Feynman Lectures Question on Figure 26-2: "It appears that action is not equal to reaction"

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SUMMARY

In the discussion regarding Feynman's Lectures, specifically Lecture 26, it is established that the law of action and reaction does not hold for two charged particles moving at right angles, as their forces do not balance, leading to a change in net momentum. This phenomenon indicates that while the momentum of the particles is not conserved, the total momentum of the system, including the electromagnetic field, remains conserved. The fourth Maxwell's equation is highlighted as crucial, stating that a magnetic field is generated by both current density and changing electric fields, which reconciles the apparent discrepancy in momentum conservation when considering the field's contribution.

PREREQUISITES
  • Understanding of Feynman's Lectures on Physics, particularly Volume 2, Lecture 26
  • Familiarity with Maxwell's equations, especially the fourth equation
  • Knowledge of electromagnetic fields and their momentum
  • Basic concepts of particle physics and momentum conservation
NEXT STEPS
  • Study the implications of the fourth Maxwell's equation on electromagnetic theory
  • Research the Poynting vector and its role in momentum conservation in electromagnetic fields
  • Explore advanced topics in particle physics related to momentum and force interactions
  • Examine case studies of charged particles in electromagnetic fields to understand practical applications
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Physics students, educators, and researchers interested in advanced electromagnetism, particularly those exploring the nuances of momentum conservation in charged particle interactions.

bob012345
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In his lecture 26 vol 2 Feynman discusses charges moving at right angles shown in fig 26-2. As Feynman states "It appears that action is not equal to reaction." He states it will be discussed further but I cannot find that. My question is what is the most updated thought on this 60 year old question? Thanks.
 
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Feynman said:
We will mention two further examples of momentum in the electromagnetic field. We pointed out in Section 26–2 the failure of the law of action and reaction when two charged particles were moving on orthogonal trajectories. The forces on the two particles don’t balance out, so the action and reaction are not equal; therefore the net momentum of the matter must be changing. It is not conserved. But the momentum in the field is also changing in such a situation. If you work out the amount of momentum given by the Poynting vector, it is not constant. However, the change of the particle momenta is just made up by the field momentum, so the total momentum of particles plus field is conserved.
 
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Thanks. I thought it might be that similar to the Feynman Disk. Anyone have any other thoughts?
 
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The fourth Maxwell’s equation states that a magnetic field is produced by a current density and by a changing electric field.
Here the changing electric field generates the component of the magnetic field which is not mentioned. After including it the action will be equal to the reaction.
 
Gavran said:
After including it the action will be equal to the reaction.

So what about momentum of the field?
 
Gavran said:
The fourth Maxwell’s equation states that a magnetic field is produced by a current density and by a changing electric field.
Here the changing electric field generates the component of the magnetic field which is not mentioned. After including it the action will be equal to the reaction.
If I understand Feynman’s argument, momentum is conserved by including the field momentum and change in the field momentum balances the forces but the forces do not balance among the charges only but the system of charges and fields.
 
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Sorry for wrong explanation in the post #4.
 
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