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It's regularly argued that in the Hamiltonian formalism, we have more freedom to choose our coordinates and that this is arguably its most important advantage.

To quote from two popular textbooks:

[S]ince there are twice as many canonical variables (q,p) as generalized coordinate, the set of possible transformations is considerably larger. This is one of the advantages of the canonical formalism. (Calkin, Lagrangian and Hamiltonian Mechanics)

"The enlargement of the class of possible transformations is one of the important advantages of the Hamiltonian treatment." (Landau-Lifshitz)

The transformations we usually care about in the Hamiltonian formalism are canonical transformations. By canonical transformations we mean all transformations that don't change the form of Hamilton's equations and fulfill

$$ \{Q, P\}_{q,p} = 1 \, , $$ where $q,p$ are the original and $Q,P$ the new phase space coordinates.

Now I've learned that one class of canonical transformations are the usual point transformations. And a second important class corresponds to the usual gauge transformation:

$$ L \to L' = L + \frac{dF(q,t)}{dt}\, . $$

Point transformations lets us choose our location coordinates freely, while gauge transformation lets us shift the conjugate momentum $$ p \to p + \frac{\partial F}{\partial q} \, . \tag{1} $$

Therefore, I was wondering if point transformation + gauge transformations are really all there is to canonical transformations. In other words, do all canonical transformations correspond to either a point transformation, a gauge transformation, or a combination of the two if we translate them into the language of the Lagrangian formalism?

Formulated more precisely: is the set of possible transformations $$ q,p \to Q,P \quad \text{such that} \quad\{Q, P\}_{q,p} = 1 \, , $$ exhausted by the transformations that correspond to point and gauge transformations in the Lagrangian formalism?

Take note that this would mean that we don't really have an enlargement of the class of possible transformations in the Hamiltonian formalism. It's only that these transformations are represented differently.


PS: In addition, we have scale transformation $L \to k L$, where $c$ is some constant factor. However, these are usually not included in the class of canonical transformations because we define them through the condition $ \{Q, P\}_{q,p} = 1 \, . $ By modifying this condition to $ \{Q, P\}_{q,p} = k \, $ we get additional transformations which don't change Hamilton's equations.

jak
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  • Why are you calling the addition of a total derivative to the Lagrangian a gauge transformation? That's not what one usually means by a gauge transformation. – ACuriousMind Feb 24 '19 at 10:34
  • Related: https://physics.stackexchange.com/q/391216/2451 Related post by OP: https://physics.stackexchange.com/q/462472/2451 – Qmechanic Feb 24 '19 at 10:39
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    @ACuriousMind this kind of transformation is called "gauge transformation" in many Classical Mechanics textbooks. See also this question of mine https://physics.stackexchange.com/q/459182/ and https://physics.stackexchange.com/q/436140/ and https://physics.stackexchange.com/q/365864/ – jak Feb 24 '19 at 10:49
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    For textbooks, see Structure and Interpretation of Classical Mechanics by Gerald Jay Sussman et. al, Theoretical Physics 2 by Nolting, Variational Principles in Classical Mechanics by Cline, $\ldots$ – jak Feb 24 '19 at 10:57
  • @ACuriousMind How would you call transformations of this kind? – jak Feb 24 '19 at 10:58
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    I have never had the need to give this transformation any other name than "adding a total derivative", but I strongly suggest to not call it a gauge transformation - once you get "actual" gauge transformations in your Hamiltonian theory (from constraints), your terminology leaves no way to distinguish the two. – ACuriousMind Feb 24 '19 at 11:03
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    @ACuriousMind A gauge transformation we can agree on is $$ \vec A \to \vec A + \nabla f , $$ where $\vec A$ is the usual vector potential used to describe electromagnetism. The conjguate momentum of a particle carrying charge $1$ in an electric field is $$\vec p = m\dot q + \vec A .$$ Hence, this conjugate momentum becomes under a gauge transformation $$ \vec p \to \vec p + \nabla f , . $$ This is exactly what happens under a transformation of the kind I described above, c.f. Eq. (1). – jak Feb 24 '19 at 12:15
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    @ACuriousMind If you disagree with this reasoning, feel free to write an answer here: https://physics.stackexchange.com/q/459182/. (I'm seriously interested...) – jak Feb 24 '19 at 12:22
  • I don't think there's anything to disagree with, it's just a name. And it is certainly the case that "true" gauge transformations in my sense can act on the Lagrangian as the addition of a total derivative (that's the definition of a (quasi-)symmetry, after all - it only changes the Lagrangian by a total derivative), as your example shows. But the converse is not true - not all transformations that act by adding a total derivative are "true" gauge transformations, since these transformations certainly include all non-gauge (quasi-)symmetries. – ACuriousMind Feb 24 '19 at 12:42

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