This question already has an answer here:

Can anyone give me an intuition of why the magnetic vector potential $A$ is sometimes interpreted as the electromagnetic momentum ?

I don't know analytical mechanics, just classical electromagnetism


marked as duplicate by Kyle Kanos, Brandon Enright, JamalS, John Rennie, Neuneck Nov 5 '14 at 8:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Who refers to the vector potential as momentum? That's nonsense, AFAIK. $\endgroup$ – ACuriousMind Nov 5 '14 at 3:07
  • $\begingroup$ Yes I'm wrong. It's sometimes interpreted as electromagnetic momentum Edited $\endgroup$ – Keith Nov 5 '14 at 3:08
  • $\begingroup$ See physics.stackexchange.com/questions/53020/… $\endgroup$ – Cheeku Nov 5 '14 at 3:15
  • $\begingroup$ I already check that one. The problem is that in the solution they show some things that I don't understand, like "canonical momentum ". I search for a more elementary answer $\endgroup$ – Keith Nov 5 '14 at 3:18
  • $\begingroup$ You might find this helpful: "Potential Momentum, Gauge Theory, and Electromagnetism in Introductory Physics" arxiv.org/abs/physics/9803023 $\endgroup$ – Alfred Centauri Nov 5 '14 at 3:36

In Lagrangian mechanics, one usually introduces the "generalized momentum" as $$ p_{\phi_i}:=\frac{\partial L}{\partial \phi_i} $$where $L$ is the Lagrangian depending on a bunch of generalized coordinates $\phi_1\dots\phi_n$. If you introduce electromagnetic forces, you can derive, that the Lagrangian in the presence of an electromagnetic field has a form like $$ L=\frac{m}{2}\left(\vec{p}-\frac{q}{c}\vec{A}\right)^2+q\Phi $$ where $A$ is your vector potential and $\Phi$ your scalar potential, therefore your generalized momentum will be shifted by $A$, i.e. if you write down the respective Newton equation, instead of $\dot{p}$ you will have this shifted by $A$.

The physical meaning behind this formal stuff is, that the electromagnetic field actually carries a momentum, which can be translated into a mechanical momentum. For example, I remember an experiment, where you put two spherical shells into a magnetic field and then turn the field off, the shells will get an angular momentum (I searched for it, didnt find it yet, will update if found).

  • 1
    $\begingroup$ Ok. Thanks for the answer. So that means that the total momentum of any particle interacting with a magnetic field is less by a factor of $A$. Why is this ? Any intuition , if the particle is on magnetostatic field is different from a changing current in time. Does this affect the energy of the particle By some other factor of $A$? $\endgroup$ – Keith Nov 5 '14 at 3:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.