The third law of thermodynamics says that at T=0 the entropy of a system must be 0. I'm confused to how to corresponds to the statistical/quantum mechanical treatment of systems, which seems to say that S doesn't have to be 0 at T=0 when degenerate energy levels are considered.

Some materials have mulitple degenerate ground states, like iron has two. So if T=0 and all atoms of iron are in the ground state, then the multiplicity of the ground state is $W_g=2$. So $S=k lnW_g=9.5654*10^{-24}J/K$, which isn't 0. Can someone clarify what my mistake is?

  • $\begingroup$ A degeneracy that is not of the form $g^{\text{system size}}$ is not relevant, as you could have redefined the entropy to make it zero at T = 0. Entropy, after all, is defined using a coarse graining procedure, you need to specify what degrees of freedom are to be kept in the macroscopic description. A few more degrees of freedom more or less doesn't matter; if you are very precise when deriving thermodynamic equations, you'll also see this when introducing external parameters of the system like volume to describe work. This also redefines the entropy, but such effects are totally negligible. $\endgroup$ Oct 17, 2015 at 16:37
  • $\begingroup$ @CountIblis Can you describe what you mean by "redefining" entropy? I was considering the Boltzmann definition of entropy, and it didn't think it was possible to redefine that based on the situation. $\endgroup$
    – Mecury-197
    Oct 17, 2015 at 16:43
  • $\begingroup$ You need to take W to be the number of physical states the system can be in that are compatible with its macroscopic specification. Here we assume that all these W states are equally likely, otherwise you need to use the Shannon formula for the entropy. The macroscopic specification will involve a small energy interval $\delta E$ in which the total energy is supposed to be, see here for details. $\endgroup$ Oct 17, 2015 at 17:07
  • $\begingroup$ Note that if you take $\delta E$ to be smaller than the spacing between the energy levels, the specification of the internal energy would uniquely define the exact physical state of the system and the entropy would equal zero. So, you clearly see that the definition of the entropy makes it, in principle, a subjective quantity. It is the amount of information that you decide not to keep in your macroscopic description of the system. But, different choices will only lead to a few degrees of freedom more or less being kept, which doesn't affect the specific entropy in the thermodynamic limit. $\endgroup$ Oct 17, 2015 at 17:10
  • $\begingroup$ The effect you are considering is due to a global symmetry of the entire system, which amounts to adding just one degree of freedom. It would be different if the ground state degeneracy would be a factor of 2 for each atom, then the residual entropy would be proportional to the system size. $\endgroup$ Oct 17, 2015 at 17:14

1 Answer 1


The third law of thermodynamics doesn't say that entropy goes to zero at T=0. You yourself gave a counter-example with your mention of degenerate ground states. The third law of thermodynamics states that the entropy settles to some constant value as temperature approaches T=0. In the case of a ground state with degeneracy g, the constant value would be $S=k_bln(g)$, where $k_b$ is the Boltzmann constant. You can see that in the case that the ground state is non-degenerate then g=1 and this equation gives S=0.


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