# Show that the Laplace-Runge-Lenz vector is conserved using poisson brackets

(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.

• The second term is incorrect. It should be constant times the unit vector in the radial direction. I hope that fixes your problem. Apr 22, 2014 at 12:13
• Sorry that's what I meant to show, I'm not sure how to illustrate the r unit vector with the formatting options. Apr 22, 2014 at 12:58
• The proof of $\{A^i,H\}_{PB}=0$ is essentially done in pt. 1 - 6 of my Phys.SE answer here. Apr 22, 2014 at 16:36

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.

• Why does this hold: "since each are functions of different variables, each must be constant?" Mar 31 at 1:08