The following infinitesimal transformation of phase space coordinates (for infinitesimal $\epsilon$) is apparently canonical (preserving Hamilton's equations and Poisson brackets):

$$ q_i' = q_i + \epsilon \frac{\partial g}{\partial p_i} $$

$$ p_i' = p_i - \epsilon \frac{\partial g}{\partial q_i} $$

where $g$ is the generator and a function of $q$ and $p$.

When computing the conditions for being canonical in terms of Poisson Brackets $\{q_i', q_j'\} = 0$, $\{p_i', p_j'\} = 0$, $\{q_i', p_j'\} = \delta_{ij}$. There seems to be second-order partial derivative terms that don't cancel. For example,

$$\{q_i', q_j'\} = \Sigma_k (\frac{\partial q_i'}{\partial q_k} \frac{\partial q_j'}{\partial p_k} - \frac{\partial q_i'}{\partial p_k} \frac{\partial q_j'}{\partial q_k}) = [(1 + \epsilon \frac{\partial ^2 g}{\partial q_i \partial p_i})(\epsilon \frac{\partial ^2 g}{\partial p_i \partial p_j}) - (\epsilon \frac{\partial ^2 g}{(\partial p_i)^2})(\epsilon \frac{\partial ^2 g}{\partial q_i \partial p_j})] + [(\epsilon \frac{\partial ^2 g}{\partial p_i \partial q_j}) (\epsilon \frac{\partial ^2 g}{(\partial p_j)^2}) - (\epsilon \frac{\partial ^2 g}{\partial p_j \partial p_i})(1 + \epsilon \frac{\partial ^2 g}{\partial q_j \partial p_j})] + \Sigma_{k \neq i,j} (\frac{\partial q_i'}{\partial q_k} \frac{\partial q_j'}{\partial p_k} - \frac{\partial q_i'}{\partial p_k} \frac{\partial q_j'}{\partial q_k}).$$

It doesn't look like these first sets of terms for $i$ and $j$ cancel and neither do the sum of the terms not with respect to $i$, or $j$. I see a similar issue with the other Poisson Brackets. Are these second-order partial derivatives all 0? I'm not sure what I am missing. If possible, any hints would be appreciated.

  • 5
    $\begingroup$ $\epsilon^2$ is second order, so you can drop those terms $\endgroup$
    – user65081
    Commented Nov 23, 2020 at 0:16
  • $\begingroup$ ohhh ok. Please feel free to post that as an answer. $\endgroup$
    – dylan7
    Commented Nov 23, 2020 at 1:14

1 Answer 1


As @WolphrameJonny said, As we are doing infinitesimal changes, We can neglect all the term which are of the higher order (greater that $1$) in $\epsilon$. So if
$$Q_j=q_j+\epsilon\frac{\partial G}{\partial p_j}+O(\epsilon^2)=q_j+\epsilon\frac{\partial G}{\partial p_j}$$ You may also be interested in the following related physics.SE posts:

Neglecting second order differentials

Rigorous underpinnings of infinitesimals in physics

Commutator of Lorentz boost generators : visual interpretation


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