I'm reading some fluid dynamics notes which are talking about a Galilean boost of the form:

$$x'=x-vt, \qquad t'=t$$

The notes immediately claim from this that the material derivative $$\frac{D}{Dt}$$ is invariant, since $$\nabla'=\nabla, \qquad \frac{\partial}{\partial t'}=\frac{\partial}{\partial t}+v \cdot\nabla$$ which works if you plug it in to the definition of the material derivative, but I don't understand why we don't have $$t'=t\implies \frac{\partial}{\partial t'}=\frac{\partial}{\partial t}.$$Can anyone clarify why this is the case?

  • 2
    $\begingroup$ Hint: Use multivariable chain rule. $\endgroup$
    – Qmechanic
    Oct 30, 2017 at 14:54

1 Answer 1


There are two frames of reference here: $(x,t)$ frame in which a fluid particle is moving at speed $v$ in X-direction, and $(x',t')$ frame which is moving with the fluid particle (so that in this frame the fluid particle is at rest). Now $t'=t$ implies $dt'=dt$, i.e. change in time is the same in both frames. However it is is not obvious that $\frac{\partial}{\partial t'}$ and $\frac{\partial}{\partial t}$ must be equal, because unlike $\{dt',dt\}$ which represents change in time itself as recorded in the two frames, $\{\frac{\partial}{\partial t'},\frac{\partial}{\partial t}\}$ is an operator that represents change in some other quantity (say, temperature of the fluid particle) with change in time as recorded in the two frames. To put it plainly, $\frac{\partial}{\partial t}$ and $dt$ are conceptually different things and therefore $\frac{\partial}{\partial t}\neq (dt)^{-1}$.

As @Qmechanic suggests the way to expand partial derivatives is by using chain rule: \begin{align} x&=x'+vt',t=t'\\ \frac{\partial}{\partial t'}&=\frac{\partial t}{\partial t'}\frac{\partial}{\partial t}+\frac{\partial x}{\partial t'}\frac{\partial}{\partial x}=\frac{\partial}{\partial t}+v\frac{\partial}{\partial x} \end{align}


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