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In the Lagrangian formulation of Classical Mechanics, we have the freedom to add a total time derivative of an arbitrary function $\Lambda$ to the Lagrangian:

$$ L \to L + \frac{d \Lambda}{dt} . $$

Does this symmetry of the Lagrangian have any particular name?

jak
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    The arbitrariness of the function $\Lambda$ is not general. It's restricted. So, if the Lagrangian is $L(q,\dot{q},t)$ then the function $\Lambda$ must not depend on $\dot{q}$ that is $\Lambda(q,t)$. – Frobenius Feb 07 '19 at 03:33

2 Answers2

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Some authors call that a gauge transformation of the Lagrangian function. Others, don't give any specific name and may object to the previous one.

For a reference for the former denomination, see for instance F. Scheck, Mechanics, Springer, 2010.

Massimo Ortolano
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    Ah yes, I've also found the term "mechanical gauge transformations" in the book "Analytical Mechanics" by Nolting. I also discovered in Tong's "Classical Dynamics" notes the comment "The Lagrangian L is not unique. We may make the transformation [...] for any function f and the equations of motion remain unchanged. [...] (As an aside: A system no longer remains invariant under these transformations in quantum mechanics. The number α is related to Planck’s constant, while transformations of the second type lead to rather subtle and interesting effects related to the mathematics of topology)." – jak Feb 19 '19 at 09:07
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    Here are some further references which explicitly call this kind of transformation a gauge transformation: Structure and Interpretation of Classical Mechanics by Gerald Jay Sussman et. al, Theoretical Physics 2 by Nolting, Variational Principles in Classical Mechanics by Cline, A Primer of Analytical Mechanics by Strocchi, Classical Dynamics by Saletan. In particular, the last one defines: "a Lagrangian undergoes a gauge transformation whenever a total time derivative is added to it." – jak Mar 01 '19 at 09:15
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  1. I would simply call the operation$^1$ $$L(q,\dot{q},\ldots, q^{(N)},t)\quad \longrightarrow \quad L(q,\dot{q},\ldots, q^{(N)},t) \quad+\quad \frac{dF(q,\dot{q},\ldots, q^{(N-1)},t)}{dt}\tag{1}$$ for "adding a total derivative term to the Lagrangian", nothing else.

  2. A quasi-symmetry transformation or a gauge transformation are by definition specified at the level of the fundamental variables of the theory (in this case, the $q$s and $t$). The operation (1) doesn't in general fulfill this.

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$^1$ The operation (1) should be amended with a prescription for the possible new boundary conditions (BCs). Note that the operation (1) [with the new BCs] may render the functional/variational derivative of the action $S$ ill-defined. However, if the functional derivative $\frac{\delta S}{\delta q}$ exists both before and after the operation (1), it is unchanged, cf. e.g. this Phys.SE post.

Qmechanic
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