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Following Counting degrees of freedom in presence of constraints, we know that there would be N-2M-S dofs if we have M 1st-class constraints and S 2nd-class constraints in N-dim phase space.

I don't know how we come up with this counting. Why does each 1st-class constraint eliminate 2 dofs and each 2nd-class constraint eliminate 1 dof?

Thanks.

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A first-class constraint eliminates 2 degrees of freedom because it, one the one hand, relates the $p_i$ and the $q^i$ with an equation, and on the other hand generates a one-parameter group of gauge transformations on the constraint surface, where all states lying in the same orbit have to be physically identified. So you lose one d.o.f. because of the constraint equation itself, and one further d.o.f. because of the gauge transformation generated.

A second-class constraint does not generate a gauge transformation, the transformation generated by it does not have physical meaning because it does not preserve the constraint surface - it maps a state on the surface to a state off the surface. So for a second-class constraint, you only have the single degree of freedom it eliminates simply by being a relation among the coordinates.

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  • $\begingroup$ Thanks for your answer and it seems to make sense. I have one further question. In case of massless spin-1 field, we have two first-class constraints: $\Pi^0 = 0$ and $\partial_i \Pi^i = 0$ without source. The gauge transformation associated with the former one acting on $A_0$ is $\delta A_0 (x) = \int \epsilon (y) [\Pi^0 (y) , A_0 (x) ] = \epsilon(x) $. In fact, this does not look like usual E&M gauge symmetry. What is wrong with my argument? @ACuriousMind $\endgroup$ – john Feb 10 '16 at 17:44
  • $\begingroup$ @john: That is indeed a bit subtle. The usual gauge transformation arises only as residual gauge freedom after imposing Gauß' law $\partial_i \pi^i = 0$. The reason why you must impose Gauß' law, but not $\pi^0 = 0$ is that Gauß' law is a secondary constraint, i.e. it is necessary for consistency of the other constraint with the equations of motion, and the Lagrangian gauge transformations generally only arise from primary first-class constraints. $\endgroup$ – ACuriousMind Feb 10 '16 at 18:02
  • $\begingroup$ I don't understand your last statement: the Lagrangian gauge transformation only arise from primary first-class constraints. In E&M case, the primary first-class constraint is $\Pi^0 = 0$, which does not generate desired gauge transformation. I guess the gauge transformation associated with $\partial_i \Pi^i = 0$ looks like $\delta A_\mu = \delta_\mu^i \partial_i \epsilon$. Here, $A_0$ does not transform at all. You see my confusion? @ACuriousMind $\endgroup$ – john Feb 10 '16 at 18:05
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    $\begingroup$ @john: Oh, I'm sorry, that was indeed an imprecise statement. This is the general, precise fact: The gauge symmetries of the usual Lagrangian action are the residual gauge transformations when all secondary constraints have been imposed. For a detailed derivation of this fact, see e.g. chapter 3 in "Quantization of gauge systems" by Henneaux/Teitelboim. $\endgroup$ – ACuriousMind Feb 10 '16 at 18:08
  • $\begingroup$ @john: For EM, you have to consider the combined transformation of both constraints on the extended action, and then find the residual transformations that preserve the secondary constraints in the sense that they do not transform the associated Lagrange multipliers. $\endgroup$ – ACuriousMind Feb 10 '16 at 18:10

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