Consider the action

$$A_{1} = \int{L(q, \dot{q})}{dt}\tag{1}$$

and the corresponding Euler–Lagrange equation

$$\frac{\partial{L}}{\partial{q}} - \frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot{q}}}\right)=0.\tag{2}$$

This equation is a general condition that $L$ fulfill. Therefore this condition you can add to the original action as Lagrange multiplier (this change has no, in principle, affect for Euler–Lagrange equation).

$$A_{2} = \int{\left[L(q, \dot{q}) + \lambda\left(\frac{\partial{L}}{\partial{q}} - \frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot{q}}}\right)\right)\right]}dt\tag{3}$$

Where $\lambda \in R$ and so the last term is

$$\frac{d}{dt}\left(\lambda\frac{\partial{L}}{\partial{\dot{q}}}\right)\tag{4} $$

This term is in the form of total derivative and so we can dropped it from the lagrangian (generates the same equation). We're getting that expression

$$A_{2} = \int{\left(L(q, \dot{q}) + \lambda\frac{\partial{L}}{\partial{q}} \right)}dt\tag{5}$$

But this lagrangian generally generates a different equation than the original lagrangian $L$.

I can't figure out where I made a mistake.


1 Answer 1


Here we assume that $\lambda$ is not a function of time, i.e. there is only one time-averaged constraint. Then OP's derivation has the following shortcomings:

  • Firstly, the new time-averaged constrained term in eq. (3) subtly changes the EOM for $q$.

    Example. Consider for simplicity the static model $L(q)~=~\frac{1}{2}q^2+\frac{1}{3}q^3$. Then the stationary points for the action (1) are $q\approx 0$ and $q\approx -1$, while the constraint yields the time-average $\langle q\rangle \approx - \frac{1}{2}$. The EOM for $q$ becomes $q^2+q\approx\lambda (2q+1)$. See also this related Phys.SE post.

  • Secondly, removing the boundary term (4) alters the EL equation for $\lambda$. More generally, boundary terms do matter if they don't vanish/aren't fixed by the pertinent boundary conditions of the theory.


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