Is it possible to view entropy as the sum of the kinetic energy of all molecules in the system? Since a system with 0 entropy would have 0 motion, and as you increase the motion of the molecules, entropy increases as well. If not, could you provide a simple counterexample in which entropy does not make sense as the sum of kinetic energy?

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    $\begingroup$ You might want to review the concept of entropy, including but not limited to the third law of thermodynamics and the statistical mechanical view of entropy. $\endgroup$ Commented Feb 8, 2017 at 0:47
  • $\begingroup$ I see the classical explanations of entropy as the 'randomness' of a system, but this seems to be a direct contradiction with the units of entropy (Joules/K). $\endgroup$
    – Andi Gu
    Commented Feb 8, 2017 at 0:54
  • $\begingroup$ @AndiGu You have randomness in Boltzmann's description of entropy. S=Boltzmann constant* lnW where W is macrostate multiplicity. But Clausius first described entropy with dS= dq/T which makes it unit J/Kwhich doesn't incorporate randomness. And Boltzmann introduced randomness to the concept of entropy(and it's a classical concept). But I think he could just do with lnW but to get it consistent with a pre-established theory Boltzmann introduced kB to the equation. $\endgroup$ Commented Feb 8, 2017 at 1:22
  • $\begingroup$ As an example, there is entropy increase upon mixing two substances together or even of mixing up sets of different objects even if both sets are already 'random'. No kinetic energy need be involved. Randomness is an unfortunate term often used to describe entropy. Its the degree of randomness that is described and what this is is not clear in everyday usage so its best totally avoided . Entropy is $S=k\ln(\Omega)$ where $\Omega$ is the number of ways of arranging objects, in molecules this means calculating the number of ways of placing then in all their possible energy levels. $\endgroup$
    – porphyrin
    Commented Feb 8, 2017 at 9:43
  • $\begingroup$ @AndiGu Your view is also a direct contradiction with the units of entropy (Joules/K). These do not match the units of energy (Joules). $\endgroup$ Commented Feb 8, 2017 at 10:09

1 Answer 1


In a crystal lattice, there are two contributions to the entropy: the vibrational entropy and the configurational entropy. The first is related to the kinetic energy; atoms with higher kinetic energies have tend to have larger displacements from equilibrium, and have a larger "footprint" in phase space, resulting in a higher entropy. The second contribution, configurational entropy, arises from disorder in the crystal lattice, namely the number of ways of arranging the atoms. In a perfect crystal, there is only one way, so the configurational entropy ($k\ln W$) will be 1.

An counterexample to the suggestion you make is any disordered solid, whether it's amorphous, a random alloy or some kind of crystal with geometric frustration. The most obvious example of this is water ice. Each water molecule occupies a lattice position and is tetrahedrally coordinated with its neighbours. In a defect-free ice crystal, the ice rules state that each molecule must accept exactly two hydrogen bonds and donate exactly two hydrogen bonds. This is quite a strong constraint, but allows orientational disorder --- the molecules can be oriented in one of six possible ways at each lattice position, resulting in an orientationally disordered structure. This will have a finite entropy even at absolute zero.

  • $\begingroup$ Thanks for the clear explanation - just one question: if 'configuration' is an aspect of entropy, how does that contribute to the J/k notion of entropy? I understand the kinetic energy part - but where does this configurational part come in in terms of energy? $\endgroup$
    – Andi Gu
    Commented Feb 8, 2017 at 23:37

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