In my introductory mechanics class, we were given a home assignment to calculate the following Poisson bracket.

In $\mathbb{R}^3$, let $\vec{r}$ and $\vec{p}$ be the generalized coordinate and the conjugate momentum, respectively. Now, let $\vec{F}$ be a smooth function in the form $$\vec{F}=f_1\,\vec{r}+f_2\,\vec{p}+f_3\,\vec{r}\times\vec{p}$$ where $f_i$ are scalar functions that only depend on $\vec{r}^2$, $\vec{p}^2$ and $\vec{r}\cdot\vec{p}$.

Denote by curly brackets $\{\cdot,\cdot\}$ the canonical Poisson bracket, i.e., $\{r_i,p_j\}=\delta_{ij}$ (Kronecker delta).

Also, denote the angular momentum by $\vec{M}=\vec{r}\times\vec{p}$, and let $\vec{n}$ be some constant vector.

Now, calculate the following: $$\{\vec{F},\vec{M}\cdot\vec{n}\}$$ Calculating this explicitly (using the Levi-Civita symbol), we get $\vec{n}\times\vec{F}$.

My question is: What is the intuition behind the identity $$\{\vec{F},\vec{M}\cdot\vec{n}\}=\vec{n}\times\vec{F}\,\,?$$

Or, rather, a more general question: What is the (geometric) intuition behind Poisson brackets?


The intuition behind OP's identity is that the Hamiltonian vector field $X_{F}=\{F ,\cdot\}$ generators (infinitesimal) transformations dual to the generator $F$.


  1. The momentum $F=\vec{p}\cdot\vec{n}$ along the direction $\vec{n}$ generates translations along the direction $\vec{n}$.

  2. The angular momentum $F=\vec{M}\cdot\vec{n}$ along the direction $\vec{n}$ generates rotations around the direction $\vec{n}$.


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.