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I am referring to another post on the same question as this post: Lagrangian $L' = L + \frac{df}{dt}$ gives the same equations of motion

In the first paragraph, the poster says:

"It is well known that when a Lagrangian $L$ is incremented by the total time derivative of a function $f$ that does not depend on the time derivatives of the generalized coordinates, the same equations of motion are obtained."

I have verified this statement and understand the derivation, but what is the intuitive implication of this? I guess the question boils down to what the Lagrangian is in relation to the equations of motion. What would change by perturbing the lagrangian by this function, $f(q_1, ...,q_n,t)$?

I am new to physics, I come from a background in math.

Qmechanic
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ty morr
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3 Answers3

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I guess the question boils down to what the Lagrangian is in relation to the equations of motion.

Yes, exactly. The clue is that the equations of motions for $q(t)$ are equivalent to the principle of least action: $$S[q,t_1,t_2] := \int_{t_1}^{t_2}L(q,\dot{q},t)dt=\text{Min}.$$

It is almost trivial to show that this action $S$ only changes by a constant, when you replace the Lagrangian $L$ by the Lagrangian $L'=L+\frac{df}{dt}$.

$$S'[q,t_1,t_2] = S[q,t_1,t_2] + f(q(t_2),t_2) - f(q(t_1),t_1)$$

So, when $S$ is minimal for motion $q(t)$, then $S'$ is minimal for the same motion $q(t)$. That means, both Lagrangians $L$ and $L'$ must lead to the same equations of motions.

Thomas Fritsch
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There is the physics SE feature of offering (in the column on the right hand side of the page) suggestions, under the heading 'Related'

In this particular case the column 'Related' offers a link to a 2011 question titled Invariance of Lagrange on addition of total time derivative of a function of coordinates and time

According to the answer to that question that invariance is in accordance with relativity of inertial motion.

Example:
The dynamics of juggling balls is independent of some constant relative velocity. You can be on the platform of a train station, or in a train moving at a constant velocity, the way the balls will move relative to you is the same. If you throw the ball straight up it will land back in your hand. This holds good for the members of the equivalence class of inertial coordinate systems.


The criterion of stationary action is that in order to find the true trajectory you find the point in variation space where the derivative of the Action $S$ (with respect to variation) is zero.

That is, in terms of stationary action the value of the Action $S$ is in itself not relevant. The relevant factor is the derivative of the Action $S$ with respect to variation.

As we know, taking a derivative is an operation that discards some information. That leaves room to reverse engineer ways in which one can modify the Lagrangian while still ending up at the same point in variation space.


For more on the concept of stationary action:

There is a 2021 answer by me about Hamilton's stationary action with a two stage derivation:
1 Derivation of the Work-Energy theorem from $F=ma$
2 Demonstration that in cases where the Work-Energy theorem holds good Hamilton's stationary action will hold good also

(Of course, the usual presentation starts with positing Hamilton's stationary action and next it is demonstrated that F=ma can be recovered from that. As pointed out by contributor Kevin Zhou (jan 16, 2020): "[...] in physics you can often run derivations in both directions.")

Cleonis
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Well, the intuitive explanation is clear and can be generalized to field theory: The functional derivative $$\frac{\delta S}{\delta\phi^{\alpha}(x)}\tag{1}$$ (at an interior point $x\in M$ of spacetime) is an object derived by making small localized variations around $x$ (away from the boundary), and its value (1) is not affected by adding a boundary term to the action $S$ (away from $x$).

For more details, see also e.g. my related Phys.SE answer here.

Qmechanic
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