5
$\begingroup$

I'm studying fluid mechanics, and I got the impression that the material derivative is nothing more than "differentiating along a path" and so I got confused on why do we need it. Basically, let $D\subset \mathbb{R}^3$ be the region containing the fluid and let $f : D\times \mathbb{R}\to \mathbb{R}$ be a time dependent function on $D$.

Suppose then $\gamma : I\subset \mathbb{R}\to D$ is a trajectory on the fluid. The material derivative of $f$ along $\gamma$ is defined by

$$\dfrac{D}{Dt}f(\gamma(t),t) = \dfrac{\partial f}{\partial t}(\gamma(t),t) + (\mathbf{u}\cdot \nabla)f(\gamma(t),t)$$

Where $\mathbf{u}$ is the spatial velocity field of the fluid. But that expression is nothing more nothing less than simply differentiating $f(\gamma(t),t)$ with respect to $t$, or better, differentiating the function $f\circ (\gamma, I)$ where $I$ is the interval in $\mathbb{R}$.

Indeed we have

$$\dfrac{d}{dt}f(\gamma(t),t) = \nabla f(\gamma(t),t)\cdot \gamma'(t) + \dfrac{\partial f}{\partial t}(\gamma(t),t) = \dfrac{D}{Dt} f(\gamma(t),t)$$

So since $\dfrac{d}{dt}= \dfrac{D}{Dt}$ why do we need the material derivative? Why do we define it, since in truth it is just the well know derivative of a composition just? What are the advantages of defining it?

EDIT: I think I got it now. When computing $\dfrac{d}{dt}$ as I said a composition is needed, in other words, we need a path. But since all information is contained in $\mathbf{u}$ we can dispose the path by defining:

$$\dfrac{D}{Dt} f(a,t) = \dfrac{\partial f}{\partial t}(a,t) + (\mathbf{u}\cdot \nabla) f(a,t)$$

And that coincides with $\dfrac{d}{dt}$ if $a = \gamma(t_0)$ for some $\gamma$ and some $t_0$. Is that really the point we are making when defining the material derivative? Is the same we would get if we had a path going through there with velocity $\mathbf{u}$ but we dispose the path since $\mathbf{u}$ already contains all needed info.

$\endgroup$
1
  • 2
    $\begingroup$ Depending on how you are solving the equations of motion, one form is preferred over the other. If you model the flow as discrete particles (Lagrangian formulation) then you solve the material derivative (and it's a coupled set of ODE's in time) whereas if you solve on a fixed domain (Eulerian formulation), you solve the second form. Some problems are better in one than the other. $\endgroup$
    – tpg2114
    Aug 30 '14 at 18:13
4
$\begingroup$

You've pretty much answered your own question; you've motivated it and derived the expression, just with a different notation.

To rephrase what you've said: the material derivative is the derivative along the path defined by the integral curves of $\vec{u}$. This is useful because this is the path taken by a small element of the fluid.

As a motivating example: imagine a fluid which is incompressible, so any given element has constant density, but is a mixture of things with different densities. This condition is summed up by $\dfrac{D\rho}{Dt}=0.$

$\endgroup$

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.