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The Third Law of Thermodynamics can be stated in various ways, one of which is:

The entropy of a perfect crystal at absolute zero is exactly equal to zero.

Is this true for only "perfect crystals" and not for (say) fluids?

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  • $\begingroup$ The third law is better formulated as a statement that no system can reach absolute zero temperature in a finite number of finite process steps. The formulation above has a number of real drawbacks, one of which is the implicit assumption that $T=0$ is a physical state, after all. The other is that it's not clear what a perfect crystal is supposed to be. How are you handling faces, edges and corners? One can't, at least not in a rational and unique way, which now leads to needing an infinite system for starters. It's a rabbit hole from there. $\endgroup$ – CuriousOne Jun 6 '16 at 5:05
  • $\begingroup$ Yes, that is my point of contention. But this statement is very common and I am sure some sort of consensus exists in the community. $\endgroup$ – noir1993 Jun 6 '16 at 5:08
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    $\begingroup$ The only material we know of that is liquid at absolute zero is helium, and that would form a Bose-Einstein condensate that is in its ground state and therefore also has zero entropy. $\endgroup$ – John Rennie Jun 6 '16 at 5:17
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    $\begingroup$ In my thermodynamics course the problem was averted by pointing out that real systems asymptotically approach vanishing entropy, but that the time it takes to get there increases exponentially (i.e. from a semi-stable glass state to an ordered state). In practice zero entropy is simply not achievable with finite systems in finite time, one can, at most, freeze out a very low entropy state and the system will, for all practical purposes, get stuck in there. $\endgroup$ – CuriousOne Jun 6 '16 at 5:17
  • $\begingroup$ Remember that even a perfect crystal has zero point energy at absolute zero, so the definition of zero entropy is a bit subtle. It basically means in the ground state. $\endgroup$ – John Rennie Jun 6 '16 at 5:18
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The hamiltonian of a perfect crystal can be approximated at low temperature as the sum of harmonic oscillator hamiltonians. In 1D we have

$$H = \sum_{i=1}^N \frac{p_i^2}{2 m} + \frac 1 2 m \omega^2 \sum_{ij} ( r_i- r_j)^2 $$

where the $ij$ sum is over nearest neighbors. It is possible to verify that the eigenvalues of this hamiltonian are

$$ E_n = \left( \frac 1 2 + n \right) 2 \hbar \omega \left| \sin \left(\frac {ka}{2} \right)\right| $$

Where $n=0,1,2,3,\dots$, $k$ is the wavevector and $a$ is the lattice spacing. There is only one fundamental state, namely the one with $n=0$.

If the fundamental state is only one, entropy must vanish. This is clear from the Boltzmann relation:

$$S= k_B \log (\Omega)$$

where $\Omega$ is the number of micostates. If there is only one possible microstate (the fundamental state), $S$ must be $0$ (because $\log(1)=0$).

But there are systems are conceivable which have more than one fundamental state, i.e. in which the fundamental state is degenerate. If the degeneracy is smaller than exponential, however, there is no real problem, since the entropy per particle $S/N$ still vanishes in the thermodynamic limit. For example, if the degeneracy is of order $N^n$ we have

$$\lim_{N\to \infty} \frac{S}{N} \propto \lim_{N\to \infty} \frac{\log(N^n)}{N} = n \lim_{N\to \infty} \frac{\log(N)}{N} = 0 $$

The real problem is when the degeneracy of the ground state is exponential, since in this case we have

$$\lim_{N\to \infty} \frac{S}{N} \propto \lim_{N\to \infty} \frac{\log(a^N)}{N} = \log(a) \neq 0$$

So the "third law" of Thermodynamics fails in systems with an exponential number of ground states.

When we (improperly) apply the definition of $S$ to non-equilibrium systems, like glasses, this phenomenon is known as residual entropy.

This is why it is necessary to specify "of a perfect crystal" when stating the third law.

Then, as mentioned as John Rennie in the comments, there are also some exceptions, like liquid Helium, which does not crystallize as $T \to 0$, but it forms a Bose-Einstein condensate. Also in this case, there is only one ground state and therefore $S(T \to 0)=0$.

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