Aprovechando que he tenido que hablar sobre el paréntesis de Poisson en un curso de Mecánica Clásica, acá comparto el cálculo del paréntesis de Poisson del vector de Laplace-Runge-Lenz (LRL) con el hamiltoniano, que -como se espera- resulta ser nulo, demostrando que el vector LRL se conserva. El camino que tomé es un tortuoso, pero funciona y ayuda a familiarizarse con las propiedades del paréntesis de Poisson. Por lo extenso de algunas ecuaciones, recomiendo leer la entrada desde un ordenador.
Sabiendo que el hamiltoniano es $\displaystyle{\mathcal{H}=\displaystyle{\frac{\mathbf{p}^2}{2\mu}-\frac{k}{r}}}$, se quiere verificar que el vector LRL, $\displaystyle{\mathbf{A}=\mathbf{p}\times\mathbf{L}-\mu{k}\displaystyle{\frac{\mathbf{r}}{r}}}$ es una constante de movimiento, calculando explícitamente el paréntesis ${\{\mathbf{A},\mathcal{H}\}}$.
Descomponiendo el vector LRL,
\begin{align}A_i&=\epsilon_{ijk}p_jL_k-\mu{k}\frac{r_i}{r}\nonumber\\[0.1in]&=\epsilon_{ijk}\epsilon_{kmn}p_jr_mp_n-\mu{k}\frac{r_i}{r}\nonumber\\[0.1in]&=\epsilon_{kij}\epsilon_{kmn}p_jr_mp_n-\mu{k}\frac{r_i}{r}\nonumber\\[0.1in]&=\left(\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}\right)p_jr_mp_n-\mu{k}\frac{r_i}{r}\nonumber\\[0.1in]&=\mathbf{p}^2r_i-(\mathbf{r}\cdot\mathbf{p})p_i-\mu{k}\frac{r_i}{r}\nonumber\\[0.1in]&=\left(\mathbf{p}^2-\mu{k}\frac{1}{r}\right)r_i-(\mathbf{r}\cdot\mathbf{p})p_i\end{align}
Así entonces, calculando el paréntesis de Poisson,
\begin{align}\{A_i,\mathcal{H}\}&=\left\{\left(\mathbf{p}^2-\mu{k}\frac{1}{r}\right)r_i-(\mathbf{r}\cdot\mathbf{p})p_i,\frac{\mathbf{p}^2}{2\mu}-\frac{k}{r}\right\}\nonumber\\[0.1in]&=\frac{1}{2\mu}\left(\left\{\mathbf{p}^2r_i,\mathbf{p}^2\right\}-\left\{(\mathbf{r}\cdot\mathbf{p})p_i,\mathbf{p}^2\right\}\right)+k\left(\left\{(\mathbf{r}\cdot\mathbf{p})p_i,\frac{1}{r}\right\}-\left\{\mathbf{p}^2r_i,\frac{1}{r}\right\}-\frac{1}{2}\left\{\frac{r_i}{r},\mathbf{p}^2\right\}\right)+\mu{k}^2\left(\left\{\frac{r_i}{r},\frac{1}{r}\right\}\right)\end{align} así, para el primer sumando,
\begin{align}\left\{\mathbf{p}^2r_i,\mathbf{p}^2\right\}&=\mathbf{p}^2\left\{r_i,\mathbf{p}^2\right\}+r_i\left\{\mathbf{p}^2,\mathbf{p}^2\right\}\nonumber\\[0.1in]&=\mathbf{p}^2\left\{r_i,\mathbf{p}^2\right\}\nonumber\\[0.1in]&=\mathbf{p}^2\,\sum_{\alpha}\left(\frac{\partial{r_i}}{\partial{r_\alpha}}\frac{\partial{\mathbf{p}^2}}{\partial{p_\alpha}}-\frac{\partial{r_i}}{\partial{p_\alpha}}\frac{\partial{\mathbf{p}^2}}{\partial{r_\alpha}}\right)\nonumber\\[0.1in]&=2p_i\mathbf{p}^2\end{align} y también
