5
$\begingroup$

Consider a theory in the Hamiltonian formalism and assume that it has constraints between canonical variables $Q, \pi$. By the Dirac terminology, the set of constraints $F_{a}(Q, \pi) \approx 0$ of the first class satisfies conditions $\lbrace F_{a}, F_{b}\rbrace_{P} \approx 0$, while the set of constraints of the second class have nonzero Poisson brackets.

Let's have massive and massless bosonic field cases with lagrangians $$ L = -\frac{1}{4}F_{\mu \nu}F^{\mu \nu} - \lambda m^{2} A^{2} , \quad \lambda_{EM} = 0, \quad \lambda_{massive} = 1. $$ For first case we have the set of the second class constraints (the second one is fake equation of motion for $A_{0}$ component) $$ \pi^{0} = \frac{\partial L}{\partial (\partial_{0}A_{0})} \approx 0, \quad F(A_{0}, \pi^{i}, j_{0}) = -\Delta A_{0} - \partial_{i}\pi^{i} + m^{2}A_{0} \approx 0,\quad \lbrace \pi_{0}(\mathbf x ), F_{b}(\mathbf y)\rbrace_{P} = -m^{2}\delta (\mathbf x - \mathbf y), $$ while for the second one we have first class constraints: $$ \pi^{0} = \frac{\partial L}{\partial (\partial_{0}A_{0})} \approx 0, \quad F(A_{0}, \pi^{i}, j_{0}) = -\Delta A_{0} - \partial_{i}\pi^{i} \approx 0,\quad \lbrace \pi_{0}(\mathbf x ), F_{b}(\mathbf y)\rbrace_{P} \approx 0. $$ Why in the first case after introducing Dirac bracket we may make the equality the constraints to zero strict (i.e., we can express $A_{0}$ as the definite function of canonical momentums and current), while in the second case the impossibility of introduction of the Dirac brackets leads to the impossibility of expression of $A_{0}$ through other canonical coordinates? I.e., how the possibility of inctoruction of the Dirac brackets changes $\approx$ to $=$?

$\endgroup$
  • $\begingroup$ Comment to question (v2): Consider providing reference/further details for $\approx$ to $=$ statement. $\endgroup$ – Qmechanic Jul 10 '14 at 14:14
1
$\begingroup$

Comment to the question (v2):

According to Ref. 1, the weak equality symbol $\approx$ usually means equality modulo all constraints:

  • primary, secondary, tertiary, $\ldots$, constraints.

  • (or in Dirac's classification) first and second class constraints.

References:

  1. M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994; p. 13.
| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, you're right. I have deleted the second sentence. $\endgroup$ – Andrew McAddams Jul 9 '14 at 17:22
1
$\begingroup$

Not sure to answer properly, but if I remember, using the Dirac bracket allowing you to get rid off the second class constraints and deal at the end only with first class constraints. And still at the end, you consider only weak equality, no =.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Hi Tom, this would make for a good comment. As an answer it's pretty incomplete. $\endgroup$ – Brandon Enright Jul 9 '14 at 18:29
  • $\begingroup$ No, it doesn't. We eliminate the second class constraints by introducing the Dirac brackets for the equation of motion, and after that can strongly set the secondary class constraints to zero. After that we can reduce the number of degrees of freedom. $\endgroup$ – Andrew McAddams Jul 10 '14 at 7:08
0
$\begingroup$

The weak equality $f \approx 0$ means, that we first must evaluate all of the Poisson brackets of the theory (the equations of motion etc.) and only after that we may set $f$ to zero. It's because the hamiltonian doesn't contain info about primary constraints (the good example is electrodynamics), and so it doesn't contain info about the secondary constraints.

Maybe, I have understood the answer to the question. We can replace the weak equality on the strong one if all dynamical brackets (Poisson bracket in the beginning) of the constraint with any (!) other function is equal to zero. As it can show, if we replace hamiltonian $H_{0}$ (which doesn't consist of info about primary constraints $f_{i}$) to the $H = H_{0} + \lambda_{a}f_{a} \approx H_{0}$, then we will have the Dirac bracket for the time evolution of each function. The Dirac bracket of all of constraints of the second kind with an arbitrary function is equal to zero, so if we can introduce this bracket we can then set all of the constraints of the second class to zero. But if there are remaining constraints of the first class, we shhould analyze our theory as gauge theory, because the first class constraints are always connected with invariance under some transformations. It may reduce our degrees of freedom.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.