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The paramagnetic/ferromagnetic phase transition is an archetypal example of a continuous (or second-order) phase transition. When the temperature $T$ approaches the Curie temperature $T_c$, the magnetization $M(T)$, which is the order parameter of the transition, continuously goes to zero.

I heard (in an informal context) that under a constant nonzero magnetic field $H$, the transition becomes first order, but I was not able to find references which clearly confirm or infirm this statement, and I am not convinced of any of the possibilities.

So my question is : is the paramagnetic/ferromagnetic transition under magnetic field, $H\neq 0$, a continuous or a first order transition ? (It would be better with a reference, but I would also like an explanation.)

Thanks in advance.

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I heard (in an informal context) that under a constant nonzero magnetic field $H$, the transition becomes first order, but I was not able to find references which clearly confirm or infirm this statement, and I am not convinced of any of the possibilities.

It doesn't work like this. At temperature $T<T_c$, where $T_c$ is the critical temperature, you have a first order phase transition in $H$ when crossing the $H=0$ line. This is due to the fact that the magnetization $M$ develops a jump discontinuity at $H=0$ for $T<T_c$. You can see this in the mean-field approximation by considering the self-consistent equation for the magnetization,

$$M = \tanh(\beta (H + 2D M))$$

where $D$ is the number of spatial dimensions of the model.

This becomes clearer when visualizing the function $M(H,T)$ (picture from Introduction to the Theory of Soft Matter: From Ideal Gases to Liquid Crystals by Jonathan V. Selinger):

enter image description here

If you follow the blue line, keeping $H$ fixed at $H=0$ and changing $T$, you will encounter a transition at $T=T_c$. During this transition, the order parameter $M$ remains continuous, while its derivative with respect to temperature, $\partial M/\partial T$, changes discontinuously. Therefore, the transition is second order.

If you follow the red line, keeping $T$ fixed at $T<T_c$, you will encounter a phase transition at $H=0$: this time, $M$ has a jump discontinuity, i.e.

$$\lim_{H\to 0} \frac{\partial M}{\partial H} = \infty$$

Therefore, we are dealing with a first order phase transition.

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  • $\begingroup$ Thanks! Follow-up: a curve parallel to the blue one on the (T,H) plane, but at non-zero H would correspond to a smooth trajectory on the manifold, i.e. a crossover (as opposed to a phase transition of any order), right? $\endgroup$ – Georg Sievelson Mar 30 '18 at 22:02
  • $\begingroup$ @GeorgSievelson I am not familiar with the exact definition of crossover. Moving along the $T$ axis at nonzero $H$ you can move from small $M$ to large $M$, but only at $T \to infty$ you will have $M=0$. I am not sure wether this works as crossover, though. $\endgroup$ – valerio Apr 3 '18 at 8:58
  • $\begingroup$ Maybe the terminology is not good, it does not seem to be unambiguously defined (see physics.stackexchange.com/q/194815/3151). What I mean is that I expect that along the T axis at nonzero H, there will be no phase transition i.e. everything will be smooth (analytic). It looks true, but I may miss a subtlety. $\endgroup$ – Georg Sievelson Apr 4 '18 at 1:26
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    $\begingroup$ @GeorgSievelson Yes, there will be no phase transition in that case, absolutely. You need some discontinuity in the order parameter or in its derivatives to have a phase transition. $\endgroup$ – valerio Apr 4 '18 at 7:47
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The Landau model for ferromagnetism has the following expression for the free energy density, as a function of temperature $T$ and magnetization $M$: $$F(T,M)=F_0(T)+\dfrac{a}{2}(T-T_C)M^2+\dfrac{b}{4}M^4+\dfrac{c}{6}M^6+\mathcal{O}(M^6)$$ First order phase transition occurs when the first derivative of $F$ (namely, the entropy) is discontinuous as $T\to T_C$, which happens in the case of a non-vanishing external magnetic field, represented by $b<0$. The stable minimum of $F$, for $T<T_C$ (and $T\sim T_c$) is: $$M_0^2=\dfrac{|b|}{2c}\left(1+\sqrt{1-\dfrac{4ac}{b^2}(T-T_C)}\right)\simeq\dfrac{|b|}{c}+\dfrac{a}{|b|}(T_C-T)$$ The first order phase transition, at $T=T_C$, is thus characterized by the discontinuity $\dfrac{|b|}{c}$ in the order parameter, which is reflected in the entropy: $$S=-\left(\dfrac{\partial F}{\partial T}\right)_M=-\dfrac{dF_0}{dT}-\dfrac{a}{2}M^2$$ $$\lim_{T\to T_C}\left[S_{T<T_C}-S_{T>T_C}\right]=-\dfrac{a|b|}{2c}$$

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    $\begingroup$ I would expect a term like $- M H$ in the Landau free energy to take into account the magnetic field. Could you explain why you take $b<0$ instead (and a $M^6 term) ? $\endgroup$ – Georg Sievelson Apr 29 '15 at 22:04
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    $\begingroup$ I am worried that your answer may be inexact. If you do the same analysis with the Landau free energy for a ferromagnet found in Landau's book, you end up with a never-vanishing order parameter which is continuous at $T_c$ (and other temperatures) when $H \neq 0$ ... There may be a subtlety somewhere, but I do not see where. $\endgroup$ – Georg Sievelson May 3 '15 at 22:47
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Maybe it is too late, but look at this lecture http://www.tcm.phy.cam.ac.uk/~bds10/phase/introduction.pdf

enter image description here

Fig. 1.3 (c)

When you change the magnetic field H at constant temperature T at $b^{'}$ the system in the spin-up state. Then you move to point $c^{'}$ at phase diagram and the system jumps discontinuity at spin-down state. So you have a first order phase transition.

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