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It is very well established how to derive conserved charges associated to the symmetries of Lagrangian using the Noether's theorem. Also in the Hamiltonian formulation, we know how to derive the charges associated to symmetries of the Hamiltonian action, using the symplectic structure etc.

My question: Suppose I have a set of equations without knowing the explicit action or Hamiltonian. I realize that the equations of motion are invariant or covariant under some transformation. Is there a way to associate conserved charges (or better conserved currents) associated to this symmetry transformation without using Lagrangian or Hamiltonian?

This question was answered here for the very special case of a system of particles where the equations of motion take a "natural form" $m_i\ddot{{\vec x}}_i=-{\vec\nabla} V(x^j)$. However, this is very limited and does not apply to a field theory or a gauge theory in general (e.g. consider general relativity without any special gauge condition).

Application A good example of this issue is the Electric-magnetic duality symmetry. Remind that Maxwell equations are invariant under the electric-magnetic duality $(E,B)\to (B,-E)$. With some effort, this symmetry can be also realized at the level of the action as shown originally by Deser and Teitelboim (here). Then using the Noether procedure, it turns out that the Noether charge corresponding to duality is helicity which is a physical observable.

In linearized GR same thing happens. The eom is duality symmetric. One can also find a duality invariant action (here)

However, the problem becomes more interesting in full GR. In this case, the Einstein equations still have an electric-magnetic duality under rotating the electric and magnetic parts of the Weyl tensor into each other (see e.g. here or section 7.2 of the book by Christodoulou and Klainerman). However, this symmetry cannot be realized at the level of the action. Indeed there is a no-go theorem by Deser et al. (here) showing that the duality symmetry cannot be realized at the level of the action. So the question is whether there is a conserved quantity that correspond to the duality symmetry in GR or not?

Qmechanic
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Ali Seraj
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    I think GR does not have the duality symmetry you are mentioning. First, the article by Deser and Seminara that you link to proves the duality symmetry that exists at the linear level cannot be extended to higher orders. Second, the article by Maartens and Basset is clear that their analysis does not apply to Einstein gravity: "Duality is an invariance only of the Bianchi identities, and not the Ricci identities, so that it does not map Einstein solutions into Einstein solutions." – Andrew Dec 06 '20 at 08:57
  • The other point is that Noether's theorem assumes a Lagrangian formulation. For equations of motion that cannot be derived from an action, I don't think there is any reason to expect a symmetry will imply the existence of a conserved quantity, in general. – Andrew Dec 06 '20 at 09:01
  • You can easily derive the continuity equation for the electric charge from maxwell equations, but I do not know of a general method –  Dec 06 '20 at 16:45

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Let me point you towards something I came across several years ago which I think might go a long way towards doing what you want. This is something that goes back to the work of Sophus Lie and was later developed further by Elie Cartan.

We of course know that Lie came up with Lie groups, which play a central role on modern physics, but the reason he invented these things was actually to investigate the role of symmetry in differential equations, and to the best of my knowledge the Lagrangian was not such a prominent tool then, so much of his work was directly in terms of differential equations.

Now, unfortunately I never finished delving into this work, so I won't be able to give an explanation here, but I can point you to this page by Peter Olver (he also wrote a book on the analysis of differential equations which I understand to be quite good).

From what I understand, Lie found a procedure to find the generators of continuous symmetries (which I note are not the same as the phase space charges given by Noether's theorem. No phase space picture is used here) of a given system of differential equations (ordinary or partial) by solving some linear algebraic equations. The problem of finding invariants would then be equivalent to solving a differential equation for the set of functions annihilated by the generator.

Richard Myers
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  • I like how your answer invites participation from the OP. – TLDR Dec 07 '20 at 13:58
  • @TLDR It is a large subject and a more detailed introduction to it would not be possible in a post here. Peter Olver's notes on the topic are rather compact and to the point, yet still manage to occupy many pages. It also depends on a certain level of comfort with differential geometry to get into. – Richard Myers Dec 08 '20 at 22:41
  • @RichardMyers Thank you for your explanations and the nice reference. However, I am still confused if symmetries of a system of differential equations necessarily correspond to conserved currents, like in case of Noether's theorem? Because as mentioned by nwolijin below, for example translation invariance of eom does not imply momentum conservation – Ali Seraj Dec 09 '20 at 14:19
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    @AliSeraj They do not. Transformations of the equations of motion are only assured to map solutions to other solutions. While this is powerful and can lead to other results, conserved charges demand symmetries of the action, which is a strictly stronger condition. In the end, a major contributor to this is the fact that action symmetries (for Noether's theorem) are demanded to be symmetries even off the equations of motion. – Richard Myers Dec 09 '20 at 19:19
  • @RichardMyers What do you think about the claim of this reference then? Phys. Rev. Lett. 78 (1997), 2869-2873. This was mentioned by someone called Eric, but his comment was unfortunately removed for some reason. – Ali Seraj Dec 10 '20 at 09:32
  • @AliSeraj I strongly suspect that what they are doing is precisely equivalent to the work of Lie and Cartan I mentioned in my answer (referencing Olver). I have not checked, but it seems their "adjoint condition" is essentially constructing a vector field tangent to the manifold of solutions and seeking vector fields (the $\Lambda$'s) which commute with it. Something along these lines anyway. Such symmetries are not the most general kind, hence my assertion that not all symmetries of the EOM lead to conservation laws remains, but of course, some do. – Richard Myers Dec 10 '20 at 19:05
  • @AliSeraj I also note that a large part of the power of Noether's theorem does not lie in the conservation, but in the connection to a symplectic structure, which non-Noether charges are not guaranteed to have (a symplectic structure is not even assured without a Lagrangian so far as I'm aware). – Richard Myers Dec 10 '20 at 19:06
  • @RichardMyers I agree and this was my original question: which symmetries of the eom lead to a conservation law and what is the procedure to do that? In the reference by Olver I did not see a discussion of conservation laws for non-Lagrangian eom. But in this reference, they do. Anyway, it is surprising that this work is not known in the physics literature. – Ali Seraj Dec 11 '20 at 06:09
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A simple comment. Consider a modification of the equation you quoted by adding friction: \begin{equation} \ddot{\vec x} = -k \dot{\vec x}-\vec\nabla V(x). \end{equation} Clearly, there is time translation invariance, however, there is no energy conservation. Similarly, you can drop $V(x)$, in this case there is also space translation that is a symmetry of EOM, but again momentum is not conserved. The bottom line is that invariance of EOM does not guarantee existence of a conserved quantity.

nwolijin
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  • Perhaps not at the level of individual orbits, but could there be conserved quantities associated with ensembles or distributions of paths? – TLDR Dec 13 '20 at 08:01
  • Your comment shows that not all symmetries of the eom are associated with conservation laws, but it doesn't show that no conserved quantities can be derived from them. Please see this reference Phys. Rev. Lett. 78 (1997), 2869-2873. – Ali Seraj Dec 13 '20 at 13:51
  • @Ali Seraj thanks for the reference. It is an interesting construction, but it does not rely (it seems to me) on symmetries of equations at all. – nwolijin Dec 13 '20 at 14:05