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We have a system that has N identical spins $n_i$, and each spin can be in state 1 or 0. The overall energy for the system is $\epsilon\sum_{i=1}^{N}n_i$.

My understanding: There is only one distinct way in which we could have 3 spins in state 1, because the spins are identical. Likewise for 4,5 or 6... spins in state 1, there will only ever be one way of choosing. So I think that the multiplicity is $N$, therefore entropy $S(E,N)=k_blog(N)$. I need to calculate the $E(T,N)$, I know $(\frac{\partial S}{\partial E})_{V,N}=1/T$ but my entropy doesn't have an E in it. How can I calculate the energy and is what Ive got so far correct?

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1 Answer 1

You can calculate $\langle E \rangle$ explicitly, through

$$\langle E \rangle = \frac{1}{Z} \sum_{\text{configurations }i} E_i \exp\left\{ -\frac{1}{T}E_i\right\}.$$ Here, $E_i$ is the energy of configuration $i$, i.e. $E_i = \epsilon \sum_j \ldots.$ In your attempt, you're focusing on the thermodynamics of the problem, but in statistical physics, that's seldom useful. You know the distribution of energy as a function of temperature and $N$ here -- at least, you know in principe how to calculate it.

If you're stuck, I can give a hint, but the calculation itself simplifies a lot if you think a little bit.

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I'm slightly confused, because the question has steps: Give an example microstate, how many distinct microstates are there for each energy level, calculate the multiplicity and hence the energy. Do we have to calculate the average energy because energy is not fixed and would that just be $N(N+1)\epsilon/2N = (N+1)\epsilon /2$ and is the $Z$ in the term you gave me the multiplicity? –  Tins Apr 21 '13 at 9:41
    
$Z$ is the partition function. You shouldn't think of $E$ as 'fixed' or not - it's a statistical observable, which depends on $T$ and $N$. The latter are the only variables in the problem. And you shouldn't give the average energy: as you see from my expression, each microstate is weighted by a factor $\exp(-E_i/T),$ so depending on $T$, some states contribute more than others. If you're stuck, you should go through the steps of your problem one by one in your question, and we can tell you if you make mistakes. –  Vibert Apr 21 '13 at 9:46
    
So far I have an example microstate would be (1,0,1,0,...,0) just any vector of N 1s and 0s. The number of distinct microstates for a fixed energy is 1 because our particles are identical. The multiplicity is the total number of microstates, which is $1+1+...+1=N+1$ Is this right so far? –  Tins Apr 21 '13 at 10:06
    
No, you need to be more careful. First you classify all possible energy levels, which are $E = 0, \epsilon, 2\epsilon, \ldots, N \epsilon.$ Then you pick one energy level, say $E = n \epsilon$ with $n < N.$ The question is then: how many microstates have energy $n \varepsilon$? If you find the general question too difficult, start with $n = 1$, $n=2$ and work your way up until you see the pattern. –  Vibert Apr 21 '13 at 10:11
    
I think that the number microstates with energy $n\epsilon=NCn$, so total number of microstates would be $2^n$, but isn't this the case for distinguishable spins, our spins are identical so would you have to divide by some factor? –  Tins Apr 21 '13 at 10:43

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