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2d Ising. Finite lattice.

  1. Aug 25, 2012 #1
    http://books.google.rs/books?id=vrc...nepage&q=Nolting Finite Ising lattice&f=false

    A finite lattice [tex]X[/tex] with so constructed boundary condition that [tex]M_s(X;T)\neq 0[/tex]

    boundary condition - all spins in the boundary are up, in Ising model [tex]S_i=1, \forall i \in \partial X[/tex]

    Wall - line that separates + and - sites.

    Two probabilities
    1) [tex]\omega_i(T)[/tex] - probability that at temperature [tex]T[/tex] site [tex]i[/tex] is occupied by spin -
    2) [tex]W_{\Gamma}[/tex] - probability that at temperature [tex]T[/tex] polygon [tex]\Gamma[/tex] exists.

    Can you tell me exactly what they suppose by polygon? There is a picture in page 247. How many polygons is on this picture?

    Also can you explain me estimation (6.54) on page 248.
     
  2. jcsd
  3. Aug 25, 2012 #2
    A polygon is nothing but a closed loop. The line forming this loop is what they call the “wall.” In Figure 6.9 there are three polygons, one encircling a single negative site, one encircling a single positive site, and one big one encircling the collection of negative sites.


    Now, consider the example Figure 6.9 where we have a [itex]6 × 6[/itex] finite lattice. Since we have fixed a boundary condition of only positive spins, there are [itex]25 (= 5 × 5)[/itex] variables, or [itex]2^{25}[/itex] possible configurations.

    Consider the site [itex]i=(x,y)=(1,1)[/itex] (with the origin [itex]= (0,0)[/itex] at the bottom left corner). In the configuration shown in Figure 6.9 this site is negative. Consider a configuration where only this site is negative and the rest of the sites are all positive. Then you would have only one polygon. Say that this configuration arises a quarter of the times out of all the possible configurations. And imagine that there exist a quarter configurations where site [itex](1,2)[/itex] will be also be negative in addition to the one at [itex](1,1)[/itex]. In this other set of configurations you would have a bigger polygon encompassing [itex](1,1)[/itex] and [itex](1,2)[/itex]. Say that the remaining half configurations have all sites positive. Now, we know that the probability [itex]w_i(T)[/itex] for [itex]i=(1,1)[/itex] is 0.5. Out of all the configurations we can have two types of polygons (say) [itex]Γ_1[/itex] and [itex]Γ_2[/itex]. And in the above example it is obvious that [itex]W_{Γ_1}(T)=0.25[/itex] and [itex]W_{Γ_2}(T)=0.25[/itex]. In this particular example the equality in Equation (6.54) would hold.

    Now, let's talk about the inequality. Consider the site [itex]i=(3,3)[/itex]. In the configuration shown in Figure 6.9 this site is positive. Therefore, this configuration is excluded while computing the probability [itex]w_i(T)[/itex], but not while computing the probability [itex]W_Γ(T)[/itex]. In some other configuration site [itex]i[/itex] will be negative; in this case this configuration is included while computing both [itex]w_i(T)[/itex] and [itex]W_Γ(T)[/itex]. I hope it is, at least qualitatively, clear why the inequality exists in the direction that is shown in Equation (6.54). If you want, you can work out a numerical example similar to the one in the previous paragraph to demonstrate this inequality.
     
    Last edited: Aug 25, 2012
  4. Aug 25, 2012 #3

    Mute

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    Homework Helper

    You can put equations in-line (like this: ##\mathcal H = \sum_{ij} J_{ij} s_i s_j##) by enclosing your latex code as shown below

    Code (Text):

    This equation, ##\mathcal H = \sum_{ij} J_{ij} s_i s_j##, would appear as an inline equation  if it weren't in the "code" environment. It would also appear inline if I did this [itex]\mathcal H = \sum_{ij} J_{ij} s_i s_j[/itex]
     
     
  5. Aug 26, 2012 #4
    Tnx a lot. I was a bit confused because they use wall and polygon, and I didn't see the difference.

    I don't understand the bold part. How can I suppose that?
     
  6. Aug 26, 2012 #5
    That was just a hypothetical example to demonstrate Equation (6.54). When you want to compute an ensemble average, you take a weighted average of all possible configurations. Now, these weights (or probabilities) are a function of temperature. Another way of stating my assumption is that I “picked” a temperature where the conditions that I described above hold. It is, however, possible (and likely) that if you fully analyze this spin system you may not find any temperature where this is true. In that case the computation of [itex]W_\Gamma(T)[/itex] and [itex]w_i(T)[/itex] is not going to be as trivial. You are going to have to account for many different possibilities; only the combinatorics gets more complicated. But the type of analysis I showed in my example can be easily extended to this more complicated system.
     
