# An infinitesmal transformation that is canonical

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.

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

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: