# Time Dependence on Landau & Lifshitz's Proof of Poisson's Bracket Canonical Invariance

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.

• 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. Commented Sep 8, 2019 at 15:47
• 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. Commented Sep 8, 2019 at 15:50
• Furthermore, the time dependence seems important to me at a first glance because the transformation itself might be time-dependent =/ Commented Sep 8, 2019 at 15:50
• 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? Commented Sep 8, 2019 at 15:53
• 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. Commented Sep 8, 2019 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). $$\Box$$

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

• +1 beautiful exposition. Commented Sep 11, 2019 at 20:38
• Is this Arnol'd or Spivak inspired? Commented Sep 12, 2019 at 1:43