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Maxwell-Boltzmann Equation (avg velocity)

  1. Oct 23, 2015 #1
    A question asked me to derive a symbolic expression for mean particle speed using the Max-Boltz equation. I know that Max-Boltz equation is a function of velocity (v).
    The Max-Boltz equation is ##f=(\frac{m}{2\pi kT})^{3/2}4\pi v^2 exp(\frac{-mv^2}{2kT})##
    Apparently the general formula for the average given a statistical function is ##\bar{v}=\int_{0}^{\infty}\frac{fvdv}{n}##
    Here is what I don't understand:

    1) Where did this formula come from? Does this formula only apply to statistical functions? What is a statistical function?
    2) It turns out that division by the number of particles (n) is unnecessary for the Max-Boltz equation. What is reasoning behind this?
    3) Why is the integration from zero to infinity? Clearly no particle can have infinite velocity...

    Thank you again!
     
  2. jcsd
  3. Oct 23, 2015 #2

    DrClaude

    User Avatar

    Staff: Mentor

    Consider a random variable with discrete values (such as a die). The expectation value is given by summing over the probability of each event times its value:
    $$
    \begin{align}
    \langle x \rangle &= \sum_i P_i x_i \\
    &= \frac{1}{6} 1 + \frac{1}{6} 2 + \frac{1}{6} 3 + \frac{1}{6} 4 + \frac{1}{6} 5 + \frac{1}{6} 6 = 3.5
    \end{align}
    $$
    The first line above is the generic equation, the second line is the specific example of a six-sided die. When the random variable is continuous, the sum becomes an integral:
    $$
    \langle x \rangle = \int f(x) x \, dx
    $$
    where ##f(x)## is the probability density function (pdf), i.e., the probability that the random variable will have a value between ##x## and ##x + dx##.

    I guess it depends on how the pdf is defined. Normally, the MB distribution will give you the speed pdf per particle, so there is no factor 1/n.

    This is a non-relativistic theory. Speed is not bounded, so you have to integrate up to infinity.
     
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