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The question was asked long time ago on this site, but not answered properly.

In classical thermodynamics, entropy is defined up to a constant. In statistical thermodynamics, there is no such freedom. What exactly is the reason for either point of view?

In classical thermodynamics, only entropy changes play a role. But why does this argument disappear in statistical mechanics?

Also several formulations of the third law allow for entropy to reach a constant value (non-zero) at lowest temperature.

Addenda

  1. The paper by Steane arxiv.org/abs/1510.02311 (he added it below) shows that the discussion is pointless, because absolute entropy values can be determined also in classical thermodynamics. He shows that classical entropy is NOT defined up to a constant. A great read.

  2. The entropy implied here is observed entropy.

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    $\begingroup$ If the ground state (the minimum energy state of the many body system at zero temperature) is unique, then there is a single microstate, implying that the entropy is zero en.wikipedia.org/wiki/Third_law_of_thermodynamics $\endgroup$
    – Quillo
    Oct 11, 2023 at 5:57
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    $\begingroup$ The statement that entropy is only defined up to a constant in classical thermodynamics is a myth. It is no more true than a similar statement about volume. To determine a volume you can measure it. The same is true of absolute entropy. See arxiv.org/abs/1510.02311 $\endgroup$ Oct 11, 2023 at 17:56
  • $\begingroup$ Isn't the idea in SE that the Questioner first gives a brief outline of each point of view, and what doubts remain? $\endgroup$ Oct 11, 2023 at 23:42
  • $\begingroup$ One first has to agree on which type of entropy one is talking about, i.e., how it is defined: physics.stackexchange.com/a/739917/247642 $\endgroup$
    – Roger V.
    Oct 12, 2023 at 8:45
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    $\begingroup$ I disagree with the question being pointless exactly because most people still believe and textbooks also propagate that very myth @AndrewSteane is referring to and so clearly has demonstrated. If you axiomatize and formulate thermodynamics along the equivalent lines of Carnot-Bronsted that is based not on energy conservation $dU=TdS+ \sum_k Y_kdX_k$ but on work conservation $SdT+\sum_k X_kdY_k=0$ then the question of $S$ being absolute or relative does not even come up. $\endgroup$
    – hyportnex
    Oct 12, 2023 at 23:28

1 Answer 1

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The key point is that statistical mechanics entropy is just a name for $$ \eta = -k_B \sum_i p_i \ln p_i \tag{1} $$ (Gibbs expression in terms of the probability of the microstates $p_i$). It is clear that if there is a macrostate such that all probabilities but one are zero $\eta=0$.

To identify $\eta$ with the thermodynamic entropy, we need to establish a link between $\eta$ and $$ S=S_0 + \int \frac{q_{rev}}{T}. $$ Such a link can be established by working with the differentials (see the accepted answer to this other question for the case of the canonical ensemble). I.e., in a closed system, we have $d \eta = dS$.

However, equality of the differentials only implies equality of the functions within an arbitrary constant. Therefore, even though equation $(1)$ is introduced without additive constants, its relation with thermodynamic entropy implies the possibility of such a constant. Of course, the constant can be chosen such that the entropy at $0$ K vanishes.

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    $\begingroup$ So one can say that statistical mechanics has no additive constant, whereas classical thermodynamics has it - correct? $\endgroup$
    – KlausK
    Oct 11, 2023 at 17:37
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    $\begingroup$ "Of course, the constant can be chosen such that the entropy at 0 K vanishes." -- although to agree with statistical mechanics, you don't necessarily want to make this choice (eg if there is a degenerate ground state). $\endgroup$
    – Andrew
    Oct 11, 2023 at 17:38
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    $\begingroup$ @Klausk In part, it is a matter of definition. It is not forbidden to add a constant. Much more important, the real contact between statistical mechanics entropy and thermodynamic entropy requires the thermodynamic limit. Any function growing less than linear with the size can be added to statistical mechanics entropy. For example, a generate ground state does not change the vanishing of entropy at zero temperature, provided the degenracy results in a contribution sublinear in N. $\endgroup$ Oct 11, 2023 at 18:28

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