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In Problem 8.15 you manually computed the energy of a particular state of a 4 x 4 square lattice. Repeat that computation, but this time apply periodic boundary conditions.

Short Answer

Expert verified

Therefore, the energy is 4J.

Step by step solution

01

Given information

In Problem 8.15 you manually computed the energy of a particular state of a 4 x 4 square lattice. Apply periodic boundary conditions.

02

Explanation

In the Ising model, a 4x4 square lattice has 16 sites. When periodic boundary conditions are applied, the energy of a certain state is given by:

Where spins are equal to:

Skν=+1Skμ=-1Siν=Siμ

Energy is equal to:

role="math" localid="1647835907601" E=-J·i,jSiν·Skν+J·i,jSiν·Skμ=-J·(-2)+J·(+2)E=4J

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Most popular questions from this chapter

Consider an Ising model of 100 elementary dipoles. Suppose you wish to calculate the partition function for this system, using a computer that can compute one billion terms of the partition function per second. How long must you wait for the answer?

At T = 0, equation 8.50 says that s¯=1. Work out the first temperature-dependent correction to this value, in the limit βn1. Compare to the low-temperature behaviour of a real ferromagnet, treated in Problem 7.64.

Modify the ising program to compute the average energy of the system over all iterations. To do this, first add code to the initialise subroutine compute the initial energy of the lattice; then, whenever a dipole is flipped, change the energy variable by the appropriate amount. When computing the average energy, be sure to average over all iterations, not just those iterations in which a dipole is actually flipped (why?). Run the program for a 5 x 5 lattice for T values from 4 down to l in reasonably small intervals, then plot the average energy as a function of T. Also plot the heat capacity. Use at least 1000 iterations per dipole for each run, preferably more. If your computer is fast enough, repeat for a 10x 10 lattice and for a 20 x 20 lattice. Discuss the results. (Hint: Rather than starting over at each temperature with a random initial state, you can save time by starting with the final state generated at the previous, nearby temperature. For the larger lattices you may wish to save time by considering only a smaller temperature interval, perhaps from 3 down to 1.5.)

In this problem you will use the mean field approximation to analyse the behaviour of the Ising model near the critical point.

(a) Prove that, when x1,tanhxx-13x3

(b) Use the result of part (a) to find an expression for the magnetisation of the Ising model, in the mean field approximation, when T is very close to the critical temperature. You should find MTc-Tβ¯,whereβ(not to be confused with 1/kT) is a critical exponent, analogous to the f defined for a fluid in Problem 5.55. Onsager's exact solution shows that β=1/8in two dimensions, while experiments and more sophisticated approximations show that β1/3in three dimensions. The mean field approximation, however, predicts a larger value.

(c) The magnetic susceptibility χis defined as χ(M/B)T. The behaviour of this quantity near the critical point is conventionally written as χT-Tc-γ , where y is another critical exponent. Find the value of in the mean field approximation, and show that it does not depend on whether T is slightly above or slightly below Te. (The exact value of y in two dimensions turns out to be 7/4, while in three dimensions γ1.24.)

Consider an Ising model in the presence of an external magnetic field B, which gives each dipole an additional energy of -μBB if it points up and +μBB if it points down (whereμB is the dipole's magnetic moment). Analyse this system using the mean field approximation to find the analogue of equation 8.50. Study the solutions of the equation graphically, and discuss the magnetisation of this system as a function of both the external field strength and the temperature. Sketch the region in the T-B plane for which the equation has three solutions.

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