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$\frac{dM}{dt} = 0$ represents a constant of motion $M.$ Why not $\frac{\partial M}{\partial t}$ represent a constant of motion $M$?

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marked as duplicate by ACuriousMind, user36790, Kyle Kanos, John Rennie, HDE 226868 Nov 2 '15 at 22:46

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$\frac{dM}{dt} = \frac{\partial{M}}{\partial{t}}+\frac{\partial{M}}{\partial{x}}\frac{d{x}}{d{t}} = \frac{\partial{M}}{\partial{t}}+v\cdot\nabla{M}$ (with no assumption on what is M) . So if $v\cdot\nabla{M} \neq0$ you can have one of $\frac{dM}{dt}$ and $\frac{\partial{M}}{\partial{t}}$ that is zero when the other is not.

$\frac{\partial{M}}{\partial{t}}=0$ means stationarity of the quantity $M$: at a given fix location in space $M$ doesn't change in time. Now, flowing particles might have their $M(x(t),t)$ changing in time, i.e. $\frac{dM}{dt} \neq 0$.

$\frac{dM}{dt} = 0$ means conservation of the quantity $M$ for the given flowing particles. Now if the flow is not stationary the value that you see at a given location x might change in time: $\frac{\partial{M}}{\partial{t}}(x) \neq 0$ .

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