Momentum of a box full of bouncing particles

We were wrong in our claim on August 25, 2023 that special relativity would break conservation of center of mass if rest masses exist. We assumed that the y velocity of the bouncing particle stays the same when the particle is pulled with a force F to the x direction. That assumption is wrong.

Let us analyze the momentum and inertia of the box full of bouncing particles.

                      --------

                         ^

                         |

                         |  vy 

                         |

                         •  m

                      --------

                        ● M  =  m / sqrt(1  -  vy² / c²)

      ^  y

      |

       -----> x

The particle m is bouncing up and down in the box. The particle M is static, with the same mass-energy as m.

Let us assume that a force F pulls M and m both to the positive x direction for a time t, giving an impulse of

       pₓ  =  F t.

What is the x velocity of each particle after the impulse?

We assume that for a particle m moving at a velocity v, the momentum p satisfies

       p²  =  v² m² / (1  -  v² / c²),

and the momentum vector p is parallel with the velocity vector v. Then

       p² -  p² v² / c²  =  v² m²

   <=>

       v²  =  p²  /  (m²  +  p² / c²).

For M, from the above formula we get the velocity

       Vₓ²  =  pₓ²  /  (m² / (1 - vy² / c²))  +  pₓ² / c²).

For the bouncing particle m, the square of the x component of the velocity is

       vₓ²  =  pₓ² / (pₓ² + py²)

                  * (pₓ² + py²)  /  (m²  +  (pₓ² + py²) / c²)

              =  pₓ²  /  (m² + (pₓ² + py²) / c²)).

Let us assume c = 1. To prove Vₓ = vₓ, we have to show

       m² / (1 - vy²)  +  pₓ²  =  m² + pₓ² + py².

   <=>

       m²  =  (m² + py²)  *  (1 - vy²).

The energy-momentum relation says

       E²  =  p²  +  m²,

and for the bouncing particle,

       E²  =  m² / (1 - vy²).

Our equation becomes

       m²  =  m² / (1 - vy²)  *  (1 - vy²),

which is true.

When a particle is x accelerated, its y momentum stays the same, but not the y velocity.

The force F pulls to the x direction. The y velocity slows down. Thus, the acceleration of the particle is not solely to the x direction, but also to the negative y direction. The acceleration is not to the direction of the force.

Conclusions

We did have a calculation error in our August 25, 2023 blog post. Calculating with velocities is error-prone in special relativity. It is better to use momenta.

On September 9, 2023 we claimed that the Lorentz transformation of the electric field Eₓ is wrong in the literature. Our claim was wrong. We erroneously assumed that the "inertia" of a particle m against a change in its velocity is γ m, like its mass-energy is.

Above we showed that the "inertia" really is γ m when the force F is normal to the velocity v of the particle. The bouncing particle behaves like a static particle whose mass M = γ m. But if the force F pulls to the direction of v, the behavior is very different. A simple example is this: we can increase the x velocity of a static particle M by any number < c, but that is not possible for a particle m which already moves to the x direction.

We will next look at the inertia of a test charge q inside the electric field of a charge Q. Does the inertia spoil the Lorentz transformation Eₓ' = Eₓ?