9
$\begingroup$

Many books say that entropy measures energy dispersal or energy spreading.

Key point 1.4: Thermodynamic processes entail spatial redistributions of internal energies, namely, the spatial spreading of energy. Thermal equilibrium is reached when energy has spread maximally; i.e., energy is distributed equitably and entropy is maximized. Thus, entropy can be viewed as a spreading function, with its symbol S standing for spreading. Although not Clausius’ motivation for using S, this can serve as a mnemonic device. Energy spreading can entail energy exchanges among molecules, electromagnetic radiation, neutrinos, and the like.

link>>>

Spreading where? Spreading in space? What kind of space?

I mean, that in my mind "the spread of energy" is some kind of "density of energy in some volume". Why then in $\mathrm dS=Q/T$ the $Q$ factor gets divided by $T$ and not by some "abstract volume measure"? Why is energy spreading measured in units $\mathrm{J/K}$ and not in $\mathrm{J/m^3}$ ?

Or, maybe, the temperature in $\mathrm S=Q/T$ is treated exactly like some kind of volume.

I can certainly think in this way:

The greater the $T$, the greater the number of "microstates" consistent with that $T$.

P.S. I'm aware about Boltzmann definition of entropy. I understand it. I just can't handle entropy in phenomenological level (classical thermo level).

$\endgroup$
5
  • 1
    $\begingroup$ Where did you see that? Without further context seems simply wrong. $\endgroup$
    – lcv
    Commented Dec 24, 2018 at 12:41
  • 3
    $\begingroup$ It is the distribution of the energy over the various microstates. $\endgroup$ Commented Dec 24, 2018 at 12:50
  • 1
    $\begingroup$ @ChesterMiller Looks like an answer to me $\endgroup$ Commented Dec 24, 2018 at 13:30
  • $\begingroup$ @ChesterMiller, is T here acts as a measure of "a number of various mictostates"? $\endgroup$
    – coobit
    Commented Dec 24, 2018 at 13:37
  • $\begingroup$ @lcv, I've added context. Hope it helps to clarify the question. Thanks. $\endgroup$
    – coobit
    Commented Dec 24, 2018 at 13:41

4 Answers 4

8
$\begingroup$

At the classical thermodynamic level, the reason people speak about spread of energy is simply rooted into the principle of maximum entropy, i.e. the property of the entropy of a compound system to increase every time an internal constraint is released and getting a maximum for unconstrained equilibrium.

As a consequence of the principle, if in an isolated system the total energy of the system is partitioned in a too asymmetric way, for example constraining this configuration by fixed, impermeable, adiabatic walls, removal of the adiabatic constraint will result in a transfer of energy, from the hottest, to the colder system. The final result at equilibrium will be that the total energy of the system will be spread over the geometry of the system in such a way to equalize the local temperatures.

One has to think directly in terms of $S$ and not of $Q/T$ because, although the latter formula is used in the operative definition of entropy, $Q$ is not a property of the system but characterizes the process of exchanging energy.

$\endgroup$
2
  • $\begingroup$ Well, as you say "the total energy will be spread across parts of the system, which were previously separated". This I understand, but why Q/T? More so, you say it yourself "Q is not a property of the system" why then "spreading" has T in it's denominator and not... well.. Volume? $\endgroup$
    – coobit
    Commented Dec 30, 2018 at 13:41
  • $\begingroup$ @coobit You have to separate two quite different things. The first one is the operational definition of entropy, and the classical approach for that is through the total sum of the $Q_{rev}/T$ contributions. Once one has defined this entropy as a thermodynamic state function, the analysis of its properties allows to extract a number of consequences. The spreading of energy being one of them. $\endgroup$ Commented Dec 30, 2018 at 13:58
2
$\begingroup$

This question is critical in the understanding of thermodynamics, and I think that few people makes it for fear of being considered fools and prefer just memorizing the book.

First, the systems theory defines a system as a set of parts standing in interrelations (Bertalanffy). This imply that systems have sub-systems and sub-systems have sub-sub-systems; also, that all systems belong to a supra-system, etc. But remember this: not in thermodynamics. In thermodynamics there are only two system types: the whole (also called the thermodynamic system) and the parts (the molecules, or also, the sub-systems).

The classical thermodynamics exercise sets two containers with gas molecules at different temperatures, then, we open a wall separating both containers and they become a single one, so the molecules start mixing.

In this case:

  • The container is the system or the whole, and the molecules are the subsystems. There it is: there are no more levels of subsystems and suprasystems. Why? Because, Clausius, Boltzmann, et.al. supposed that things are a physical fact of the universe, and this might not be the truth (to be explained at the end). Anyway, we need this false truth in order to calculate the thermodynamic variables, and this false truth simplifies things.

