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Maxwell-Boltzmann Distribution and Probability

  1. Mar 18, 2014 #1
    In the book 'Macroscopic and Statistical Thermodynamics' they derived the Maxwell-Boltzmann distribution by maximizing entropy using lagrangian multipliers with constants ##\alpha## and ##\beta##.

    The final result is given as:

    [tex] \frac{\overline {n_j}}{g_j} = e^{-\beta \epsilon_j}e^{-\alpha}[/tex]

    where ##\overline {n_j}## is the occupation number and ##g_j## is the number of states of jth energy level.

    After solving for ##e^{-\alpha} = \frac{N}{V}\left(\frac{h^2}{2\pi mkT}\right)^{\frac{3}{2}}## and integrating density of states to find ##g_j = \frac{V}{h^3} 4\pi p^2 dp##:

    We obtain the maxwell-boltzmann distribution:

    [tex]\overline {n_j} = \overline {n_{(p)}} dp = N 4\pi \left(\frac{\beta}{2\pi m}\right)^{\frac{3}{2}} p^2 e^{\frac{-p^2}{2mkT}} dp[/tex]

    I obtain the correct speed distribution ##\propto p^2 e^{\frac{-p^2}{2mkT}}##, but What is the probability/fraction of finding a particle with momentum p?



    In Blundell's Book, a shorter approach is taken using gibbs' expression for entropy to find the boltzmann probability:

    20jqmtx.png

    Here's the earlier reference to equation (4.13):

    mt5ls7.png
     
    Last edited: Mar 18, 2014
  2. jcsd
  3. Mar 19, 2014 #2
    Since the momentum p is a continuous quantity, looking for the probability of finding a particle with exactly p is pretty meaningless. One usually says that, if [itex]f(p)[/itex] is the Maxwell-Boltzman distribution then the probability of find a particle with momentum between p and p+dp is:
    $$
    P_{[p,p+dp]}=f(p)dp.
    $$
    For a general interval [itex]p\in[p_1,p_2][/itex] the probability is simply:
    $$
    P_{[p_1,p_2]}=\int_{p_1}^{p_2}f(p)dp.
    $$
     
  4. Mar 19, 2014 #3
    Yes I get all that, but what is ##f_{(p)}## from ##\frac{\overline {n_j}}{g_j} = e^{-\beta \epsilon_j}e^{-\alpha}##?
     
  5. Mar 19, 2014 #4
    Well, I suppose that [itex]\bar{n}_j=\bar{n}(p)dp[/itex] is the average number of particles with momentum between p and p+dp. As you can see from your expression it is equal to the total number of particles [itex]N[/itex] times something. By definition, that "something" is your distribution [itex]f(p)[/itex].
     
  6. Mar 19, 2014 #5
    Just divide by N.
    [tex]f=\frac n N[/tex]
     
  7. Mar 19, 2014 #6
    Ok, so the momentum distribution is:

    [tex]\overline {n_j} = \overline {n}_{(\vec {p})} d^3p = N \left(\frac{\beta}{2\pi m}\right)^{\frac{3}{2}} e^{\frac{-p^2}{2mkT}} d^3\vec{p}[/tex]

    By changing the dp's to dv's, thus the velocity distribution is:

    [tex]\overline {n_j} = \overline {n}_{(\vec {v})} d^3v = N \left(\frac{\beta m}{2\pi }\right)^{\frac{3}{2}} e^{\frac{-p^2}{2mkT}} d^3\vec{v}[/tex]

    For a particular direction ##v_x##, simply just take one part of 'dv':

    [tex]\overline {n_j} = \overline {n}_{(\vec {v_x})} dv_x = N \left(\frac{\beta m}{2\pi }\right)^{\frac{1}{2}} e^{\frac{-p_x^2}{2mkT}} d\vec{v}_x[/tex]

    With ##f_{(\vec{v})} = \frac{n_j}{N} = \left(\frac{\beta m}{2\pi}\right)^{\frac{3}{2}} e^{\frac{-p^2}{2mkT}}## and

    ##f_{(\vec{v_x})} = \left(\frac{\beta m}{2\pi}\right)^{\frac{1}{2}} e^{\frac{-p_x^2}{2mkT}}##

    To change from velocity to speed distribution simply expand the ##d^3\vec{p}## to ##4\pi p^2 dp## or ##d\vec{v_x} = 4\pi v^2_x dv_x##.
     
    Last edited: Mar 19, 2014
  8. Mar 19, 2014 #7
    At a first glance I would say that all the factors are correct. So, yes!
     
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