Diffusion with an even number of reflecting boundaries

1 vote
39 views a) This is the behaviour I expect from both concentrations at t= 0. Based on the theory of diffusion I would say that as t goes on, the Gaussian distributions will spread out and after a large number of collisions both concentrations will end up spreading out more or less evenly throughout the whole volume. Please let me know if this reasoning is wrong.

A doubt came to my mind as I imagine the same scenario if we had absorbing boundaries instead of reflecting ones. Therefore, does really matter the kind of boundary we are dealing with when we are trying to predict how the two concentrations (in this case) are going to spread out as time flows?

EDITED:

My initial idea about the identical scenario regarding either absorbing boundaries or reflecting ones was wrong. That is because after a long time in presence of absorbing boundaries, the concentration will end up being zero.

b) Based on what I have said before, I would expect the limiting case to be that both concentrations were equal to each other.

EDITED:

Actually I was wrong as there is only one distribution of concentration, which tends to be uniform after a long time (as explained at a)).

I have been looking for a deeper explanation and I came up with the idea that at the boundaries the current is zero:

$$j =-D\frac{\partial c(+-a/2,t)}{\partial x} = 0$$

Which means that c(x,t) has vanishing derivatives. How could this fact help to explain the limiting behaviour of the concentration after a long time?

c)
This household potential: As we are dealing with the distribution of particles in thermal equilibrium, the first thing that came to mind when I read potential was the Boltzmann equation:

$$n = ke^{-V(x)/KT}$$

EDITED:

I am coming up with the Boltzmann distribution because it can tell us the distribution of molecules. If the potential energy is known as a function of distance, then the proportion of them at different distances is given by this law. And we do know the potential, which is the square-well. So in the equilibrium concentration I should be able to check whether with the square-well potential we get:
- When $|x| \le \frac{a}{2}$ we get a constant (the value of the concentration in equilibrium between the boundaries).
- When $|x| > \frac{a}{2}$ we get zero (outside the barriers).
But how can I check this?

d)
I came across an explanation. It is not detailed and I barely understand it. Please explain it if you see that it makes sense: (I do not know why pasteboard did not work. I had to use Imgur):

https://imgur.com/a/8TLgeRe

edited Nov 17, 2018

1 vote

(a)

Your prediction is correct. With reflecting boundaries the limiting concentration tends to a uniform distribution. With absorbing boundaries the limiting concentration tends to zero.

The type of boundary makes a difference. The starting concentration is the same, the end concentration is different, so there must be an increasing difference in between the two cases. One similarity is that in both cases the final distribution is uniform : with reflecting boundaries it is a uniform non-zero value, with absorbing boundaries it is a uniform zero value.

The question asks you to explain the behaviour at the boundaries. The concentration at the reflecting boundary rises steadily to the final limiting value. I interpreted this as asking about $\frac{\partial c}{\partial t}$. However, perhaps it is asking about $\frac{\partial c}{\partial x}$, prompting you to state that $\frac{\partial c}{\partial x}=0$ as you suggested.

(b)

The limiting concentration is $N/a$, which should be marked on your sketch. It is $N/a$ because concentration in 1D is number of particles per unit length : there are $N$ particles spread uniformly over a length of $a$.

Yes that is a good point : $\frac{\partial c}{\partial x}=0$ at reflecting boundaries. This is not true for higher derivatives.

Why $\frac{\partial c}{\partial x}=0$ at a reflecting boundary

Consider first the distribution without the RH reflecting boundary. Close to any point the distribution is approximately linear, eg $c=A(\frac{a}{2}-x)+B$ near the RH boundary position $x=+\frac{a}{2}$. The reflection of this concentration in the RH boundary is $c'=A(x-\frac{a}{2})+B$. When the reflecting boundary is replaced, the concentration to the left of the RH boundary is the sum of the unreflected distribution and its reflection, ie $c'+c=2B$, which is a constant.

It is not obvious to me how this condition could explain the limiting distribution. I do not think it does explain it. However, it does fulfill the requirement for a reflecting boundary.

Why the limiting distribution is uniform

Remove the boundaries. Divide the x axis into an infinite number of intervals or boxes of length $a$ to the left and right of the box at the origin between $x= \pm a/2$.

As the initial distribution evolves it spreads into the adjacent boxes, becoming broader and flatter as well as smaller in height. The highest concentration is always in the central box, and concentration falls monotonically in both directions outside this box. After a long time the distribution is far broader than the box width $a$. In each box the concentration is approximately constant, with a small linear decrease or increase to right and left of centre respectively.

When the LH and RH reflecting boundaries of the central box are replaced, the distribution in the central box becomes the sum of the distributions in all adjacent boxes. Distributions in odd numbered boxes are reversed before they are added. Because there are approximately equal numbers of boxes with decreasing/increasing concentrations, and approximately equal numbers of odd and even boxes, the linear increasing/decreasing contributions tend to cancel out more exactly as the number of boxes increases. The constant contributions all add up to another constant function, ie a uniform distribution, which is equal in area to the area under the spreading Gaussian function.

