5
$\begingroup$

The first definition of entropy given by Clausius is I believe this $$S=Q/T$$ It is as I understand a common fact to understand entropy and maybe often teach it as a measure of disorder through the statistical definition of Boltzmann or Gibbs( depending on the ensemble) $$S=k\lnΩ$$ My question depending entropy, after some searching (look at A MODERN VIEW OF ENTROPY by Frank L. LAMBERT ) is this:

Is the physical meaning of entropy to be understood only in statistical terms as disorder because of the change in the statistical weights $Ω,$ or by looking to the thermodynamics as well, move to a definition of entropy as energy dispersal? In other words, conceive the physical meaning of entropy as a dispersal of the energy inside (or maybe at some points outwards) the system under consideration, where dispersal stands for a more wide allocation through the interior parts of the system( classical or quantum mechanical).

$\endgroup$
5

3 Answers 3

1
$\begingroup$

Thermodynamic entropy is not a definition. Contrary to the sickeningly numerous sources stating that it is, entropy is NOT disorder. Entropy is also NOT a measure of how much we know about the physical system. In a sentence, entropy is a measure of the number of distinct energy microstates consistent with the total energy and the physical constraints of the system. This is a property unique to each system which could care less who measures it (in fact it cannot be measured, only entropy changes can be measured).

$\endgroup$
2
  • 13
    $\begingroup$ " entropy is a measure of the number of distinct energy microstates consistent with the total energy and the physical constraints of the system." That's only true if energy is the only constrained quantity. More generally, entropy is the set of states consistent with all constraints imposed on the system, including energy, volume, etc. (and temperature and pressure are the Lagrange multipliers associated with those constraints). $\endgroup$
    – DanielSank
    Jul 12, 2017 at 21:35
  • $\begingroup$ And is the idea of a microstate well defined? $\endgroup$
    – TLDR
    Oct 16, 2021 at 23:20
1
$\begingroup$

S=Q/T and S=klnΩ do not contradict each other. To understand this, first you need to understand S=Q/T does not describe the absolute entropy. It describe the entropy change. For example, when 1J is extracted from a block of stone at 300K, the stone's entropy is reduced by S = Q/T = 1J/300K. (Strictly speaking, we should use dS = dQ/T, as the temperature of the stone may change. To get the total change of S, we need to integrate over the entire energy extraction process: $\Delta$S = ${\int} dQ/T$ ).

Now, microscopically, when you extract 1J from the stone, it cools down a little bit (lower temperature). As temperature changes the phonon (lattice vibration in the stone) distribution (a Bose-Einstein distribution since phonons are bosons), the available microscopic states for the lattice vibration (Ω) becomes less, and the statistical equation $S=k ln Ω$ will give you the same result of reduced entropy (or strictly, $\Delta$S = ${\int} dQ/T = k \ln Ω_{after} - k \ln Ω_{before}$ ).

In conclusion, the key to reconcile the two equation is to understand the microscopic (statistical) meaning of temperature - while classically it measures how hot an object is; microscopically it describes the probability distribution of particles (or quasi-particles such as phonons).

$\endgroup$
1
$\begingroup$

It's not a case of either/or; it is both/and.

To clarify, let's compare with another quantity: internal energy. In thermodynamic terms, internal energy is that quantity which is a function of state and whose change between states (for a closed system) is equal to the work required to move between those states under conditions of thermal isolation. In microscopic terms, the internal energy is the sum of the kinetic and field energies of the parts of the system. These two definitions do not contradict one another and it does not help our understanding to claim that one of them is one the right one 'really' and other a mere hanger-on. They are mutually consistent statements about a physical property, both of which bring insight and illumination.

Now for entropy similar things can be said. You can define it as a function of state whose change between states (for a closed system) is equal to the integral of $dQ/T$ if the change is accomplished reversibly. You can also define entropy as $-k_{\rm B} \sum_i p_i \ln p_i$ or as $k_{\rm B} \ln \Omega$. The statistical definitions require some understanding of what we are counting by the index $i$ and the probabilities $p_i$, and what we mean by a set of microstates and constraints. But the statistical and thermodynamic statements do not contradict one another (far from it), nor is one the right one 'really' and the other a mere hanger-on or embarrassing relative. They are mutually consistent statements about a physical property, both of which bring insight and illumination.

$\endgroup$

Your Answer

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

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