$\frac{dM}{dt} = 0$ represents a constant of motion $M.$ Why not $\frac{\partial M}{\partial t}$ represent a constant of motion $M$?
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7Possible duplicate of What is the difference between implicit and explicit time dependence e.g. $\frac{\partial \rho}{\partial t}$ and $\frac{d \rho} {dt}$? – ACuriousMind Nov 02 '15 at 16:46
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What is M? What is the physical situation? Without that info, it's hard to help. (But see the link in the comment above.) – march Nov 02 '15 at 16:57
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Can you please tell how M will matter? Some examples , if you can give? – Syed Jaffri Nov 02 '15 at 17:05
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I'm taking M, for example say a function of x,v and t. i.e. M=M(x,v,t) – Syed Jaffri Nov 02 '15 at 17:22
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Why did dM/dt=0 represent M as a constant of motion ,and not del(M)/del(t)=0? – Syed Jaffri Nov 02 '15 at 17:23
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I,m not asking when they are equal or not – Syed Jaffri Nov 02 '15 at 17:24
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$$M = M\left( {t,{x_1},...,{x_n}} \right)$$ $$\frac{{dM}}{{dt}} = \frac{{\partial M}}{{\partial t}}\frac{{dt}}{{dt}} + \sum\limits_{i = 1}^n {\frac{{\partial M}}{{\partial {x_i}}}\frac{{d{x_i}}}{{dt}}} $$ – Nov 02 '15 at 19:30
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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: $\frac{\partial{M}}{\partial{t}}(x) \neq 0$ .
Fabrice NEYRET
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