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(I realise similar Phys.SE questions already exist but there is no answer with the Poisson bracket notation, I'll take this down if someone lets me know I should have commented in the existing question.)

I am trying to show that the Poisson bracket between the Hamiltonian and the Laplace-Runge-Lenz vector vanishes, i.e.

$$\left\{H,A\right\}_{PB}=0$$

where $\vec{A} = \left(p \times L\right) - m k\cdot \hat{r}$, and the Hamiltonian is for an orbit is given by, $$H = \frac{m}{2}\left(\frac{dr}{dt}\right)^2 + \frac{k}{r}$$

I have been trying to use tensor notation to write out the cross product term and the fundamental Poisson brackets but am not having any luck.

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  • $\begingroup$ The second term is incorrect. It should be constant times the unit vector in the radial direction. I hope that fixes your problem. $\endgroup$ – suresh Apr 22 '14 at 12:13
  • $\begingroup$ Sorry that's what I meant to show, I'm not sure how to illustrate the r unit vector with the formatting options. $\endgroup$ – user12800 Apr 22 '14 at 12:58
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    $\begingroup$ The proof of $\{A^i,H\}_{PB}=0$ is essentially done in pt. 1 - 6 of my Phys.SE answer here. $\endgroup$ – Qmechanic Apr 22 '14 at 16:36
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Recall $A^2 = m^2k^2+2mEL^2$. We can first prove the magnutude is constant:

$$[A^2,H] = [m^2k^2+2mEL^2,H] = [2mEL^2,H]$$

$$ [A^2,H] = 4mEL[L,H] = 4mEL \dot{L} $$

Since $\dot{L} = 0 $, this implies $\dot{A} = 0$. Not we jus need to show the components are constant. Notice:

$$\vec{A} = <yL_z-mk,xL_z-mk,0>$$

Thus $A_x = A_x(y)$ and $A_y=A_y(x)$. Since $A_x^2+A_y^2= constant$ and each are functions of different variables, each must be constant.

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