  • The first law deals with energy conservation. Where? In the whole, as the sum of the energy of the parts. The total energy of the parts (internal to each part, internal to each molecule) is not considered (now you understand why avoiding the existence of sub-sub-systems seems a good idea to simplify things).

  • The above principle implies that thermodynamics requires the denial of the energy and mass relativistic equivalence. If a molecule would be polymerized (split into atoms + energy [the energy of the atomic bounding]), then, a sub-system is being converted into sub-sub-systems + energy, and that would add a large complexity to thermodynamics (in fact, that's a real problem when thermodynamics is applied to chemistry). So, the first law deals only with the conservation of energy of the whole, not the conservation of energy of the parts.

  • The second law, (my definition of entropy and the second law, coherent with Boltzmann or Clausius: "energy tends to propagation"), deals with the dispersal of the energy within the whole, that is, the energy dispersal or propagation across the parts or sub-systems. Then, we don't talk about the entropy of the molecules, the atoms, the quarks, or the supra-system (the room, or the universe). We always talk about the entropy of the system. The target where entropy is calculated is always the system, in this configuration of a unique "system made by indivisible parts". It is the entropy of the system that always grows.

When someone says that the entropy of a refrigerator (the whole) decreases at the expense of the context (e.g. the kitchen room, which is the supra-system), it is a sort of naive application of Clausius' formulas. The intention is to say that naturally, if the fridge is cold and the exterior is hot, then, the fridge tends to increase its temperature in order for entropy to grow. But for entropy to decrease, the energy dispersal between the apples, bananas and grapes (the subsystems) in the fridge would not disperse, but concentrate into each fruit-thing. What is happening is that the fridge is just losing energy, and transferring it to the environment. What we are talking about is not the entropy of the refrigerator, but the entropy of the kitchen (or the system made by the interaction kitchen -- refrigerator). Then, the error is to say that the entropy of the refrigerator decreases. The correct statement is to say that the entropy of a kitchen reduces when a refrigerator is active (energy is concentrated outside the refrigerator, which is an amazing human achievement!). But at the expense of the entropy of the suprasystem: the universe. In order to assess the second law, the energy must conserve, and that's a problem in open systems. It's contradictory to assess entropy changes in a system which energy is changing.

  • The third law states that entropy is $0$ when temperature is $0K$. This implies the negation of Einstein's energy-mass equivalence. In fact, here we are saying that entropy is zero (zero energy dispersal) in the whole, the gas container, or the kitchen. We are stating that entropy (whatever it means, which is related to the energy) is zero within a portion of matter. Our subjective, fragile, human-rational perception is tracing the line between mass and energy (and quantum physicists come to the rescue: "wait! there is still ground state energy!!!, so don't say that entropy is zero, say, perhaps that entropy is... a constant!" (which evidently contradicts the essence of the 3rd law)), assuming that energy and mass are completely different physical features. Well, they are not. But we need this false truth in order to use the other two laws.

  • Finally, the zeroth law (temperature is a transitive relation) just formalizes the feeling that is temperature, in order to be used as a physical fact. Arieh Ben-Naim states that temperature should be expressed in energetic units. I've always considered this as a good idea, but, well, in the meantime we're stucked with Kelvin.

Hope this answers to your question, but of course, sets new questions (that's why the book-memorizing-without-asking-obvious-questions approach is wrong: science has problems; science can only be perfect for fools!). The following link addresses the false truth presented above. In simple terms, the false truth is accepting that our perception is the truth, but that's not correct. Macrostates correspond to our perception, and microstates to that which we cannot perceive. But thermodynamics is not strict with such separation: the microstatic level includes several facts that correspond to our perception; mainly, the existence of things or solid objects (e.g. molecules as objects that have absolutely no internal energy) as physical facts. We know that things are just manifestations of our perception, not real physical facts.