This is an averaging process. The distribution is continuously expanded and chopped into an increasing number of slices, and these slices are added together. Every part of the distribution gets mixed with every other part. The differences, such as peaks and troughs, are averaged out.

(c)

I do not see why the author thinks that this problem is the same as drift-diffusion in a square potential well. It is not the same. There is no drift here. The peaks in concentration remain at their initial positions, they do not move.

Why drift does not affect the final distribution

If the initial concentration drifts as well as diffuses then there is no difference in the final distribution, which is again uniform. The reason is that drift does not prevent the distribution from spreading out. It continues to diffuse at the same rate. Also, the peak concentration is confined to the central box; the concentration decays to left and right of the centre just as it did without drift.

The same averaging process as above takes place : the distribution is chopped up into a larger and larger number of slices, which are added together - outer, middle and inner portions of the distribution get mixed together and lose their individual features.

The Boltzmann Equation

The Boltzmann Equation does not seem to be of any use in this problem. It relates particle numbers $n$ and energy levels. In an infinite square well the particle KE does not change because the particles cannot gain enough energy to escape. The kinetic energy of each particle is also fixed : either by definition, or because all collisions which the particle make are elastic.

(In the Random Walk model of diffusion, ideal particles change direction at random, without any cause, and keep the same speed. However, real diffusing particles only change direction because of causes such as collisions. These are either collisions with invisible particles, such as smoke particles colliding with much smaller air molecules, or collisions with each other. This leads to a distribution of speeds, such as the Maxwell-Boltzmann Distribution. However, for most purposes each diffusing particle is assumed to have the same root-mean-square speed.)

Another scenario is continuous boundaries, in which a particle which exits across the left boundary re-enters immediately across the right boundary. The approach to equilibrium is not quite the same as with reflecting boundaries, but the limiting distribution is identical.

(d)

I do not understand what this is asking for. You already have an expression for $g_n(x,t)$ from earlier. What else is there to be done here?

Why an infinite number of Gaussians is required to model the boundary conditions

The explanation you found seems to be as follows :

Start with a diffusing Gaussian distribution at $x=0$. If you place an identical 2nd Gaussian distribution at $x=+a$ then the RH boundary at the midpoint $x=+a/2$ will always satisfy the boundary condition $\frac{\partial c}{\partial x}=0$ because of symmetry. However, the bc does not hold at $x=-a/2$.

If you add a 3rd Gaussian at $x=-a$ then the LH boundary at $x=-a/2$ is not symmetrical, and neither now is the RH boundary, so the bc does not hold at either boundary.

However, adding more pairs of Gaussians either side of the centre makes the boundaries at $x=\pm a/2$ look more symmetrical. The number of Gaussians either side differs by only one. The final 'odd' Gaussian is so far away that only the tail of this Gaussian overlaps the boundary, so the effect it makes on the bc gets smaller and smaller as it gets further away from the centre.

In the limit, with an infinite comb of Gaussians, both boundaries at $x=\pm a/2$ are symmetrically placed because each has an infinite number of Gaussians on either side of it, every one with its mirror image. There can be at most one extra Gaussian on the opposite side of the centre, but this is so far away that is has an insignificant effect on the bc $\frac{\partial c}{\partial x}=0$, which will again hold.

answered Oct 29, 2018 by (28,448 points)
edited Oct 31, 2018
I updated my question. Note:

a) It is clear.
a) I worked with $\frac{\partial c}{\partial x}$. I think the graphs look quite similar whether we use $\frac{\partial c}{\partial x}$ or $\frac{\partial c}{\partial t}$ (in this case).

b) Why is $N/a$ the limiting concentration? It has to be obvious but I do not see it (thank you for your explanation about the uniformity of the limiting distribution).

c) Correct me if I am mistaken Sr but I think we are dealing with a system of non-interacting particles. Why did you come up with the statement that all collisions are elastic? Are you referring to the particles-boundaries collisions?

You are right. There is no particle with enough velocity to overcome the potential barrier. Therefore, the Boltzmann law cannot help us out here...

I am interested in what you wrote about drift and diffusion: 'if the initial concentration drifts as well as diffuses then there is no difference in the final distribution, which is again uniform'. It can explain the match between b) and c). May you delve into this?

d)Thank you  for the explanation.
How can they change direction at random without any apparent cause? Newton told us there has to be an external force exerted on the particles for triggering such a change. However, the potential inside the square well is zero. Therefore $F = - \frac{dU}{dx} = 0$. So if there is no external force, how can we explain such a phenomenon?
I have read that $\frac{\partial c}{\partial x}=0$ at a reflecting boundary because there is no current passing through it (it reflects every particle). This sound convincing but I wanted to know your thoughts about it.