# Counting degrees of freedom in presence of constraints

In a $N$ dimensional phase space if I have $M$ 1st class and $S$ 2nd class constraints, then I have $N-2M-S$ degrees of freedom in phase space. How can I calculate the degrees of freedom in configuration space?

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The number of physical degrees of freedom (DOF) or dynamical variables is simply the number of generalized positions whose evolution is given by a second order in time differential equation. Using the OP's notation, the number of DOF is $${1\over 2}(N-2M-S)$$ For instance, in electrodynamics the phase-space is six-dimensional $\{A_i,F_{0i}\}_{i=1}^3$ and the Gauss law is a first class constraint. Thus $N=6,\, M=1, S=0$. So that there is two DOF corresponding to the two polarizations of electromagnetic waves or the two photon's helicities.

One can take an alternative and equivalent point of view in which the phase-space consists of $\{A_{\mu},F_{0\mu}\}_{\mu=0}^3$ and besides the Gauss law one has the first class constraint $F_{00}\approx0$ (the symbol $\approx$ is read "weakly zero" and means zero when the constraints are verified, you may perfectly write $=$) which Poisson commutes with the Gauss law and both are therefore first class constraints. Then $N=8,\, M=2, S=0$ and the number of DOF is still two, of course.

In the case of the gravitational field, the counting of DOF is analogous. The phase-space consists of $\{h_{ab},p_{ab}\}_{a=1,b=1}^{a=3,b=3}$, with $h_{ab}$ the components of the spatial metric and $p_{ab}$ their conjugated momenta. The four $(0,\mu)$ Einstein equations are not dynamical equations —since they do not contain second order temporal derivatives— but first class constraints. Hence $N=12,\, M=4,\, S=0$ so that the number of DOF is two corresponding to the two polarizations of gravitational waves.

However, consider the case of the Procca field (a vectorial field of mass $m$). Now the phase-space consists of $\{A_{\mu},F_{0\mu}\}_{\mu=0}^3$ and there are two constraints $\partial_i\, F_{0i}=m^2A_0$ —I am considering a theory with no matter fields besides the vectorial field, if one added other fields, then there would be a density of charge $\rho$ in the right hand side— which reduces to the Gauss law when $m=0$ and $F_{00}=0$ like in the electromagnetic case. However, now due to the mass term, the two constraints do not Poisson commute, thus the constraints are second class. Hence $N=8,\, M=0, S=2$ and the number of degrees of freedom is three corresponding to the three helicities of a massive vectorial particle.

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With $M$ 1st class constraints there should be imposed $M$ gauge-fixing conditions.

So the dimension of the physical phase space$^1$ is $N-2M-S$.

The dimension of the physical configuration space$^2$ is $\frac{N-2M-S}{2}$.

In other words, there are $\frac{N-2M-S}{2}$ physical degrees of freedom (d.o.f.), cf. e.g. this Phys.SE post.

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$^1$ Phase space is the space of generalized positions and momenta.

$^2$ Configuration space consists of generalized positions.

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