https://philosophy.stackexchange.com/a/57718/23407

$\endgroup$
5
  • $\begingroup$ The entropy of system doesn't, in general, go to zero as temperature go to zero. See Nernst's theorem $\endgroup$
    – Alexander
    Commented Dec 29, 2018 at 9:40
  • $\begingroup$ Your interpretation of zero law is also kind of a tongue-twist. The zero'th law states equivalence of equilibrium systems. Then you invent some parameter to quantify it and call it temperature. The equivalence is not under debate, but the definition of temperature is. See this article. When referring to obscure physics opinions ("Arieh Ben-Naim and other physicists"), please refer also to the consensus, in order not to distort the opinions of people new to the field. $\endgroup$
    – Alexander
    Commented Dec 29, 2018 at 9:53
  • $\begingroup$ @Alexander: Entropy not 0 at 0K: correct. The 3rd law allows the calculation of the absolute value of entropy, which is complex to do when energy flows between sub-systems. At $0K$, energy stops flowing, so it is the point that we can use to define a conventional magnitude (whatever value is ok, as long as it supports further calculation). The definition of temperature was improved (temperature is a transitive relation), whilst temperature is originally a feeling (a perception), of course, subject to debate. Removed "other physicists". Thanks. $\endgroup$
    – RodolfoAP
    Commented Dec 29, 2018 at 9:59
  • $\begingroup$ In the whole, as the sum of the energy of the parts. The total energy of the parts (internal to each part, internal to each molecule) is not considered Hm.. Do you say that "whole energy = energy of parts" and then say the opposite ( =The total energy of the parts is not considered = there is no energy of the parts"? Sorry, I can't understand. $\endgroup$
    – coobit
    Commented Dec 30, 2018 at 13:45
  • $\begingroup$ @coobit, a molecule is made by matter, which ultimately has an equivalence to energy (independently of the fact that that it is technically difficult to convert matter to energy). But in thermodynamics, matter has no equivalence to energy. Only the kinetic energy of molecules is considered, not the potential energy (e.g. energy types like the Morse potential or the equivalence mass -- energy of the atoms in the molecule). $\endgroup$
    – RodolfoAP
    Commented Dec 31, 2018 at 9:08
0
$\begingroup$

Rob: regarding your comment: "entropy will be equal to total energy is NOT dimensionally consistent"...your comment is entirely correct. I should have said that BH entropy is equal to the total radiation energy (confined to the Schwarzschild surface area) divided by BH temperature (which is constant for all BH's). But Q/T will nevertheless increase (but never decrease) as a BH gets bigger.

$\endgroup$
0
$\begingroup$

The idea that entropy (or increase of entropy) measures energy spreading is just wrong. It is unfortunate that this idea has gotten traction in recent years and now even appears in some textbooks (thanks largely to its promulgation by Frank Lambert).

It is easy to construct a counterexample: consider two boxes of equal volume containing ideal gas. Box 1 has many fewer particles and a slightly higher temperature than box 2 ($N_1 << N_2, T_1 > T_2$), so that the higher temperature box has less internal energy. When the boxes are put in thermal contact, which way will the heat flow? The second law says it will flow from high T to low T, that is, from box 1 to box 2. That means that the energy is not spreading out, but becoming more concentrated.

More colloquially, proponents of the "entropy measures spreading" idea (including the Leff article you cited) often invoke a cup of hot soup cooling down, in which case the increase of entropy does correspond to energy spreading out into the ambient air. But what if you replace the cup of hot soup with a glass of ice water? In that case the ice water will absorb heat from the air. One might object that this is really "spreading out" of energy from the air into the glass, but since the heat capacity of the water is so much higher than the air's, the water starts out with more thermal energy content, and then gets heated by the lower-energy room air. Entropy causes the energy to concentrate in the glass, where it was higher to begin with - there is no sense in which this can be understood as spreading of energy.

$\endgroup$
4
  • $\begingroup$ So, what entropy measures then? Your opinion, please. $\endgroup$
    – coobit
    Commented Jan 24, 2023 at 11:20
  • $\begingroup$ heat will from box 1 to box 2. That means that the energy is not spreading out, but becoming more concentrated. But Box` has fewer particles, thus giving away enrgy to a more particled Box2 the energy spreads from fewer particles to more particles, thus "the spreading". Yeah, Box1 has lower internal enegry, but it is more concentrated in the tiny amount of particles which make up the box1, no? $\endgroup$
    – coobit
    Commented Jan 24, 2023 at 11:25
  • $\begingroup$ @coobit Energy goes from where it is low density (box 1) to high density (box 2). That is not spreading, but concentrating. (If you think going from low N to high N means it is spreading, then take instead the case where $T_1 < T_2$. Then the energy goes from high N to low N.) $\endgroup$
    – pwf
    Commented Jan 25, 2023 at 1:16
  • $\begingroup$ As indicated in your question, the entropy of a state measures the number of independent microstates associated with that state. IMO that's always a reliable way to think about it. The reason it's associated with temperature is because, all else being equal, a system at higher temperature will experience a smaller relative increase in entropy than one at lower temperature for a given heat input. $\endgroup$
    – pwf
    Commented Jan 25, 2023 at 1:25

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.