# Difference between $dM/dt$ and $\partial M/\partial t$ [duplicate]

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

• – ACuriousMind Nov 2 '15 at 16:46
• 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 2 '15 at 16:57
• Can you please tell how M will matter? Some examples , if you can give? – Syed Jaffri Nov 2 '15 at 17:05
• I'm taking M, for example say a function of x,v and t. i.e. M=M(x,v,t) – Syed Jaffri Nov 2 '15 at 17:22
• Why did dM/dt=0 represent M as a constant of motion ,and not del(M)/del(t)=0? – Syed Jaffri Nov 2 '15 at 17:23

$\frac{dM}{dt} = \frac{\partial{M}}{\partial{t}}+\frac{\partial{M}}{\partial{x}}\frac{d{x}}{d{t}} = \frac{\partial{M}}{\partial{t}}+v\cdot\nabla{M}$ (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: $\frac{\partial{M}}{\partial{t}}(x) \neq 0$ .