Are the first order phase transitions always associated with a latent heat?

Is the first order ferromagnetic transition below the critical temperature associated with latent heat?

For example, the transition of ferromagnetic configuration with all its spins aligned up to a ferromagnetic configuration with all its spins aligned down at $T=T_F<T_c$, when the magnetic field changes from $H\to -H$, is a first order transition. Is this also associated with latent heat?

If yes, how do we calculate it? $Q=T_F\Delta S$, but I think, $\Delta S=0$ because both up-aligned and down-aligned configurations have same entropy. I'm not quite sure that $\Delta S=0$ because if $\Delta S=0$ how can it be discontinuous? But for this transition to be first order, entropy should be discontinuous according to the Ehrenfest criterion.

• Seems it's a rather unusual system to consider for phase transitions. Are these two 'phases', the initial spin aligned up phase and the final spin aligned phase, even truly different 'phases'? They appear to be essentially identical to each other except for the direction of their oriented spins. And if we were to regard them as being different phases, then it seems that one could argue that there are an infinite number of phases in this spin system since each and every different angle orientation of the oriented spin system would be a separate phase. – Samuel Weir Jul 27 '17 at 22:05
• @SamuelWeir This is quite a famous example of first order magnetic phase transition. – SRS Jul 27 '17 at 22:20
• Yes, I believe that I've seen it presented before, but never thought deeply about it, at least not in regards to issues such as latent heats and the number of ordered phases that are associated with such a system. – Samuel Weir Jul 27 '17 at 22:31
• I don't think the entropy must be discontinuous for the transition to be first order. According to Wikipedia the transition is first order if there is a discontinuity in the first derivative of the free energy "with respect to some thermodynamic variable". Here the discontinuity is in the magnetization, which is the derivative of the free energy with respect to the external field. – user8153 Jul 28 '17 at 18:22
• Goldenfeld's "Lectures on Phase Transitions and the Renormalization Group" says that "Ehrenfest proposed that phase transitions could be classified as 'nth order' if any nth derivative of the free energy with respect to any of its arguments yields a discontinuity at the phase transition" (p. 14). I don't think that from this classification it follows that a first order transition must have non-zero latent heat. – user8153 Jul 28 '17 at 18:56

This is a very nice question, but to answer that, we should first clarify the OP in detail.

-- First scenario: Abrupt change

One may consider a scenario in which a ferromagnetic material prepared in an ordered state (eg., with $\langle m \rangle = \uparrow$) is suddenly subjected to a finite “opposite” external magnetic field (ie., antiparallel to the initial magnetisation) while the temperature is kept fixed below the critical temperature.

I think this abrupt scenario cannot be called a phase transition in the common sense. Such a procedure will induce an abrupt change in the properties of the system (esp. its ground-state). Free energy will be discontinuous itself, so it is a “zeroth-order/discontinuous phase transition”, if you wish. Phase transitions are usually defined as processes in which a slow variation of a thermodynamic variable (usually, temperature) leads to drastic changes in the thermodynamic phase of a system. Abrupt changes are trivial in this regard: we know a priori that they induce qualitative changes in the system. In our case, if one changes the external magnetic field abruptly, and waits long enough to let the system relax to its equilibrium, one will see that the magnetisation of the system is reversed to be parallel to the final external field; the system adapts itself to the applied field if we allow it to relax (see diagram below).

                               *************
apply H     ** complicated **
initial phase  ———————————>  **  relaxation **  ———>  final phase
**   process   **
*************


Notice that the devices of equilibrium statistical mechanics cannot deal with the intervening inherently non-equilibrium relaxation process. They can only explain the equilibrium initial and final (ordered) phases. In our scenario, we implicitly assume that we wait long enough for the system to relax.

So let's consider a quasi-static (slow) scenario. But, to be more quantitative, we need a simple mean-field analysis.