  7. Aug 26, 2012 #6
    Ok. Tnx a lot for your help. I have just one more question. This ##|\Gamma|## (tnx Mute :)) is lenght of ##\Gamma## is that like correlation lenght? In book is defined like wall units. I don't understand that. What that means? Suppose we build ##\Gamma## from ##2m## vertical and ##2n## horisontal units. ##|\Gamma|## is then equal to what and why? Can I have, for example ##1## vertical, and ##1## horisontal unit?
     
  8. Aug 26, 2012 #7
    [itex]|\Gamma|[/itex] is the perimeter of the polygon. In the example in Figure 6.9, [itex]|\Gamma|[/itex] for the two small polygons is 4, and that for the large polygon is 14. This is measured in units of the lattice constant; or maybe it is more appropriate to say the lattice constant of the "dual" lattice (in this example they are the same, and so it doesn't matter). Say you have two adjacent and aligned spins, and one of the spin flips, then a "wall" is created between those spin sites. This represents a "gapped" excitation in your system, where the value of that gap is [itex]2J[/itex]. But as described in the text you cannot just have an isolated segment of the wall sitting there. The curve has to close in on itself and form a loop (or polygon). Therefore the energy of the system, above the ground state (which is all positive spins), is [itex]2J|\Gamma|[/itex].
     
  9. Aug 26, 2012 #8
    How you get those numbers ##4,4,14?##
     
  10. Aug 26, 2012 #9
    The number of [itex](+)[/itex]-[itex](-)[/itex] links across the wall
     
  11. Aug 26, 2012 #10
    If I watch small polygons. ##+/-## signs have 4 nearest neighbour ##-/+## sign. They don't have the same spins across the wall. So do I watch nearest neighbours? So for big polygon I have 13 spins nn with oposite sign and one + spin which is in small polygon so ##13+1=14##. Is that logic?
     
  12. Aug 26, 2012 #11
    No, in the big polygon there are 14 nn spins. I think you missed one. Note that the spin at (1,4) has 3 links. Are you sure you didn't count only 2 links for that one?
     
  13. Aug 26, 2012 #12
    Maybe I don't understand what do you think when you say links.
    If ##(0,0)## is left boton then spin (-) is in ##(1,1)##. What is 4 links for him?

    ##(0,0) - (row,column)##

    So links of ##(1,1)## are ##(1,1)\leftrightarrow(1,0), (1,1)\leftrightarrow(0,1),(1,1)\leftrightarrow(1,2),(1,1)\leftrightarrow(2,2)##
    Right?

    Links of ##(3,3)## are ##(3,3)\leftrightarrow(3,4), (3,3)\leftrightarrow(3,2),(3,3)\leftrightarrow(4,3),(3,3)\leftrightarrow(2,3)##
    Right?

    For big polygon links are
    ##(2,2)\leftrightarrow(2,1), (2,2)\leftrightarrow(1,2),(3,2)\leftrightarrow(3,1),(4,1)\leftrightarrow(4,0),(4,1)\leftrightarrow(5,1),(4,2)\leftrightarrow(5,2),(4,3)\leftrightarrow(5,3),(4,4)\leftrightarrow(5,4)...##

    Also when this lattice and dual lattice wouldn't be the same.
     
  14. Aug 26, 2012 #13
    The last one is incorrect (but, I think you made a typo). It should be ##(1,1)\leftrightarrow(2,1)##

    Yes, this is right.

    ##(2,2)\leftrightarrow(2,1), (2,2)\leftrightarrow(1,2),(3,2)\leftrightarrow(3,1),(4,2)\leftrightarrow(4,1),(4,2)\leftrightarrow(5,2),(4,3)\leftrightarrow(5,3),(4,4)\leftrightarrow(5,4),(4,4)\leftrightarrow(4,5),## ##(3,4)\leftrightarrow(3,5),(2,4)\leftrightarrow(2,5),(1,4)\leftrightarrow(1,5),(1,4)\leftrightarrow(0,4),(1,4)\leftrightarrow(1,3),## ##(2,3)\leftrightarrow(1,3)##

    One example: triangular and honeycomb lattices are dual to each other.
     
  15. Aug 26, 2012 #14
    Yes. Mistake in writing. I'n bit tired. Tnx a lot for your help. Tomorrow I'm gonna read again text from the book.
     
  16. Sep 3, 2012 #15
    I read this again what tejas777 wrote and I still have a problem with relation
    [tex]\omega_i(T)\leq \sum_{\Gamma\supset i}W_{\Gamma}(T)[/tex]
    ##\omega_i## is probability that spin in site ##i## is down. ##W_{\Gamma}## is probability that site ##i## is in polygon ##\Gamma##. Can you show inequality in rigoruos way?
    And one more question. Why spontaneous magnetisation of finite lattice without boundary condition must be equal to zero.
     
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