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.


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.


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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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

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