The Material Derivative in Cartesian Coordinates

Recently, we've had some controversy about some terms in a momentum equation in one of the papers that one of my colleagues is studying. Indeed, for compressible and non constant viscosity fluids, it is not straightforward to come up with the equations of motion as this situation is rarely the case in our daily engineering investigations.

As the problems we are dealing become more complicated (in fact, we add the complication as we solve the problem one step at a time. For example, we start with an inviscid incompressible model, then we add viscosity, then we add compressibility to the inviscid model, then add viscosity to the compressible model etc…).

So anyway, I decided to once and for all settle down the issue and put together a convenient and easy to understand post.

## The Material Derivative

First, let us start by going over the material derivative one more time. As you most probably know, there are two reference frames that can be used in studying fluid motion; namely, the Lagrangian and Eulerian frames.

In the Lagrangian description, each particle in the fluid is followed as if it were a "rigid body". This means that each particle has a unique identifier or tag. This description is akin to rigid body dynamics and is a great way of describing a relatively small number of particles. However, when dealing with a fluid, there is a very large number of particles thus rendering a Lagrangian description of a fluid flow problem very tedious and intractable.

In the Eulerian description, flow properties (velocity, pressure, temperature, etc…) are defined as a function of space and time. This means that instead of tagging each particle of the flow, the viewer fixes a volume in space and identifies the flow properties in that region of space. Therefore, it does not matter which particle passes through the volume since that particle will assume the flow property of that point in space.

The natural way of defining the velocity or acceleration is based on a Lagrangian description because it is the easiest (and that's how we've done it in dynamics in high school)! When it comes to fluid flow, the acceleration of a fluid particle as seen in the Eulerian reference is different from the Lagrangian description. This is due to the fact that as the particle moves about, its velocity and position change as well. In Mathematical terms, the particle velocity is

(1)
\begin{align} \mathbf{V}\left[\mathbf{r}(t), t \right] \equiv \mathbf{V}\left[x(t),y(t),z(t), t \right] \end{align}

of course, the velocity is a vector

(2)
\begin{align} \mathbf{V}\left[\mathbf{r}(t), t \right] \equiv u\left[\mathbf{r}(t), t \right]\mathbf{i} + v\left[\mathbf{r}(t), t \right]\mathbf{j} + w\left[\mathbf{r}(t), t \right]\mathbf{k} \end{align}

Now the acceleration is obtained by differentiating the velocity with respect to time. But since the velocity is a function of the position, and the position is a function of time, then we have to use implicit or chain rule differentiation as follows

(3)
\begin{align} \frac{{\rm d}\mathbf{V}}{{\rm d}t} = \frac{\partial \mathbf{V}}{\partial t} + \frac{\partial \mathbf{V}}{\partial x}\frac{{\rm d}x}{{\rm d} t} + \frac{\partial \mathbf{V}}{\partial y}\frac{{\rm d}y}{{\rm d} t} + \frac{\partial \mathbf{V}}{\partial z}\frac{{\rm d}z}{{\rm d} t} \end{align}

which yields the following expression for the acceleration of ANY fluid particle as seen by an observer in an Eulerian reference.

(4)
\begin{align} \frac{{\rm d}\mathbf{V}}{{\rm d}t} = \frac{\partial \mathbf{V}}{\partial t} + u \frac{\partial \mathbf{V}}{\partial x} + v \frac{\partial \mathbf{V}}{\partial y} + w \frac{\partial \mathbf{V}}{\partial z} \end{align}

If we were in a Lagrangian frame, then the velocity is only a function of time because in that reference, the observer is riding on the particle as it flows through space.

page revision: 3, last edited: 06 Dec 2010 06:09