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Suppose from the Hamiltonian I got the Primary constraints $$(\Phi_m,\Phi)$$ And $\dot \Phi_m$ , $\dot \Phi$ leads to secondary constraints $$(\gamma_m,\gamma)$$ respectively. Now if the commutation of $\Phi_m$ with $\gamma_m$ is non zero (i.e they both are second class) does it terminate the commutation chain there?

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    $\begingroup$ Can you please explain your notation $(\Phi_m,\Phi)$ ? $\endgroup$ – user10001 Sep 15 '12 at 6:20
  • $\begingroup$ $(\Phi_m,\Phi)$ are like the momenta.For Lagrangian L, $\Phi_m=\frac{\partial L}{\partial \dot q}$. Same goes for $\Phi$ $\endgroup$ – aries0152 Sep 15 '12 at 8:06
  • $\begingroup$ Then i think notation $(\Phi_1,\Phi_2)$ would be more appropriate. $m$ may be confused for a variable subscript. $\endgroup$ – user10001 Sep 15 '12 at 8:24
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OP wrote(v1):

Now if the (Poisson) commutation of $\Phi_m$ with $\gamma_m$ is non zero (i.e they both are second class) does it terminate the (Poisson) commutation chain there?

No, not necessarily. The rest of the $4\times 4$ matrix $\{\chi^i,\chi^j\}$ of the $4$ constraints still contains unspecified entries.

Finally, don't forget to check if there are tertiary constraints, quaternary constraints, etc.

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To follow the process of deducation, \dot{\kappa}={\kappa,H_T}, to find whether there are some other new constraints. If there are no any new constraints, you can be taken as you are at the end.

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