2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 $\Lambda(q,t)$. $\endgroup$ – Frobenius Feb 7 at 3:33
5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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)." $\endgroup$ – JakobH Feb 19 at 9:07
  • 1
    $\begingroup$ 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." $\endgroup$ – JakobH Mar 1 at 9:15
2
$\begingroup$
  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.

--

$^1$ 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.

$\endgroup$

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.