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Mean field approximation and entropy

  1. Apr 12, 2016 #1

    CAF123

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    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. jcsd
  3. Apr 12, 2016 #2

    Charles Link

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    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.
     
  4. Apr 13, 2016 #3

    CAF123

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    Hi Charles Link,
    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!
     
  5. Apr 13, 2016 #4

    Charles Link

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    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
  6. Apr 13, 2016 #5

    CAF123

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    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?
     
  7. Apr 13, 2016 #6

    Charles Link

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    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.
     
  8. Apr 13, 2016 #7

    CAF123

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    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.
     
  9. Apr 13, 2016 #8

    TSny

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    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.
     
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