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Negative Temperatures (Two level systems)

  1. May 31, 2007 #1

    Dox

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    Hello everybody!

    1. The problem statement, all variables and given/known data

    I'm interested in a statistical system with N particles which possible energies are, [tex] 0,\epsilon>0[/tex].

    a) Find the entropy of the system.

    b) Find the most probable [tex]n_0[/tex] and [tex]n_1[/tex], and find the mean square fluctuations of these quantities.

    c) Find the temperature as a function of [tex]U[/tex], and show it can be negative.

    d) What happens when a system of negative temperature is allowed to exchange heat with a system of positive temperature?





    2. Relevant equations

    [tex]W(n_0,n_1)=\frac{N!}{n_0!n_1!}[/tex]

    [tex]S=k\ln \Omega, [/tex] with [tex]\Omega=\sum'_{\{n\}} W(n_0,n_1).[/tex]

    [tex]\frac{1}{T}=\frac{\partial S}{\partial U}.[/tex]


    3. The attempt at a solution

    I try the following,
    [tex]N= n_0+n_1[/tex] and [tex]U=n_1 \epsilon_1-n_0\epsilon_0= n_1\epsilon_1.[/tex], so

    [tex]W=\frac{N!}{(N-n_1)! n_1!}\;\Rightarrow \; n_1^*=\frac{N}{2}.[/tex]
    That should answer the question (b).

    Meanwhile, by using the Stirling's formula,

    [tex]S=k\left(N\ln N -(N-n_1)\ln (N-n_1) - n_1\ln n_1,[/tex]
    I used [tex]N=n_1+n_2[/tex] for deriving the temperature, but I got

    [tex]\frac{1}{T}= \frac{k}{\epsilon_1}\ln\left(\frac{n_1+n_0}{n_1}\right),[/tex] which is not negative, :uhh:

    Finally, I guess that if I allow a sut up with negative temperature interact with a standard thermodynamical system the second law breaks... Am I right??
     
  2. jcsd
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