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Free path ambiguity

  1. Oct 30, 2005 #1
    We have an expression f(x)dx =aexp(-ax)dx
    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.

  2. jcsd
  3. Oct 30, 2005 #2

    Physics Monkey

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    If the probability that a particle travels a distance in (x,x+dx) before colliding is [tex] p(x) dx = a e^{-a x} dx [/tex], then the most probable value is x=0. Also, the mean free path is defined as [tex]
    \int^\infty_0 a x e^{-ax} dx = 1/a
    as it must be. But to say "most particles collide immediately" needs more qualification. What is the number of particles that travel a distance between less than or equal to the mean free path? The answer is just
    N(x\leq 1/a) = N \int^{1/a}_0 a e^{-ax} dx = N (1 - e^{-1}) = .632 N
    thus only somewhat more than half have a free path less than the mean free path. It doesn't therefore make sense to say that most of the particles collided almost immediately, in fact almost no particles "collide immediately". You can easily check, for instance, that if you define "colliding immediately" as having a free path less than one thousandth of the mean free path, then only about 1 in a thousand particles "collide immediately".
    Last edited: Oct 30, 2005
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