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Particle Statistics

  1. Dec 14, 2011 #1
    Hallo, I posted this in General Math, and I decided to post it here also because this room seems more appropriate. The formulas and part of the text are quoted from "Klimontovich - Statistical theory of non-equilibrium processes in a plasma":

    Let [itex]N_{a}(\textbf{x},t) =\Sigma_{i=1,N_{a}}\delta(\textbf{x}-\textbf{x}_{ai})[/itex] be the phase density of particles of species [itex]a[/itex] and [itex]f_{N}[/itex] the distribution function of the coordinates and momenta of the all [itex]N=\Sigma_{a} N_{a}[/itex] particles of the system respectively.

    The statistical average of [itex]N_{a}[/itex] is then

    [itex]\overline{N_{a}( \textbf{x},t )}[/itex]=[itex]\int\sum_{i=1,N_{a}}\delta(\textbf{x}-\textbf{x}_{ai})f_{N}
    \prod_{a}d^{6}\textbf{x}_{a1}...d^{6}\textbf{x}_{a_{N_{a}}}[/itex]

    and since all the particles of one kind are identical

    =[itex] N_{a} \int\delta(\textbf{x}-\textbf{x}_{a1})f_{N}
    \prod_{a}d^{6}\textbf{x}_{a1}...d^{6}\textbf{x}_{a_{N_{a}}}[/itex]

    If we define

    [itex]f_{a}(\textbf{x}_{a1},t)=V \int f_{N}d^{6}\textbf{x}_{a2}...d^{6}\textbf{x}_{a_{N_{a}}}
    \prod_{b\neq a}d^{6}\textbf{x}_{b1}...d^{6}\textbf{x}_{b_{N_{b}}}[/itex] where [itex]V[/itex] is the volume of the particle, then we can write

    [itex]\overline{N_{a}}( \textbf{x},t ) = n_{a} f_{a}(\textbf{x},t)[/itex] where [itex]n_{a}[/itex] is the mean concentration of particles of the kind [itex]a[/itex]

    Up to here everything seems ok. He now tries to connect the mean values of the products of the phase densities [itex]N_{a},N_{b}[/itex] in the following way, where my problems come:

    Splitting the double sum

    [itex]\Sigma_{i=1,N_{a}}\Sigma_{j=1,N_{b}} \delta(\textbf{x}-\textbf{x}_{ai}) \delta(\textbf{x}'-\textbf{x}_{bj}) [/itex]

    into the two parts (why???)

    [itex]\Sigma_{i=1,N_{a}}\Sigma_{j=1,N_{b}}\delta(\textbf{x}-\textbf{x}_{ai})\delta(\textbf{x}'-\textbf{x}_{bj}) [/itex]

    (for xaixbj when a=b)

    +

    [itex] \delta_{ab}\Sigma_{j=1,N_{a}} \delta(\textbf{x}-\textbf{x}_{ai}) \delta(\textbf{x}-\textbf{x}')[/itex]

    we obtain, neglecting unity when compared with [itex]N_{a}[/itex] (when do we compare unity with [itex]N_{a}[/itex] ???)

    [itex]\overline{N_{a}( \textbf{x},t )N_{b}( \textbf{x}',t)}=n_{a}n_{b}f_{ab} ( \textbf{x},\textbf{x}',t)+\delta_{ab}n_{a}\delta( \textbf{x}-\textbf{x}')f_{a}(\textbf{x},t) [/itex]

    where [itex]f_{ab}(\textbf{x}_{1a},\textbf{x}_{1b},t)=V^{2} \int f_{N}d^{6}\textbf{x}_{a2}...d^{6}\textbf{x}_{a_{N_{a}}}d^{6}\textbf{x}_{b2}...d^{6}\textbf{x}_{b_{N_{b}}}\prod_{c \neq a,b}d^{6}\textbf{x}_{c1}...d^{6}\textbf{x}_{c_{N_{c}}}[/itex]

    So, please, can anyone explain me the logic behind this?


    Thank you very much in advance,
    Kaniello
     
  2. jcsd
  3. Dec 24, 2011 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    It would be hard for a person only familiar with mathematics to interpret this excerpt from a book on physics. For example, I don't know the the [itex] \delta [/itex] is an indicator function, a Dirac [itex] \delta [/itex], or something else and I certainly don't know what [itex] \delta_{ab} [/itex] or x' represents. You'll probably get a better answer by posting in a section of the forum that deals with statistical physics.

    If you want help from a mathematician, I suggest that you give a link that explains the physics that is going on. For example, one link I found about "phase density" is the PDF http://www.google.com/url?sa=t&rct=...sg=AFQjCNFYEck6SnQDMfBDreU8TobxNNzY4A&cad=rja
     
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