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Entropy of ensemble of two level systems

  1. Aug 5, 2017 #1
    1. The problem statement, all variables and given/known data
    The fundamental equation of a system of [itex]\tidle{N}[/itex] atoms each of which can exist an atomic state with energy [itex]e_u[/itex] or in atomic state [itex]e_d[/itex] (and in no other state) is
    [tex]
    F= - \tilde{N} k_B T \log ( e^{-\beta e_u} + e^{-\beta e_d} )
    [/tex]
    Here [itex]k_B[/itex] is Boltzmann's constant [itex]\beta = 1/k_BT[/itex]. Show that the fundamental equation of this system, in entropy representation, is
    [tex]
    S = NR \log \left (\frac{1+Y^{e_d/e_u}}{Y^Y} \right)
    [/tex]
    where
    [tex]
    Y= \frac{U-\tilde{N}e_u}{\tilde{N}e_d-U}
    [/tex]

    Hint: introduce [itex] \beta =1/k_BT[/itex] and show that [itex]U = F + \beta \partial F / \partial \beta = \partial ( \beta F) /\partial \beta [/itex]. Also, for definiteness, assume [itex] e_u < e_d[/itex] and note that [itex]\tilde{N}k_B = NR[/itex].


    2. Relevant equations
    [tex]
    S = -\partial F / \partial T = -(\partial F /\partial \beta )(\partial \beta /\partial T) = (\partial F / \partial \beta)( \beta /T)
    [/tex]

    3. The attempt at a solution
    I showed the hint but will not write out the solution here. Then we have
    [tex]
    U = \partial ( \beta F) /\partial \beta = \tilde{N} \frac{e_u e^{-\beta e_u} + e_d e^{-\beta e_d}}{ e^{-\beta e_u} + e^{-\beta e_d}}[/tex]
    from which I get [itex]
    e^{\beta( e_u - e_d)} = Y [/itex].

    Also,
    [tex]
    S = \tilde{N}k_B \log ( e^{-\beta e_u} + e^{-\beta e_d} ) +\tilde{N} \beta k_B \frac{e_u e^{-\beta e_u} + e_d e^{-\beta e_d}}{ e^{-\beta e_u} + e^{-\beta e_d}}
    [/tex]
    which I rewrote as
    [tex]
    NR\log (1+Y) - NR\beta e_u + NR \beta \frac{e_u +e_d Y}{1+Y} = NR \log (1+Y)+ NR \beta \frac{Y(e_d-e_u)}{1+Y} = NR \log (1+Y) - NR \frac{Y}{Y+1} \log Y = NR \log \left ( \frac{1+Y}{Y^{Y/Y+1}} \right)
    [/tex]
    which is not (at least obviously) the same, but I cannot see my mistake. Help would be appreciated.
     
  2. jcsd
  3. Aug 10, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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