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I Motion of a particle from its momentum distribution

  1. Jan 9, 2018 #1
    Hi all,

    I recently learnt the concept of Maxwell's speed distribution and became interested in how to use similar momentum distributions to study the probabilistic motion of a classical free particle. I have done some of my own reading on probabilities and distributions (no formal lessons yet) and with some help, tried to derive a probability density for a classical particle. Could someone look through my work and point out mistakes in my understanding?

    The classical equation of motion for a free particle is (##m = 1##),
    $$X = x_0 + pt$$
    where ##x_0## and ##pt## are random variables.
    Let ##F## denote Cumulative Distribution Function and ##f## for Probability Density
    $$F_X = \int_{-\infty}^{\infty} F_{pt} (a - y) f_{x_0}(y)dy$$
    $$f_X = \int_{-\infty}^{\infty} f_{pt} (a - y) f_{x_0}(y)dy$$
    Since this is a classical particle we talk about, then ##f_{x_0}(y) = \delta (y - x_0)## since it has ##P = 1## of being at ##x_0##
    This gives,
    $$f_X = f_{pt} (a - x_0) $$
    If I know the momentum probability density ##f_p##, I could use the fact that
    $$F_p = P(p<a) = \int_{-\infty}^{a} f_p(b) db$$
    $$F_{pt} = P(pt<a) = P(p<\frac{a}{t}) = \int_{-\infty}^{\frac{a}{t}} f_p(c) dc$$
    and do a quick change of variables, compare integrands (both with limits ##-\infty## to ##a##) that would give,
    $$f_{pt}(b) = \frac{1}{t} f_p(b/t)$$
    Finally,
    $$f_X = f_{pt} (a - x_0) = \frac{1}{t} f_p( \frac{a - x_0}{t})$$
    Which gives me a probability density of a free particle , ##f_X##

    Does the above work? Many thanks in advance!
     
  2. jcsd
  3. Jan 15, 2018 at 11:25 AM #2

    Stephen Tashi

    User Avatar
    Science Advisor

    What does it mean to say "##pt##" is a random variable?

    If we treat ##pt## as a single random variable with a distribution independent of time then as time passes the distribution of ##p## would have to vary in order to keep the distribution of the product ##pt## the same. Is that the situation you want to model?

    It amounts to saying that the position of the particle (at any given time whatsoever) has the same distribution but the distribution of its velocity (and momentum) changes with respect to time in a special way.

    (I'm assuming you don't wish to treat time ##t## as a random variable.)



    I interpret this to mean:
    ##F_X(a) = \int_{-\infty}^{\infty} F_{pt}(a-y) f_{x0}(y) dy ##
    ##f_X(a) = \int_{-\infty}^{\infty} f_{pt} (a - y) f_{x_0}(y)dy##

    which makes sense if the distribtion of ##pt## is independent of time and the disribution of velocity is not, but that strikes me as a strange assumption for physical process.


    What does "##x_0##" denote? Previously, you said ##x_0## denoted a random variable, but now you are treating ##x_0## as some constant.

    Perhaps you are expressing the thought that if ##f_X(x_0)## is the probability density for the particle being at a position ##x_0## then it is also the probability density for the particle really being at ##x_0##. It is unnecessary to say this. That is implicit in defining the probability distribution ##f_X(x)## for position.
     
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