5
$\begingroup$

In the third law of thermodynamics, entropy goes to zero or to a constant value at vanishing absolute temperature. The change of entropy also goes to zero.

The third law is valid in the thermodynamic limit, which means, it is valid in the case that the system is infinitely large. In contrast, in the real world, systems are of finite size.

Is entropy continuous in real life, when systems have finite size? And is the third law of thermodynamics correct for finite systems?

EDIT: What happens for nanoparticles with 10, 100, 1000 or 10000 atoms?

$\endgroup$

3 Answers 3

6
$\begingroup$

You seem to be interested in the case of mesoscopic systems. Here you will have to be precise about the preparation of the ensemble (and you will easily leave the realm of equilibrium thermodynamics if you're not careful).

Entropy is always the property of an ensemble not of a system. We get away with ignoring this in the macroscopic case because:

  • We look at the system at timescales large compared to the scale of time evolution of the internal degrees of freedom. So we can describe the systems by the time average and assume it is the same as the ensemble average (which holds for "reasonable" systems).

  • We look at very large systems, so we can average over small cells of the system, and so again describe our system as an ensemble with some average properties.

If you don't have this, e.g. because you have 1000 atoms in a harmonic trap, then, it's all about preparation. The questions are then ones like: How does the system interact with the environment (particle exchange, energy exchange)? Was the system prepared by taking part of a much larger system that was in equilibrium, and does it, thereby, "freeze" the equilibrium distribution of it?

If you do something like microcanonical preparation of the system (which you, of course, can't really do exactly experimentally), again and again – then you have a microcanonical ensemble, and this will have a well defined entropy, and this entropy will only take on certain discrete values. However, your system will no longer have a well defined temperature (because temperature can only be defined if entropy can be assumed to be continuous).

On the other hand, if you have a single two-level system connected to a heat bath (preparation in the canonical ensemble), and prepare this again and again, the entropy of your ensemble of two level systems is well defined again and continuously depends on the temperature of your heat bath. In this case, the system has the temperature of the heat bath by definition (!). But a single measurement of the two level system won't tell you the entropy – you have to measure again and again (as the entropy is a property of the ensemble not of the system).

As soon your system is small and there is no microcanonical preparation and no mechanism to equlibrate with a large system either – then equilibrium thermodynamics will totally break down. You can easily put five atoms in a trap with an arbitrary initial probability distribution, but you can't expect the laws of thermodynamics to hold then.

$\endgroup$
0
5
$\begingroup$

Real-world systems are composed of an extremely large number of atoms - the thermodynamic description is safely applied to anything containing of the order of Avogadro number of atoms, $N_A\sim 10^{24}$, where discreteness is beyond anything that our measurement instruments could detect.

Moreover, even with the perfect measurement instruments, we have to deal with the fact that whatever is our system, it is only a part of a larger and truly infinite world, the interaction with which washes out any discrete effects.

$\endgroup$
2
  • $\begingroup$ I clarified the question with an edit. $\endgroup$
    – KlausK
    Feb 1, 2023 at 15:31
  • $\begingroup$ @KlausK Why do you think a theory should apply outside of range of its applicability? $\endgroup$
    – Roger V.
    Feb 1, 2023 at 15:35
2
$\begingroup$

It depends on what kind of entropy you are computing. If the problem is computing an entropy for $N$ particles occupying a $M$ number of energy states, then your multiplicity will be taking discreete values, and your entropy will be discrete as well.

If it is a problem of computing entropy for $N$ particles distributed in $(\vec{x}, \vec{p})$ phase space, multiplicity will be continuous, and the entropy will be continuous too.

Finally, if you consider magnetic particles, which have two- or three-dimensional magnetization vector, then you Hamiltonian, your partition function, and, hence, the entropy will all be continuous.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.