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A Negative T for a spin 1 system in the canonical ensemble

  1. May 20, 2017 #1
    I'm interested in an apparent inconsistency with the result for negative temperatures for a spin 1 system of N particles.

    The partition function of such a system is

    \begin{equation}
    Z=(1+2\cosh(\beta \,\epsilon))^{N}
    \end{equation}
    where each particle can be in one of three energy states: [itex]E=-\epsilon,\,0,\,\epsilon.[/itex]

    The average energy of the system is

    \begin{equation}
    \langle E \rangle=-\frac{2N\epsilon\,\sinh(\beta\,\epsilon)}{1+2\cosh(\beta\epsilon)}.
    \end{equation}

    Let me define [itex]\mathcal{E}=\langle E \rangle/N\epsilon[/itex]. If I'm to write [itex]\beta=1/k_{_{B}}T[/itex] in terms of this dimensionless energy [itex]\mathcal{E}[/itex], I'll arrive at the equation

    \begin{equation}
    \mathcal{E}+(\mathcal{E}+1)e^{\beta B}+(\mathcal{E}-1)e^{-\beta B}=0.
    \end{equation}

    I tried to solve this as a quadractic equation, obtaining

    \begin{equation}
    \beta = \frac{1}{B}\ln\left[\frac{-\mathcal{E}+\sqrt{4-3\mathcal{E}^{2}}}{2}\right].
    \end{equation}

    The first thing to observe is that we must have

    \begin{equation}
    \mathcal{E}^{2}<\frac{4}{3}
    \end{equation}

    for a real temperature to exist. Now, for a solution with negative temperature to exist, we have to have that

    \begin{equation}
    -\mathcal{E}+\sqrt{4-3\mathcal{E}^{2}}<2
    \end{equation}
    \begin{equation}
    -\mathcal{E}<1
    \end{equation}

    If [itex]\mathcal{E}>0[/itex] this relation is always satisfied. If [itex]\mathcal{E}=-|\mathcal{E}|[/itex], then the condition reads [itex]|\mathcal{E}|<1[/itex]. This is the part I don't get. If [itex]|\mathcal{E}|>1[/itex] then we have positive temperatures. But this means that

    \begin{equation}
    |\langle E \rangle| > N\epsilon
    \end{equation}

    and [itex]N\epsilon[/itex] is the energy the system has if all N particles are in identical alignment with respect to the external magnetic field. It's the maximum energy possible. How can the average temperature be larger than the maximum possible?


    Thank you very much.
     
  2. jcsd
  3. May 25, 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.
     
  4. May 26, 2017 #3
    Your equation (3) is not quadratic. The average energy depends on the temperature, so the coefficients in equ 3 are not constants.
     
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