My question is in reference to Landau's Vol. 1 Classical Mechanics. On Page 6, the starting paragraph of Article no. 4, these lines are given:

If an inertial frame $К$ is moving with an infinitesimal velocity $\mathbf\epsilon$ relative to another inertial frame $K'$, then $\mathbf v' = \mathbf v+\mathbf \epsilon$. Since the equations of motion must have the same form in every frame, the Lagrangian $L(v^2)$ must be converted by this transformation into a function $L'$ which differs from $L(v^2)$, if at all, only by the total time derivative of a function of co-ordinates and time.

1) Doesn't this hold for same frame? Why is Landau changing the Lagrangian of frame $K$, $L$, to $L'$ with the change satisfying this condition? So, how can he assume that the action would be minimum for the same path in $K'$ as there was in $K$? In two frames the points $q_1$ and $q_2$ aren't same which are at $t_1$ and $t_2$.

2) How did he assume that this is the one and only way to change the Lagrangian without changing the path of least action? Can we prove this?

With respect to first question, I feel that there is something fundamentally amiss from my argument as the Lagrangian is dependent only on magnitude of velocity, so $q_1$ and $q_2$ won't matter. I have made an explanation myself that since the velocity is changed infinitesimally, it should essentially be the same path governed by the previous Lagrangian, the path it took with constant velocity $v$. But, still I am not convinced. The argument isn't concrete in my head. Please build upon this argument or please provide some alternative argument.

I know the question (1) and argument above are very poorly framed but I am reading Landau alone without any instructor and so have problems forming concrete ideas.


2 Answers 2


Consider the principle of least action $\delta S = 0$, where the action $S$ is given by: $$S= \int_{t_1}^{t_2}L(\mathbf{q}(t),\dot{\mathbf{q}}(t),t)\ \text{d} t$$ from this we can derive Lagrange's Equations. Now let $G\in C^1(t_1,t_2)$, that is $G$ is an at least once differentiable function on $(t_1,t_2)$. Then $$S' = \int_{t_1}^{t_2}\left(L(\mathbf{q}(t),\dot{\mathbf{q}}(t),t) +\dfrac{\text{d}G}{\text{d}t}\ \right) \text{d} t=\int_{t_1}^{t_2}L(\mathbf{q}(t),\dot{\mathbf{q}}(t),t)\ \text{d} t + \Delta G = S + \text{const}$$ Thus $\delta S' = \delta S$ and $L(\mathbf{q}(t),\dot{\mathbf{q}}(t),t) +\dfrac{\text{d}G}{\text{d}t}$ is an equivalent Lagrangian of the system and will produce all the same dynamics.


Even if you change frames, the physics is still the same and the particle will follow the same path, no? And there is certainly more than one way to change the Lagrangian without affecting the path of least action--add any combination of total time derivatives to it.

When I said it would follow the same path, I meant the same path after you take into account the fact that you shifted frames. If $q_1$ and $q_2$ label the same point even after you shift frames (so that in new coordinates $q_1=q^{new}_1−ϵt$ and etc.) then the particle will be at $q_2$ at $t_2$ if it was at $q_1$ at $t_1$.

I mean that if the particle starts at time $t_1$ at $q_1$ it DOES end up at $q_2$ at time $t_2$ provided you take into account how the points look different because of the new frame. The path taken by the particle must be the same; physics doesn't depend on the inertial frame you are in and this is the point Landau is making. If you think that in frame K' the particle doesn't end up at $q_2$ at time $t_2$ then it is purely because the points are labeled differently in this frame. It also has nothing to do with v being infinitesimal; that doesn't matter.

As for the other question, you can also multiply the whole Lagrangian by a constant. That's kind of obvious though. Basically you need the kinetic energy term, and the only way you can further modify it is by adding terms right? If you multiplied the Lagrangian by a non-constant term, for instance, the form of the kinetic energy term would change. Then he rules out what terms you can add.

Maybe if you go to this page and look under the section "Is the Lagrangian unique?" it will help:


Basically you change the Lagrangian by adding terms, and it can be proved that ONLY if this term is a total time derivative of a function of coordinates and time that the action is still extremized. In other words, no, it is not possible to keep the path of least action by adding any other function.

  • $\begingroup$ There isn't an absolute path in Galilean mechanics, the paths ARE different in the two frames where $x' = x -Vt$ $\endgroup$ Apr 11, 2014 at 20:15

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