0
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ Hint: Use multivariable chain rule. $\endgroup$
    – Qmechanic
    Oct 30 '17 at 14:54
5
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.