Elimination of velocities from momenta equations for singular Lagrangian

Question

this doubt is related to Generalized Hamiltonian Dynamics paper by Dirac.

Consider the set of $n$ equations : $p_i$ = $∂L/∂v_i$,

(where $v_i$ is $q_i$(dot) = $dq_i/dt$, or time derivative of $q_i$)($L$ is the lagrangian, $q$ represent degrees of freedom in configuration space)

Now Dirac says : "If the $n$ quantities $∂L/∂v_i$ on the right-hand side of the given equations are NOT independent functions of the velocities, we can eliminate the $v$'s from (the above-given) set of equations and obtain one or more equations: $\phi_j(q, p) = 0$, ($j = 1, 2, ... , m$ if there are $m$ such constraints)"

Could anyone please explain how this comes about? I can't understand how the $v$'s can be necessarily eliminated, and if the $p$'s are not all independent, then we can simply obtain relations like $\sum_i a_ip_i = 0$ (where $a$'s are non-zero coefficients), not involving $q$'s at all.

Thanks (and apologies in case I'm missing something obvious).

Answer 1

Maybe an example:

A particle moving in 2 dimensions has a Lagrangian

$$L = \frac{\dot{x}^2 +\dot{y}^2}{2} $$

So $$p_x = \frac{\partial L}{\partial \dot{x}} = \dot{x}$$ $$p_y = \frac{\partial L}{\partial \dot{y}}=\dot{y}$$

Suppose it's constrained to move on a circle $x^2+y^2=R^2$

Now there is a constraint between the p's which you can get from differentiating the constraining circle, namely $$x\dot{x}+y\dot{y}=0$$ This is a constraint, but not of the type you are talking about, since the Lagrangian is still regular.

To obtain a Lagrangian which is singular rather than regular, we require c onstraints which result in the vanishing of the Hessian matrix $\frac{\partial^2L}{\partial \dot{q}_i \partial \dot{q}_j}$. This means that the Legendre transform (sometimes called the Floer map) from the tangent bundle to the cotangent bundle (phase space) $$\mathcal{FL} : TQ \rightarrow T^{*}Q$$ given by $$(q_i,\dot{q}_i) \rightarrow \left(q_i, p_i=\frac{\partial L}{\partial \dot{q}_i}\right)$$

is not invertible. It's image is restricted by a bunch of constraint functions. (Caveat, assuming we're restricted to a neighbourhood where rank of Hessian is constant).

For example, for the following Lagrangian $$L=\frac{1}{2}(\dot{x}^2+\dot{y}^2)+\dot{x}\dot{y}+4x\dot{y}+2x^2+4xy$$

the Hessian determinant is easily seen to vanish. The generalized momenta are $$p_x=\dot{x}+\dot{y}$$ $$p_y=\dot{x}+\dot{y}+4x$$ You can then eliminate $\dot{x}$ and $\dot{y}$ from these relations to find your constraint equation.

(Edited to provide example appropriate to the OP's question)

Answer 2

One of the basic examples of singular actions with this type of constraint is the relativistic particle:

$$ S =mc^2 \int \mathrm d\tau \sqrt { \eta_{\alpha \beta} \dot{x}^{\alpha}\dot{x}^{\beta}} = m c^2\int \mathrm d\tau \sqrt{{\dot{x}_0}^2 -{\dot{x}_i}^2 }$$

The canonical momenta:

$$p_{\alpha} = m c^2\frac{\eta_{\alpha\beta} \dot{x}^{\alpha}}{\sqrt { \eta_{\alpha \beta} \dot{x}^{\alpha}\dot{x}^{\beta}}}$$

are not functionally independent as they satisfy the constraint equation:

$$\eta^{\alpha\beta} p_{\alpha}p_{\beta} = m^2 c^4$$

which is equivalent to one of the possibilities

$$\phi = p_0 \pm \sqrt{m^2c^4 + p_i^2} = 0$$

Now, the method of the elimination of the redundant degree of freedom according to Dirac is to find any additional constraint $\pi$ (gauge fixing) such that

$$ \{ \pi, \phi \} = 1$$

A natural choice is performed by choosing

$$ \pi = t - \tau = 0$$

This choice corresponds to choosing the time coordinate as the parameter along the trajectory. In this simple example we can eliminate the time coordinate velocity from the action explicitly:

The action becomes:

$$ S =m c^2\int \mathrm dt \sqrt{1 -{\dot{x}_i}^2 }$$

Now, this theory is uncontrained, the canonical momenta

$$p_i = -m c^2\frac{\dot{x}_i}{\sqrt{1 -{\dot{x}_i}^2 }}$$

are functionally independent, and the Hamiltonian is given by:

$$H = p_i \dot{x}_i - L = \sqrt{m^2 c^4 + p_i^2}$$

which is the well known relativistic energy.