\begin{align}\left\{(\mathbf{r}\cdot\mathbf{p})p_i,\mathbf{p}^2\right\}&=\left(\mathbf{r}\cdot\mathbf{p}\right)\left\{p_i,\mathbf{p}^2\right\}+p_i\left\{(\mathbf{r}\cdot\mathbf{p}),\mathbf{p}^2\right\}\nonumber\\[0.1in]&=p_i\left\{(\mathbf{r}\cdot\mathbf{p}),\mathbf{p}^2\right\}\nonumber\\[0.1in]&=p_i\,\sum_\alpha\left(\frac{\partial}{\partial{r_\alpha}}(\mathbf{r}\cdot\mathbf{p})\frac{\partial}{\partial{p_\alpha}}\mathbf{p}^2-\frac{\partial}{\partial{p_\alpha}}(\mathbf{r}\cdot\mathbf{p})\frac{\partial}{\partial{r_\alpha}}\mathbf{p}^2\right)\nonumber\\[0.1in]&=2p_i\mathbf{p}^2\end{align}
Entonces podemos ir reduciendo el vector LRL a
\begin{equation}\{A_i,\mathcal{H}\}=k\left(\left\{(\mathbf{r}\cdot\mathbf{p})p_i,\frac{1}{r}\right\}-\left\{\mathbf{p}^2r_i,\frac{1}{r}\right\}-\frac{1}{2}\left\{\frac{r_i}{r},\mathbf{p}^2\right\}\right)+\mu{k}^2\left(\left\{\frac{r_i}{r},\frac{1}{r}\right\}\right)\end{equation}
Para el siguiente sumando,
\begin{align}\left\{(\mathbf{r}\cdot\mathbf{p})p_i,\frac{1}{r}\right\}&=(\mathbf{r}\cdot\mathbf{p})\left\{p_i,\frac{1}{r}\right\}+p_i\left\{(\mathbf{r}\cdot\mathbf{p}),\frac{1}{r}\right\}\nonumber\\[0.1in]&=(\mathbf{r}\cdot\mathbf{p})\,\sum_\alpha\left(\frac{\partial{p_i}}{\partial{r_\alpha}}\frac{\partial{r^{-1}}}{\partial{p_\alpha}}-\frac{\partial{p_i}}{\partial{p_\alpha}}\frac{\partial{r^{-1}}}{\partial{r_\alpha}}\right)+p_i\,\sum_\alpha\left(\frac{\partial}{\partial{r_\alpha}}(\mathbf{r}\cdot\mathbf{p})\frac{\partial{r^{-1}}}{\partial{p_\alpha}}-\frac{\partial}{\partial{p_\alpha}}(\mathbf{r}\cdot\mathbf{p})\frac{\partial{r^{-1}}}{\partial{r_\alpha}}\right)\nonumber\\[0.1in]&=(\mathbf{r}\cdot\mathbf{p})\frac{r_i}{r^3}+\frac{p_i}{r}\nonumber\\[0.1in]&=\left\{\mathbf{p}^2r_i,\frac{1}{r}\right\}\nonumber\\[0.1in]&=\mathbf{p}^2\left\{r_i,\frac{1}{r}\right\}+r_i\left\{\mathbf{p}^2,\frac{1}{r}\right\}\nonumber\\[0.1in]&=\mathbf{p}^2\,\sum_\alpha\left(\frac{\partial{r_i}}{\partial{r_\alpha}}\frac{\partial{r^{-1}}}{\partial{p_\alpha}}-\frac{\partial{r_i}}{\partial{p_\alpha}}\frac{\partial{r^{-1}}}{\partial{r_\alpha}}\right)+r_i\,\sum_\alpha\left(\frac{\partial}{\partial{r_\alpha}}\mathbf{p}^2\frac{\partial{r^{-1}}}{\partial{p_\alpha}}-\frac{\partial}{\partial{p_\alpha}}\mathbf{p}^2\frac{\partial{r^{-1}}}{\partial{r_\alpha}}\right)\nonumber\\[0.1in]&=2\frac{r_i}{r^3}(\mathbf{r}\cdot\mathbf{p})\end{align} y también
