This is a very simple problem (PR-2.pdf), but interpreting the result seems to be quiet difficult. Please, leave a comment with your opinion.

Consider two particles A and B in translation with uniformly accelerated vertical motion in a frame S (X,Y,T) such that the segment AB with length L remains always parallel to the horizontal axis X (X_A = 0, X_B = L). If we assume that the acceleration vector (0, E) is constant and we take the height of both particles to be defined by the expressions Y_A = Y_B = 0.5 ET², we have that the vertical distance between A and B in S is always (see fig. in PR - 2.pdf):

1) Y_B - Y_A = 0

If S moves with constant velocity (v, 0) with respect to another reference s(x,y,t) whose origin coincides with the origin of S at t = T = 0, inserting the Lorentz transformation for A (Y = y, T = g(t - vx_A/c²), xA = vt) into Y_A= 0.5 ET² and the Lorentz transformation for B (Y = y, T = g(t - vx_B/c²), x_B = vt + L/g) into Y_B= 0.5 ET² we get that the vertical distance between A and B in s(x,y,t) is:

2) y_B - y_A = 0.5 E (L²v²/c⁴- 2Lvt/c²g)

which shows us that, at each instant of time "t" the distance y_B - y_A is different despite being always constant in S (eq.1). As we know that the classical definition of translational motion of two particles is only possible if the distance between them remains constant, we conclude that in s the two particles cannot be in translational motion despite being in translational motion in S.

More information in:

Preprint A BRIEF OVERVIEW ON THE CONSEQUENCES OF CONSIDERING A LOCALL...

PR-2.p
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Quantum Optics and Quantum Information

7th Feb, 2022

Stam Nicolis

University of Tours

Of course the problem may be solved within Newtonian mechanics quite trivially, by realizing that the only way the segment can remain (a) horizontal and (b) of fixed length is if the initial conditions are y_A(0)=y_B(0), v_A,x(0)=v_B,x(0) and v_A,y(0)=v_B,y(0).

Therefore it doesn't make sense demanding that the segment can satisfy conditions (a) and (b) if the initial conditions are different. It suffices to write Newton's equations along x and y for the two particles to show this.

Then Newton's equations ensure that y_A(t)=y_B(t) and that |x_A(t)-x_B(t)|=|x_A(0)-x_B(0)| which every first year student learns in the first few weeks.

Newton's equations, also, imply that a Galilean transformation along x is consistent with these relations, since the force along x is zero...

And the statement that the vertical distance could change is, just, wrong. There's just no way it could change; it can be shown to be a constant of motion, by writing the equation satisfied by the relative distance, y_A(t)-y_B(t), namely d²(y_A(t)-y_B(t))/dt²=0, since the acceleration is constant along y and the same for both particles-and both particles are assumed to have the same initial velocity along y.

In fact it's, also, trivial to see from Newton's equations, that, if the two particles are subject to the same force along x, the segment will remain horizontal and of fixed length; it will just be accelerating along x. Motion along x is independent of motion along y...

The equation for y_A(t)-y_B(t) isn't invariant under transformations to a non-inertial frame; in Newtonian mechanics these would be transformations of time that aren't linear functions of time. But that's no surprise. (And Lorentz transformations are transformations that are linear in the coordinates.)