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Deriving conserved charges from the equations of motion

Physics Asked by Ali Seraj on December 13, 2020

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" $m_iddot{{vec x}}_i=-{vecnabla} V(x^j)$. 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?

2 Answers

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.

Answered by Richard Myers on December 13, 2020

A simple comment. Consider a modification of the equation you quoted by adding friction: begin{equation} ddot{vec x} = -k dot{vec x}-vecnabla 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.

Answered by nwolijin on December 13, 2020

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