Gabriel Maia
- 70
- 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.
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.