# Feynman's Nobel classical electrodynamics action

by johne1618
Tags: action, classical, electrodynamics, feynman, nobel
 PF Gold P: 1,165 I think it should be easy if you just replace $X^i$ by $(ct_i,\mathbf r_i)$ 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.