\begin{align}\left\{\frac{r_i}{r},\mathbf{p}^2\right\}&=r_i\left\{\frac{1}{r},\mathbf{p}^2\right\}+\frac{1}{r}\left\{r_i,\mathbf{p}^2\right\}\nonumber\\[0.1in]&=-2r_i\frac{1}{r^3}(\mathbf{r}\cdot\mathbf{p})+\frac{2}{r}p_i\end{align} entonces se tiene
\begin{align}k&\left(\left\{(\mathbf{r}\cdot\mathbf{p})p_i,\frac{1}{r}\right\}-\left\{\mathbf{p}^2r_i,\frac{1}{r}\right\}-\frac{1}{2}\left\{\frac{r_i}{r},\mathbf{p}^2\right\}\right)=k\left[\frac{p_i}{r}+(\mathbf{r}\cdot\mathbf{p})\frac{r_i}{r^3}-2(\mathbf{r}\cdot\mathbf{p})\frac{r_i}{r^3}-\frac{1}{2}\left(\frac{2}{r}p_i-2(\mathbf{r}\cdot\mathbf{p})\frac{r_i}{r^3}\right)\right]=0\end{align} por lo que la derivada temporal del vector LRL se reduce a
\begin{equation}\{A_i,\mathcal{H}\}=\mu{k}^2\left\{\frac{r_i}{r},\frac{1}{r}\right\}\end{equation} que evidentemente es nulo, por tanto se concluye que
\begin{equation}\mathbf{\dot{A}}=\left\{\mathbf{A},\mathcal{H}\right\}=\mathbf{0}\end{equation} y en efecto $\mathbf{A}$ es constante de movimiento.
En Mathematica es sencillo programar el paréntesis de Poisson para verificar lo anterior. Las siguientes líneas hacen esa tarea para cada componente.
Sabiendo que el hamiltoniano es $\displaystyle{\mathcal{H}=\displaystyle{\frac{\mathbf{p}^2}{2\mu}-\frac{k}{r}}}$, se quiere verificar que el vector LRL, $\displaystyle{\mathbf{A}=\mathbf{p}\times\mathbf{L}-\mu{k}\displaystyle{\frac{\mathbf{r}}{r}}}$ es una constante de movimiento, calculando explícitamente el paréntesis ${\{\mathbf{A},\mathcal{H}\}}$.
Descomponiendo el vector LRL,
\begin{align}A_i&=\epsilon_{ijk}p_jL_k-\mu{k}\frac{r_i}{r}\nonumber\\[0.1in]&=\epsilon_{ijk}\epsilon_{kmn}p_jr_mp_n-\mu{k}\frac{r_i}{r}\nonumber\\[0.1in]&=\epsilon_{kij}\epsilon_{kmn}p_jr_mp_n-\mu{k}\frac{r_i}{r}\nonumber\\[0.1in]&=\left(\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}\right)p_jr_mp_n-\mu{k}\frac{r_i}{r}\nonumber\\[0.1in]&=\mathbf{p}^2r_i-(\mathbf{r}\cdot\mathbf{p})p_i-\mu{k}\frac{r_i}{r}\nonumber\\[0.1in]&=\left(\mathbf{p}^2-\mu{k}\frac{1}{r}\right)r_i-(\mathbf{r}\cdot\mathbf{p})p_i\end{align}
Así entonces, calculando el paréntesis de Poisson,
\begin{align}\{A_i,\mathcal{H}\}&=\left\{\left(\mathbf{p}^2-\mu{k}\frac{1}{r}\right)r_i-(\mathbf{r}\cdot\mathbf{p})p_i,\frac{\mathbf{p}^2}{2\mu}-\frac{k}{r}\right\}\nonumber\\[0.1in]&=\frac{1}{2\mu}\left(\left\{\mathbf{p}^2r_i,\mathbf{p}^2\right\}-\left\{(\mathbf{r}\cdot\mathbf{p})p_i,\mathbf{p}^2\right\}\right)+k\left(\left\{(\mathbf{r}\cdot\mathbf{p})p_i,\frac{1}{r}\right\}-\left\{\mathbf{p}^2r_i,\frac{1}{r}\right\}-\frac{1}{2}\left\{\frac{r_i}{r},\mathbf{p}^2\right\}\right)+\mu{k}^2\left(\left\{\frac{r_i}{r},\frac{1}{r}\right\}\right)\end{align} así, para el primer sumando,
