Main question

When we talk about symmetry operation in classical mechanics, do we necessarily mean transformations on the configuration space (e.g. translations, rotations etc) or could it also be nontrivial, nonlinear transformations on the phase space (e.g., canonical transformations (CT))? I note that CTs for a system with $n$ degrees of freedom form the symplectic group ${\rm Sp}(2n,{\rm R})$, and they preserve the form of Hamilton's equations (though not necessarily the Hamiltonian).

Comments on what I am (not) asking, terminology etc

  • I want to make a distinction between symmetry and symmetry operation. Transformations such as translations, rotations etc form groups and in quantum mechanics, it is customary to call them symmetry operations whether or not a system has those symmetries/invariances. I use the word symmetry operation, with that connotation. By that token, since CTs form a group, the question is whether they are also symmetry operations like rotations, translations etc.

  • Translations defined as $q\to q+{\rm const}$ is defined on the configuration space, which then automatically implies $\dot{q}\to \dot{q}$. Therefore, $\dot{q}\to \dot{q}$ is not really a separate transformation. I am not talking about those. I know that those are symmetry operations. I am enquiring whether nontrivial transformations on phase space, in particular, canonical transformation are symmetry operations. To start with, they have to be defined on the phase space (not configuration space) and $q$'s and $p$'s in a nontrivial way.

  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/98714/2451 , physics.stackexchange.com/q/461879/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 22, 2020 at 13:30
  • 1
    $\begingroup$ Does this answer your question? How are symmetries precisely defined? $\endgroup$ Jun 22, 2020 at 13:32
  • $\begingroup$ Not really. The answer there does not lift my confusion. I have added a line to express my confusion. $\endgroup$ Jun 22, 2020 at 13:44
  • $\begingroup$ If memory serves, the canonical transformations typically take advantage of the sympletic symmetry that naturally exists in the Hamiltonian. $\endgroup$
    – Cort Ammon
    Jun 24, 2020 at 1:33
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    $\begingroup$ @mithusengupta123 Contrary to Note 1 of your post, the answer here (physics.stackexchange.com/a/98785/19976) does not exclude transformations on phase space. (physical) Symmetries can be transformations on any of the objects in any mathematical model that leave the physics the model predicts invariant. If e.g. you're formulating mechanics via phase space and hamiltonians, you can consider transformations on phase space. If you formulate mechanics via paths through configuration space and an action principle, then transformations on the set of paths can be considered. $\endgroup$ Jun 27, 2020 at 3:36

3 Answers 3


I think the answer you are looking for is that any transformation of the phase space that leaves the action stationary is considered a symmetry of the system at hand. A system is typically completely specified by functions that determine paths on the phase space of the system. So there is no real reason to restrict yourself to the configuration space only when defining what a symmetry is.

Thus, you are not, in principle, restricted to transformations of just the coordinates. Where any other answers may have referred to transformations of the configuration space only, this would certainly be a symmetry, but may not capture all the symmetries of the system.

For example, another poster has given the example of a free particle; this system certainly admits the symmetry $q \rightarrow q + \textrm{const},$ but it is clear the system also admits the symmetry $\dot{q} \rightarrow \dot{q} + \textrm{const}$.

In answer to your follow-up points, a general canonical transformation (CT) is in essence a reformulation of the dynamics of the system. The only difference between canonical transformations and transformations of the configuration space is that in the former the momenta are treated on equal footing to the coordinates. But there's no reason to call such transformations a symmetry any more than calling the transformation to polar coordinates a symmetry of the system.

The generalized nature of CTs does, however, mean that transformations of the system generated by a symmetry can be expressed as a canonical transformation, but there is no claim that every CT represents a symmetry. If you perform a CT, then you re-express the system in terms of new $(Q^k,P_k)$, and the dynamics is governed by a new function $K(Q^k, P_k, t)$, which is the Hamiltonian in the transformed phase space. This "Kamiltonian" and the Hamiltonian will, in general, differ by a total derivative that can be thought of as generating the CT. Only if the Hamiltonian is invariant under the (infinitesimal) generating function can we say that the generating function is a constant of the motion, and that the system therefore has a corresponding symmetry. A general CT will change the form of the Hamiltonian in the new phase space, so it cannot be so understood to be a symmetry of the system.

Goldstein (Classical Mechanics) covers this and more in some detail in chapter 9; I would recommend giving it a read if you can.

  • $\begingroup$ I have added a summary in the question, in response to your answer. $\endgroup$ Jun 26, 2020 at 2:28
  • $\begingroup$ Dear @Zorawar Would you tolerate a few more follow-up questions about your answer? I understand this answer is three years old :-( $\endgroup$ Jun 23, 2023 at 17:26
  • $\begingroup$ @Solidification I think it's more pertinent that I am three years older... I can't promise I have the time to answer, to be frank. And you probably ought to write these as a separate question anyway. $\endgroup$
    – Zorawar
    Jun 24, 2023 at 18:54

Symmetry on classical mechanics are coordinate transformation that doesn't change the motion equation, for example, the lagrangian (in this approach you must use the configuration space) of particle-free is:

$ L=\frac{1}{2}mv^{2} $

applying Euler-Lagrange Equations, we have:

$\frac{d}{dt}(mv^{i})=0 $

if you change the coordinates moving the origin $x=x'+\delta$, where $\delta$ is a constant, the motion equation now is:

$\frac{d}{dt}(mv'^{i})=0 $

you can see together expression have the same form. With the Hamilton approach, the Hamilton Equations don't change the form, in this context is where you must use the phase space. A canonical transformation is a coordinate transformation defined on phase space that not change the Poisson Bracket.

Generally all about symmetries in quantum mechanics is true for classical mechanics, the difference is that 'the motion equations' are hardier solve than the classical theory and use Lie Groups is 'easier'.


If a distinction is needed between transformations of the configuration space preserving the equations of motion and phase space symmetries that do not have obvious restriction to the configuration space one could use the terms explicit symmetries vs. hidden symmetries

As a reference, the following review:

  • Frolov, V. P., Krtouš, P., & Kubizňák, D. (2017). Black holes, hidden symmetries, and complete integrability. Living Reviews in Relativity, 20(1), 1-221, doi:10.1007/s41114-017-0009-9.

makes this distinction in this passage:

… However, the contrary is not true: not every phase space symmetry can be easily reduced to the configuration space. Symmetries which have the direct counterpart on the configuration space will be called the explicit symmetries, those which cannot be reduced to the configuration space transformation are called the hidden symmetries.

For example, for particle performing geodesic motion in curved spacetime explicit continuous symmetries of the configuration space are transformations generated by Killing vector fields, these are the isometries of the spacetime.

Killing tensors of rank $s \ge 2$ provide examples of hidden symmetries of such system: a Killing tensor $k^{a_1 a_2 … a_s} $ corresponds to an integral of motion $K=k^{a_1 a_2 … a_s} p_{a_1} p_{a_2} … p_{a_s}$ that is a monomial in momentum components $p_a$. This conserved quantity generates a symmetry of the phase space, but its projection on spacetime depends explicitly on particle's momenta, so it is not a pure spacetime transformation.


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