# Mean field approximation and entropy

1. Apr 12, 2016

### CAF123

1. The problem statement, all variables and given/known data
Consider a D dimensional Ising model with N sites, defined by the Hamiltonian $$\mathcal H = -J \sum_{\langle i j \rangle} \sigma_i \sigma_j - h \sum_i \sigma_i$$ where the sum extends over nearest neighbours and each spin variable $\sigma_i = \pm 1$. For a given spin configuration we denote the number of spins up and spins down by $N_+$ and $N_-$ respectively. The magnetisation is defined by $m=(N_+ - N_-)/N$

a) Using Stirling's approximation, show that the entropy of the system can be written as $$S/N = \log 2 - \frac{1}{2} (1+m) \log(1+m) - \frac{1}{2} (1-m) \log(1-m)$$ and that in the mean field approximation $$f(T,m) = -\frac{1}{2} Jzm^2 + \frac{1}{2} T((1+m) \log(1+m) - (1-m) \log(1-m)) - T\log 2$$

2. Relevant equations
$$S = \ln \Omega,$$ in units of $k_B=1$

3. The attempt at a solution
There are N sites and on each site, the spin can be up or down. So isn't the total number of spin configurations (or number of microstates) just $2^N$? This would give $S = N \log 2$, the first term in that expansion but I don't see how the other terms come about.

Thanks!

2. Apr 12, 2016

This doesn't seem like an introductory homework, but in any case I got the first part to work by using $\Omega=N!/((N_+)!(N_-)!)$ . The number of states is apparently given by the number of possible combinations that can create a given $m$. It's an exercise in algebra, but their answer for S is correct.

3. Apr 13, 2016

### CAF123

Ah ok. I tried this expression for $\Omega$ and I did not seem to get the correct result - in particular where would the factor of log 2 come from?

Thanks!

4. Apr 13, 2016

The factor of $\ln 2$ (it's actually $N \ln 2$ ) is what is left over after a lot of terms cancel. If I remember correctly $N_+=N(1+m)/2$ and $N_-=N(1-m)/2$ There are several places where you use $N=N_+ + N_-$ and when you take $\ln(\Omega)$ , Stirling's formula is used 3 times. (twice in the denominator). If you are careful with the algebra, I think you will get the same result. (I think the $N \ln(2)$ comes from the 1/2's in the denominators of $N_+$ and $N_-$, but I'd need to look again at my calculations to be sure.)

Last edited: Apr 13, 2016
5. Apr 13, 2016

### CAF123

I see. thanks! The only thing I don't understand is why would the number of states be counted in this way? If I just count the number of spin configurations then the total number of states would be $2^N$ if I can distinguish between the spins. Then this would yield an entropy with only the first term present. Any ideas?

6. Apr 13, 2016

I had Statistical and Thermal physics (we used the book by F. Reif), quite a number of years ago. If the spins are in a lattice, I think your way of counting them might be appropriate, but I am certainly no expert. I picked the method of the combinations simply because it then resulted in the book's answer.

7. Apr 13, 2016

### CAF123

I see, thanks anyway! ok maybe we can await an answer from someone else, @TSny maybe? I thought my spins were on a lattice no? I imagined the Ising model in D dim to be like a hypercubic with spins placed on the nodes of the cubes.

8. Apr 13, 2016

### TSny

The spins are on a lattice.

2N represents the total number of possible micro states. You want the entropy corresponding to a particular value of magnetization m. Only a subset of the 2N possible micro states have the magnetization m. Charles gave you an expression for the number of micro states Ω in the subset. If you use his expression you should get the desired expression for S/N corresponding to a magnetization m.