Negative Temperatures (Two level systems)

1. May 31, 2007

Dox

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, $$0,\epsilon>0$$.

a) Find the entropy of the system.

b) Find the most probable $$n_0$$ and $$n_1$$, and find the mean square fluctuations of these quantities.

c) Find the temperature as a function of $$U$$, 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

$$W(n_0,n_1)=\frac{N!}{n_0!n_1!}$$

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

$$\frac{1}{T}=\frac{\partial S}{\partial U}.$$

3. The attempt at a solution

I try the following,
$$N= n_0+n_1$$ and $$U=n_1 \epsilon_1-n_0\epsilon_0= n_1\epsilon_1.$$, so

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

Meanwhile, by using the Stirling's formula,

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

$$\frac{1}{T}= \frac{k}{\epsilon_1}\ln\left(\frac{n_1+n_0}{n_1}\right),$$ 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??