First-class constraint: Difference between revisions

Revision as of 11:07, 9 September 2019

Revision as of 01:38, 12 September 2019

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From this consistency condition, we immediately get the [[First class constraints#Constrained Hamiltonian dynamics from a Lagrangian gauge theory|secondary constraint]]

:{{math|''r''²−''R''²{{=}}0}} .

<math>\begin{align}

By the same reasoning, this constraint should be added into the Hamiltonian with an undetermined (not necessarily constant) coefficient {{mvar|u}}₂. At this point, the Hamiltonian is

0&=\{H,p_\lambda\}_\text{PB}\\

&=\sum_{i}\frac{\partial H}{\partial q_i}\frac{\partial p_\lambda}{\partial p_i}-\frac{\partial H}{\partial p_i}\frac{\partial p_\lambda}{\partial q_i}\\

&=\frac{\partial H}{\partial \lambda}\\

&=\frac{1}{2}(r^2-R^2)\\

&\Downarrow\\

0&=r^2-R^2

\end{align}</math>

By the same reasoning, thisThis constraint should be added into the Hamiltonian with an undetermined (not necessarily constant) coefficient {{mvar|u}}_2,. At this point,enlarging the Hamiltonian isto

:<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>

LikewiseSimilarly, from this secondary constraint, we getfind the tertiary constraint,

<math>\vec{p}\cdot\vec{r}=0</math>,

<math>\begin{align}

by demanding, for consistency, that <math>\{r^2-R^2,\, H\}_{PB} = 0</math> on-shell.

0&=\{H,r^2-R^2\}_{PB}\\

&=\{H,x^2\}_{PB}+\{H,y^2\}_{PB}+\{H,z^2\}_{PB}\\

&=\frac{\partial H}{\partial p_x}2x+\frac{\partial H}{\partial p_y}2y+\frac{\partial H}{\partial p_z}2z\\

&=\frac{2}{m}(p_xx+p_yy+p_zz)\\

&\Downarrow\\

<math>0&=\vec{ p}\cdot\vec{ r}=0</math>,

\end{align}</math>

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

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where {{mvar|u}}₁, {{mvar|u}}₂, and {{mvar|u}}₃ 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.

We keep turning the crank, demanding this new constraint have vanishing [[Poisson bracket]]

The tertiary constraint's consistency condition yields

:<math>

0=\{\vec{p}\cdot\vec{r},\, H\}_{PB} = \frac{p^2}{m} - mgz+ \lambda r^2 -2 u_2 r^2 = 0.

</math>

However, now, this is ''not'' a quaternary constraint, but a condition which fixes one of the undetermined coefficients. In particular, it fixes

We might despair and think that there is no end to this, but because one of the new Lagrange multipliers has shown up, this is not a new constraint, but a condition that fixes the Lagrange multiplier:

:<math>

u_2 = \frac{\lambda}{2} + \frac{1}{r^2}\left(\frac{p^2}{2m}-\frac{1}{2}mgz \right).

</math>

Plugging this into our Hamiltonian gives us (after a little algebra)

<math>

H = \frac{p^2}{2m}(2-\frac{R^2}{r^2}) + \frac{1}{2}mgz(1+\frac{R^2}{r^2})+u_1p_\lambda+u_3\vec p \cdot\vec r

</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.