Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

I noticed that a calculus of variations problem is just an integral over a differential form. Therefore, I would think it would be possible to formulate the Euler-Lagrange equations using exterior calculus. However, I do not know of how to reconcile the notion of a functional derivative with say an exterior derivative.

share|cite|improve this question

The correct context for this problem is on infinite jet bundle and using the variational bicomplex.

The main reference for this is Ian Anderson's work:

share|cite|improve this answer

Sometimes, you can. If a functional is stationary on a particular path, it means that "around that path", up to first order, the value of the functional does not change. Intuitively, "on the path, the functional does not depend on the path". Yes, it sounds weird. But it simply means that along the path, and only there, a certain 1-form is exact! Let me make an example: the action in physics.

The functional is:

$$ S= \int L\,dt = \int (p\dot q -H)\,dt.$$

$ dS = L\,dt = p\,dq - H\,dt $ is the 1-form you are talking about.

If there is a line along which it is exact (so that writing $dS$ makes sense), that line must be an extremum!

That form is closed if:

$$ \frac{\partial p}{\partial t} = -\frac{\partial H}{\partial q} .$$

Does this ring a bell, speaking of Euler-Lagrange? This is (almost) Hamilton's equation. If there is a line on which this equation is satisfied, then that line is an extremum for the action.

The converse, unfortunately, is not always true. For example, the form could be nowhere closed. But there will be again an extremum, in general!

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.