I'm reading Landau & Lifshitz's Mechanics and, at a certain point when discussing canonical transformations, they prove that Poisson brackets are canonical invariants.

The proof starts with Landau & Lifshitz stating that, since time appears as a parameter in the canonical transformations

$$p_i = \frac{\partial F}{\partial q_i}, \; P_i = - \frac{\partial F}{\partial Q_i}, \; H' = H + \frac{\partial F}{\partial t},\tag{45.7}$$

$$p_i = \frac{\partial \Phi}{\partial q_i}, \; Q_i = \frac{\partial \Phi}{\partial P_i}, \; H' = H + \frac{\partial \Phi}{\partial t},\tag{45.8}$$

it is sufficient to prove that the Poisson brackets are canonical invariants for quantities that do not depend explicitly on time, i.e., proving that if $\frac{\partial f}{\partial t} = \frac{\partial g}{\partial t} = 0$, then $$[f,g]_{p,q} = [f,g]_{P,Q}\tag{45.9}.$$ Why is this the case? It seems to me that he is storing the time dependence in the coordinates and momenta instead of the functions $f$ and $g$ themselves, but this would prove the result only for those coordinates that leave $f$ and $g$ without explicit time-dependence.

  • $\begingroup$ Why would the time dependence matter in working out a formula that involves derivatives with respect to $p$ and $q$ and then changing to $P$ and $Q$ variables - if you work it out directly, which is the alternative method they indicate to prove this, you can also see why time dependence wont affect things. $\endgroup$
    – bolbteppa
    Sep 8 '19 at 15:47
  • $\begingroup$ In his argument it is relevant, because he will then consider $g$ as the Hamiltonian of some fictitious system and use the expression for the total time derivative of $f$ in that system. Since such derivative can only depend on the dynamics ($g$), but not on the coordinates, the Poisson brackets must be canonical invariants. $\endgroup$ Sep 8 '19 at 15:50
  • $\begingroup$ Furthermore, the time dependence seems important to me at a first glance because the transformation itself might be time-dependent =/ $\endgroup$ Sep 8 '19 at 15:50
  • $\begingroup$ How is adding a partial derivative with respect to $t$ on the right going to change anything in the paragraph between 45.9 and 45.10? $\endgroup$
    – bolbteppa
    Sep 8 '19 at 15:53
  • $\begingroup$ It is not clear to me yet whether it changes or not. I mean, I could pick coordinates such that $\frac{\partial f}{\partial t} = 0$, but I am still unsure why such a transformation would change $\frac{df}{dt} = 0$ precisely in the way neccessary for the brackets to remain invariant. $\endgroup$ Sep 8 '19 at 16:05

Obviously L&L's conclusion (45.9) is correct as always, but their 2nd proof in the paragraph between eqs. (45.9) & (45.10) seems not entirely convincing, although they do mention all the important ingredients along the way. In contrast, their 1st proof method by direct calculation is a straightforward exercise and quite convincing. In this answer, we extend the (non-degenerate) Poisson bracket construction to arbitrary (i.e. not necessarily canonical) coordinate systems to obtain a deeper understanding of this beautiful geometric gadget.

  1. Foliation. Consider a $(2n+1)$-dimensional codimension-1 foliation $M$ with a global time coordinate $t$, meaning that in an overlap $U\cap U^{\prime}$ between two (foliation-adapted) coordinate neighborhoods $U,U^{\prime} \subseteq M$, the coordinate transformation $(z,t)\to (z^{\prime},t^{\prime})$ is of the form $$ \begin{align} z^{\prime J}&~=~f^J(z,t), \qquad J~\in~\{1, \ldots, 2n\} ,\cr t^{\prime}&~=~t. \end{align}\tag{A} $$ The chain rule implies that the bases for vector fields and 1-forms transform as $$ \begin{align} \frac{\partial}{\partial z^J}&~=~\sum_{K=1}^{2n}\frac{\partial z^{\prime K}}{\partial z^J}\frac{\partial}{\partial z^{\prime K}}, \qquad J~\in~\{1, \ldots, 2n\} ,\cr \frac{\partial}{\partial t}&~=~\frac{\partial}{\partial t^{\prime}}+\sum_{K=1}^{2n}\frac{\partial z^{\prime K}}{\partial t}\frac{\partial}{\partial z^{\prime K}}, \end{align}\tag{B} $$ and $$ \begin{align} \mathrm{d}z^{\prime J}&~=~\sum_{K=1}^{2n}\frac{\partial z^{\prime J}}{\partial z^K}\mathrm{d}z^K+\frac{\partial z^{\prime J}}{\partial t}\mathrm{d}t, \qquad J~\in~\{1, \ldots, 2n\} ,\cr \mathrm{d}t^{\prime}&~=~\mathrm{d}t, \end{align}\tag{C} $$ respectively.

