Poisson Brackets of Classical Particle in Radial Potential on Plane

Theorem
Let $P$ be a classical particle embedded in a 2-dimensional Euclidean manifold.

Let the real-valued functions $\map r t$, $\map \theta t$ denote the position of $P$ in polar coordinates, where $t$ is time.

Suppose, the potential energy of $P$ depends only on $r$.

Then $P$ has the following Poisson brackets:

Proof
The standard Lagrangian of $P$ in polar coordinates is:


 * $L = \dfrac 1 2 m \paren { {\dot r}^2 + r^2 {\dot \theta}^2 } - \map U r$

The canonical momenta are:


 * $p_r = \dfrac {\partial L} {\partial \dot r} = m \dot r$


 * $p_\theta = \dfrac {\partial L} {\partial \dot \theta} = m r^2 \dot \theta$

The Hamiltonian associated to $L$ in canonical coordinates reads:


 * $H = \dfrac {p_r^2} {2 m} + \dfrac 1 2 \dfrac {p_\theta^2} {m r^2} + \map U r$

Then:

$\sqbrk {r, p_r} = \paren {\dfrac {\partial r} {\partial r} \dfrac {\partial p_r} {\partial p_r} - \dfrac {\partial p_r} {\partial r} \dfrac {\partial r} {\partial p_r} } + \paren {\dfrac {\partial r} {\partial \theta} \dfrac {\partial p_r} {\partial p_\theta} - \dfrac {\partial p_r} {\partial \theta} \dfrac {\partial r} {\partial p_\theta} } = 1$

$\sqbrk {\theta, p_\theta} = \paren {\dfrac{\partial \theta} {\partial r} \dfrac {\partial p_\theta} {\partial p_r} - \dfrac {\partial p_\theta} {\partial r} \dfrac {\partial \theta} {\partial p_r} } + \paren {\dfrac {\partial \theta} {\partial \theta} \dfrac {\partial p_\theta} {\partial p_\theta} - \dfrac {\partial p_\theta} {\partial \theta} \dfrac {\partial \theta} {\partial p_\theta} } = 1$

$\sqbrk {r, H} = \paren {\dfrac {\partial r} {\partial r} \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} { m r^2 } + \map U r} } {\partial p_r} - \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial r} \dfrac {\partial r} {\partial p_r} } + \paren {\dfrac {\partial r} {\partial \theta} \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial p_\theta} - \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2}  + \map U r} } {\partial \theta} \dfrac {\partial r} {\partial p_\theta} } = \dfrac {p_r} m$

$\sqbrk {p_r, H} = \paren {\dfrac {\partial p_r} {\partial r} \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial p_r} - \dfrac {\partial \paren {\frac {p_r^2}{2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial r} \dfrac {\partial p_r} {\partial p_r} } + \paren {\dfrac {\partial p_r} {\partial \theta} \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial p_\theta} - \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2}  + \map U r} } {\partial \theta} \dfrac {\partial p_r} {\partial p_\theta} } = -\dfrac {\partial U} {\partial r}$

$\sqbrk {\theta, H} = \paren {\dfrac {\partial \theta} {\partial r} \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial p_r} - \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial r} \dfrac {\partial \theta} {\partial p_r} } + \paren {\dfrac {\partial \theta} {\partial \theta} \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial p_\theta} - \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial \theta} \dfrac {\partial \theta} {\partial p_\theta} } = \dfrac {p_\theta} {m r^2}$

$\sqbrk {p_\theta, H} = \paren {\dfrac {\partial p_\theta} {\partial r} \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial p_r} - \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial r} \dfrac {\partial p_\theta} {\partial p_r} } + \paren {\dfrac {\partial p_\theta} {\partial \theta} \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial p_\theta} - \dfrac {\partial \paren {\frac {p_r^2} {2 m} + \frac 1 2 \frac {p_\theta^2} {m r^2} + \map U r} } {\partial \theta} \dfrac {\partial p_\theta} {\partial p_\theta} } = 0$