1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

General thermo questions [Thermal average occupancy]

  1. Nov 5, 2009 #1
    Howdy,

    Just studying for a test, need to clear something up and I can't find it in any of my books.

    My question is in regards to [itex]N[/itex], which to me seems like it is the same as [itex]<N>[/itex] also known as the thermal average occupancy. This quantity represents the thermal average number of the orbitals in the system while in thermal and diffusive contact with a reservoir. In such a domain, we want to use the grand partition function:

    [tex]z=\sum_{ASN}e^{-\beta(N\mu-\varepsilon_s)}=\sum_{ASN}\lambda^Ne^{(-\beta\varepsilon_s)}[/tex]

    where
    [tex]\beta=\frac{1}{K_bT}[/tex], [tex]\lambda=e^{\beta\mu}[/tex]

    And the following definitions for [itex]<N>[/itex]:

    [tex]<N>=\frac{1}{z}\sum_{ASN}Ne^{-\beta(N\mu-\varepsilon_s)}[/tex]

    and

    [tex]<N>=\lambda\sum_{S}e^{-\beta\varepsilon_s}[/tex]

    My question: What is the connection between the last two equations for [itex]<N>[/itex]? Thanks yall

    IHateMayonnaise
     
  2. jcsd
  3. Nov 6, 2009 #2
    There are two ways of getting ensemble averages: you can either take an average weighted by the coefficients in the partition function


    Say, if

    [tex] Z = \sum_{\mathrm{states}} \rho, \hspace{0.5cm} \mathrm{then} \hspace{0.5cm} \langle N \rangle = \frac{1}{Z} \sum_{\mathrm{states}} N \rho. [/tex]

    The second way that I know of to calculate averages is to derivate the thermodynamic potential twice. Your second formula looks like it might be something of the sort, but I'm really not very sure.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: General thermo questions [Thermal average occupancy]
  1. Thermal average (Replies: 0)

Loading...