1
$\begingroup$

The form-invariance of the Lagrange equations implies the existence of a function $\ A( q_k, t)$ so that

$\ \begin{equation} L' (q_k, v_k, t) -L(q_k, v_k, t) = \frac d {dt} A( q_k, t) \end{equation}$

Is there any way of proving the converse? That is, asuming the existence of the funcion $A$ prove the form-invariance of the Lagrangre equations.

$\endgroup$
2
$\begingroup$

An elegant proof of the converse is obtained by using the action formulation. Recall that for a given Lagrangian $L$, the action is a functional of paths defined as follows: \begin{align} S_L[q] = \int_{t_a}^{t_b} dt\, L_q(t) \end{align} Now suppose that two Lagrangians differ by a total time derivative; \begin{align} L'_q(t)-L_q(t) = \frac{dA_q}{dt}(t), \end{align} then we immediately find that \begin{align} S_{L'}[q] - S_{L}[q] = A_q(t_b) - A_q(t_a) \end{align} If, in addition, $A$ is a local function (see Definition of Local Function) of paths $q$, then this implies that \begin{align} \delta S_{L'}[q] - \delta S_L[q] = 0 \end{align} for all variations keeping the endpoints of paths fixed. Therefore, for all such variations, $q$ is a stationary point of $S_{L'}$ if and only if $q$ is a stationary point of $S_L$. Now, if $L$ and $L'$ are local functions that depend only on $q$, it's first derivative, and time, namely if \begin{align} L_q(t) &= L(q(t), \dot q(t), t) \\ L'_q(t) &= L'(q(t), \dot q(t), t), \end{align} then a path $q$ is a stationary point of the action for all such variations if and only if it satisfies the corresponding Euler-Lagrange equations. Combining these facts, we see that if $q$ satisfies the EL equations for $L$, then it is a stationary point of $S_L$, so it is also a stationary point of $S_{L'}$ and therefore satisfies the EL equations for $L'$, as desired.

Addendum.

We show that the variation of a function $A$ that is local in paths $q:[t_a,t_b]\to \mathbb R$ and their first derivatives vanishes at the endpoints $t_a$ and $t_b$ for all variations that keep path endpoints fixed.

Let a function $A$ that is local in paths $q$ and their time derivatives be given. Namely, $A$ is such that there exists a function $\alpha$ for which \begin{align} A_q(t) = \alpha(q(t), \dot q(t), t) \end{align} for all paths $q$. Now let a path $q$ defined on $[t_a, t_b]$ be given, and let $\delta q$ be a variation of $q$, then if $\delta A_q$ denotes the variation of $A_q$ induced by $\delta q$, then we have \begin{align} \delta A_q(t) = \frac{\partial \alpha}{\partial q}(q(t), \dot q(t), t)\delta q(t) + \frac{\partial\alpha}{\partial\dot q}(q(t), \dot q(t),t) \delta\dot q(t) \end{align} Now, for a sufficiently smooth variation that keeps the endpoints $q(t_a), q(t_b)$ of the path fixed, one has \begin{align} \delta q(t_a) &= \delta q(t_b)=0\\ \delta \dot q(t_a) &= \delta \dot q(t_b) = 0, \end{align} which you should be able to convince yourself of (see https://physics.stackexchange.com/a/93290/19976 for mathematical details of variations that will help) from which it follows that \begin{align} \delta A_q(t_a)= \delta A_q(t_b) = 0. \end{align} as desired.

$\endgroup$
  • $\begingroup$ Simply perfect. Thank you. Just a detail: I've read somewhere that the function A could depend on velocities, but we would need to know the final velocities. I don't understand what this means. In the third step of your prove the two A functions cancel each other because the q coordinates don't change from tb to ta. But why shouldn't the velocities change? $\endgroup$ – carllacan Jan 30 '14 at 1:43
  • $\begingroup$ @carllacan Sure thing. See the addendum for a proof that $A$ can depend on velocities. The terms $A_q(t_a)$ and $A_q(t_b)$ don't cancel each other, it is their variations that cancel each other provided the variation we are considering keeps path endpoints fixed and provided $A$ is local. I'm not sure what "need to know their final velocities" means in this case. $\endgroup$ – joshphysics Jan 30 '14 at 2:15
  • 1
    $\begingroup$ That last link is the most illuminating thing I read about mechanics in a month. I cannot thank you enough, is the first explicit definition of $\delta q$ I've seen. $\endgroup$ – carllacan Jan 30 '14 at 2:48
  • 1
    $\begingroup$ @carllacan I can totally identify with your frustration, and it makes me happy to hear you've benefited in that way from the other post. It took years until I was finally able to piece together all of the mathematically vague stuff one finds in mechanics books. The nice thing is that once you understand what the heck a variation is in that context, it carries over to everywhere else in physics (like classical field theory). $\endgroup$ – joshphysics Jan 30 '14 at 3:00

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.