1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear algebra application: entropy

  1. Oct 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider a linear chain of N atoms. Each atom can be in 3 states (A,B,C) but an atom is state A cannot be next to an atom in state C. Find the entropy per atom as N approaches infinity.

    Accomplish this by defining the 3-vector [tex] \vec{v}^{j} [/tex] to be the number of allowed configurations of the j-atom chain ending in type A, B, C. Then show that [tex] \vec{v}^{j} = \textbf{M}\vec{v}^{j-1}[/tex]. Then [tex] \vec{v}^{j} = \textbf{M}^{j-1}\vec{v}^{1}[/tex]. Show that in the limit of large N, the entropy per atom is dominated by the largest eigenvalue of M, and is given by [tex] k ln(1 + \sqrt{2})[/tex].


    3. The attempt at a solution

    For the first j-atom chains, it is evident that

    [tex] \vec{v}^{1} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} [/tex], [tex] \vec{v}^{2} = \begin{bmatrix} 2 \\ 3 \\ 2 \end{bmatrix} [/tex], [tex] \vec{v}^{3} = \begin{bmatrix} 5 \\ 7 \\ 5 \end{bmatrix} [/tex]

    which implies that

    [tex] \textbf{M} = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} [/tex]

    Right now I am having trouble with the first part: show that [tex] \vec{v}^{j} = \textbf{M}\vec{v}^{j-1}[/tex]. It is easy to show for specific cases using the vectors I have determined above, but I am confused on how to generalize this relation.
     
  2. jcsd
  3. Oct 15, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok, let v(n)_A, v(n)_B and v(n)_C be number of n atom chains ending in A, B and C (the three components of your column vectors). Then to get v(n+1)_A you take any n atom chain ending in A or B (not C) and add an A. So v(n+1)_A=v(n)_A+v(n)_B. Now do v(n+1)_B and v(n+1)_C. Aren't those linear equations the same as v(n+1)=Mv(n)?
     
  4. Oct 15, 2008 #3
    Okay, so let

    [tex] \vec{v}^{j-1} = \begin{bmatrix} v^{j-1}_{A} \\ v^{j-1}_{B} \\ v^{j-1}_{C} \end{bmatrix} [/tex]

    where [tex] v^{j-1}_{A} [/tex] is the number of configurations that in end in A for j-1 atoms, and the same for B and C. Then,

    [tex] \vec{v}^{j} = \textbf{M}\vec{v}^{j-1} = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix} v^{j-1}_{A} \\ v^{j-1}_{B} \\ v^{j-1}_{C} \end{bmatrix}= \begin{bmatrix} v^{j-1}_{A} + v^{j-1}_{B} \\ v^{j-1}_{A} + v^{j-1}_{B} + v^{j-1}_{C} \\ v^{j-1}_{B} + v^{j-1}_{C} \end{bmatrix} [/tex]
     
  5. Oct 15, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure. Doesn't that express the condition "A cannot be next to C"?
     
  6. Oct 15, 2008 #5
    Thanks for helping me out of my stupor. That was ridiculously easy. I found the eigenvalues 1, 1+sqrt(2), and 1-sqrt(2). I used these to construct the diagonal matrix D given by

    [tex] \textbf{D} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 + \sqrt{2} & 0 \\ 0 & 0 & 1 - \sqrt{2} \end{bmatrix} [/tex]

    Then I used the basis vectors of the eigenspace to construct the orthogonal matrix Q:

    [tex] \textbf{Q} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{2} & \frac{1}{2} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{2} & \frac{1}{2} \end{bmatrix} [/tex]

    So,

    [tex] \textbf{M}^{N} = \textbf{Q}^{T}\textbf{D}^{N}\textbf{Q} [/tex]

    For large N, I can ignore the eigenvalues 1 and 1 - sqrt{2}, and construct [tex] \textbf{M}^{N} [/tex] from the large eigenvalue 1 + sqrt{2}, which shows that the entropy per atom is dominated by the largest eigenvalue, right? But how does this show that [tex] W \approx 1 + \sqrt{2} [/tex]?
     
  7. Oct 15, 2008 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    What's your definition of entropy for this system?
     
  8. Oct 15, 2008 #7
    S = k ln W

    where k is boltzmann's constant.
     
  9. Oct 15, 2008 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok. What's W?
     
  10. Oct 15, 2008 #9
    It's the total number of configurations, which is the sum of the components of the vector [tex] \vec{v}^{j} [/tex]. But the sum of these components does not necessarily equal the eigenvalue. If I recall from earlier, it's off exactly by a factor of 3/2. Does this really matter for large N? This factor results in an overall difference of about 30% for the value the logarithm term compared with the given approximation.
     
    Last edited: Oct 15, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Linear algebra application: entropy
  1. Linear Algebra (Replies: 1)

Loading...