Feynman's Nobel classical electrodynamics action

In summary, Feynman discusses the electrodynamic action between particles in his Nobel lecture, and mentions that it can be expressed in both 4-vector and 3-vector forms. He asks for someone to re-express it for a pair of particles in 3-vector form, noting that the Darwin Lagrangian is based on a universal time while the Feynman action uses proper time for each particle. The resulting action for the Darwin Lagrangian is a double integral over two trajectories, while the Feynman action uses just coordinates and velocities at a common time.
  • #1
johne1618
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In his Nobel lecture Feynman describes an electrodynamic action between a set of particles (equation 1, one third way thru lecture):

http://www.nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html

The action is in 4-vector form.

I wonder if someone could do me a favour and re-express the action just for a pair of particles in 3-vector form?

I presume the action is related to the Darwin Lagrangian. The main difference between them is that in the Feynman action each particle has its own proper time whereas the Darwin lagrangian is expressed in a universal time.

Thanks,

John
 
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  • #2
I think it should be easy if you just replace [itex]X^i[/itex] by [itex](ct_i,\mathbf r_i)[/itex] in the integral and play a little with the formula. However, the resulting action will contain double integral over two trajectories, while the action for the Darwin Lagrangian is just an integral of certain function over coordinate time. The motions described by the two are quite different in general; the motion due to the action with Darwin Lagrangian can be thought of as best approximation to the motion implied by the Feynman action, possible with action that uses just coordinates and velocities at common time.
 

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