# First order phase transition in a classical system

I've never liked discontinuous quantities in classical physics, so I find the discontinuity in heat capacity weird.

My question is, do first order phase transitions ever really exist? Or are our discontinuous experimental $C_v$ vs. $T$ graphs just really steep curves? Discontinuous for all practical and theoretical purposes.

I understand that the theoretical model of a pure component system has a first order phase transition - and therefore a discontinuity in $C_p$. But its theoretically impossible to have a pure component (the chemical potential for any impurity in $A$ is infinite for a pure $A$). So effectively we only ever have multicomponent systems.

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– Arnold Neumaier Nov 26 '12 at 16:44

## 3 Answers

In classical thermodynamics (the best verified physical theory we have), phase transitions are true discontinuities, obtained from a continuous thermodynamic potential by taking derivatives (or even second derivatives for heat capacities) at points where these do not exist.

Thus there is nothing weird about them. It is no more weird than that the derivative of the absolute value function has a jump at zero.

First order phase transitions also appear in impure materials, though at slightly different volume, pressure and temperature than in the pure case.

Note that the discontinuities are inherited from statistical mechanics (Lee-Yang theorem).

 Note that thermodynamics applies to matter regarded as a continuum. Once one looks at atoms or molecules inside a system one has entered in the realm of statistical mechanics, a game with different rules. In general, as one looks at any physical system in more detail, the previously useful description starts to become approximate if not invalid. In particular, questions about continuity or differentiability lose their meaning (or regain it in a very different way).

[Edit2] Traditional thermodynamics in its usual axiomatic form (e.g., Callen) is valid for finitely extended matter. From a microscopic point of view, thermodynamics is usually justified in terms of statistical mechanics, based on the thermodynamic limit of infinite volume. But this is necessary only if the derivation is based on the microcanonical or canonical ensemble. In the grand canonical ensemble, thermodynamics follows without a thermodynamic limit (see Chapter 9 of my book http://lanl.arxiv.org/abs/0810.1019). In the original derivation of the Lee-Yang theorem (for ferromagentism), there is no phase transition without a thermodynamic limit, as the number of particles in an Ising system is bounded. However, real fluids treated from the most basic level, relativistic quantum field theory, have no upper bound on the number of particles, so that the original Lee-Yang statement no longer applies.

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Very interesting Arnold. My question, I guess, is more about the seam between theory and reality. I can imagine a theoretical system having a discontinuity at it's nth derivative, just like I can imagine a set of (classical) atoms standing perfectly still and achieving 0K. What happens when I add features to the model? Let's say I have a finite set of (classical) atoms at 0K in a rigid box larger than their crystal volume. I start to heat the atoms. Will the jump in Cv be perfectly discontinuous? Won't vacancies (holes) enter the solid and act like a "second component"? – user1512321 Nov 19 '12 at 16:59
@user1512321: Holes or other impurities act like an admixture of a second substance, if they can be treated collectively. But discontinuity is a matter of how closely you describe a system. The separating surface between two phases is a manifold in thermodynamics but gets increasingly rough as you magnify it, until the manifold description loses its meaning. See also the edit to my answer. – Arnold Neumaier Nov 19 '12 at 17:19
Thank you. Sorry for being nitpicky, but I've always liked the "seams" problems in science. – user1512321 Nov 19 '12 at 17:29

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This is wrong. First-order phase transitions, in the sense of being discontinuous curves, only exist if you take the limit of a system of infinite size. Real systems that we measure are not of infinite size, they are merely very large indeed, and consequently any $C_V$ vs $T$ curve for a real system is not discontinuous but just very steep, as the OP suggests. – Nathaniel Nov 26 '12 at 11:40

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@Nathaniel: Even continuity is an idealization. Without idealization, there is ''really'' only a physics we do not understand including quantum gravity), and all our current physics is only for all practical purposes. The standard model holds only for practical purposes, and so does statistical mechanics. Questions about what is ''really'' the case become therefore unanswerable, unless one settles on a particular basis. The notion of a phase trasition is well-defined only in thermodynamics, hence this is the correct basis. I had already mentioned in my answer what happens on lower levels. – Arnold Neumaier Nov 26 '12 at 16:31
show 4 more comments

In a sense you are right. The heat capacity only becomes discontinuous for a system of infinite extent. For all others it is continuous.

But that's a theoretical concern only. At the theoretical point of discontinuity the slope of the heat capacity is infinite. Pick any finite value for the slope, no matter how large and I can find a finite system where the slope is larger than that.

As for "impurities", there is no problem handling multicomponent systems. Chemists do it all the time. The results are only slightly more complex than for single component systems and my remarks about the phase transition above are still true.

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I think you understood my question. I know that one can always get a system with a steeper slope, just like you can (theoretically) get arbitrarily close to 0K, but not 0K. – user1512321 Nov 19 '12 at 16:45

"My question is, do first order phase transitions ever really exist?"

Yes they certainly do. Most of phase transitions are first order. I would not estimate in percents with confidence, but my feeling is that more than 90% of all phase transitions are the transitions of the first order. That is the answer from the experimental point of view.

Discussing it from the theoreticl point of view, I cannot see, why the fact that the body has a final size brings you to a conclusion that its capacitance should be continuous. The body in thermodynamics is indeed treated as a finite, as soon as one of the thermodynamic variables is, say, volume, or a number of particles. Let me just remind you that the two phases, 1 and 2, may be characterized by their free energies, F_1, and F_2. As soon as the free energy has the variables: V, T and N, it describes a body of a finite volume. The free energies under discussion are two different functions. This is important to understand. In some cases they may be slightly different, but there are certainly transitions where they essentially differ from one another. It depends upon the transition under study. In the transition point, however, the free energies of the phases are equal: F_1=F_2, but not their derivatives. Nothing strange that the second derivatives of these two different functions are not equal. It is, in contrast, not natural to expect them to be equal.

There is a different source of a perplexity in the case of the first order transitions. It is that each experiment is always performed during a certain time, while the transition itself has its characteristic time, or few characteristic times. If the dynamic of the transition is slow, the time of the experiment duration may be not enough for relaxation. It is often the case, if the transition is diffusive, but may be met also in other cases. This kinetic nature may "wash out" the curve C_v=C_v(t) and it may seem to be continuous. This is however, the kinetic effect only.

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