If I have a given hamiltonian $H$ and some $f$ which claims to be an integral of the motion and I have this identity $$\frac {d}{dt} = \{\;,H\} + \frac {\partial}{\partial t}$$ where $\{\;,H\}$ is the Poisson bracket. I actually don't understand the difference between $\frac {d}{dt}$ and $\frac {\partial}{\partial t}$ in this case. If $f$ is that integral, then $\{f,H\} + \frac {\partial f}{\partial t} = 0$. What is the intuition behind it?
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Crossposted from https://math.stackexchange.com/q/2277214/11127 – Qmechanic May 12 '17 at 02:49
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Possible duplicates: https://physics.stackexchange.com/q/9122/2451 and links therein. – Qmechanic May 12 '17 at 02:51
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I think the very detailed answer you got on MS on the difference between $\partial f/\partial t$ and ${\rm d}f/{\rm d}t$ should have clarified part of this question. I just want to add that if $f = f({\bf q}(t), {\bf p}(t),t)$ then
\begin{eqnarray} \frac{{\rm d}f}{dt} &=& \frac{\partial f}{\partial q^\alpha}\frac{{\rm d}q^\alpha}{{\rm d}t}+ \frac{\partial f}{\partial p_\alpha}\frac{{\rm d}p_\alpha}{{\rm d}t} + \frac{\partial f}{\partial t} \\ &\stackrel{\rm Hamilton~ Eq.}{=}& \frac{\partial f}{\partial q^\alpha}\frac{\partial H}{\partial p_\alpha} - \frac{\partial f}{\partial p_\alpha}\frac{\partial H}{\partial q^\alpha} + \frac{\partial f}{\partial t} \\ &\stackrel{\rm Poisson~bracket}{=}& \{f, H\} + \frac{\partial f}{\partial t} \end{eqnarray}
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