-- Mean-field analysis

To consider the transition more quantitatively, let's model a ferromagnetic material with a nearest-neighbour Ising model ($S = \frac{1}{2}$), and “solve” it within the mean-field approximation. The derivations of the mean-field Hamiltonian and partition function are given in modern statistical physics textbooks and will not be repeated here (see eg., Schwabl, F. “Statistical Mechanics” (2010) [WCat]). The dimensionless mean-field free energy density reads

$$f(h, T; m) := \frac{1}{T_c} \frac{F}{N} = \frac{1}{2} m^2 - \frac{T}{T_c} \ln \big( 2\cosh( \mathcal{M}_h ) \big)$$

where $\mathcal{M}_h := \frac{T_c}{T} ( h + m )$, $h$ is the rescaled external magnetic field, $h := h_{ext}/T_c$, $m$ is the mean-field, $T_c$ is the critical temperature for spontaneous magnetisation, and $T$ is the temperature. Note that we use natural units where $k_B = 1 = \hbar$.

To determine the mean-field $m$, we minimize the free energy and obtain the self-consistent mean-field equation,

$$m = \tanh( \mathcal{M}_h ) ~.$$

The nature of the solutions for $T < T_c$ differs from those for $T > T_c$: ie., in absence of external fields, $h \rightarrow 0$, we have a finite spontaneous magnetisation ($m \neq 0$) when $T < T_c$, but no spontaneous magnetisation ($m = 0$) when $T > T_c$.

From the free energy, one can readily obtain the entropy,

$$s := \frac{S}{N} = - \frac{1}{N} \frac{\partial F}{\partial T} \big\vert_{V, h} = \ln \big( 2 \cosh(\mathcal{M}_h) \big) - m \, \mathcal{M}_h ~,$$

specific heat at constant volume, $c_V$,

$$c_V := \frac{C_V}{N} = \frac{1}{N} T \frac{\partial S}{\partial T} \Big\vert_{V, h} = -\frac{1}{N} T \frac{\partial^2 F}{\partial T^2} \Big\vert_V = - \frac{T \mathcal{M}_h^2}{1 - \frac{T}{1 - m^2}} ~.$$

and magnetic susceptibility, $\chi_h$,

$$\chi_h := \frac{\partial m}{\partial h} = - \frac{1}{1 - \frac{T}{1 - m^2}} ~.$$

To obtain the relations above, we have used a rescaled temperature, $T \mapsto \frac{T}{T_c}$, and plugged in the mean-field equation and its temperature derivative,

$$\frac{\partial m}{\partial T} = \frac{\mathcal{M}_h}{1 - \frac{T}{1 - m^2}} ~,$$

when necessary.

According to the Ehrenfest classification (see this figure), the order of a transition can be determined from the discontinuities in the derivatives of the free energy (see Side Remarks).

-- Second scenario: Quasi-static change

One may consider a variation of the first scenario where, at a fixed $T < T_c$, the strength of the magnetic field is slowly (quasi-statically) varied from a positive value to some negative value. We observe that in this case, $m$ has a jump at $h = 0$ (see figure below); therefore, a discontinuous (1st-order) phase transition happens when $h$ crosses 0.

For concreteness, let's suppose that we prepare the system in a preferred magnetised state (say, $\uparrow$) with a tiny external field $h_\uparrow$ in a temperature $T < T_c$, and then we slowly vary the magnitude of the field to a finite value $h_\downarrow$ in the opposite direction. Then we calculate the free energy and entropy in the initial and final phases and changes thereof:

$$\Delta f := f(h_\downarrow, T) - f(h_\uparrow, T) \\ \Delta s := s(h_\downarrow, T) - s(h_\uparrow, T) ~;$$

from this we obtain the exchanged heat in the process,

$$\Delta q = T \Delta s ~.$$

First order phase transitions are always associated with a latent heat?

If the initial and final applied fields have the same magnetisation, since the entropy is an even function of $m$, then there will be no entropy change and hence, no heat exchange, $\Delta q = 0$, as mentioned by OP. Here, it is important to note the presumptions behind statements like “first-order phase transitions are associated with a latent heat”. The measurement scenario presumed for this statement is that one varies the temperature quasi-statically and measures the thermodynamic quantities as temperature crosses a certain value $T_\ast$. In the case of a “first-order transition”, varying the temperature across $T_\ast$ will incur a finite latent heat. Thus, that is not a statement about arbitrary external fields, like magnetisation. However, there is still an analogy.