\begin{align}\left\{\mathbf{p}^2r_i,\mathbf{p}^2\right\}&=\mathbf{p}^2\left\{r_i,\mathbf{p}^2\right\}+r_i\left\{\mathbf{p}^2,\mathbf{p}^2\right\}\nonumber\\[0.1in]&=\mathbf{p}^2\left\{r_i,\mathbf{p}^2\right\}\nonumber\\[0.1in]&=\mathbf{p}^2\,\sum_{\alpha}\left(\frac{\partial{r_i}}{\partial{r_\alpha}}\frac{\partial{\mathbf{p}^2}}{\partial{p_\alpha}}-\frac{\partial{r_i}}{\partial{p_\alpha}}\frac{\partial{\mathbf{p}^2}}{\partial{r_\alpha}}\right)\nonumber\\[0.1in]&=2p_i\mathbf{p}^2\end{align} y también
\begin{align}\left\{(\mathbf{r}\cdot\mathbf{p})p_i,\mathbf{p}^2\right\}&=\left(\mathbf{r}\cdot\mathbf{p}\right)\left\{p_i,\mathbf{p}^2\right\}+p_i\left\{(\mathbf{r}\cdot\mathbf{p}),\mathbf{p}^2\right\}\nonumber\\[0.1in]&=p_i\left\{(\mathbf{r}\cdot\mathbf{p}),\mathbf{p}^2\right\}\nonumber\\[0.1in]&=p_i\,\sum_\alpha\left(\frac{\partial}{\partial{r_\alpha}}(\mathbf{r}\cdot\mathbf{p})\frac{\partial}{\partial{p_\alpha}}\mathbf{p}^2-\frac{\partial}{\partial{p_\alpha}}(\mathbf{r}\cdot\mathbf{p})\frac{\partial}{\partial{r_\alpha}}\mathbf{p}^2\right)\nonumber\\[0.1in]&=2p_i\mathbf{p}^2\end{align}
Entonces podemos ir reduciendo el vector LRL a
\begin{equation}\{A_i,\mathcal{H}\}=k\left(\left\{(\mathbf{r}\cdot\mathbf{p})p_i,\frac{1}{r}\right\}-\left\{\mathbf{p}^2r_i,\frac{1}{r}\right\}-\frac{1}{2}\left\{\frac{r_i}{r},\mathbf{p}^2\right\}\right)+\mu{k}^2\left(\left\{\frac{r_i}{r},\frac{1}{r}\right\}\right)\end{equation}
Para el siguiente sumando,
\begin{align}\left\{(\mathbf{r}\cdot\mathbf{p})p_i,\frac{1}{r}\right\}&=(\mathbf{r}\cdot\mathbf{p})\left\{p_i,\frac{1}{r}\right\}+p_i\left\{(\mathbf{r}\cdot\mathbf{p}),\frac{1}{r}\right\}\nonumber\\[0.1in]&=(\mathbf{r}\cdot\mathbf{p})\,\sum_\alpha\left(\frac{\partial{p_i}}{\partial{r_\alpha}}\frac{\partial{r^{-1}}}{\partial{p_\alpha}}-\frac{\partial{p_i}}{\partial{p_\alpha}}\frac{\partial{r^{-1}}}{\partial{r_\alpha}}\right)+p_i\,\sum_\alpha\left(\frac{\partial}{\partial{r_\alpha}}(\mathbf{r}\cdot\mathbf{p})\frac{\partial{r^{-1}}}{\partial{p_\alpha}}-\frac{\partial}{\partial{p_\alpha}}(\mathbf{r}\cdot\mathbf{p})\frac{\partial{r^{-1}}}{\partial{r_\alpha}}\right)\nonumber\\[0.1in]&=(\mathbf{r}\cdot\mathbf{p})\frac{r_i}{r^3}+\frac{p_i}{r}\nonumber\\[0.1in]&=\left\{\mathbf{p}^2r_i,\frac{1}{r}\right\}\nonumber\\[0.1in]&=\mathbf{p}^2\left\{r_i,\frac{1}{r}\right\}+r_i\left\{\mathbf{p}^2,\frac{1}{r}\right\}\nonumber\\[0.1in]&=\mathbf{p}^2\,\sum_\alpha\left(\frac{\partial{r_i}}{\partial{r_\alpha}}\frac{\partial{r^{-1}}}{\partial{p_\alpha}}-\frac{\partial{r_i}}{\partial{p_\alpha}}\frac{\partial{r^{-1}}}{\partial{r_\alpha}}\right)+r_i\,\sum_\alpha\left(\frac{\partial}{\partial{r_\alpha}}\mathbf{p}^2\frac{\partial{r^{-1}}}{\partial{p_\alpha}}-\frac{\partial}{\partial{p_\alpha}}\mathbf{p}^2\frac{\partial{r^{-1}}}{\partial{r_\alpha}}\right)\nonumber\\[0.1in]&=2\frac{r_i}{r^3}(\mathbf{r}\cdot\mathbf{p})\end{align} y también
\begin{align}\left\{\frac{r_i}{r},\mathbf{p}^2\right\}&=r_i\left\{\frac{1}{r},\mathbf{p}^2\right\}+\frac{1}{r}\left\{r_i,\mathbf{p}^2\right\}\nonumber\\[0.1in]&=-2r_i\frac{1}{r^3}(\mathbf{r}\cdot\mathbf{p})+\frac{2}{r}p_i\end{align} entonces se tiene
\begin{align}k&\left(\left\{(\mathbf{r}\cdot\mathbf{p})p_i,\frac{1}{r}\right\}-\left\{\mathbf{p}^2r_i,\frac{1}{r}\right\}-\frac{1}{2}\left\{\frac{r_i}{r},\mathbf{p}^2\right\}\right)=k\left[\frac{p_i}{r}+(\mathbf{r}\cdot\mathbf{p})\frac{r_i}{r^3}-2(\mathbf{r}\cdot\mathbf{p})\frac{r_i}{r^3}-\frac{1}{2}\left(\frac{2}{r}p_i-2(\mathbf{r}\cdot\mathbf{p})\frac{r_i}{r^3}\right)\right]=0\end{align} por lo que la derivada temporal del vector LRL se reduce a
\begin{equation}\{A_i,\mathcal{H}\}=\mu{k}^2\left\{\frac{r_i}{r},\frac{1}{r}\right\}\end{equation} que evidentemente es nulo, por tanto se concluye que
\begin{equation}\mathbf{\dot{A}}=\left\{\mathbf{A},\mathcal{H}\right\}=\mathbf{0}\end{equation} y en efecto $\mathbf{A}$ es constante de movimiento.
En Mathematica es sencillo programar el paréntesis de Poisson para verificar lo anterior. Las siguientes líneas hacen esa tarea para cada componente.
(*Definición paréntesis de Poisson*)
PB[u_, v_, q_Symbol, p_Symbol] := D[u, q] D[v, p] - D[v, q] D[u, p]
(*Definiciones*)
p = {p1, p2, p3}; r = {r1, r2, r3};
P = FullSimplify[Norm[p], p1 > 0 && p2 > 0 && p3 > 0]
R = FullSimplify[Norm[r], r1 > 0 && r2 > 0 && r3 > 0]
A := Cross[p, Cross[r, p]] - r/R
H = P^2/2 - 1/R
(*El vector LRL es ortogonal al momento angular*)
A . Cross[r, p] == 0 // Simplify
(*El vector LRL se conserva*)
Simplify[Sum[{PB[A[[1]], H, r[[j]], p[[j]]], PB[A[[2]], H, r[[j]],
p[[j]]], PB[A[[3]], H, r[[j]], p[[j]]]}, {j, 1, 3}]]
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