Poisson Brackets

Is this quantity conserved under canonical transformations? It is! As long as you have $\eta$ a canonical coordinate, then you don’t need to specify $\eta$ or $q,p$ on the brackets.

Suppose we have $\eta \to \zeta$. So, $MJ{M}^T=J$.

Then, $[U,V{]}_{\eta }={\frac{\partial U}{\partial \eta }}^{T}J\frac{\partial V}{\partial \eta }={\frac{\partial U}{\partial \eta }}^T{M}^{T}JM\frac{\partial V}{\partial \eta }$. Note, $\frac{\partial V}{\partial {\eta }_i}=\sum _j\frac{\partial V}{\partial \partial {\zeta }_j}\frac{\partial {\zeta }_j}{\partial {\eta }_i}\Rightarrow \frac{\partial V}{\partial \eta }=M^T\frac{\partial V}{\partial \zeta }$. Then, $[U,V{]}_{\eta }={\frac{\partial U}{\partial \zeta }}^{T}MJ{M}^T\frac{\partial V}{\partial \zeta }=\frac{\partial U}{\partial \zeta }J\frac{\partial V}{\partial \zeta }=[U,V{]}_{\zeta }$.

Suppose, $U$ and $V$ are canonical coordinates, $[q_i,q_j]=0=[p_i,p_j]$. For, $[q_i,p_j]={\delta }_{ij}=-[p_i,q_j]$. So, $[\eta ,\eta ]=J$. Thus, you can use this to determine if a set of coordinates are canonical or not.

1. Anticommutivity: $[U,V]=-[V,U]$.
2. $[U,U]=0$.
3. Bilinearity: $[aU+bV,W]=a[U,W]+b[V,W]$.
4. Libenitz Rule: $[UV,W]=U[V,W]+[U,W]V$.
5. Jacobi Identity: $[U,[V,W]]+[V,[W,U]]+[W,[U,V]]=\varnothing$. This will be important to find the constants of motion.

Looking ahead: $[U,H]=0$ and $U$ does not explicitly depend on time, $U$ is a constant of the motion. Note: $\frac{dU}{dt}={\frac{\partial U}{\partial \eta }}^{T}J\frac{\partial H}{\partial \eta }+\frac{\partial U}{\partial t}=[U,H]+\frac{\partial U}{\partial t}$.

For, $U\to {\eta }_i$, ${\dot{\eta }}_i=[{\eta }_i,H]\Rightarrow \dot{\eta }=[\eta ,H]={\left(\frac{\partial \eta }{\partial \eta }\right)}^{T}J\frac{\partial H}{\partial \eta }=J\frac{\partial H}{\partial \eta }$. For $U\to H$, $\frac{dH}{dt}=\frac{\partial H}{\partial t}$. Thus, if $H$ does not explicitly depend on time, $H$ is a constant of the motion.

If we have $U(q,p,t)$ how do we determine if it is a constant of the motion? So, $\frac{dU}{dt}=0\Rightarrow [U,H]=-\frac{\partial U}{\partial t}$.

Then, if $U$ and $V$ are both constants of the motion, then $\alpha U+\beta V$ is a constant of the motion. Further, if $U,V$ do not depend on time explicitly, $[U,V]=\mathbb{C}$. Hence, $[U,V]$ is a constant of the motion.

Additionally, you can show that $U,V$ are both constants of the motion, this still holds. I.e. $[U,V]=\mathbb{C}$ is still a constant of the motion.

$$\begin{array}{rl}\frac{d[U,V]}{dt}& =[[U,V],H]+\frac{\partial [U,V]}{\partial t}\\ & =-[H,[U,V]]+\frac{\partial [U,V]}{\partial t}\\ & =[U,[V,H]]+[V,[H,U]]+\frac{\partial [U,V]}{\partial t}\\ & =[U,-\frac{\partial V}{\partial t}]+[V,\frac{\partial U}{\partial t}]+\frac{\partial [U,V]}{\partial t}\\ & =-[U,\frac{\partial V}{\partial t}]+[V,\frac{\partial U}{\partial t}]+\frac{\partial [U,V]}{\partial t}\\ & =-[U,\frac{\partial V}{\partial t}]+[V,\frac{\partial U}{\partial t}]+[U_t,V]+[U,V_t]\\ & =0.\end{array}$$

Suppose we have a function driven by a single parameter in the phase space: $U(\alpha )$. Suppose $A$ and $B$ are two points in the phase space with canonical coordinates. Thus, we can think of this as a finite canonical transformation. So, you can take infinitesimal steps of this parameter $\alpha$ to reach $B$ from $A$. So, $\zeta =\eta +\delta \eta$. Hence, we have some generator $\delta \eta =\epsilon J\frac{\partial G}{\partial \eta }$. Recall, $[\eta ,H]=J\frac{\partial H}{\partial \eta }$. So, $\delta \eta =\epsilon [\eta ,G]$. Consider $\epsilon \to dt$ and $G\to H$. So, $\delta \eta =dt[\eta ,H]=dt\dot{\eta }=d\eta$. Thus, the Hamiltonian drives the system along its trajectory. Hence, the motion of the system is a canonical transformation.

Define $dU=U(B)-U(A)$ as the change in the system under the canonical transformation, for $B$ an infinitesimal point away from $A$. So, $dU=U(\eta +\delta \eta )-U(\eta )$. Then, ${\frac{\partial U}{\partial U}}^T\delta \eta ={\frac{\partial U}{\partial \eta }}^T\epsilon J\frac{\partial G}{\partial \eta }$. Remark: in this case we can change $\epsilon \to \alpha$. So, $dU==\epsilon {\frac{\partial U}{\partial \eta }}^{T}J\frac{\partial G}{\partial \eta }=\epsilon [U,G]$. So, for $\alpha \to t$, $\frac{dU}{dt}=[U,G]$. In general, $\frac{dU}{d\alpha }=[U,G]$. $U(\alpha )=U(A)+\alpha \frac{dU}{d\alpha }(A)+\mathcal{O}({\alpha }^2)$. So, $\alpha \to t$ and $G\to H$. Then, $U(t)\Rightarrow \frac{dU}{dt}=[U,H]$. Hence, $U(t)=U(0)+t\frac{dU}{dt}(0)+\mathcal{O}(t^2)=U(0)+t[U,H](0)+\mathcal{O}(t^2)$.

$H=\frac{p^2}{2m}-max$.

Let $U=x$. Then,

$$\begin{array}{rl}x(t)& =x(0)+t\frac{dx}{dt}(0)+\frac{t^2}{2}\frac{d{x}^2}{d{t}^2}+\mathcal{O}(t^3)\\ & =x(0)+t[x,H](0)+\frac{t^2}{2}\frac{d{x}^2}{d{t}^2}+\mathcal{O}(t^3)\\ & =x(0)+t\left(\frac{p}{m}\right)(0)+\frac{t^2}{2}\frac{d{x}^2}{d{t}^2}+\mathcal{O}(t^3)\\ & =x(0)+v(0)t+\frac{t^2}{2m}\frac{dp}{dt}+\mathcal{O}(t^3)\\ & =x(0)+v(0)t+\frac{t^2}{2m}(0-(-ma))+\mathcal{O}(t^3)\\ & =x(0)+v(0)t+\frac{a{t}^2}{2}+\mathcal{O}(t^3)\\ & =x(0)+v(0)t+\frac{a{t}^2}{2}.\end{array}$$