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The third law of thermodynamics can be stated as:

$$\lim_{T \to 0 \ \mathrm{K}}S = 0 $$

Now, from statistical mechanics: $S = k_B \log W $ where $W$ is the number of microstates. Notice that if $W=1$, then $S=0$. In other words, is the third law a consequence of the fact that systems tend to a single microstate as temperature drops to $0$?

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  • $\begingroup$ Perhaps absolute zero would imply zero kinetic energy of particles (no motion at all) so the system could not change its microstate. $\endgroup$
    – AVS
    Commented Jan 14, 2023 at 20:43
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    $\begingroup$ @AVS temperature zero does not correspond in general to zero kinetic energy. First of all, because of quantum mechanics, dominant close to $T=0$, which allows for the zero-point kinetic energy. Moreover, there are thermodynamic systems, like radiation, where the concept of particles with some kinetic energy becomes meaningless. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Jan 15, 2023 at 7:25

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Nernst's law is usually expressed as stating that as $T\to 0$ the entropy of the system approaches a constant. This is more like saying that the heat capacity goes to zero and is more general than just saying that $S(T=0)=0$ because the former also includes the ground state degeneracies, see the wikipedia article for such examples.

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The most general statement of the third law is that as $T \rightarrow 0$, system's entropy approaches a unique value, independently of the value of any other thermodynamic variable. Since thermodynamic entropy is defined within an additive constant, it is always possible to put $$ S(T=0)=0. $$

Apparently, the microcanonical formula from Statistical Mechanics: $$ S= k_B \log W,\tag{1} $$ where $W$ is the number of microstates compatible with a given macrostate at fixed energy, volume, and number of moles would imply the non-degeneracy of the ground state once we translate the limit for vanishing temperature into the limit for energy approaching the ground state.

However, such a conclusion is not consistent with the fact that formula $(1)$, as all the similar formulae of statistical mechanics, does correspond to the thermodynamic entropy only at the thermodynamic limit, i.e., in the limit of a large system. Indeed, a degenerate ground state may be compatible with equation $(1)$ and with the third law, provided the degeneracy of the ground state does not grow with the system size so fast to give a contribution to the entropy per particle (or per volume).

Therefore, the correct statement should be the third law is a consequence of the fact that systems tend to a ground state whose degeneracy $g$ is sub-extensive as temperature drops to zero. Formally, we have to require that $$ \lim_{T \rightarrow 0^{+}} g(E,V,N)/V=0 $$ whenever $E,V,N$ go to infinity by keeping constant their ratios. That is a milder condition than the non-degeneracy of the ground state.

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  • $\begingroup$ Could you provide a good reference for the last part of your answer? Thanks in advance! $\endgroup$
    – Tobias Fünke
    Commented Jan 15, 2023 at 17:02
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    $\begingroup$ @TobiasFünke The first reference I recall saying something explicitly on such an issue is Di Castro and Raimondi "Statistical Mechanics and Applications in Condensed Matter" Cambridge University Press, 2015, section 4.2. However, I think that knowing what to look for, similar observations should be around also elsewhere. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Jan 15, 2023 at 22:24

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