# A quicker way to verify that a function is a constant of motion?

I have three particles that we can indicate with $\alpha$ ($\alpha$=0,1,2), they are identified by the $r^i_\alpha$ coordinates and $p^\beta_j$ conjugata momenta ($\beta=0,1,2$ and $i,j=1,2,3$).

I have this transformation: $$\overrightarrow r_\alpha -> \overrightarrow R_\alpha + \overrightarrow \epsilon$$

$$\overrightarrow p_\alpha -> \overrightarrow P_\alpha=\overrightarrow p_\alpha$$

and I have to demonstrate that $\sum p_\alpha$ is a constant of motion.

The Hamiltonian of the system is $$H=\frac{\overrightarrow p_0^2}{2m}+\frac{\overrightarrow p_1^2}{2m}+\frac{\overrightarrow p_2^2}{2m}-2V(\overrightarrow r_1- \overrightarrow r_0)+V(\overrightarrow r_2-\overrightarrow r_1)$$

I have calculated Poisson bracket $[H, \sum p_\alpha]$ that is equal to zero.

Considering that the generating function of the transformation is:

$$F_2=\sum \overrightarrow r_\alpha \cdot \overrightarrow P_\alpha+\overrightarrow \epsilon \cdot \sum \overrightarrow P_\alpha$$ I'd like to know if there is a quicker way to answer to the question.

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Related question by OP: physics.stackexchange.com/q/66717/2451 and links therein. – Qmechanic Jun 1 '13 at 21:59