Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Weiss molecular field theory

  1. Oct 21, 2017 #1

    MathematicalPhysicist

    User Avatar
    Gold Member

    This question is more about the maths than the physics.

    So I am reading the textbook by Bergersen and Plischke, and they get the following:

    $$m= \tanh [ \beta (qJm+h)]$$

    where ##m## is the magnetization, ##q## is the number of nearest neighbours of site ##0##, ##J## and ##h##are the coefficients in the Hamiltonian: ##H = -J\sum_{<ij>} \sigma_i \sigma_j -h \sum_i \sigma_i##;

    For the question, they write that ##m(h,T) ## satisfies: ##m(h,T) = -m(-h,T)##;
    but I tried to showed this and I didn't succeed.

    Here's my attempt:

    $$-m(-h,T) = -\tanh [ \beta(-qJm(-h,T)-h) ] = m(-h,T)$$

    I used the fact that ##\tanh(-x) = -\tanh(x)##.

    Am I wrong?
    Did they mean something else here?

    Thanks.
     
  2. jcsd
  3. Oct 21, 2017 #2
    In your original expression for [itex] m[/itex] is seems it is a function of [itex] q,J, \beta ,h[/itex] and there is also another [itex]m[/itex] inside this expression. Why are there two m's? And I also don't see the T dependence from your statement [itex] m \rightarrow m(h,T)[/itex], either in your m expression or in the Hamiltonian, unless it's somehow derived from J.
     
  4. Oct 22, 2017 #3

    MathematicalPhysicist

    User Avatar
    Gold Member

    The ##m## in the LHS and the one in the RHS are the same or at least that's what I think is the case.

    As for the ##T## dependence, as is well known in statistical mechanics textbooks we denote by ##\beta := \frac{1}{k_B T}## where ##k_B## is Boltzmann constant and ##T## is the temperature.
     
  5. Oct 22, 2017 #4
    Its confusing to think about if it is the same m on both sides. But maybe they mean:
    [itex]
    -m(-h,T) = -\tanh(\beta[qJ(-m(-h,T))-h]) = -\tanh(-\beta[qJm(-h,T)+h])=\tanh(\beta[qJm(-h,T)+h])=m(h,T,m(-h,T))
    [/itex]

    In other words, the equality is [itex] -m(-h,T,m(-h,T))=m(h,T,m(-h,T))[/itex]
     
  6. Oct 22, 2017 #5

    MathematicalPhysicist

    User Avatar
    Gold Member

    Seems so, thanks.

    Quite recursive here.
     
  7. Oct 23, 2017 #6

    DrClaude

    User Avatar

    Staff: Mentor

    The choice of the orientation of the axis along which spin is measured is arbitrary. Therefore, the solution should be invariant to that choice of orientation. Changing ##h## to ##-h## corresponds to inverting which spin state is the ground state and which is the excited state, so the magnetization should change sign, but not amplitude.

    If ##m## is a solution of
    $$
    m= \tanh [ \beta (qJm+h)]
    $$
    then let ##\bar{m}## be the result when both ##m## and ##h## change sign:
    $$
    \begin{align*}
    \bar{m} &= \tanh [ \beta (qJ(-m)-h)] \\
    &= \tanh [ -\beta (qJm+h)] \\
    &= -m
    \end{align*}
    $$
    which is consistent with the assumption that ##-m## is a solution to the equation when the substitution ##h \rightarrow -h## is made, so ##-m(-h,T) = m(h,T)##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Weiss molecular field theory
  1. Mean field theory (Replies: 7)

  2. Molecular Orbitals (Replies: 4)

Loading...