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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1$\begingroup$ Can you please explain your notation $(\Phi_m,\Phi)$ ? $\endgroup$– user10001Sep 15, 2012 at 6:20
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$\begingroup$ $(\Phi_m,\Phi)$ are like the momenta.For Lagrangian L, $\Phi_m=\frac{\partial L}{\partial \dot q}$. Same goes for $\Phi$ $\endgroup$– aries0152Sep 15, 2012 at 8:06
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$\begingroup$ Then i think notation $(\Phi_1,\Phi_2)$ would be more appropriate. $m$ may be confused for a variable subscript. $\endgroup$– user10001Sep 15, 2012 at 8:24
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2 Answers
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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.
answered Sep 14, 2012 at 21:03
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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.
