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.

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    $\begingroup$ @ACuriousMind this kind of transformation is called "gauge transformation" in many Classical Mechanics textbooks. See also this question of mine physics.stackexchange.com/q/459182 and physics.stackexchange.com/q/436140 and physics.stackexchange.com/q/365864 $\endgroup$
    – jak
    Feb 24, 2019 at 10:49
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    $\begingroup$ 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, $\ldots$ $\endgroup$
    – jak
    Feb 24, 2019 at 10:57
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    $\begingroup$ 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. $\endgroup$
    – ACuriousMind
    Feb 24, 2019 at 11:03
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    $\begingroup$ @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). $\endgroup$
    – jak
    Feb 24, 2019 at 12:15
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    $\begingroup$ @ACuriousMind If you disagree with this reasoning, feel free to write an answer here: physics.stackexchange.com/q/459182. (I'm seriously interested...) $\endgroup$
    – jak
    Feb 24, 2019 at 12:22


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