2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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