Feynman's Nobel classical electrodynamics action


by johne1618
Tags: action, classical, electrodynamics, feynman, nobel
johne1618
johne1618 is offline
#1
Dec29-12, 06:09 AM
P: 355
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_priz...n-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
Phys.Org News Partner Physics news on Phys.org
Information storage for the next generation of plastic computers
Scientists capture ultrafast snapshots of light-driven superconductivity
Progress in the fight against quantum dissipation
Jano L.
Jano L. is offline
#2
Dec29-12, 06:13 PM
P: 1,027
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.


Register to reply

Related Discussions
classical electrodynamics Advanced Physics Homework 2
Classical Electrodynamics Science & Math Textbook Listings 5
classical electrodynamics problem Advanced Physics Homework 3
Jackson's Classical Electrodynamics Advanced Physics Homework 6
Richard Feynman, 1965 Physics Nobel Prize Winner General Physics 7