[Sergio_Prats (OP)]

The 4/3 problem in classical electrodynamics can be explained here: https://www.feynmanlectures.caltech.edu/II_28.html [nofollow]

It says that, for the electromagnetic field generated by a spherical surface with radius a, moving at speed v, the mass calculated from the EM field energy is less than the mass calculated from the EM field momentum, which is the mass that would be necessary to make the field mass move at the same speed than the particle, v. While the mass that comes from the energy is m_e=(1/2)*e²/(a*c²), the momentum is p=(2/3)*e²/(a*c²v), hence  from the relation (2/3)/(1/2) comes the name of 4/3 problem.

What I noticed is that the field mass-energy is not only moving because of its momentum but also because of the interaction with the charge, E·J which is draining energy from the frontal part of the sphere and moving it to the back which is equivalent to a backward velocity and the sum of that effect plus the extra velocity of the momentum for the previous mass U_e brings to the correct speed v. The effect is similar to be cycling on a road full of round stones in which 1/4 of your pedaled the back wheel finds slippery stones which make you to lose your impulse, you need to pedal faster to keep the desired speed.

I have written this 3-page document explaining this with a bit more detail that can be found here:

https://www.slideshare.net/SergioPL81/adding-a-shift-term-to-solve-the-43-problem-in-classical-electrodinamics [nofollow]

Any comments will be appreciated.