Statistical Mechanics: classical Heisenberg Modelby eXorikos Tags: classical, heisenberg, mechanics, model, statistical 

#1
Jul2711, 12:34 PM

P: 249

1. The problem statement, all variables and given/known data
You have a latice of particles that all have spin 1, but they can change the direction of their spin so constraint [itex]\leftS_j\right=1[/itex]. There is only interaction with the closest neighbours so we have the following hamiltonian: [itex]H = J \sum_{\left\langle ij \right\rangle} \vec{S_i} \cdot\vec{S_j}  \vec{h} \cdot \sum^{N}_{j = 1} \vec{\vec{S_j}}[/itex] Choose a good orderparameter to treat this in the molecular field approximation. Calculate the selfconsistent equation for this order parameter and determine the spontaneous magnetisation for [itex]T<T_c=Jq/3k_b[/itex]. 2. Relevant equations [itex]Z=\int_{\leftS_1\right=1}\cdots \int_{\leftS_N\right=1} d^3 S_1 \cdots d^3 S_N \exp{\left(\beta H\right)}[/itex] [itex]M = \frac{1}{\beta} \nabla_h ln Z[/itex] 3. The attempt at a solution As order parameter I pick [itex]\vec{M} = \sum_j \vec{S_j}[/itex] and than I approximate the hamiltonian with q nearest neighbors by [itex]H = \frac{Jq}{2N} \left(\sum^N_{j=1} \vec{S_j}\right)^2  \vec{h} \cdot \sum^{N}_{j = 1} \vec{\vec{S_j}}[/itex] This gives [itex]Z=\int_{\leftS_1\right=1}\cdots \int_{\leftS_N\right=1} d^3 S_1 \cdots d^3 S_N \exp{\left(\frac{\beta Jq}{2N} \left(\sum^N_{j=1} \vec{S_j}\right)^2 + \beta \vec{h} \cdot \sum^{N}_{j = 1} \vec{\vec{S_j}} \right)}[/itex] But I can't manage the integral. How do I calculate this integral? The rest I presume is correct? 



#2
Aug211, 12:41 PM

P: 249

Anybody?




#3
Aug1711, 08:41 AM

P: 249

*Bump*




#4
Aug2211, 10:53 PM

HW Helper
P: 1,391

Statistical Mechanics: classical Heisenberg Model[tex]Z=\int_{\leftS_1\right=1}\cdots \int_{\leftS_N\right=1} d^3 S_1 \cdots d^3 S_N \exp\left[\left(\frac{\beta Jq\vec{M}}{2N} + \beta \vec{h}\right) \cdot \sum^N_{j=1} \vec{S_j}\right].[/tex] Treating M as independent of the S_i, the different sites are no longer coupled and you can now perform the integrals. If you then compute M by [itex]M = \beta^{1}\nabla_h \ln Z[/itex], you will find the right hand side has M in it  this is your selfconsistent equation for M. (Well, you have an equation for each component of M). Analyze the selfconsistent equation(s) and show that at some temperature there is a phase transition (i.e., a new solution to the self consistent equation appears). 



#5
Sep211, 12:21 PM

P: 249

Am I on the right track trying to use spherical coordinates or is there another way of calculating the integral more easy? Because then I can keep R=1 and integrate both angles respectively from 0 to pi and 0 to 2pi.




#6
Sep311, 03:43 AM

P: 249

Set [tex]\vec{h}^{eff}=\frac{\beta Jq\vec{M}}{2N} + \beta \vec{h}[/tex] [tex]Z=\int_{\leftS_1\right=1}\cdots \int_{\leftS_N\right=1} d^3 S_1 \cdots d^3 S_N \exp\left[\vec{h}^{eff} \cdot \sum^N_{j=1} \vec{S_j}\right][/tex] [tex]= \prod_{j=1}^{N} \int_{\leftS_j\right=1} d^3S_jexp\left(\vec{h}^{eff} \cdot \vec{S_j}\right)[/tex] Now trying to calculate it seems easier to work with spherical coordinates because of the condition that [itex]\leftS_j\right=1[/itex], but I'm guessing there is some kind of technique I don't know about for calculating this integral. Because doing the dot product in spherical coordinates will make life hard. 



