The third law of thermodynamics states that the entropy of a perfect homogeneous solid is zero at absolute zero temperature.

The reason to ask for "perfect and homogeneous" is that such a solid has a ground state that is not degenerate.

But here is the question. A perfect crystal can be oriented in various directions in space. All these orientation are degenerate states. Why are the different states not counted? Shouldn't they lead to a non-zero entropy at absolute zero temperature?

  • 1
    $\begingroup$ You are typically interested in the thermodynamic limit ie the limit of infinite number of atoms in your crystal. Entropy is extensive, but rotation invariance only adds an intensive term, which is negligible in the thermodynamic limit. $\endgroup$
    – LPZ
    Jan 26 at 12:50
  • $\begingroup$ You can independently rotate and modify the solid (or any system). For the von Neumann entropy you have for product states that $S(\rho_A \otimes \rho_B) = S(\rho_A) + S(\rho_B)$. So without actual maths, just as intuition: Doesn't counting the rotated states yield nothing more than an additive constant that is the same in every system? (With some $S(\rho_A)$ for the rotation, modulo detailed maths and an equidistribution of uncountably many states...) Is that the intensive addition you are referring to, @lpz? $\endgroup$
    – kricheli
    Jan 26 at 13:24
  • $\begingroup$ Yes that’s what I meant @kricheli. Returning to the original question, i’ll also add that you could get an additional extensive term if you could rotate each atom independently in the lattice. Similar reasonings can give you residual entropy of some solids like ice. $\endgroup$
    – LPZ
    Jan 26 at 15:12
  • $\begingroup$ @lpz I did not understand why different orientations do not count. Could you explain it in a little more detail? Is it because the number of possible orientations is negligible compared to the number of atoms? Feel free to answer "officially"! $\endgroup$
    – KlausK
    Jan 26 at 16:09
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    $\begingroup$ @KlausK Place the crystal on the table, then move around and look at it from a different angle. Its orientation relative to you has changed, should your movement affect entropy, or any property of the crystal for that matter? $\endgroup$
    – Themis
    Jan 26 at 20:21

1 Answer 1


The entropy definition you speak of is based on the numbers of microstates that realise a given macrostate. Now, what exactly is a micro- and macrostate is arbitrary to some extent and mostly changes what question you want your entropy to answer. However, I would argue that at least one of the following applies:

  • Different orientations of the crystal are different macrostates. Even if your crystal is as perfect a sphere as it can be (and thus a sphere on the macroscopic level), you can distinguish orientations, e.g., with X-ray crystallography. Thus I would consider crystal orientation to be a macroscopic property.

  • Different orientations of the crystal are the same microstate. After all, it’s still the very same crystal. This particularly applies if the crystal is floating in a vacuum devoid of external fields such as gravitation and electromagnetism: Here, you cannot distinguish between rotating the crystal and the observer. (See also Themis’ comment.)

  • If your crystal is not floating in a vacuum devoid of external fields such as gravitation and electromagnetism, the different rotations are not equivalent anymore. Some orientation of the crystal is gravitationally or electromagnetically favoured and it will turn until it achieves that orientation.


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