In our scenario, when the magnetic field is changed slowly from $h_\ast - \delta h$ to $h_\ast + \delta h$, with a small $\delta h > 0$, then the internal energy of the system suddenly changes by a finite value $\sim m \, \delta h + h_\ast \, \delta M$, due to the discontinuity of $m$ at $h_\ast$. This amount of energy is absorbed from/released to the external field -- this is essentially the work performed by the external field on the system which leads to a change in the internal energy as $du = đ q + dw$. This is similar to the usual case when temperature is varied near the transition point and a latent heat $\Delta q$ is absorbed from/released to the reservoir.

-- Side Remarks

(I) Is this scenario truly a first-order phase transition?

The fact that this scenario is indeed a first-order phase transition, can be seen by an analysis of singularities in thermodynamic quantities. First let's consider the magnetisation. The mean-field equation can be expanded around $h = 0$ in an ordered phase where $\frac{T}{T_c} \lesssim 1$ and $m \sim \mathcal{O}(1)$ is finite. Then we obtain the approximate mean-field equation,

$$m \sim \tanh(m/T) + \frac{h}{T} ( 1 - \tanh^2(m/T) ) ,$$

where only the leading linear term in $h$ is kept (“linear response” approximation).

Since $m$ and $T$ are $\mathcal{O}(1)$, we can replace the $\tanh$ functions with their asymptotic values; namely,

$$\tanh(m/T) \sim \text{sign}(m) \sim \text{sign}(h) = \pm 1 ~;$$ The second approximation is valid since the sign of $h$ determines the sign of $m$ (see above). From this, we readily get the non-analyticity in $m(h)$ near $h = 0$ (in the ordered phase when $T < T_c$):

$$m \sim \text{sign}(h) ~.$$

By an analogous method, we can see that entropy has also a singularity:

$$s = - \frac{\partial f}{\partial T} \sim \text{const.} - \frac{h}{T} (m \, \frac{\partial m}{\partial h}) ~.$$

The derivative, $\frac{\partial m}{\partial h}$ behaves like a $\delta$-function, so we can “model” it by a Lorentzian of infinitesimal width $\varepsilon$,

$$\frac{\partial m}{\partial h} = \lim_{\varepsilon \rightarrow 0} \frac{1}{h^2 + \varepsilon^2} ~,$$

where unnecessary factors are dropped in the Lorentzian. Then, the entropy behaves as

$$s \sim \lim_{\varepsilon \rightarrow 0} \frac{h \, \text{sign}(h)}{h^2 + \varepsilon^2} = \lim_{\varepsilon \rightarrow 0} \frac{| h |}{h^2 + \varepsilon^2} = - \frac{1}{|h|} ~.$$

So, $s(h)$ behaves non-analytically as $-\frac{1}{|h|}$ in the vicinity of $h = 0$ (in the ordered phase when $T < T_c$) (see figure below). This completes the picture of the discontinuous phase transition.

(II) The Python code to compute and visualize all of these is accessible here.

• I'm sorry, but I don't really see how this addresses the question. 1) OP is not making any assumption about the change being "sudden", and I don't see how the speed at which $H$ is changed is relevant: we take the initial (equilibrium) state, we change the magnetic field in whatever way we want, we wait the system to equilibrate again and we measure the thermodynamic quantities that we want. 2) The mean field analysis is irrelevant here. I suppose OP knows the Ising model, and in any case the question is not about mean field theory. – valerio Feb 9 '18 at 23:23
• To conclude, the main point is not addressed, i.e.: for 1st order transition we expect a latent heat. In this example, the transition is 1st order but $\Delta S=0$, so there is no latent heat. How is this possible? This is the question that should be addressed. – valerio Feb 9 '18 at 23:24
• @valerio : Thanks for your comments; those are indeed appreciated -- I'll modify the answer accordingly. However, I partially disagree with you. My answer is meant first to emphasise the hidden presumptions in statements like "1st-order phase transitions are associated with a latent heat", instead of saying tacitly yes/no to the question in OP. So (1) one needs to explicate the scenario for measurement (eg., application of the fields), and (2) to ascertain that the scenario is indeed a phase transition and of 1st order; hence, the mean-field analysis. – AlQuemist Feb 10 '18 at 12:09
• I just think there are too many unnecessary details and the latent heat issue is not addressed directly enough. Then, I also don't agree with your statement that if the change in $H$ is abrupt this is not a 1st order phase transition.Like I tried to say in the first comment, I think the speed at which $H$ is varied is not really relevant here since we are assuming that the system is at equilibrium before and after the transition. It is widely accepted that this is a 1st order phase transition, and none of my sources talks about the speed at which $H$ is varied.But of course I could be wrong. – valerio Feb 10 '18 at 12:27
• @valerio : (i) The speed of changes is very important in such systems. In the context of phase transitions, actually there is always a presumption of slow change of the applied field. For abrupt changes, the dynamics of the system can be much more complicated that we can consider here. (ii) The details (eg. mean-field) provide a solid ground for the statements made. So, I believe they are needed for such a fundamental question and the following discussions. – AlQuemist Feb 10 '18 at 12:58

It depends on how you do the phase transition. Consider a gas in an insulated box. You can cause a phase transition to happen by just reducing the volume, with no heat input required. Your example with the magnet is similar, since we're only changing $H$, which is analogous to $V$. So in this sense there is no latent heat, i.e. the phase transition itself can be performed without it.

However, in the standard use of the term "latent heat", we imagine performing a phase transition by changing the temperature, while e.g. fixing the volume for the gas example. By the definition of a first-order phase transition $$\frac{\partial F}{\partial T} \bigg|_{V}$$ is discontinuous, which means that $$C = - T \frac{\partial^2 F}{\partial T^2} \bigg|_{V}$$ contains a delta function, i.e. a latent heat.

Technically, there could be zero latent heat in weird situations, because a first-order phase transition only has to have some derivative of $F$ be discontinuous, and $F$ depends on multiple variables. It could be that $\partial F / \partial T$ is miraculously perfectly continuous, which would require a perfectly horizontal line in a phase diagram, but this is definitely not the generic case, and I don't think it can occur for "normal" matter.

• Interesting point of view. However, about the gas example: let's say that we perform the transition at constant $T$ by changing the volume: we can then calculate $T\Delta S$, where $\Delta S$ is the entropy difference between the initial and final state. This is the latent heat of the transition, or am I wrong? – valerio Feb 8 '18 at 9:12
• @valerio92 If you expand the gas slowly and adiabatically, the change in entropy is exactly zero. The point is that there are multiple ways of crossing a phase transition, and putting in a latent heat is just one of them. That’s something all of us (3 answers) agree on, I think. I’m highlighting this particular point in case it’s what OP was confused about. – knzhou Feb 8 '18 at 9:43
• So the conclusion is that we have a latent heat only if the transition is first order and if it is in tempearture, i.e. if $T$ is the control parameter. – valerio Feb 8 '18 at 12:04
• @valerio92 Yup! If you think of $F$ as some abstract function in $(T, V)$ space, you can always see the latent heat is "there" from the graph of $F$. But you don't have to "pay" it as you cross the phase transition, you can just pay it later. (It's even more extreme for the liquid-gas transition because you can just go around the phase transition barrier entirely.) – knzhou Feb 8 '18 at 15:26
• @valerio92 Yeah, not many sources talk much about first-order phase transitions at all, since they're in some sense "messier" than higher-order ones. I liked the discussion here (lecture 3). It doesn't say much about latent heat directly. Sources that only deal with first-order transitions briefly probably focus on latent heat because it's such a strong experimental signature. I'll edit with more details tonight! – knzhou Feb 8 '18 at 16:07

I agree with @user8153 that I don’t think entropy has to be discontinuous for the phase transition to be first order. The free energy for this case is $F=-MdH$ since the other parameter temperature is constant. At the phase transition, the first order derivative of free energy, $M = -\frac{\partial F}{\partial H}$ is discontinuous.