Content deleted Content added
Expanding article |
→Examples: Merge from Second class constraints |
||
Line 134:
The last two terms are a linear combination of the Gaussian constraints and we have a whole family of (gauge equivalent)Hamiltonians parametrized by ''f''. In fact, since the last three terms vanish for the constrained states, we can drop them.
==Second class constraints==
In a constrained Hamiltonian system, a dynamical quantity is '''second class''' if its Poisson bracket with at least one constraint is nonvanishing. A constraint that has a nonzero Poisson bracket with at least one other constraint, then, is a '''second class constraint'''.
See [[first class constraints]] or [[Dirac bracket]] for the preliminaries.
===An example: a particle confined to a sphere===
{{Disputeabout|"The second class constraints and Hamiltonian given in this example"|date=March 2009}}
Before going on to the general theory, let's look at a specific example step by step to motivate the general analysis.
Let's start with the [[action (physics)|action]] describing a [[Newtonian]] particle of [[mass]] m constrained to a surface of radius R within a uniform [[gravitational field]] ''g''. When one works in Lagrangian mechanics, there are several ways to implement a constraint: one can switch to generalized coordinates that manifestly solve the constraint or one can use a Lagrange multiplier.
In this case, the particle is constrained to a sphere, therefore the natural solution would be to use angular coordinates to describe the position of the particle instead of Cartesian and solve the constraint in that way (the first choice). For didactic reasons, instead, consider the problem in Cartesian coordinates with a Lagrange multiplier term.
The action is given by
<math>S=\int dt L=\int dt \left[\frac{m}{2}(\dot{x}^2+\dot{y}^2+\dot{z}^2)-mgz+\frac{\lambda}{2}(x^2+y^2+z^2-R^2)\right]</math>
where the last term is the [[Lagrange multiplier]] term enforcing the constraint.
Of course, we could have just used different [[coordinates]] and written it as
<math>S=\int dt \left[\frac{mR^2}{2}(\dot{\theta}^2+\sin^2(\theta)\dot{\phi}^2)+mgR\cos(\theta)\right]</math>
instead, but let's look at the former coordinatization.
The [[conjugate momentum|conjugate momenta]] are given by
<math>p_x=m\dot{x}</math>, <math>p_y=m\dot{y}</math>, <math>p_z=m\dot{z}</math>, <math>p_\lambda=0</math>.
Note that we can't determine <math>\dot{\lambda}</math> from the momenta.
The [[Hamiltonian mechanics|Hamiltonian]] is given by
<math>H=\vec{p}\cdot\dot{\vec{r}}+p_\lambda \dot{\lambda}-L=\frac{p^2}{2m}+p_\lambda \dot{\lambda}+mgz-\frac{\lambda}{2}(r^2-R^2)</math>.
We can't eliminate <math>\dot{\lambda}</math> at this stage yet. We are here treating <math>\dot{\lambda}</math> as a shorthand for a function of the [[symplectic space]] which we have yet to determine and ''not'' an independent variable. For notational consistency, define <math>u_1=\dot{\lambda}</math> from now on. The above Hamiltonian with the <math>p_\lambda</math> term is the "naive Hamiltonian". Note that since, on-shell, the constraint must be satisfied, one cannot distinguish between the naive Hamiltonian and the above Hamiltonian with the undetermined coefficient, <math>\dot{\lambda}=u_1</math>, on-shell.
We have the [[primary constraint]] p<sub>λ</sub>=0.
We require, on the grounds of consistency, that the [[Poisson bracket]] of all the constraints with the Hamiltonian vanish at the constrained subspace. In other words, the constraints must not evolve in time if they are going to be identically zero along the equations of motion.
From this consistency condition, we immediately get the [[First_class_constraints#Constrained_Hamiltonian_dynamics_from_a_Lagrangian_gauge_theory|secondary constraint]]
r<sup>2</sup>-R<sup>2</sup>=0.
By the same reasoning, this constraint should be added into the Hamiltonian with an undetermined (not necessarily constant) coefficient <math>u_2</math>. At this point, the Hamiltonian is
:<math>
H = \frac{p^2}{2m} + mgz - \frac{\lambda}{2}(r^2-R^2) + u_1 p_\lambda + u_2 (r^2-R^2)
</math>
And from the secondary constraint, we get the tertiary constraint
<math>\vec{p}\cdot\vec{r}=0</math>,
by demanding on the grounds of consistency that <math>\{r^2-R^2,\, H\}_{PB} = 0</math> on-shell. Again, one should add this constraint into the Hamiltonian since on-shell no one can tell the difference. Therefore, so far, the Hamiltonian looks like
:<math>
H = \frac{p^2}{2m} + mgz - \frac{\lambda}{2}(r^2-R^2) + u_1 p_\lambda + u_2 (r^2-R^2) + u_3 \vec{p}\cdot\vec{r},
</math>
where <math>u_1</math>, <math>u_2</math>, and <math>u_3</math> are still completely undetermined. Note that frequently all constraints that are found from consistency conditions are referred to as "secondary constraints" and secondary, tertiary, quaternary, etc. constraints are not distinguished.
The tertiary constraint's consistency condition yields
:<math>
\{\vec{p}\cdot\vec{r},\, H\}_{PB} = \frac{p^2}{m} - mgz+ \lambda r^2 -2 u_2 r^2 = 0.
</math>
This is ''not'' a quaternary constraint, but a condition which fixes one of the undetermined coefficients. In particular, it fixes
:<math>
u_2 = \frac{\lambda}{2} + \frac{1}{r^2}\left(\frac{p^2}{2m}-\frac{1}{2}mgz \right).
