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Landau Free Energy -- Phase transistions

  1. Apr 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider the Landau free energy ##\mathcal{L}=-hm+r_1 t m^2+Cm^3+s_0 m^4## with an additional ##m^3## term. We consider zero magnetic field case ##h=0## and ##s_0>0##.
    a.) Please show that there are two critical temperatures ##t*## and ##t_1##. When ##t<t*##, a second minimal of free energy (not the one at m=0) appears, and when ##t<t_1##, the second minimal has smaller free energy than the minimal at m=0.

    b.) Comparing with the second order phase transition, please show that there is no symmetry breaking or ergodicity breaking for the first order phase transition. Why is symmetry breaking related to the second order phase transition, but not the first order phase transition?
    2. Relevant equations
    Nothing

    3. The attempt at a solution
    So the first thing I did was to find all the minimals.

    ##\frac{\partial \mathcal{L}}{\partial m}=0 \rightarrow m=-\frac{3C}{8s_0} \pm \frac{1}{2}\sqrt{(\frac{3C}{4s_0})^2-\frac{2r_1 t}{s_0}} =m_{\pm}##

    In order for these minimals to exist, we require the expression under the square root to be positive, thus we get.

    ##t<\frac{s_0}{2 r_1}\left(\frac{3C}{4s_0}\right)^2=t*##

    Thus as long as ##t<t*## there will be two additional minimals, besides the minimal at m=0. However, these minimals will be less than the minimal at m=0 until a certain value of ##t##. In order to find the value of ##t## that equates all the minimals together, we do the following.

    ##\mathcal{L}(m=m_1)<\mathcal{L}(m=0) \rightarrow r_1t m^2+Cm^3+s_0m^4<0 \rightarrow m=\frac{-C}{2s_0} \pm \frac{1}{2}\sqrt{(\frac{C}{s_0})^2-\frac{4r_1 t}{s_0}}##

    We again require the expression under the square root to be positive, this gives us,

    ##t<\frac{s_0}{4r_1}\left(\frac{C}{s_0}\right)^2=t_1##

    Thus for ##t<t_1##, the minima will be less than the minima at m=0.

    So now I believe I found the expressions for ##t_1## and ##t*##, my issue comes with dealing with the phase transitions.

    My understanding is, when the absolute minimum changes from one value at ##m=0##, to two values at ##m=m_{\pm}## suddenly, then this indicates a phase transition.

    My confusion comes into determining what exactly is the first order and second order phase transitions. I believe this is a first order phase transition because ##m_{min}## changes abruptly, and the first derivative of the free energy corresponds to ##m_{min}##, so therefore the discontinuity occurs at the first derivative of the free energy and this is a first order phase transition.

    I do not know how to find the second order phase transition, and how to determine if symmetry is broken. The only thing I can think of is to plug the values of ##m_{min}## into ##\mathcal{L}## and see if both minimums are at the same value. Any help would be appreciated!!
     
  2. jcsd
  3. Apr 29, 2015 #2

    king vitamin

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    Yes, a first-order phase transition is where [itex]m_{min}[/itex] shifts discontinuously at [itex]t^*[/itex]. In contrast, the second-order phase transition requires that [itex]m_{min}[/itex] is continuous at the critical temperature (though its derivative will not be). Since [itex]m_{min} = 0[/itex] for [itex]t > t^*[/itex], for a second-order phase transition we require that [itex]m_{min} = 0[/itex] right at [itex]t = t^*[/itex], even as approached from below. So you should be able to find the points where the transition is second order by plugging your value of [itex]t^*[/itex] into your values for the minima of the broken phase and demand that the minima be zero. These will only occur for special values of C and r1.
     
  4. Apr 29, 2015 #3
    Thank you for your reply!

    So when I plug in ##t^*## into the expression for ##m_{min}## the square root term vanishes (as it should, since the value for ##t^*## comes directly from requiring that the term under the square root is greater than zero). and I am left with...

    ##m_{min}=\frac{-3C}{8s_0}=0##

    Is this correct? This means that the only way it can be satisfied is when ##C=0##, meaning no ##m^3## term to begin with. This also tells me that at ##C\ne 0##, the symmetry is broken as there is only one minimum and it is non-zero. This makes sense as the ##m^3## term is odd and will bring about non-symmetrical properties upon sign change of ##m##.

    However, it's derivative with respect to C is continuous, a constant. So maybe I am misunderstanding something : \
     
    Last edited: Apr 29, 2015
  5. Apr 29, 2015 #4

    king vitamin

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    The mathematics looks correct (I know that you should have gotten no second-order transitions unless C=0).

    But if [itex]C \neq 0[/itex], the original potential is not symmetric under [itex]m \rightarrow -m[/itex] in the first place, so you're not breaking any symmetry by having an expectation value. In contrast, for [itex]C = 0[/itex], you do have the symmetry, but your system will choose one of the minima breaking it.
     
  6. Apr 29, 2015 #5
    So I am re-writing this problem to make sure I fully get it, and I believe I should be plugging in the value ##t_1## into the expression for the minima since this is the point in which the transition occurs. (##t^*## is the point where the other two minima emerge, but still ##m=0## is the equilibrium since it is the absolute minima). When I do this I get the following expression.

    ##m_{min}=-\frac{3C}{8s_0} \pm \frac{1}{4\sqrt{2}}\frac{|C|}{s_0}##

    Since this expression includes an absolute value (from taking the square root), the derivative of this term is discontinuous at C=0, indicating a second order phase transition at this point (and only at C=0).

    However, I do not understand why symmetry breaking is only related to the second order phase transition and I am not sure how to show that symmetry is not broken for the first order transition especially since the potential isn't symmetric in the first place unless C=0. :\

    Thank you for your help again btw! I appreciate the responses.
     
  7. Apr 29, 2015 #6

    king vitamin

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    I don't understand why you're taking a derivative with respect to C? The first/second order designation refers to the derivatives of the free energy with respect to [itex]m_{min}[/itex] - whether it changes discontinuously or its derivative does.
     
  8. Apr 29, 2015 #7
    I don't have a solid explanation, but I figured that in order to take the derivative of ##m_{min}##, it would have to be with respect to one of the changing variables. Here is why I am doing this, not sure if it makes any sense though.

    In class we did the standard example of magnetism using a free energy without a ##m^3## term and one of the expressions we got was (after taking h=0)
    ##m_{min}=\sqrt{-t}##. We then showed a plot of ##m_{min}## vs ##t## and he said that the expression ##\sqrt{-t}## is continuous, but it's derivative is not meaning it is second order. We then took ##h<0,h>0## and ##h=0## and got an ##m_{min}## that looked like a step function when plotted (the plot was ##m_{min}## vs ##h##). He then said that this expression is discontinuous, meaning it is first order.

    This seems like the same sort of problem except instead of a parameter of ##t## or ##h## it is ##C## and ##r_1##.
     
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