Noether's theorem for discrete symmetry

[Dalor]

I am wondering if it existes some discret version of the Noether symmetry for potential with discrete symmetry (like $C_n$ ).

The purpose is to describe the possible evolution of the phase space over the time without having to solve equations numerically (since even if the potential may have some symmetry solving the equation may be long and exhibe some chaotic behavior).

Thank's

 

[Dalor]

So no one has any insight for that ? Even a small one ?

 

[fresh_42]

So no one has any insight for that ? Even a small one ?

No, I am not sure about it, but I cannot imagine it. Noether's theorem is a result of a variation process, continuously changing differential coordinates. It is a statement about Lie groups, i.e. analytical groups - originally called continuous groups. So major ingredients simply break away in any discrete case. Maybe one can consider different connection components or coverings to map discrete behavior, but that's more a guess than an insight.

 

[Dalor]

Thank you for the answer, yes I came to the same conclusion it does not seems easy to go from continuous to discret case. And may be Noether Theorem is not the correct approach.

But I said Noether Theorem at first because I am interesting in concervation law in my Hamiltonian.

Maybe there is other "easy" way to find invariant quantity along the trajectory in hamiltonian mechanic ? (This is for classical case not quantum one).
And by considering symmetry we could maybe conclude some interesting things on the solution even if the full solution is not found.

I am interesting in this because even if I know that the solution exhib chaotic solution it does not mean than the full phase space is reachable from any initial condition.

 

[fresh_42]

Maybe lattice theory or lattice gauge theory can offer insights, both dealing with Lie groups.

 

[Orodruin]

To me this discussion is the clearest in Hamiltonian mechanics. A constant of motion is some function $S(q,p)$ of your generalised coordinates that satisfies $\dot S = 0$. From the equations of motion, this derivative is given by $\dot S = \{S,H\}$ and therefore $\{S,H\} = 0$ for a constant of motion. Due to the anti-symmetry of the Poisson bracket, this also means that $\{H,S\} = 0$, which means that $H$ does not change under the canonical transformations generated by $S$, i.e., those canonical transformations are a one-parameter set of transformations that are a symmetry of the Hamiltonian. In other words, you can go both ways, continuous symmetry ⇔ constant of motion.