Time evolution of Galilean boost I was introduced the generator of Galilean boost $K=mx-pt$.
I was given an Hamiltonian with several particles: $H=\sum_i \frac{p_i^2}{2m_i}+V(|x_i-x_j|)$ where the potential only depends on the relative position between the particles.
I was asked to compute the poisson brackets $\{K,H\}$ and show that the general time evolution formula is given by :
$$\frac{d K}{d t}=\frac{\partial K}{\partial t}+\left\{K, H\right\}.$$
I managed to derive the time evolution formula but I feel I am doing something wrong. First the Poisson brackets:
\begin{align}
\{K,H\}&= \frac{\partial K}{\partial x}\frac{\partial H}{\partial p}-\frac{\partial K}{\partial p}\frac{\partial H}{\partial x}\\
&=p+t\frac{ \partial V}{ \partial x}\\
&=p-t\frac{d p}{d t}.\\ 
\end{align}
I was told that $\frac{d K}{d t}=0$, which I do not understand why. I found:
$$\frac{d K}{d t}=-t\frac{d p}{d t}\;(\,\neq 0\;?)$$
Finally, I read that the Poisson bracket $\{K,H\}$ describes
the effect of an infinitesimal boost on the system’s energy. This eludes my understanding. I could show that the Hamiltonian changes when I perform a change of coordinates $x'=x-v_0t$, but I do not know how to link it with the Poisson brackets.
I find this an interesting topic, and if someone could help clarify my understanding with some explanations I would be very grateful. Any remark or link to references is always appreciated.
 A: I'm not sure WP would help you.
Your comment clarification is crucial. It indicates your potential is invariant under a translation, whether that translation is proportional to the time, or not. Without loss of generality, take two particles, at positions x and y.  You then have
$$
(\partial_x  + \partial_y) V(|x-y|)= ((\partial_x  + \partial_y)|x-y|) ~~V'(|x-y|) =0,
$$
since the left parenthesis vanishes,
$$
(\partial_x  + \partial_y)\sqrt {(x-y)^2}=0.
$$
That is, for your expression K really indicating $\sum_i K_i$ as well as the PBs,  you get
$$
\partial_t K + t \{ K,H\}= t(\partial_x  + \partial_y) V(|x-y|)=0.
$$
An infinitesimal transformation generated by K is
$$
\delta x_i = v_0 \{ x_i , K \}= -v_0 t, \qquad \delta p_i = v_0 \{ p_i , K \}= -mv_0 \\  \leadsto ~~~  \delta H = -v_0 \sum_i (p_i+ t\partial_i V) = -v_0 \sum_i p_i .
$$
This reflects the corresponding term in the WP Lie algebra, involving the Hamiltonian and the common Galilean boost.
Note the invariant $mH-P^2/2$, so that
$$\delta (mH-P^2/2) =0.$$
