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In classical relativistic Hamiltonian mechanics there is a so-called "no-interaction theorem" (see, for example, this article for a proof). Roughly, it states that if we have an $N$-body mechanical system hamiltonian $H$ in $d>1+1$, such that one can construct boosts/translations/rotations generators obeying Poincare algebra together with $H$ then one can make a canonical transformation of variables $p,q$ such that all the generators assume there free particle form.

There is a certain assumption in the proof of the theorem that I don't understand the meaning of: we assume that "coordinates of the particles $q^a$ transform correctly under the inhomogeneous Lorentz group". This is expressed in terms of 3 relations on Poisson brackets with Poincare generators; the most mysterious one is $$ [q^a, K] = q^a [q^a, H] $$ with $K$ being the boost operator. What is the meaning of these conditions? Why it is natural to demand that?

Reason I'm asking is that in the few examples of relativistic $N$-body systems in 2D that I know (Ruusenars-Schneider model in particular) this relations do not hold, although it looks like the relevant variables are precisely particle coordinates. It is stated in their original article here that this is the exact reason why they were able to find a non-trivial $N$-body relativistic system.

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  • $\begingroup$ In that paper $a$ labels the particles (1 to n) and their index $i$ labels space dimensions (1 to 3), so (1.2) says $\{q^a_i, K_j\}=q^a_i\{q^a_j, H\}$. But (1.2) contradicts (3.1) in that paper, so I guess we'll never know. $\endgroup$
    – user21299
    Commented Feb 9, 2021 at 23:51
  • $\begingroup$ Well, there're just some typos in that article here and there. Still, this requirement should also make sense in 2d, when space dimension index doesn't matter; if one could at least clarify the meaning of this requirement in this case, which is of particular interest to me, it would be very helpful. $\endgroup$ Commented Feb 10, 2021 at 5:42
  • $\begingroup$ No idea -- I'm also mystified $\endgroup$
    – user21299
    Commented Feb 16, 2021 at 2:41

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