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Is it right that if a proton is trapped in a magnetic field it is forced to do a circular motion of a distinct radius.Now, if a quark inside that proton does a rotation inside the proton volume can the quark trajectory in the magnetic field be described by a helicoidal curve shown on the picture? I am interested in the problem that arises from the differences in the lengths of the simple circle of the proton center of mass trajectory and the quarks' helicoidal curve when one rotation in the magnetic field has elapsed,which implies that the velocities of the proton center of mass and the distinct quark are different...So if a proton is accelerated to reach 99.9% of the speed of light inside that magn. field the quark speed should be slightly higher as the helicoidal-circular curve is slightly longer than the pure circular curve?

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    $\begingroup$ Is it valid to even consider a classical trajectory for a quark? $\endgroup$ – BioPhysicist May 5 at 18:15
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    $\begingroup$ @BioPhysicist you can't even talk properly about single quarks inside a proton, unless you are probing with high enough energies $\endgroup$ – FrodCube May 5 at 18:30

The trajectories of charged tracks in a magnetic field are given by the solution of classical electromagnetism equations. If the velocity of the particle is perpendicular to the magnetic field, a circle is closed. If there is an angle to the field a spiral is induced by momentum conservation in that direction.

Electrons , pions etc, though, are quantum mechanical entities when interacting with each other and need quantum mechanical solutions for the interaction, in macroscopic dimensions, where h_bar, the Planck constant is commensurate to zero. Charged quantum particles also obey classical equations in macroscopic dimensions, this can be seen in the bubble chamber pictures here.

The quantum mechanical interactions are calculated using quantum field theory and the data up to now fit the calculations with great accuracy. The predictions of the theory are on the probability of finding a particle at (x,y,z,t), so an accumulation of data with the exact same conditions has to be made, in order to check theory with experiment.

In bound quantum mechanical states, there are no trajectories. The electrons around a nucleus, the protons inside the nucleus, the quarks inside the proton do not move in trajectories/orbits but in orbitals, probability loci. A single electron measured about a nucleus will just give an (x,y,z,t) , a point in the plots of the link.

There is no way a macroscopic magnetic field can affect bound state particles, because any effect has to happen with elementary particle interactions. The magnetic field interacting with the proton macroscopically turning it into a circle expends all the possible interaction energy in making the proton turn. There is no field penetrating the proton, so it does not even make sense to ask about quark-external-magnetic-field interactions. The only magnetic interactions the quark may see are internal, due to the magnetic moments of the nuclei in the nucleus.

  • $\begingroup$ I have read about energy levels,qantum numbers, radial and angular nodes and probability distribution in the case eg. of H-atom... but if 'gamma' for relativistic mass for a proton is same nevertheless if taking in account the velocity of the center of mass or 'expected non stationary' quarks does it imply they are not moving at all inside there? $\endgroup$ – Janko Bradvica May 6 at 9:15
  • $\begingroup$ in bound states the notion of motion is superseded by "probability of being at (x,y,z,t)" . The classical definition of velocity has no meaning for bound states. $\endgroup$ – anna v May 6 at 10:35

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