</math>
Now that there are new terms in the Hamiltonian, one should go back and check the consistency conditions for the primary and secondary constraints. The secondary constraint's consistency condition gives
:<math>
\frac{2}{m}\vec{r}\cdot\vec{p} + 2 u_3 r^2 = 0.
</math>
Again, this is ''not'' a new constraint; it only determines that
:<math>
u_3 = -\frac{\vec{r}\cdot\vec{p}}{m r^2}.
</math>
At this point there are no more constraints or consistency conditions to check.
Putting it all together,
:<math>H=\left(2-\frac{R^2}{r^2}\right)\frac{p^2}{2m} + \frac{1}{2}\left(1+\frac{R^2}{r^2}\right)mgz - \frac{(\vec{r}\cdot\vec{p})^2}{mr^2} + u_1 p_\lambda</math>.
When finding the equations of motion, one should use the above Hamiltonian, and as long as one is careful to never use constraints before taking derivatives in the Poisson bracket then one gets the correct equations of motion. That is, the equations of motion are given by
:<math>
\dot{\vec{r}} = \{\vec{r}, \, H\}_{PB}, \quad \dot{\vec{p}} = \{ \vec{p},\, H\}_{PB}, \quad \dot{\lambda} = \{ \lambda,\, H\}_{PB},
\quad \dot{p}_\lambda = \{ p_\lambda, H\}_{PB}.
</math>
Before analyzing the Hamiltonian, consider the three constraints:
:<math>
\phi_1 = p_\lambda, \quad \phi_2 = r^2-R^2, \quad \phi_3 = \vec{p}\cdot\vec{r}.
</math>
Notice the nontrivial [[Poisson bracket]] structure of the constraints. In particular,
:<math>
\{\phi_2, \phi_3\} = 2 r^2 \neq 0.
</math>
The above Poisson bracket does not just fail to vanish off-shell, which might be anticipated, but even on-shell it is nonzero. Therefore, <math>\phi_2</math> and <math>\phi_3</math> are '''second class constraints''' while <math>\phi_1</math> is a [[first class constraint]]. Note that these constraints satisfy the regularity condition.
Here, we have a symplectic space where the Poisson bracket does not have "nice properties" on the constrained subspace. But [[Paul Dirac|Dirac]] noticed that we can turn the underlying [[differential manifold]] of the [[symplectic space]] into a [[Poisson manifold]] using a different bracket, called the [[Dirac bracket]], such that the Dirac bracket of any (smooth) function with any of the second class constraints always vanishes and a couple of other nice properties.
If one wanted to canonically quantize this system, then, one needs to promote the canonical Dirac brackets not the canonical Poisson brackets to commutation relations.
Examination of the above Hamiltonian shows a number of interesting things happening. One thing to note is that on-shell when the constraints are satisfied the extended Hamiltonian is identical to the naive Hamiltonian, as required. Also, note that <math>\lambda</math> dropped out of the extended Hamiltonian. Since <math>\phi_1</math> is a first class primary constraint it should be interpreted as a generator of a gauge transformation. The gauge freedom is the freedom to choose <math>\lambda</math> which has ceased to have any effect on the particle's dynamics. Therefore, that <math>\lambda</math> dropped out of the Hamiltonian, that <math>u_1</math> is undetermined, and that <math>\phi_1 = p_\lambda</math> is first class, are all closely interrelated.
Note that it would be more natural not to start with a Lagrangian with a Lagrange multiplier, but instead take <math>r^2-R^2</math> as a primary constraint and proceed through the formalism. The result would the elimination of the extraneous <math>\lambda</math> dynamical quantity. Perhaps, the example is more edifying in its current form.
===Example: Proca action===
Another example we will use is the [[Proca action]]. The fields are <math>A^\mu = (\vec{A},\phi)</math> and the action is
:<math>S = \int d^dx dt \left[ \frac{1}{2}E^2 - \frac{1}{4}B_{ij}B_{ij} - \frac{m^2}{2}A^2 + \frac{m^2}{2}\phi^2\right]</math>
where
:<math>\vec{E} \equiv -\nabla\phi - \dot{\vec{A}}</math>
and
:<math>B_{ij} \equiv \frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j}</math>.
<math>(\vec{A},-\vec{E})</math> and <math>(\phi,\pi)</math> are [[canonical variables]]. The second class constraints are
:<math>\pi \approx 0</math>
and
:<math>\nabla\cdot\vec{E} + m^2 \phi \approx 0</math>.
The Hamiltonian is given by
:<math>H = \int d^dx \left[ \frac{1}{2}E^2 + \frac{1}{4}B_{ij}B_{ij} - \pi\nabla\cdot\vec{A} + \vec{E}\cdot\nabla\phi + \frac{m^2}{2}A^2 - \frac{m^2}{2}\phi^2\right]</math>.
1, N. K. Falck and A. C. Hirshfeld, 1983, "Dirac-bracket quantisation of a constrained nonlinear system: the rigid rotator", Eur. J. Phys. 4 p. 5. {{doi|10.1088/0143-0807/4/1/003}} (Note that the form of the quantum momentum in this paper is dubious.)
2, T. Homma, T. Inamoto, and T. Miyazaki, 1990, "Schrödinger equation for the nonrelativistic particle constrained on a hypersurface in a curved space ", Phys. Rev. D 42, p. 2049. http://prd.aps.org/abstract/PRD/v42/i6/p2049_1. (Tote that the Hamiltonian suggested by the authors from second form of the constraint, (i.e., the time derivative of the <math> f(x)=0 </math> ), is not completely compatible with the formalism of Hamiltonian mechanics.)
==See also==
| |||