#7
Sep411, 01:27 PM

HW Helper
P: 1,391

[tex]Z_j= \int_0^\pi d\theta \int_0^{2\pi} d\phi~\sin^2\theta \exp\left[ h_x^{eff}\cos\phi\sin\theta + h_y^{eff}\sin\phi\sin\theta + h_z^{eff}\cos\theta\right][/tex] This integral, however, is still rather hard to do. I think what may be easier to due is use [itex]h^{eff} \cdot S = h^{eff}S\cos\gamma = h^{eff}\cos\gamma[/itex], where [itex]\gamma[/itex] is the angle between the vector [itex]h^{eff}[/itex] and [itex]S[/itex]. [itex]\gamma[/itex] runs from 0 to [itex]\pi[/itex]. There's still another angle [itex]\phi[/itex] which runs over 0 to 2*Pi, but the integrand no longer depends on it so you just get a factor of 2*Pi. Now, I'm afraid I don't remember the Jacobian off the top of my head, so you'll have to work that out. [tex]Z_j= 2\pi \int_0^{\pi} d\gamma~\mathcal J(\gamma) \exp\left[h^{eff}\cos\gamma\right][/tex] I don't know if the resulting integral will be doable. It may be expressible in terms of some sort of Bessel function (or a modified Bessel function). 



#8
Sep411, 03:58 PM

P: 249

That's all way to complicated to solve in an exam of 4 ours together with two other questions. This was a question of a previous exam so it shouldn't be so convoluted. Thanks for your help!




#9
Sep611, 11:32 AM

HW Helper
P: 1,391

If I were in an exam situation and I just wanted to find the critical temperature, I would probably try assuming that there is a disordered phase in which the total magnetization was zero, and then choose [itex]\mathbf{h} = h_z \hat{z}[/itex]. Near the transition the z component of the magnetization is probably the one that would acquire a nonzero value, so the integral you would have to do is
[tex]Z_j= 2\pi \int_0^{\pi} d\theta~\sin^2\theta \exp\left[h^{eff}_z\cos\theta\right][/tex] This apparently can be evaluated in terms of a modified Bessel function, but unless you've memorized that it probably won't be too handy on an exam. So, what I would do now is take a derivative with respect to h_z. This brings down a factor of cos(theta) and a beta, so you get [tex]M_z = 2\pi \int_0^{\pi} d\theta~\sin^2\theta \cos\theta\exp\left[h^{eff}_z\cos\theta\right],[/tex] which still isn't doable. However, again, you just want the transition, so we can do two things: 1) set h_z = 0 now. From the Ising model, you should have some intuition that there is no transition for finite h. The transition you're looking for is a zerofield transition. 2) Assume M_z is small and expand the exponential. You'll find you have to expand the exponential to 3rd order in M_z, as the zeroth and second order terms vanish, and you can't have just the first order term or you'd have M = stuff*M. Doing the expansion (quickly  you'll have to check my work) yields something that looks like [tex]M_z = \frac{2\pi\beta Jq}{2N}I_1 M_z + 2\pi\left(\frac{\beta Jq}{2N}\right)^3I_2 M_z^3[/tex] where the I's are integrals that you could do because they're just trigonometric. Anywho, we don't even care about I2. Cancelling a factor of M_z you get something like stuff*M_z^2 = 1  other stuff. Now, the 1 other stuff obviously has to be positive, otherwise you have no solution other than M_z = 0 (which we divided out). So, solve (1  other stuff) = 0 for T to find the transition temperature. I'm don't think I got the exact result you quoted in your first post, but it'll be close, and for a timeconstrained test it would hopefully get you most of the points for the problem. By the way, I might redefine your order parameter as [itex]M = (1/N)\sum_j S_j[/itex]. That might get rid of N in your equations. 


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