Question

A particle of dust is bombarded by air molecules and follows a zigzag path at constant speed v. Assuming each step has a length d.... What is the length of the displacement vector after N steps where N is large? Assume that each step is taken in a random direction on the plane. A particle of dust is bombarded by air molecules and follows a zigzag path at constant speed v. Assuming each step has a length d.... What is the length of the displacement vector after N steps where N is large? Assume that each step is taken in a random direction on the plane. @Physics

Answer 1

You probably mean the average displacement, because the actual displacement in any one instance varies depending on the exact pattern of bombardment.

The average DISPLACEMENT, meaning the vector from the starting position to the ending position, is zero. You can easily understand this by noting that the system is perfectly symmetric.

The mean squared displacement, however, meaning the distance from the starting point to the ending point, is equal to d times the square root of N.

Google "random walks in 2 dimensions" for more information. This is an extensively studied field, so you may have to wade through a lot of sophisticated math you don't want.

The probability distribution of distances from the start to the end is a Gaussian.

Answer 2

Awesome info, thanks!

Answer 3

Is there any simple way to find the mean squared displacement using vectors for example?

Answer 4

Sure. The mean squared displacement <r^2> = <x^2 + y^2>, where r is the vector from the start to the end, x and y are its x and y components, and the angle brackets (< and >) mean "take the average overall many patterns of bombardment."

Because the system is symmetric, we can see that <x^2> = <y^2>, so we can focus in on <x^2>, the mean squared displacement along the x axis. Now, the total displacement along the x axis is the result of N steps. Let us call the displacement that happens in step i x_i. Then:

<x^2> = <(x_1 + x_2 + x_3 + .... x_N)^2>

Multiplying out the term in parens, we get:

= <x_1^2 + x_1 x_2 + .... + x_2^2 + x_2 x_1 + .... + x_N^2 >

That is, we get N terms where the displacement at a given step (say step 1, or 5, or N) is squared, plus lots of other terms where we average the displacement at one step multiplied by the displacement at another.

Because what happens at one step is not correlated with what happens at another, we can write terms like this the following way:

<x_1 x_2> = <x_1><x_2>

That is, in this case, the average of the displacement in step 1 times the displacement in step 2 is equal to the average displacement in step 1 time s the average displacement in step 2. We can only do this when the two quantities have no connection to each other, are uncorrelated. (It's like saying the average of your grade multiplied by somebody else's grade equals the product of your average grades -- if and only if your grades are not correlated, e.g. you do not study together.)

Now we can get down to brass tacks. <x_1> is zero, because at step 1 the displacement is just as likely to be left (in the minus x direction) as right (in the plus x direction). So it averages to zero. <x_2> is also zero, and so on, so *every* term that has a product of the displacement at one time multiplied by the displacement at another is zero.

We're left with N terms like <x_1^2> + <x_2^2> .... All these terms are identical, because the mean squared displacement at step 6 has to be just the same as it is at step 1. So we only need to calculate the mean square displacement at one step. Let's call that sx^2. What this tells us is that the mean square x displacement is N sx^2.

Now we do the same for the mean square y displacement, and we get a similar result, it is equal to N sy^2, where sy^2 is the mean square y displacement in one step. So the mean square vector displacement <r^2> = N (sx^2 + sy^2>.

Now we just use the fact that by definition sx^2 + sy^2 = d^2, where d is the size of the step. On each step, the length of the step is fixed, all that can vary is the direction. That means the mean square x step size plus the mean square y step size has to equal the mean square step size, which is d^2.

End result, <r^2> = N d^2.

That may strike you as a lot of complicated reasoning, and it is, but unfortunately the direct math way to prove the statement is even worse. You have to start by writing down all the possible paths. If you do that, you'll find that the probability distributio for the length of the path follows the binomial distribution, and you can then generalize that, for a very large number of steps, to the Gaussian probability distribution. That involves some mildly tricky math. Then you can directly calculate the mean square displacement, but that involves a little calculus.