PHZ3400-11 Problem Set 3: Difference between revisions

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''' Problem 2 Gap to low temperature excitations'''
''' Problem 2 - Gap to low temperature excitations'''


At <math>T=0</math>, our equation of state in zero external field gives fully saturated magnetization <math>m=1</math> (consider the positive solution). At very small but finite temperature, there is a very small temperature correction of the form
At <math>T=0</math>, our equation of state in zero external field gives fully saturated magnetization <math>m=1</math> (consider the positive solution). At very small but finite temperature, there is a very small temperature correction of the form


<math>\delta m(T) = 1- m(T) \propto \exp \{-E_g /T \} </math>. Determine the energy of the gap to low-temperature excitations, in terms of the interaction<math>J</math>, and the coordination number <math>z</math>. How does it compare to the energy to perform a single spin-flip from the ground state (all spins aligned)?
<math>\delta m(T) = 1- m(T) \propto \exp \{-E_g /T \} </math>. Determine the energy of the gap <math>E_g </math>to low-temperature excitations, in terms of the interaction <math>J</math>, and the coordination number <math>z</math>. How does it compare to the energy to perform a single spin-flip from the ground state (all spins aligned)?




''' Problem 2 - Hysteresis '''
''' Problem 3 - Hysteresis '''


Consider the same equation os state for <math>T = 0.5 Jz < T_c</math>. Solve numerically the equation os state starting at large field (say <math>h = 3</math>, and plot the solution as a function of <math>h</math>. You should find that the magnetization, which is close to <math>m=1</math> for large field, decreases as the field is rediced, but remains finite at <math>h=0+</math>. Continue plotting it by further reducing the field, until you reach the (negative) [http://en.wikipedia.org/wiki/Hysteresis coercive field] <math>h_{-}</math>, where the positive solution can no longer be found.  Repeat the calculation starting at very large negative fields and finding the negative magnetization solution, which can be found all the way to the (positive) coercive field <math>h_{+}</math>. In the field range between the coercive fields, there are two solutions and we find the [http://en.wikipedia.org/wiki/Hysteresis hysteresis loop], which is typically seen at any first-order transition.
Consider the same equation os state for <math>T = 0.5 Jz < T_c</math>. Solve numerically the equation os state starting at large field (say <math>h = 3</math>, and plot the solution as a function of <math>h</math>. You should find that the magnetization, which is close to <math>m=1</math> for large field, decreases as the field is rediced, but remains finite at <math>h=0+</math>. Continue plotting it by further reducing the field, until you reach the (negative) [http://en.wikipedia.org/wiki/Hysteresis coercive field] <math>h_{-}</math>, where the positive solution can no longer be found.  Repeat the calculation starting at very large negative fields and finding the negative magnetization solution, which can be found all the way to the (positive) coercive field <math>h_{+}</math>. In the field range between the coercive fields, there are two solutions and we find the [http://en.wikipedia.org/wiki/Hysteresis hysteresis loop], which is typically seen at any first-order transition.

Latest revision as of 16:49, 13 February 2011

Problem 1 - Critical Behavior of a Ferromagnet

a) Show that the mean-field equation of state for a ferromagnetic Ising model is given by (use units where ; ):

.

b) Consider , and solve this equation numerically (using Mathematica, MAPLE, or FORTRAN), and plot . At , you will find three solutions. Pick the one with positive magnetization and plot it as a function of temperature. Show the printout of your computer plot as a part of the homework.

c) By expanding the RHS in powers of , to cubic order, solve analytically for the magnetization close to . Plot the expression you obtain on the same graph as the full numerical solution (dashed lines). The two curves should agree only close to

d) Now choose a small but finite external magnetic field . Solve the above equation numerically, again picking the positive solution. You should find that transition is "rounded" and the magnetization remains finite at any temperature.

e) Now chose , and expand the above equation to linear order in , and to cubic order in . Solving for as a function of , show that at the critical point

, where the critical exponent .


Problem 2 - Gap to low temperature excitations

At , our equation of state in zero external field gives fully saturated magnetization (consider the positive solution). At very small but finite temperature, there is a very small temperature correction of the form

. Determine the energy of the gap to low-temperature excitations, in terms of the interaction , and the coordination number . How does it compare to the energy to perform a single spin-flip from the ground state (all spins aligned)?


Problem 3 - Hysteresis

Consider the same equation os state for . Solve numerically the equation os state starting at large field (say , and plot the solution as a function of . You should find that the magnetization, which is close to for large field, decreases as the field is rediced, but remains finite at . Continue plotting it by further reducing the field, until you reach the (negative) coercive field , where the positive solution can no longer be found. Repeat the calculation starting at very large negative fields and finding the negative magnetization solution, which can be found all the way to the (positive) coercive field . In the field range between the coercive fields, there are two solutions and we find the hysteresis loop, which is typically seen at any first-order transition.