Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the Lagrangian for a charged particle in an electromagnetic field

$$L = \frac{1}{2}mu^2 - q(\phi - \frac{\vec{A}}{c}\cdot \vec{u})$$

the energy of the particle is contained in the kinetic term, the rest being interaction terms of the particle with the electromagnetic field. If it's travelling at some velocity, then it will generate its own magnetic field and therefore possess a magnetic self energy, so which term in $L$ contains this?

share|cite|improve this question

The interaction term is only with an external field. To add the self interaction you should add field terms, but this requires lagrangian density formalism.

share|cite|improve this answer

Let the speed-of-light be $c=1$. OP's action

$$\tag{1} S[{\bf r}]~=~\int \! dt ~ L({\bf r}(t), \dot{\bf r}(t),t),$$

where the Lagrangian is

$$\tag{2} L({\bf r}(t), \dot{\bf r}(t),t)~=~\frac{1}{2}m\dot{\bf r}(t)^2 - q(\phi({\bf r}(t),t) - {\bf A}({\bf r}(t),t)\cdot \dot{\bf r}), $$

treats a non-relativistic charged particle's position ${\bf r}$ as an active variable and the electromagnetic $4$-potential $(\phi,{\bf A})$ as a passive background variable. Variation of this action wrt. the active variables leads to Newton's 2nd law for the charged particle with a resulting force given by the Lorentz force.

This action (1) does not describe the full electromagnetic theory with Maxwell's equations. For that one needs to add the well-known $F_{\mu\nu}F^{\mu\nu}$ term in the action, cf. the answer by Angel Joaniquet Tukiainen.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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