We have an expression f(x)dx =aexp(-ax)dx(adsbygoogle = window.adsbygoogle || []).push({});

where x is the free path, and 1/a is the mean free path. f(x)dx gives us the probability that a particle's free path lies between x and x+dx. The formula is derived by assuming that the probability of a collision occuring in a distance dx= adx. Now assume that the probability of NO collision occuring in a distance x is p(x). Therefore:

p(x+dx) = p(x)*(1-adx)

Also, p(x+dx) = p(x) +(dp/dx)dx

If we equate these we find that dp/dx = -ap

Therefore p(x)=Aexp(-ax)

But the probability of a particle not colliding in 0 distance is 1, therefore A=1. Therefore p(x) = exp(-ax). Our expression gives us the probability that a particle has not collided in distance x. Therefore to find the probability that it collides in interval (x, x+dx), we use p(x)*adx = aexp(-ax)dx = f(x)dx

n.b the free path is defined as being the distance a particle travels before its first collision.

We were then asked to find "the most probable value of the free path". Therefore we can look at the max. value of p(x), or f(x) which is a constant multiple of p(x). This gives us the max value at x=0, which makes good physical sense, as we earlier defined p(x) as being the prob. of no collision occurring in distance x, and if x =0, then no collision can occur, so p(0)=1. however, we also defined the free path as being the distance a particle travelled before it did in fact definitely collide, so if x=0 is most probable value, it would mean that most particles collided immediatley. These are obviously two irreconcilable conclusions that we are drawing, and I can't see which one of them is wrong and why. It probably has something to do with ambiguous definitions of the free path. I would be grateful if somebody could point out where I was going wrong.

Thanks

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# Homework Help: Free path ambiguity

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