# Diffusion interpetation

1. Jan 26, 2010

### mordechai9

Something has been bothering me recently regarding a diffusion problem. Consider a source of particles in a box, diffusing outwards. Set the boundary conditions so that the density is zero at the edge of the box.

We can solve this problem in "steady state" and we find essentially a parabolic (or x^2) dependence of the density inside the box, dropping away from the source. However, since the density is zero at the boundaries of the box, the mass flux there is zero (rho*v*A). What's going on here? How can we have a steady state problem with mass clearly leaving the box without any mass flow actually going out of the box?

Does this simply mean that the mass interpretation is incorrect for this example?

2. Jan 26, 2010

### Pythagorean

I don't see any evidence that mass is "clearly leaving the box" from what you've said here. Would you extrapolate on where this information comes from? You said the flux there is zero, so nothing is passing through the boundaries.

3. Jan 26, 2010

### mordechai9

Ah, well, you have a "source" inside the box. Thus mass is being created (or injected) into the box, and unless it's flowing out, you couldn't have a steady state problem.

4. Jan 26, 2010

### Pythagorean

In diffusion problems, I always took source to mean where the material is all contained, dense to a point in the box. Not that material is actually being injected into the box, just that that's the point the atoms are diffusing from.

So if I spray perfume in one corner of the room and then leave the room and close the door, the place where I sprayed the perfume is the source. No additional perfume is being injected, it's just diffusing from that point until it is evenly distributed throughout the room.

5. Jan 26, 2010

### mordechai9

Right, well, that can't be steady state though -- the perfume would just diffuse to fill the room, and then you wouldn't have a source term in the equation any more. The source term has to act like a steady input of material for this problem to be steady state.

6. Jan 27, 2010

### Mapes

Whoa! Just because some curve has a value of zero at some point doesn't mean its slope is also zero (and the slope corresponds to the magnitude of flux).

7. Jan 27, 2010

### mordechai9

No... this is incorrect. The mass flux is given by rho*V*A, and it is the value of the density (rho) that shows up, not the derivative (d(rho)/dx).

8. Jan 27, 2010

### Pythagorean

I believe you're using Fick's First Law, which assumes steady-state:

So yeah, you're saying that phi (in this equation) is zero, but that doesn't mean that d(phi)/dx is. You have to solve the differential equation for phi still and get it out of that differential form. Then you can apply your boundary conditions.

9. Jan 27, 2010

### Mapes

The mass flux into any infinitesimal element is $\rho AV$, but the mass flux out is $\rho AV+\frac{\partial}{\partial x}(\rho AV)dx$. $\rho$ is not uniform. Even though it is zero at the boundary, it is nonzero an infinitesimal distance away, and thus the mass flux is nonzero. Does this make sense?

10. Jan 27, 2010

### mordechai9

Pythagorean - Fick's law doesn't assume steady state... You get steady state just by assuming all time derivatives are equal to zero.

Mapes - Actually, the mass flow out is just $$\int \rho V \cdot da$$, where you integrate over the surface of the box. We're not talking about the flow into and out of a differential element, we're talking about the flow into/out of a closed surface, and that's all. If rho is zero at the boundary, you don't have flow in or out, end of story.

11. Jan 27, 2010

### Mapes

So when you ignore our interpretation and use your own interpretation, you find a paradox, as you pointed out in your first post. What does that tell you?

EDIT: Or, looking at it another way, what does the outward velocity need to be outside the box to maintain the density of particles at the outer edge of the box at zero? To get arbitrarily close to zero density, the velocity needs to grow very large. The product is nonzero. The flux just inside and just outside the box is equal and nonzero.

Last edited: Jan 27, 2010