Can this experiment break Lorentz symmetry?

[Ibix]

For the two projectiles with different absolute velocities, this predicts:
$$\Delta t_1 \approx \frac{d}{V+u}, \quad \Delta t_2 \approx \frac{d}{V-u}$$

But, as I have already pointed out twice, those two $d$ need to be different because the distances over which the forces apply are only equal in the lab frame. This is true even in Newtonian mechanics, let alone relativity.

Furthermore, you need to transform the various fields and currents from the lab frame into the frame where the lab is moving and recalculate the forces; they will not be equal and opposite. And you appear to be assuming a Galilean transform of velocity.

 

[Asaad-Hamad]

But, as I have already pointed out twice, those two $d$ need to be different because the distances over which the forces apply are only equal in the lab frame. This is true even in Newtonian mechanics, let alone relativity.

Furthermore, you need to transform the various fields and currents from the lab frame into the frame where the lab is moving and recalculate the forces; they will not be equal and opposite. And you appear to be assuming a Galilean transform of velocity.

You are absolutely applying the standard relativistic methodology correctly, and I understand your point about the transformation of forces and distances.

However, the core purpose of this thought experiment is to test the foundational Principle of Relativity itself. The principle asserts that the laws of physics are the same in all inertial frames, which guarantees the symmetry you are describing.

This experiment is constructed from a Lorentzian perspective, which posits:

1. A rest frame exists.
2. Lorentz contraction and time dilation are real physical effects.
3. The Principle of Relativity is an apparent symmetry, maintained by these physical compensations.

From this viewpoint, the question is not how to transform forces and distances between frames, but whether a local measurement can detect an asymmetry in a process that those physical effects might not perfectly conceal.

The experiment deliberately uses a fixed, platform-defined distance $d$ and identical trigger mechanisms. The hypothesis is that the duration of the force application over this distance is a physical process whose rate is set by absolute time, not by the platform's time-dilated clocks.

If this hypothesis were correct, then even after accounting for the different field configurations in the rest frame, the measured impulse $J = \int F dt$ would be different for the two projectiles when measured by a clock that somehow tracks absolute time. This difference would manifest as a break in symmetry within the lab frame.

Of course, the standard response—which I expect is correct—is that any measuring device on the platform, including clocks and force sensors, is governed by the same laws and will themselves be affected in a way that perfectly compensates, always measuring $J_1 = J_2$ and preserving the symmetry. This experiment is designed to test the limits of that perfect compensation.

In essence, the experiment asks: "Is the duration of this specific physical process truly self-contained within the lab frame, or is it externally dictated by a preferred frame of reference?"

 

[Ibix]

However, the core purpose of this thought experiment is to test the foundational Principle of Relativity itself.

Then you cannot use Newtonian or Einsteinian formulas in your analysis, since both use the principle of relativity.

 

[Asaad-Hamad]

Then you cannot use Newtonian or Einsteinian formulas in your analysis, since both use the principle of relativity.

That is the critical point, my argument hinges on the partitioning of energy into 'spark' and 'passive' components.

1. The spark energy ($W = F \cdot d$) is the small, measurable energy we input locally in the platform frame.
2. The passive energy is the vast, undetectable (in the platform frame) kinetic energy inherent in the object's absolute motion ($\frac{1}{2}mV^2$).

The key insight is that the same spark energy input $W$ produces a different change in absolute velocity ($\Delta v$) depending on the initial absolute velocity, due to the quadratic kinetic energy law:
$$\Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) = W$$
This means:
$$\Delta v = \sqrt{v_i^2 + \frac{2W}{m}} - v_i$$
For a large $v_i$ (e.g., $V+u$), the same $W$ produces a smaller $\Delta v$ than for a small $v_i$ (e.g., $V-u$).

Therefore, the kinematic formula $\Delta t \approx d / v_{avg}$ is not just a mathematical tool; it reflects this physical reality. The projectile with higher absolute velocity ($V+u$) spends less absolute time under force, receives less impulse, and thus gains less velocity from the spark energy than its counterpart.

The apparent symmetry in the platform frame is an illusion maintained by the 'passive energy' transfer, which ensures the relative velocities remain equal and opposite. But the mechanism of how that symmetry is maintained—through different interaction durations and impulses—would reveal the absolute motion.

from rest frame perspective, we only measure the exerted energy in the inertial frame and we think this is what caused the huge change in the absolute kinetic energy ( masking it by energy is frame dependent) where the truth is almost all the change in the kinematic energy came from the passive energy transfer, the light-mass forward-moving projectile pushed the heavy platform with vast kinetic energy slightly rearward and gained huge energy.

 

[Ibix]

But you are still using Newtonian formulae, which means you accept the principle of relativity. So you are not testing the principle of relativity.

What you are actually doing is mis-applying Newtonian physics and getting into a mess. For example this:

The key insight is that the same spark energy input $W$ produces a different change in absolute velocity ($\Delta v$) depending on the initial absolute velocity, due to the quadratic kinetic energy law:
$$\Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) = W$$

is a common misconception that stems from failing to consider the work done by the third law pair force of the force you are calculating. Once you account for that you will find that the same input energy produces the same delta-V. At the moment you are simply ignoring some energy and trying to come up with a fantastic explanation of where the energy you are ignoring is.

 

[Nugatory]

@Asaad-Hamad We can answer questions, but we can’t make you like the answer.

The questions in the original post and repeated in #11 have been answered, so this thread is closed.