# Non-holonomic constraints in Dirac-Bergmann theory

The Dirac-Bergmann algorithm effectively isolates the physical degrees of freedom of a system, by changing from Poisson brackets $\{\cdot,\cdot\}_\mathrm{PB}$ to Dirac brackets $\{\cdot,\cdot\}_\mathrm{D}$.

Quick overview: let $\chi_i\approx0$ be the constraints. We demand that $\chi_i=\chi_i(q,p)$ and write $M_{ij}\equiv \{\chi_i,\chi_j\}_\mathrm{PB}$ for every second class constraint. Finally, the Dirac brackets are given by $\{\cdot,\cdot\}_\mathrm{D}=\{\cdot,\cdot\}_\mathrm{PB}-\{\cdot,\chi_i\}M^{ij}\{\chi_j,\cdot\}$.

My question: if we have a velocity-dependent constraint, $\chi_0=\chi_0(q,\dot q)$, and $p=p(\dot q)$ is not invertible (singular Legendre transform), then the Poisson bracket $\{\chi_0,\cdot\}_\mathrm{PB}$ is not defined. Does this mean it is impossible to define $\{\cdot,\cdot\}_\mathrm{D}$?

• A "constraint" in the sense of Dirac-Bergmann is a function of $q$ and $p$, something that "carves out" a constraint surface in the phase space. $\dot{q}$ never enters, since it is not a coordinate of the phase space. If you have some "constraint" that depends on $\dot{q}$, you're not in a theory on the phase space. – ACuriousMind Mar 10 '16 at 20:08
• But what if we start off with a lagrangian and some constrains, which are velocity-dependent? Can't we switch into hamiltonians and try to impose this constrain there? – AccidentalFourierTransform Mar 10 '16 at 20:13
• I don't think so. The very nature of Hamiltonian constraints is that they correspond to singularities in the Legendre transform and also gauge symmetries of the Lagrangian (action). They do not come from Lagrangian constraints. – ACuriousMind Mar 10 '16 at 20:17
• Hamiltonian constrains usually correspond to Legendre singularities and gauge symmetries, right. At least, this is what naturally arises in practical applications. But in principle, we could have other constrains that we want to impose for some reason (in the same way we sometimes impose constrains in lagrangian systems: to constrain motion into a plane and the like). My point is: constrains do not come from anything in particular: they come from what we want to impose. Singularities and gauge symmetries are one possibility, but we can impose other constrains if we want, right? – AccidentalFourierTransform Mar 10 '16 at 20:24
• Yes, but that is not what the Dirac-Bergmann method is built for (and there is in general no method to deal with arbitrarily strange constraints you might dream up, as far as I know). When one says "constrained Hamiltonian mechanics" or "Dirac-Bergmann recipe", one means that the constraints are functions on the phase space. – ACuriousMind Mar 10 '16 at 20:29

$$L(Q,\dot{Q},t)~=~L_0(q,\dot{q},t) + \lambda^a \chi_a (q,\dot{q},t),$$
with non-holonomic velocity-depending constraints $\chi_a=\chi_a (q,\dot{q},t)$; Lagrange multipliers $\lambda^a$; and where we have introduced the shorthand notation
$$Q^I~=~\{q^i; \lambda^a\}, \qquad P_I~=~\{p_i; \pi_a\}.$$
Provided that the theory is well-posed and consistent, we can in principle still apply the Dirac-Bergmann recipe to the extended configuration space of $Q^I$-variables in order to perform a (possible singular) Legendre transformation to achieve the corresponding Hamiltonian $H(Q,P,t)$ and possible constraints, of first and/or second class, and finally derive the Dirac bracket in the extended phase space.