# Eulerian description

1. Dec 17, 2008

### maze

I am having some trouble understanding some things about the Eulerian description of a continuum.

Suppose we have a fluid that is continuously deforming and moving in time. If x are the spatial coordinates that the fluid is passing through, t is time, and p is a scalar function p(x,t), for example density, then for a path $\gamma(t)$ through space the chain rule gives,

$$\frac{d}{dt}\rho = \frac{\partial \rho}{\partial t} + \dot{\gamma} \cdot \nabla \rho$$

for the derivative of p in along the path $\gamma$.

So far so good. But then, multiple sources I have read make the jump to the following statement without justification: In general (not on a path),

$$\frac{d}{dt}\rho = \frac{\partial \rho}{\partial t} + v \cdot \nabla \rho$$

where v is the velocity of the fluid at that point x and time t. This seems nonsensical though. If p(x,t) really describes the exact density at a spatial point x and time t, then shouldn't we simply have $\frac{d\rho}{dt} = \frac{\partial \rho}{\partial t}$? In the Eulerian description, isn't x just a static point in space? Plus, what is the fluid velocity v doing in there?

I've been stuck on this for several hours, and I'm pretty sure there's an important but subtle point I'm missing.

2. Dec 17, 2008

### Andy Resnick

Eulerian coordinates co-move with the material. As a result, when evaluating how the material deforms or flows, the problem essentially reduces to a comparison of how the local coordinate system has deformed. Usually, to avoid confusion, authors write out the 'original' coordinates in capital letters: (X,Y,Z) and the deformed coordinates in lowercase: (x,y,z). Also, the material derivative is written as D/Dt rather than d/dt. Unfortunately, a side-effect (IMO) is entirely too many different symbols.

So, in your question- in the Eulerian description, X is static, but x is not, so d[f(x,t)]/dt must be expanded by the chain rule, resulting in the material derivative.

Does that help? It's worthwhile to spend the time understanding this.

3. Dec 17, 2008

### maze

So, ok let me see if I understand. Would it be a correct interpretation to say that the material derivative of some function p, (Dp/Dt)(x,t), measures the rate of change of p along the flow of a particular particle crossing through x at time t?

For example, suppose you drop a bunch of thermometers into a river, all over the place. Then as time progresses the thermometers will flow with the river, tracing out trajectories in space. Further, you could make a graph of temperature against time for each thermometer, as it moves around. Then pick a point in the river x and a time t, and find the closest thermometer (they are all over the place so this should be no problem (-: ). The material derivative at x,t is the slope of the closest thermometer's temperature vs time graph, at time t.

Written another way, pick some x,t. If the deformation is 1-1, then this x,t corresponds to a unique reference place X. The particle starting at X traces out a path $\gamma$ with velocity v, starting at X and passing through x at time t. We have already seen that the rate of change of p as measured by an observer traveling along a path $\gamma$ is given by,

$$\frac{D}{Dt}\rho = \frac{\partial \rho}{\partial t} + \dot{\gamma} \cdot \nabla \rho$$

but here $\dot{\gamma}=v$, so we have

$$\frac{D}{Dt}\rho = \frac{\partial \rho}{\partial t} + v \cdot \nabla \rho$$

Is this the right way of thinking?

4. Dec 17, 2008

### Andy Resnick

I think you are close. The function p varies both in space and time, say 'the weather', for example. Then the first term of
$$\frac{D}{Dt}\rho = \frac{\partial \rho}{\partial t} + v \cdot \nabla \rho$$

corresponds to you sitting on the ground and the weather changing. The second term corresponds to you getting in a car or airplane and going somewhere else.

Going with your thermometer analogy, becasue the thermometers are moving in the river, asking how T(x,t) varies means keeping track both of how x varies (x = x(X,t)) as well as t.

How's that?

5. Dec 19, 2008

### maze

Thank you, I understand now.