}$ for $K^{\prime}$ and $f(r,v)=ma$ for $K$. The form of the law is the same in both frames. $\endgroup$ Commented Dec 28, 2023 at 14:44

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