# Calculate integral of motion condition with Poisson brackets

Problem statement: The Hamiltonian of a system is given by the formula:

\begin{equation*} H = \frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2} + V(r,\theta). \end{equation*}

Under what condition is $f=p_\theta^2$ an integral of motion?

Attempted solution:

In order for $f$ to be an integral of motion, according to Poisson theorem, is that:

\begin{equation*} [f, H] =0, \end{equation*}

where $[,]$ is the Poisson bracket.

Therefore:

\begin{align*} \left[\frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2} + V(r,\theta), p_\theta^2\right] &= 0 \Leftrightarrow \\ \left[\frac{p_r^2}{2m}, p_\theta^2\right] + \left[\frac{p_\theta^2}{2mr^2}, p_\theta^2\right] + \left[V(r,\theta), p_\theta^2\right] &= 0. \end{align*}

But how do I proceed from here on?

EDIT1: Regarding the first term:

\begin{align*} \left[\frac{p_r^2}{2m},p_\theta^2\right]&=\frac{1}{2m}\left[p_r^2,p_\theta^2\right]=\frac{1}{2m}[p_r p_r, p_\theta^2]=\frac{1}{2m}\left(p_r[p_r,p_\theta^2]+p_r[p_r,p_\theta^2]\right)\\ &=\frac{1}{m}[p_r,p_\theta^2]= \frac{1}{m}\left(p_\theta[p_r, p_\theta]+p_\theta[p_r,p_\theta]\right) = \frac{2p_\theta[p_r,p_\theta]}{m} \end{align*}

So it boils down to what is $[p_r, p_\theta]$ equal to ? I'm not sure whether it can be answered directly without resorting to the definition of Poisson bracket.

The coordinates in our problem are $q_i = \{r, \theta\}, p_i = \{p_r, p_\theta\}$. Therefore:

\begin{align*} [p_r, p_\theta] &= \sum_{i=1}^2 \left(\frac{\partial p_r}{\partial q_i}\frac{\partial p_\theta}{\partial p_i} - \frac{\partial p_r}{\partial p_i}\frac{\partial p_\theta}{\partial q_i}\right)\\ &= \left(\frac{\partial p_r}{\partial r}\frac{\partial p_\theta}{\partial p_r} - \frac{\partial p_r}{\partial p_r}\frac{\partial p_\theta}{\partial r}\right) + \left(\frac{\partial p_r}{\partial \theta}\frac{\partial p_\theta}{\partial p_\theta} - \frac{\partial p_r}{\partial p_\theta}\frac{\partial p_\theta}{\partial \theta}\right)\\ \end{align*}

And now I'm stuck again.

The way I understand it $p_r, p_\theta$ are two components of momentum along $r, \theta$ respectively. Does that imply that $p_r$ does not depend on $\theta$ and that it only depends on $r$ ? If the answer is yes, then the result for the first term is zero.

EDIT 2: Based on the answers in Hamiltonian formalism all the canonical variables are taken independent to each other, therefore $[p_r, p_\theta]=0$. Similarly it may be shown that the 2nd term is also zero.

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I think it is pretty obviously in your last line. So what are the results of the commutator $[p_r,p_\theta]$ and $[p_\theta,p_\theta]$? Note that $p_\theta$ is not commute with $\theta$. Then you should get the answer.
And "What is the identity to get from $[A^2,B^2]$ to $[A,B]$?" :) – Alex Nelson Feb 6 '14 at 21:46
Regarding $[p_r, p_\theta]$ please see my latest question. Regarding $[p_\theta, p_\theta]$ how did you get rid of $r^{-2}$ ? Shouldn't it be $[p_\theta/r^2, p_\theta]$ ? – Zet Feb 7 '14 at 2:17
@Zet In Hamiltonian formulation, $p_r$ and $p_\theta$ are treated as independent variables, so the derivative with repest to others canonical variable would be zreo. – hwlau Feb 7 '14 at 3:15