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.
Answer
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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