  2. Contact manifold. Assume furthermore that the manifold $M$ is endowed with a contact 1-form class $[\Theta]$, with (an atlas of locally defined) representatives $$\Theta~=~\vartheta - H \mathrm{d}t, \qquad \vartheta~=~\left. \Theta \right|_{\mathrm{d}t=0}~=~\sum_{J=1}^{2n} \vartheta_J\mathrm{d}z^J. \tag{D}$$ In the overlap $U\cap U^{\prime}$ between two (foliation-adapted) coordinate neighborhoods $U,U^{\prime} \subseteq M$, the 1-form is allowed to differ by an exact 1-form $$\Theta~=~\Theta^{\prime}+\mathrm{d}F,\tag{E} $$ i.e. we allow for a finite abelian gauge transformation with generating function$^1$ $F$. The $\Theta$-components wrt. the two coordinate neighborhoods are related as $$ \begin{align} \vartheta_J -\frac{\partial F}{\partial z^J}~&=~\sum_{K=1}^{2n} \frac{\partial z^{\prime K}}{\partial z^J} \vartheta^{\prime}_K, \cr H + \frac{\partial F}{\partial t}~&=~H^{\prime}-\sum_{K=1}^{2n} \frac{\partial z^{\prime K}}{\partial t} \vartheta^{\prime}_K.\end{align}\tag{F} $$

  3. Presymplectic 2-form. Consider the exterior derivative $$ \mathrm{d}~=~d+ \mathrm{d}t\frac{\partial}{\partial t} , \qquad d~=~\left. \mathrm{d} \right|_{\mathrm{d}t=0}~=~\sum_{J=1}^{2n} \mathrm{d}z^J\frac{\partial}{\partial z^J}, \tag{G} $$ and define a (globally defined) closed 2-form $$\Omega~=~ \mathrm{d}\Theta~=~\omega+ \mathrm{d}t\wedge\left( \frac{\partial \vartheta}{\partial t} +\mathrm{d}H\right),\tag{H} $$ $$d\vartheta~=~\omega~=~\left. \Omega \right|_{\mathrm{d}t=0}~=~\frac{1}{2}\sum_{J,K=1}^{2n} \omega_{JK}\mathrm{d}z^J\wedge\mathrm{d}z^K. \tag{I}$$ Note that $\Omega$ is independent of the representative $\Theta$ for the class $[\Theta]$ and independent of the coordinate system. Interestingly, the object $\omega$ depends on coordinate system, cf. eq. (C), but one may check that the components $\omega_{JK}$ transform covariantly $$ \omega_{IJ} ~=~\sum_{K,L=1}^{2n} \frac{\partial z^{\prime K}}{\partial z^I} \omega^{\prime}_{KL}\frac{\partial z^{\prime L}}{\partial z^J}, \qquad I,J~\in~\{1, \ldots, 2n\} \tag{J} $$ under a coordinate transformation (A). The transformation law (J) is essentially equivalent to what L&L mean by the phrase "time appears as a parameter". Geometrically it reflects the foliation.

  4. Poisson bracket. We now impose that $\omega$ is non-degenerate, i.e. invertible. Define the Poisson bracket via the inverse structure $$ \{f,g\}~:=~ \sum_{J,K=1}^{2n}\frac{\partial f}{\partial z^J} (\omega^{-1})^{JK} \frac{\partial g}{\partial z^K}. \tag{K} $$ The Jacobi identity is a consequence of the closedness relation $$d\omega~=~\left. \mathrm{d}\Omega \right|_{\mathrm{d}t=0}~=~0.\tag{L}$$ It is straightforward to check that the Poisson bracket (K) is invariant under coordinate transformations (A). This fact essentially yields the sought-for conclusion (45.9) as we shall see in the next section.

  5. Canonical transformations. Finally, recall L&L's definition of a canonical transformation $$\Theta~=~\Theta^{\prime}+\mathrm{d}F, $$
    $$\Theta~=~\sum_{i=1}^n p_i \mathrm{d}q^i - H \mathrm{d}t, \qquad \Theta^{\prime}~=~\sum_{i=1}^n p^{\prime}_i \mathrm{d}q^{\prime i} - H^{\prime} \mathrm{d}t^{\prime}. \tag{45.5b}$$ Both the un-primed and primed coordinate systems $z=(q,p)$ and $z^{\prime}=(q^{\prime},p^{\prime})$ are canonical/Darboux coordinates $$ \omega~=~\sum_{i=1}^n \mathrm{d}p_i \wedge \mathrm{d}q^i, \qquad \omega^{\prime}~=~\sum_{i=1}^n \mathrm{d}p^{\prime}_i \wedge \mathrm{d}q^{\prime i}, \tag{M}$$ i.e. they share the same canonical components $$\omega_{JK}~=~\begin{bmatrix} \mathbb{0} & -\mathbb{1}\cr \mathbb{1} & \mathbb{0}\end{bmatrix}_{2n\times 2n}, \tag{N}$$ and hence they lead to the same Poisson bracket (K). $\Box$


$^1$ The generating functions must satisfy a consistency condition on triple overlaps of 3 coordinate neighborhoods.

  • $\begingroup$ +1 beautiful exposition. $\endgroup$ Sep 11 '19 at 20:38
  • $\begingroup$ Is this Arnol'd or Spivak inspired? $\endgroup$
    – DanielC
    Sep 12 '19 at 1:43

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