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I've been reading the Postulates of Classical Thermodynamics, and I haven't found anywhere to be said that the Absolute Entropy of a system has to be a positive number.

The third one states that the Temperature, $dU/dS$, must be Positive, since $S$ is monotone increasing with $U$, and that as $U(S)$ approaches flatness, $S$ approaches zero. But this doesn't mean that $S$ can't be negative. Consider for instance the function $U = kS^3$, which is monotone increasing, and where $S$ approaches zero as $dU/dS$ approaches zero from both sides, meaning $S$ could be negative.

Im pretty sure $S$ can't be negative, but I'd like to see the actual proof or Postulate that says so in CLASSICAL (not Statistical) Thermodynamics. Maybe it's included in the definition of entropy and I missed it?

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    $\begingroup$ Just a comment about temperature. You can have negative temperature if you for instance make the majority spins in a magnet point oposite to an external field. Adding more energy here actually decreases the entropy. $\endgroup$ – Mikael Fremling Jul 9 '16 at 23:15
  • $\begingroup$ Huh. Doesn't that violate the third postulate? $\endgroup$ – Juan Perez Jul 10 '16 at 2:33
  • $\begingroup$ Well, the postulate does really only consider systems with infinite number of degrees of freedom. Whenever you have only a finite number of degrees of freedom you should be able to get negative temperature. Note however that for these systems there is still a one-to-one correspondence between $U$ and $T$. $\endgroup$ – Mikael Fremling Jul 10 '16 at 8:51
  • $\begingroup$ The third law of thermodynamics only applies to systems at zero temperature - it doesn't apply to (or rule out) systems at negative temperature. $\endgroup$ – tparker Jul 10 '16 at 16:29
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In classical physics, the absolute entropy is not well-defined - only entropy differences matter, so shifting your entropy by an overall additive constant doesn't matter. So entropies can indeed be negative, just as energies can, and that doesn't say anything at all about your system, just about your choice of coordinates.

For example, the microcanonical ensemble for a system with $d$ degrees of freedom is given by $$\Omega(E) = \int \frac{d^dx\, d^dp}{h^d} \delta(H(\{p,x\}) - E) \delta E.$$

In quantum mechanics, $h$ is Planck's constant, but in classical mechanics it can be anything you want - it's just there to get the units right. So if you decide to make $h$ large enough, then $\Omega(E)$ will become less than 1, and so the entropy $S(E) = k_B \ln \Omega$ will be negative.

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  • $\begingroup$ Thanks! So entropy can be negative. Though the other answer says otherwise. In any case, with quantum mechanics and Planck's constant, S should be positive then? $\endgroup$ – Juan Perez Jul 10 '16 at 2:38
  • $\begingroup$ @JuanPerez The other answer and I are talking about slightly different notions of entropy. Graham Reid is talking about the "Gibbs entropy," which is indeed always positive. I'm talking about the "Boltzmann entropy," which can be either positive or negative. Both notions are useful in different contexts. $\endgroup$ – tparker Jul 10 '16 at 2:41
  • $\begingroup$ Thanks again. That Gibbs entropy was the one I'm more familiar with. in the end, I guess looking at things through statistical mechanics is more insightful than looking at them from classical thermodynamics $\endgroup$ – Juan Perez Jul 10 '16 at 2:45
  • $\begingroup$ @JuanPerez Well, it actually depends in the quantum case. For a finite system, the entropy of a certain energy eigenvalue is just the logarithm of the energy degeneracy, which is clearly positive or zero. But in the thermodynamic limit, the quantum Boltzmann entropy is actually still only defined up to an additive constant, because of the freedom to choose the width $\delta E$ of the energy window. But for a fixed choice of $\delta E$, taking the semiclassical limit of $\Omega(E)$ yields the expression above with $h$ being Planck's constant. $\endgroup$ – tparker Jul 10 '16 at 2:47
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    $\begingroup$ I did, but it says that it only is publicly shown if my reputation is more than 15 $\endgroup$ – Juan Perez Jul 10 '16 at 14:13
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A better definition of Entropy is given by the statistical mechanics definition of the Gibbs Entropy:

$ S = -k_B \sum{p_i \ln{p_i}}$

Where $S$ is the entropy, $k_B$ is Boltzman's constant and $p_i$ is the probability of the $\mathrm{i^{\mathrm{th}}}$ microstate to be occupied. For any probabilities less than 1, the quantity is clearly always positive. It can be shown that this definition is consistent with the classical thermodynamic entropy

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